Chemistry Handbook
Part 1
Department of Chemistry and Chemical Biology
McMaster University
2015
R.S. Dumont
Acknowledgments
I am grateful to Pippa Lock for her reviews of some of the chapters of this book. I also thank
John Tromp, David Brock and Peter Kruse for helpful comments.
1
Table of Contents
1.
2.
3.
The Mole
9
1.1 Counting Atoms and Molecules
Example 1.1 How many helium atoms?
Equation 1.1 Moles from mass and molar mass
Example 1.2: How many water molecules?
1.2 Balancing Chemical Reactions
Example 1.3: Balance the following chemical reactions
1.3 Stoichiometry and Limiting Reactant
Example 1.4: Limiting reactant and yield
1.4 Density
Equation 1.2: Mass from volume and density
Example 1.5: Mass of 1 mL cube of copper
Example 1.6: Mass of 1 m3 cube of copper
1.5 Solutions
Equation 1.3: Moles from concentration and volume
Example 1.7: How much HCl?
Example 1.8: Water volume added to achieve dilution
1.6 Titration and Solution Stoichiometry
Example 1.9: Titration of NaOH solution
Example 1.10: Titration of Na2CO3 solution
Problems
Solutions
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The Ideal Gas Law
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2.1 The Gas Laws
Equation 2.1: The ideal gas law
Example 2.1: Molar mass of a gas
Example 2.2: Temperature change accompanying volume change
Example 2.3: Pressure change accompanying volume change
Example 2.4: Temperature change accompanying pressure change
Example 2.5: Yield of HCl
2.2 Partial Pressure
Equation 2.2: Sum of partial pressues
Example 2.6: The mole fraction of water in saturated air
Problems
Solutions
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Electronic Structure
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3.1 Light
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Equation 3.1. Speed of light
Figure 3.1. Travelling wave
Example 3.1. Frequency range of visible light
Figure 3.2. The electromagnetic spectrum
Equation 3.2. Photon energy
Figure 3.3. Blackbody radiation
3.2 Photoelectric Effect
Equation 3.3. The photoelectric effect
Figure 3.4. The photoelectric effect
Figure 3.5. Electron kinetic energy vs frequency
Example 3.2. The photoelectric effect
3.3 Quantum Mechanics and the Hydrogen Atom
Equation 3.4. Energy levels of hydrogen
2
4.
5.
Equation 3.5. Transitions of hydrogen
Example 3.3. Transitions of hydrogen
3.4 Orbitals of Hydrogen
Figure 3.6. The energy levels of hydrogen
Table 3.1. Numbers of orbitals and states of hydrogen
Example 3.4. Spin and orbital quantum numbers
Figure 3.7. Orbital shapes
Table 3.2. Orbital shapes
3.5 The Periodic Table
Equation 3.6. Energy levels of hydrogen-like ions
Figure 3.8. Ground state electronic configuration of H, He and Li
Figure 3.9. Ground state electronic configuration of Be, B and C
Figure 3.10. Aufbau principle
Example 3.5. Ground state electron configurations of atoms
Figure 3.11. The periodic table block structure
Problems
Solutions
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Periodic Properties
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4.1 Atomic/Ionic Radius
Figure 4.1. Trends in atomic radius
Example 4.1. Order according to radius
Table 4.1. Atomic radii of main group elements
Table 4.2. Ionic radii of main group elements
4.2 Ionization Energy
Figure 4.2. Trends in ionization energy
Figure 4.3. Ionization energy across the second row
Table 4.3. Ionization energies of main group elements
Example 4.2. Order according to ionization energy
Successive ionization energies
Example 4.3. Successive ionization energies
4.3 Electron Affinity
Figure 4.4. Trends in electron affinity
Example 4.4. Order according to electron affinity
Table 4.4. Electron affinities of main group elements
4.4 Chemical Properties
Figure 4.5. Trends in electronegativity
Table 4.5. Electronegativies
Problems
Solutions
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Bonding
56
5.1 Types of Bonding
Ionic
Covalent
Metallic
Example 5.1: Type of bonding
5.2 Lewis Structures
Figure 5.1: Finding the Lewis structure of F2O
Figure 5.2: Finding the Lewis structure of CO2
Figure 5.3: Lewis structure of NO2
Example 5.2: Draw Lewis structures
Figure 5.4: Lewis structures of BeF2 and SnCl2
Figure 5.5: Lewis structures of BF3
Figure 5.6: Lewis structure of BF4‒
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3
Formal Charge
Figure 5.7: Finding the formal charges in CO2
Figure 5.8: Lewis structures of SCN‒, OCN‒ and CO
Figure 5.9: Finding formal charges in HSO4‒
Figure 5.10: Charged-minimized Lewis structure of HSO4‒
Figure 5.11: Lewis structure of NO3‒
Figure 5.12: Resonance structures of NO3‒
Figure 5.13: Resonance structures of HSO4‒
Example 5.3: Average formal charges and bond orders
Example 5.4: Lewis structures, average formal charge and bond order
5.3 Molecular Shape and Polarity
Figure 5.14: Two electron domains – linear geometry
Figure 5.15: Three electron domains – trigonal planar geometry
Figure 5.16: Trigonal planar methanal
Figure 5.17: Bent nitrous acid
Figure 5.18: Four electron domains – tetrahedral geometry
Figure 5.19: Tetrahedral CCl4
Figure 5.20: Trigonal pyramidal NH3 and SO32‒
Figure 5.21: Five electron domains – trigonal bipyramidal geometry
Figure 5.22: Trigonal bipyramidal SbF5
Figure 5.23: Seesaw shaped SF4
Figure 5.24: T-shaped BrCl3
Figure 5.25: Linear I3‒
Figure 5.26: Six electron domains – octahedral geometry
Figure 5.27: Octahedral SF6
Figure 5.28: Square pyramidal IF5
Figure 5.29: Square planar XeF4
Figure 5.30: T-shaped XeF3‒
Molecular Polarity
Figure 5.31: Bent water – showing net dipole
Figure 5.32: Tetrahedral, non-polar SiCl4
Example 5.5: VSEPR class and shape – polar or non-polar
5.4 Aqueous Solutions
Figure 5.33: Solvation of sodium and chloride in aqueous solution
Example 5.6: Electrolyte solutions
Example 5.7: Pairs of compounds forming solutions
Problems
Solutions
6.
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Types of Reactions
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6.1 Solubility and Precipitation Reactions
Solubility
Table 6.1. Solubility of common salts
Example 6.1: Which salts are soluble?
Precipitation Reactions
Example 6.2: Write net ionic equations
Example 6.3: Identify a precipitation reaction
6.2 Brønsted-Lowry Theory of Acid-Base Reactions
Example 6.4: Identify acid and base in a given reaction
6.3 Strong and Weak Acids and Bases
Table 6.2. Strong acids and a few weak acids
Table 6.3. Strong bases and a few weak bases
Table 6.4. Acid and base strength of conjugate pairs
Example 6.6: Ordering acids and bases according to strength
6.4 Types of Acids and Bases
Binary Acids
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4
Oxoacids and Oxoanions
Table 6.4. Oxoacids and oxoanions
Organic Acids and Bases
Table 6.5. Some simple organic acids and bases
6.5 Oxidation-Reduction Reactions
Oxidation Numbers
Example 6.7: Assign oxidation numbers
Example 6.8: Which reactions are redox reactions?
Table 6.6. Some simple organic acids and bases
6.6 Balancing Redox Reactions
Acid Solution
Base Solution
Example 6.9: Balance the redox reactions in acid or base
Disproportionation Reactions
6.7 Visible Reactions
Example 6.10: Which reactions produce a visble change
Problems
Solutions
7.
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Chemical Equilibrium
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7.1 Dynamic Equilibrium
7.2 The Equilibrium Constant
Equation 7.1: Activity
Equation 7.2: Equilibrium constant, K
Example 7.1: Expressions for K
Example 7.2: Determining unknown concentration
Equation 7.3: K for a sum of reactions
7.3 The Reaction Quotient
Example 7.3: Net forward or reverse reaction
7.4 Shift in Equilibrium – le Châtelier's Principle
Example 7.4: Shift in equilibrium
Example 7.5: Hydrogen and iodine
7.5 Solubility
Example 7.6: Solubilities of BaSO4 and CaF2
Equation 7.4: Solubility in terms of Ksp
The Common Ion Effect
Example 7.7: Common ion effect
7.6 Acid-base Equilibrium
Autoionization of Water
Equation 7.5: Autoionization of water equilbrium
Equation 7.6: pH and pOH
Equation 7.7: pKw = pH + pOH
Acid Dissocation Equilibrium
Example 7.8: pH of a weak acid solution
Example 7.9: pH of a dilute weak acid solution
Figure 7.1: Percent dissociation of a weak acid
Base Association Equilibrium
Example 7.10: pH of concentrated and dilute weak base solutions
7.7 Strength of Acids and Bases
Equation 7.8: pKa and pKb
Table 7.1: Acid pKa and base pKb values
Equation 7.9: pKa + pKb = pKw
Strength of Binary Acids
Strength of Oxoacids
Strength of Bases
Example 7.11: Order scids and bases according to strength
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5
8.
9.
Problems
Solutions
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150
Thermochemistry
157
8.1 System and Properties - Energy
8.2 Heat
Equation 8.1: Heat from ∆T, mass and specific heat capacity
Example 8.1: Heat flow into water
Example 8.2: Heat flow from copper to water
8.3 Work
Figure 8.1: Gas in cylinder with piston
Equation 8.2: Work for a constant pressure process
Example 8.3: Compression – work done on a gas
Example 8.4: Work done on or by a gas
8.4 Energy Conservation – the 1st Law of Thermodynamics
Equation 8.3: 1st law of thermodynamics
Example 8.5: Heating and compressing a gas
8.5 Enthalpy
Equation 8.4: Definition of enthalpy
Equation 8.5: Heat for a constant volume process
Equation 8.6: Heat for a constant pressure process
Example 8.6: Calorimetry measures enthalpy change
Example 8.7: Using Hess’ Law
Equation 8.7: ∆H from enthalpies of formation
Figure 8.2: Graphical depiction of Hess' law for formation reactions
Example 8.8: ∆H from enthalpies of formation
Equation 8.8: ∆H from bond enthalpies
Figure 8.3: Graphical depiction of Hess' law for bond dissociations
Example 8.9: ∆H from bond enthalpies
8.6 Lattice Enthalpy and the Born-Haber Cycle
Equation 8.9: ∆Hf of sodium chloride
Equation 8.10: Lattice enthalpy of sodium chloride
Figure 8.4: The Born-Haber cycle for sodium chloride
Example 8.10: Enthalpy of formation of zinc bromide
Equation 8.11: Coulomb’s law
Example 8.11: Order ionic solids according to lattice enthalpy
Problems
Solutions
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Entropy and Gibbs Free Energy
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9.1 Spontaneous Processes
Figure 9.1. Spontaneous mixing of gases
9.2 Entropy
Equation 9.1. The Boltzmann formula
Equation 9.2. The second law of thermodynamics
Figure 9.2. Three types of spontaneous processes
Properties of Entropy
Equation 9.3. Thermodynamic definition of entropy
Figure 9.3. The molar entropy of water versus temperature
Equation 9.4. Entropy and enthalpy of fusion relation
Equation 9.5. Entropy and enthalpy of vaporization relation
Standard Molar Entropy
Figure 9.4. The intramolecular motions of NO, NO2 and N2O4
Example 9.1. Order according to molar entropy
Entropy of Reaction
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6
10.
Equation 9.6. ∆S from absolute entropies
Example 9.2. ∆S from absolute entropies
Example 9.3. Sign and magnitude of ∆S
9.3 The Second Law of Thermodynamics
Equation 9.7. ∆Ssurr and ∆Hsys
Equation 9.8. ∆Suniv > 0
Example 9.4. ∆Suniv for formation of water
9.4 Gibbs Free Energy
Equation 9.9. Definition of Gibbs free energy
Equation 9.10. ∆G > 0
Equation 9.11. ∆G in terms of ∆H and ∆S
Table 9.1 Spontaneity of exothermic and endothermic processes
Equation 9.12. The crossover temperature
Standard Gibbs Free Energy of Reaction
Equation 9.13. ∆G in terms of Gibbs free energies of formation
Figure 9.5 Temperature dependence of ∆G
Example 9.5 Spontaneity of decomposition of N2O4(g)
9.5 Gibbs Free Energy and Equilibrium
Equation 9.14. ∆G under non-standard conditions
Equation 9.15 ∆G and the equilibrium constant
Equation 9.16 ∆G = R T ln(Q/K)
Equation 9.17 K from ∆G
Figure 9.6. Reaction quotient and Gibbs free energy
Example 9.6. Spontaneity of hydration of calcium sulfate
9.6 Coupled Reactions
Figure 9.7. Condensation of phosphoric acid and ADP to form ATP
Problems
Solutions
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Electrochemistry
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10.1 Electron Transfer Reactions
Figure 10.1. Direct reaction of zinc and copper II
10.2 Electrochemical Cells
Figure 10.2. The Zinc-Copper Cell
Cell Diagrams
Example 10.1. Cell diagrams
10.3 Standard Cell Potential
Equation 10.1. Ecell and ∆G
Equation 10.2. Ecell° and ∆G°
Oxidation and Reduction Half-Cell Potential
Figure 10.3. The standard hydrogen electrode
Equation 10.3. Ecell° is a difference of reduction potentials
Table 10.1. Standard reduction potentials
Example 10.2. Standard cell potentials from reduction potentials
Example 10.3. Spontaneous cell reactions
Combining Reduction Potentials
10.4 Cell Potential and Equilibrium
Equation 10.4. Ecell° in terms of K
Equation 10.5. K in terms of Ecell°
Non-standard Conditions and the Nernst Equation
Equation 10.6. Ecell under non-standard conditions – Nernst equation
Figure 10.4. Ecell versus Q
Example 10.4. Ecell under non-standard conditions
Measuring Concentrations with Electrochemical Cells
Example 10.5. Molar solubility of AgBr from cell potential
10.5 Concentration Cells
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Equation 10.7. Concentration cell potential
Example 10.6. Concentration cell potentials
The pH Meter
Figure 10.5. A pH meter
Concentration Cells in Living Systems
10.6 Corrosion
Figure 10.6. Corrosion of iron under a water droplet
Figure 10.7. Cathodic protection and enhanced corrosion
Example 10.7. Choosing cathodic protection agents
10.7 Batteries
Figure 10.8. The dry cell
Problems
Solutions
11.
Math Skills
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11.1 Scientific Notation
Example 11.1: Scientific notation
Table 11.1. Units of Distance
11.2 Dimensional Analysis
Example 11.2: How many Cu atoms are there in a penny?
11.3 Solving for a variable
Example 11.3: New concentration after dilution
Example 11.4: Two equations and two unknowns
11.4 Logarithms
Periodic Table of the Elements
Thermodynamic Properties
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8
1.
The Mole
Early chemists discovered that substances combine in certain combinations to form new
substances. For example, 32 g (grams) of oxygen combines with 4 g of hydrogen to form
36 g of water. Also, 32 g of oxygen combines with 12 g of carbon to form 44 g of carbon dioxide.
Eventually people realized that substances are made of atoms, and that atoms combine in
specific ratios. The various combinations of masses that were observed to react is explained
when we account for different atoms having different masses.
The numbers of atoms encountered in everyday life is very large. Even a tiny quantity of a
substance consists of a vast number of atoms. It is inconvenient to constantly talk about such
large numbers of atoms when discussing chemical reactions – or substances in general. This
approach to dealing with large numbers is actually quite common. For example, when discussing
the national debt of the United States, we hear numbers like 14.3 trillion dollars. This number is
expressed in units of trillions of dollars. This is much more convenient than writing the number as
$14300000000000.
It turns out that there are typically a lot more atoms than there are dollars of national debt. Thus,
the unit for numbers of particles used in chemistry is much bigger than 1 trillion. It is
NA = 6.022×1023 mol−1, the Avogadro constant. The unit itself is called the mole (or mol). If we
have 1 mol of hydrogen atoms, we have 6.022×1023 hydrogen atoms.1 This number arises
because chemists wanted 1 mol of hydrogen atoms (the lightest atoms) to have a mass of about
1 g. They achieve this by setting the mass of 1 mol of carbon-12 atoms to exactly 12 g (carbon12 is the common isotope of carbon). This convention defines the value of the Avogadro
constant.
1.1 Counting Atoms and Molecules
The masses of 1 mol of each type of atom2 are given in the periodic table. With this data, we can
determine how many atoms are in a sample by measuring the mass of the sample.
Example 1.1: If a sample of Helium has a mass of 6.600 g, then how many helium atoms are in
the sample?
Approach: Find mol of He. Convert to number of atoms.
The molar mass of helium (i.e., the mass of 1 mol of helium atoms) is expressed as 4.003 g
mol-1. You find this in the periodic table. Therefore, 6.500 g of helium consists of
6.600 g
 1.6488 mol of helium 3
4.003 g mol1
To get the number of helium atoms not in units of mol, we simply multiply by NA = 6.022×1023
atoms mol−1. The Avogadro constant has units of “particles” per mole. Here, the particle is a
helium atom. Thus, there are
1
All data and physical constants are taken from D.R. Lide (ed.), CRC Handbook of Chemistry
and Physics, 93rd Ed. (CRC Press, Boca Raton, 2012), unless otherwise indicated.
2
Molar masses found in the periodic table are averages of isotope masses, weighted according
to natural abundance.
3
The final digit in 1.6488 is written as a subscript to indicate that we are carrying the extra digit
(the second 8) into the next step of the calculation. It is written as a subscript because it is not a
significant figure. Carrying the extra figure reduces rounding errors during the steps of a
calculation.
9
1.6488 mol × 6.022×1023 atoms mol‒1 = 9.929×1023 atoms
Observe how the units cancel out, leaving the desired unit of atoms:
1.6488 mol × 6.022×1023 atoms mol−1 = 9.929×1023 atoms
Dividing the mass of a sample of atoms by the molar mass of the atom gives the amount of the
substance in mol.
amount (in mol) =
mass (in g)
Molar mass (in g mol1 )
n
m
M
1.1
Multiplying by NA then gives the number of atoms. We can use the same formula to count the
number of molecules in a sample.
Example 1.2: How many water molecules are in 20.0 g of water?
Approach: Find number of moles of water. Convert to number of molecules.
mass of water
Molar mass of water
20.0 g

 1.110 mol
18.015 g mol1
amount of water (mol) =
This is the amount of water. To answer the question, we must multiply by the Avogadro
constant.
1.110 mol × 6.022×1023 molecules mol‒1 = 6.68 ×1023 molecules
The Avogadro constant has units of “particles” per mole. Here, the particle is a water
molecule. The “particle” can be any unit we need.
1.2 Balancing Chemical Reactions
Chemical reactions are happening all the time. Some are slow, and barely noticeable. Others
are fast. Some of these involve the release of energy in the form of heat, light and/or sound.
Chemical reactions are conversions of one set of compounds – the reactants – into another set of
compounds – the products. They readily occur when there are stable products that can be
reached through some rearrangement of the atoms of the reactants. A principal feature of any
chemical reaction is that the total number of each type of atom, among reactants, must equal the
same tally for the products. Atoms are neither created nor destroyed in a chemical reaction (this
is the Law of Conservation of Mass). This allows us to balance chemical reactions and use the
associated stoichiometric coefficients to determine the expected amounts of the various products.
Example 1.3: Balance the following chemical reactions:
(a) Mg(s) + Cl2(g) → MgCl2(s)
(b) Ca(s) + O2(g) → CaO(s)
(c) CH4(g) + O2(g) → CO2(g) + H2O(g)
(a) This chemical equation is already balanced. There are two chlorine atoms and one
magnesium on each side of the equation.
10
(b) Calcium is already balanced – one Ca on each side. To balance oxygen, we need two
mol of CaO on the right. This changes the calcium balance, which is fixed by having two mol
of Ca on the left. The result is
2 Ca(s) + O2(g) → 2 CaO(s) .
It is also acceptable to balance this reaction using ½ as the coefficient of O2(g) – i.e.,
Ca(s) + ½ O2(g) → CaO(s) .
(c) Carbon is already balanced. To balance H we need two mol H2O on the right. Now we
need two mol O2 on the left two balance the four mol O on the right – two from CO2 and two
from 2 H2O. The result is
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) .
This method is called balancing by inspection. There are reactions that are more challenging to
balance. Often these are oxidation-reduction reactions, which are balanced with the half-reaction
method – see Chapter 6.
1.3 Stoichiometry and Limiting Reactant
Knowing the amounts of substances is important because the coefficients in a chemical reaction
give the relative amounts of the substances that react and are produced. For example, the
balanced chemical equation
2 H2(g) + O2(g) → 2 H2O(l)
tells us that the amount of hydrogen needed to completely consume a sample of oxygen is twice
the amount of oxygen in the sample. It also tells us that the amount of water produced equals the
amount of hydrogen consumed.
Sometimes the relative amounts of reactants actually used in a reaction are not the same as the
stoichiometric coefficients. In such cases we must determine the reactant that is completely
consumed (the limiting reactant). The other reactant(s) is(are) in excess.
Example 1.4: Suppose we carry out the above reaction with 30.0 g of O2 and 4.00 g of H2.
(a) Which reactant is the limiting reactant?
(b) How much of the other reactant remains at the end of the reaction?
(c) What mass of water is produced?
Approach: Find the number of moles of each reactant. Use reaction stoichiometry to
determine the limiting reactant.
30.0 g
 0.9375 mol
31.9988 g mol1
4.00 g
amount of H 2 =
 1.984 mol
2.01588 g mol1
amount of O 2 =
(a) To completely consume 1.984 mol of H2 requires 1.984/2 = 0.9920 mol of O2. However,
there is not enough O2 (only 0.938 mol). Therefore, O2 is the limiting reactant. The 0.938
mol of O2 is consumed completely, and some H2 remains unreacted.
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(b) The amount of H2 consumed is twice the amount of O2 consumed. 1.876 mol of H2 is
consumed, leaving 1.984 − 1.876 = 0.108 or 0.11 mol of H2 unreacted.
(c) Since the amount of H2 consumed equals the amount of H2O produced, 1.876 mol of H2O
is produced.
2 H2(g)
+
Before
1.984 mol
After
0.108 mol
O2(g)
→
2 H2O(l)
0.938 mol 0 mol
limiting reactant
0
1.876 mol
We are asked for the mass of H2O produced. This is the theoretical yield, determined by
calculation based on amount of limiting reagent used.
mass of water = amount of water  Molar mass of water
 1.876 mol  18.015 g mol1  33.80 g
1.4 Density
Sometimes we are given the volume of a substance and not its mass. To determine the amount
of the substance, we need to know the density of the substance. The density of a substance is
the mass per unit volume. For example, the density of water under ordinary conditions is about
1.00 g mL−1. If we know the volume of a substance, then we can determine its mass from its
density.
mass = volume  density
or
m=Vd
1.2
Example 1.5: The density of copper is 8.94 g mL−1. What is the mass of a cube of copper
1.00 cm × 1.00 cm × 1.00 cm ? Note that 1 mL = 1 cm3.
Approach: Find cube volume. Use density and volume to solve for mass.
First, we need the volume of the cube in mL
1.00 cm × 1.00 cm × 1.00 cm = 1.00 cm3
The mass of the copper cube is m = V x d,
1.00 mL × 8.94 g mL-1 = 8.94 g
Sometimes more elaborate unit conversions are required.
12
=
1.00 mL
Example 1.6: What is the mass of a cube of copper 1.00 m × 1.00 m × 1.00 m ?
Again, we need the volume of the cube in mL.4 This time, we will use the definition of the
litre, 1 L = 1 dm3 = 103 mL. [Note that 1 dm = 10−2 m].
V = (1.00 m)3
=
(10.0 dm)3
=
= 10.03 dm3
= 1.00 × 103 dm3 = 1.00 × 103 L
1.00 × 103 L × 103 mL/L
=
106 mL
The mass of the copper cube is
106 mL × 8.94 g mL−1 = 8.94 × 106 g = 8.94 × 103 kg
1.5 Solutions
Sometimes we are given a volume of a solution. We can determine the amount of a dissolved
substance, if we know the concentration.
amount = volume × concentration
or, n = c V
1.3
Example 1.7: How much HCl (mol) is in 250. mL of a solution with HCl concentration, [HCl],
equal to 2.00 mol L−1?
Approach: Use concentration and volume to find mol.
amount of HCl (mol) = 0.250 L × 2.00 mol L−1 = 0.500 mol
Note the conversion of volume in mL to volume in L because the concentration is given in
mol L−1.
The concentration of a substance is the number of moles per unit volume. Dissolved species can
be very dilute (low concentration) or highly concentrated (high concentration).
A dilute solution can be prepared from a concentrated solution by adding pure water. This is
called dilution.
Example 1.8: What volume of pure water (in mL) must be added to 100. mL of a [NaOH] = 2.00
mol L−1 sodium hydroxide solution to make a solution with [NaOH] = 0.500 mol L−1?
Approach: Find mol NaOH. Use final concentration to find final volume. Take difference
between final and initial volumes.
The amount of NaOH (mol) in the initial and final solutions is the same – only pure water is
added. Therefore, since n = cV,
Amount of NaOH (mol) = amount initial = ciVi = amount final = cfVf
(You may recognize this as c1V1 = c2V2)
amount of NaOH = 0.100 L  2.00 mol L1 = V f  0.500 mol L1
where Vf is the final solution volume.
4
Alternatively, we could convert the density to units of g m−3.
13
Vf =
0.100 L  2.00 mol L1
 0.400 L
0.500 mol L1
The volume of water added is the change between the final and the initial volumes:
V  V f  Vi = 0.400  0.100 L = 0.300 L
Therefore, 300. mL of pure water must be added to perform the dilution.
1.6 Titration and Solution Stoichiometry
A strong acid such as HCl will react completely with a base. The acid is said to neutralize the
base. We can detect when the base is completely consumed by adding a small amount of an
acid-base indicator such as phenolphthalein to the base solution. Phenolphthalein makes the
solution pink until all the base is consumed, at which point the solution turns colorless.
An HCl solution with known concentration can be used to determine the unknown concentration
of a base solution.
Example 1.9: Suppose 221.6 mL of a standard HCl solution with [HCl] = 1.007 mol L−1 are
required to completely neutralize the NaOH in 100.0 mL of a sample solution of unknown NaOH
concentration. NaOH and HCl react according to
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
What is the NaOH concentration of the sample solution?
Approach: Find mol NaOH. Use NaOH volume and mol to determine concentration.
Since the reaction is 1:1, for complete reaction, the amount of HCl (mol) added must
equal the amount of NaOH (mol) in the sample solution. Therefore,
amount of NaOH  amount of HCl  0.2216 L 1.007 mol L1  0.22315 mol
This amount of NaOH was present in 100.0 mL of the sample solution. Therefore, the
NaOH concentration of the sample solution was c = n /V,
[NaOH] 
0.22315 mol
 2.232 mol L1
0.1000 L
Example 1.10: Suppose 178.1 mL of a standard HCl solution with [HCl] = 1.007 mol L−1 are
required to completely neutralize the Na2CO3 in 100.0 mL of a sample solution of unknown
Na2CO3 concentration. Na2CO3 and HCl react according to
Na2CO3(aq) + 2 HCl(aq) → 2 NaCl(aq) + H2O(l) + CO2(g)
What is the Na2CO3 concentration of the sample solution?
Approach: Find mol Na2CO3. Use volume and mol to find concentration.
Take note of the stoichiometry of the balanced equation. The amount of HCl added must
be twice the amount of Na2CO3 in the sample solution. Therefore,
14
amount of Na 2 CO3 =
=
1
 amount of HCl
2


1
 0.1781 L  1.007 mol L1 = 0.089673 mol
2
and
[Na 2 CO 3 ] 
0.08967 3 mol
 0.8967 mol L1
0.1000 L
Problems:
1.1 How many grams of calcium oxide, CaO, can be produced from the reaction of 4.20 g of
calcium metal and 1.60 g of oxygen gas?
1.2 (a) How many water molecules are in an Olympic-sized swimming pool (volume 2500 m3)?
The density of water is 1.00 g/mL. (b) If each water molecule were replaced by a solid cube
with volume 1.00 cm3, what volume swimming pool (in km3) would be required to hold all the
cubes?
1.3 What mass of hydrogen is required to produce 40.0 mL of water (density = 1.00 g mL-1) in the
hydrogen fuel cell reaction:
2 H2(g) + O2(g) → 2 H2O(l)
1.4 What is the concentration of a KOH solution, if 107 mL are required to neutralize 250 mL of
1.09 mol L-1 nitric acid (HNO3) solution?
1.5 What volume of 1.21 mol L−1 LiOH solution is required to neutralize 305 mL of 0.511 mol L−1
H2SO4 solution?
15
Solutions:
1.1
First, write a balanced reaction:
2 Ca(s) + O2(g) → 2 CaO(s)
Initial amount of calcium:
n  Ca  
m  Ca 
M  Ca 
=
4.20 g
 0.1048 mol
40.08 g / mol
=
1.60 g
 0.0500 mol
31.9988 g / mol
Initial amount of oxygen:
n  O2  
m  O2 
M  O2 
Since O2 and Ca react in a 1:2 ratio, O2 is the limiting reactant – there is less than 0.0524
mol (the amount needed to consume the entire 0.1048 mol of Ca).
Amount of CaO produced = 2 × (amount of O2) = 2 x 0.500 mol = 0.100 mol
Mass of CaO produced = (0.100 mol) × (56.08 g/mol) = 5.61 g
1.2
(a) Because the density is expressed in g mL-1, we must convert m3 to mL. Since 1 m =
102 cm,
2500 m3 = 2500 × (102 cm)3 = 2500 × 106 cm3 = 2.500 × 109 mL
Mass of water = volume × density = 2.500 × 109 mL × 1.00 g / mL = 2.500 × 109 g
Amount of water =
n  H 2O  
m  H 2O 
M  H 2O 
=
2.500 109 g
 1.387 7  108 mol
18.015 g / mol
Multiply by the Avogadro constant to get the number of water molecules:
1.3877 ×108 mol × 6.022 ×1023 molecules mol−1 = 8.3567 ×1031 molecules
(b) The total volume of 8.3567 ×1031 cubes is
8.3567 ×1031 cubes × 1.00 cm3/cube = 8.3567 ×1031 cm3
= 8.3567 ×1031 cm3 (10−5 km/cm)3 = 8.357 ×1016 km3
This volume corresponds to giant cube more than half of 1 million kilometers along each
side.
16
1.3
Mass of water produced = volume × density = 40.0 mL × 1.00 g /mL = 40.0 g
Amount of water produced:
n  H 2O  
m  H 2O 
M  H 2O 
=
40.0 g
 2.220 mol
18.015 g / mol
From the balanced chemical equation,
2 H2(g) + O2(g) → 2 H2O(l),
we see there is a 1:1 ratio of hydrogen consumed to water produced. Therefore, 2.220
mol of hydrogen is required.
Mass of H2 required = (2.220 mol) × (2.016 g mol-1) = 4.48 g
1.4
First write the balanced chemical equation.
KOH(aq) + HNO3(aq) → KNO3(aq) + H2O(l)
We see that KOH and HNO3 react in a 1:1 ratio. Therefore, the amount of KOH added
must equal the amount of HNO3 in the nitric acid solution. Therefore,
amount of KOH = amount of HNO3 = 0.250 L × 1.09 mol L−1 = 0.2725 mol
[KOH] = (amount of KOH) / (volume of KOH solution) = 0.2725 mol / 0.107 L
= 2.55 mol L−1
is the concentration of the KOH solution.
1.5
First write the balanced chemical equation.
2 LiOH(aq) + H2SO4(aq) → Li2SO4(aq) + 2 H2O(l)
We see that LiOH and H2SO4 react in a 2:1 ratio. Therefore, the amount of LiOH added
must equal twice the amount of H2SO4 in the sulfuric acid solution. Therefore,
amount of LiOH = 2 × amount of H2SO4
= 2 × (0.305 L × 0.511 mol L-1) = 0.3117 mol
volume of LiOH solution = (amount of LiOH) / [LiOH] = 0.3117 mol / 1.21 mol/L
= 0.258 L = 258 mL
is the volume of the LiOH solution required.
17
2.
The Ideal Gas Law
2.1 The Gas Laws
At low pressures all gases have the same behavior in the following sense:
1) With pressure and temperature fixed, the volume of the gas is proportional to the number of
particles (atoms in the case of a monatomic gas such as Ne, or molecules in the case of a
molecular gas such as CO2). The proportionality constant is the same for all gases. This is
Avogadro’s Law:
V = constant × n
2) If temperature is increased, with pressure fixed, the volume increases in proportion to the
temperature (provided the temperature is measured on the absolute scale). This is Charles’
Law:
V = constant × T
3) If pressure is increased, with temperature fixed, the volume decreases inversely with the
pressure. This is Boyle’s law:
V = constant / P
4) If temperature is increased, with volume fixed, the pressure increases in proportion to the
temperature. This law follows from Charles’ and Boyle’s Laws:
P = constant × T
Note that each of these laws has a different constant that depends on the values of the fixed
properties.
These observations are encapsulated in the ideal gas law:
PV  nRT
2.1
where R = 8.3145 J mol-1 K-1 = 0.083145 L bar mol-1 K-1 is the gas constant. Gases that obey this
law are said to be ideal. The law is valid at low pressure for all gases because, under these
conditions, the atoms or molecules are so far apart that forces between them have no significant
effect on the pressure of the gas. Note that 1 bar = 100 kPa is the standard IUPAC unit of
pressure. It is just slightly smaller than atmospheric pressure at sea level, 1 atm = 101.3 kPa.
The first observation allows us to determine the molar mass of a gas by measuring the mass of a
known volume of the gas at known temperature and pressure.
Example 2.1: Suppose a 1.000 L sample of a gas at 1.000 bar pressure and 293.0 K has a
mass of 1.807 g. What is the molar mass of the gas?
Approach: Solve for the amount of gas. Divide mass by amount to give the molar mass of
the gas.
The amount of gas is determined, from PV = nRT, by
PV
1.000 atm  1.000 L

RT 0.083145 bar mol1K 1  293.0 K
 0.041048 mol
n
18
The molar mass is determined from
molar mass 
mass
1.807 g

amount 0.041048 mol
 44.02 g mol1 .
The gas is likely CO2. Carbon dioxide has molar mass, 44.01 g mol‒1.
The second observation: Changes in volume indicate changes in temperature – if pressure is
fixed.
Example 2.2: A balloon of gas increases in volume from 10.0 L to 12.8 L at 1.000 bar pressure.
If the gas was initially at 295 K, what is its final temperature?
Approach: The amount of gas does not change. Equate the initial and final amounts.
Cancel out R and the pressure which also does not change.
The pressure and the amount of gas are constant. Therefore,
n
PVi PV f

RTi RT f
or
Vi V f

Ti T f
or
Tf 
Vf
Vi
Ti 
12.8 L
 295 K
10.0 L
 378 K
The third observation: Changes in volume indicate changes in pressure – if temperature is fixed.
Example 2.3: A gas-filled balloon increases in volume from 10.0 L to 15.6 L at 298 K. If the gas
was initially at 1.00 bar pressure, what is its final pressure?
Approach: Equate the initial and final amounts. Cancel out R and the temperature which
also does not change.
The temperature and the amount of gas are constant. Therefore,
19
n
PV
PV
i i
 f f
RT
RT
or
PV
i i  Pf V f
or
Vi
10.0 L
Pi 
1.00 bar
15.6 L
Vf
Pf 
 0.641 bar
The fourth observation: Changes in pressure indicate changes in temperature – if volume is fixed.
Example 2.4: A gas is held in a cylinder with fixed volume of 10.0 L. The pressure at is
observed to increase from 1.00 bar to 15.0 bar. If the gas was initially at 298 K, what is its final
temperature?
Approach: Equate the initial and final amounts. Cancel out R and the volume which also
does not change.
The volume and the amount of gas are constant. Therefore,
n
Pf V
PV
i

RTi RT f
or
Pi Pf

Ti T f
or
Tf 
Pf
Pi
Ti 
15.0 bar
 298 K
1.00 bar
 4470 K
Because the volume of a gas is proportional to amount (mol), we are sometimes faced with
reaction stoichiometry questions that provide volume data for reactant gases.
Example 2.5: H2 and Cl2 react according to
H2(g) + Cl2(g) → 2 HCl(g)
Suppose 3.00 L of hydrogen gas is mixed with 4.00 L of chlorine gas, at 1.00 bar pressure and
298 K, then ignited. How much HCl is produced?
Approach: Determine the amount of the limiting reactant. The reaction stoichiometry then
determines the amount of product.
Under the same conditions of pressure and temperature, a smaller volume means a smaller
amount. Since the gases combine in a 1:1 ratio, the gas with the smaller volume is the
limiting reactant. The limiting reactant is H2 since it has the lesser volume.
20
amount of H 2 consumed = n 
PV
RT
1.00 bar  3.00 L
0.083145 bar mol1K 1  298 K
 0.1211 mol

According to the reaction stoichiometry, the amount of HCl produced is twice this amount;
i.e.,
amount of HCl = 2 × 0.1211 mol = 0.242 mol
2.2 Partial Pressure
When we have a mixture of gases in a vessel, each gas contributes to the total pressure in the
vessel – each gas has its own partial pressure. The total pressure in the vessel is the sum of the
partial pressures of all gases in the vessel. For example, if we have a mixture of gases, A, B, C,
…, the total pressure would be expressed as:
PTOTAL = PA + PB + PC + …
2.2
The partial pressure of any gas in a mixture can be determined if we know the total pressure of
the mixture and the mole fraction of the gas of interest. The mole fraction, x, of any gas in a
mixture is its proportion with respect to the whole quantity. For example, if we have our mixture
of gases A, B and C, and we know that we have 2.0 mol A, 1.5 mol B and 1.5 mol C, then the
mole fraction of gas A is:
xA = 2.0 mol / (2.0 + 1.5 + 1.5) mol = 2.0/5.0 or 0.40
If the total pressure of the mixture under a given set of conditions were 1.4 bar, then the partial
pressure of gas A would be the total pressure multiplied by the mole fraction of gas A:
PA = PTOTAL x xA = (1.4 bar) (0.40) = 0.56 bar
Example 2.6: The vapor pressure of water at 298 K is 24 torr (1 bar = 750 torr). This is the
partial pressure of water in air saturated with water at 298 K. What is the mole fraction of water in
1.00 bar air at 298 K with 100% humidity (i.e., the air is saturated with water)?
Approach: Mole fraction is proportional to partial pressure, and we know the total
pressure – i.e., 1.00 bar.
Total air pressure = 1.00 bar x 750 torr bar−1 = 750 torr
xwater = 24 torr / 750 torr = 0.0320
The mole fraction of water vapor in air with 100% humidity is 0.032.
21
Problems:
2.1 (a) Determine the density of He, N2 and Ar at 1 bar pressure and 298 K.
(b) Which gas can be used to displace air in a cylinder, by pouring the gas into the cylinder?
2.2 What volume (in L) of CO2 is consumed by reaction with 1.00 kg of CaO at 298 K and 1.00
bar pressure?
CaO(s) + CO2(g) → CaCO3(s)
2.3 Hydrochloric acid reacts with aluminum to produce hydrogen gas:
2 Al(s) + 6 HCl(aq) → 2 AlCl3(s) + 3 H2(g)
An impure sample of aluminum contains an unknown amount of Al and other impurities.
Suppose 110 mL of H2, at 1.00 bar and 298 K, are collected from the reaction with HCl of all
the aluminum in the 3.65 g impure sample. What percentage (by mass) of the sample is Al?
Assume that the other components of the sample do not react with HCl.
2.4 Because the reaction in question 2.3 is carried out in aqueous solution, any gas collected
from such a reaction will be a mixture of the product gas – in this case, hydrogen – and water
vapor. Gases collected bubbling out of an aqueous are saturated with water. Thus, the 110
mL of H2 collected in question 2.3 must be the dry volume – i.e., the volume after all of the
water is removed (e.g., with a dessicant). Do question 2.3 again, but this time, suppose that
the 110 mL of gas collected has not been dried – it contains water vapor. You must correct
either the pressure or the volume of H2 in order to get the actual amount (mol) of H2.
2.5 Hydrogen peroxide decomposes into water and oxygen.
2 H2O2(aq) → 2 H2O(l) + O2(g)
What volume of oxygen is liberated from the complete decomposition of 100. mL of a 3.00%
hydrogen peroxide solution, at 1.00 bar and 293 K? The density of the hydrogen peroxide
solution is 1.01 g mL−1.
22
Solutions:
2.1
(a) These three gases are essentially ideal under ordinary conditions. Therefore,
n
P
1.00 bar


 0.04036 mol L1
V RT 0.083145 bar mol 1 K 1  298 K
for all three gases. However, the gases have different molar masses, and
consequently different densities. For example
d = m/V = nM/V
m  He 
nM  He 
n
M  He 
V
V
V
 0.04036 mol L1  4.002 g mol1  0.162 g L1
density  He  
=

Similarly,
density  N 2   0.04089 mol L1  M  N 2 
 0.04036 mol L1  28.013 g mol1  1.13 g L1
and
density  Ar   0.04089 mol L1  M  Ar 
 0.04036 mol L1  39.948 g mol1  1.61 g L1
.
(b) Denser gases can displace less dense gases. Air is mostly nitrogen, with 20%
oxygen (which is a little denser). Only argon can displace air by pouring it into a cylinder.
Nitrogen will not pour through air – it will mix slowly. Helium pours upwards because the
surrounding air quickly displaces this less dense gas, forcing it to the ceiling and beyond.
2.2
Because molar masses are expressed in g mol−1, we must convert kg to g:
1.00 kg = 1.00 × 103 g
Amount of calcium oxide consumed:
n  CaO  
m  CaO 
M  CaO 
=
1.00  103 g
 17.83 mol
56.08 g mol1
From the balanced chemical equation,
CaO(s) + CO2(g) → CaCO3(s),
the amount of CO2 consumed = amount of CaO consumed = 17.83 mol
23
At 298 K and 1.00 bar pressure, this corresponds to the volume of CO2,
nRT 17.83 mol  0.083145 bar mol1K 1  298 K

1.00 bar
P
 429 L
V
2.3
To determine the amount of aluminum reacted, we need to calculate the amount of
hydrogen gas:
PV
1.00 bar  0.110 L

RT 0.083145 bar mol1K 1  298 K
 0.004440 mol
n
From the reaction stoichiometry, we see that three mol of H2 are produced for every 2
mol of Al.
Therefore, the amount of Al = (2/3) × 0.004440 mol = 0.002960 mol.
The mass of Al = 0.002960 mol × 26.9815 g/mol = 0.07986 g.
Since the mass of the sample was 3.65 g, the percentage Al (by mass) is
0.07986 g
100%  2.19%
3.65 g
2.4
We must correct the volume, 110 mL, or pressure, 1.00 bar, of the collected gas to
account for the water contribution. It is easier to correct for pressure since we are given
the partial pressure of water in the collected gas – it is 24 torr or
24 torr / 750 torr bar−1 = 0.0320 bar
The total pressure is 1.00 bar. This is the sum of the partial pressures of hydrogen and
water. Therefore, the partial pressure of hydrogen is
1.00 bar − 0.0320 bar = 0.968 bar
Now, we can follow the solution of question 2.3 with the partial pressure of hydrogen set
to 0.968 bar rather than 1.00 bar. This by-passes determining the corrected collected
volume – i.e., determining the dried volume of collected gas.5
The amount of hydrogen gas:
0.968 bar  0.110 L
PV

RT 0.083145 bar mol1K 1  298 K
 0.00430 mol
n
The amount of Al = (2/3) × 0.00430 mol = 0.00286 mol.
5
The dried volume of product gas is just
0.968 bar
 110 mL  1.06  10 2 mL
1.00 bar
24
The mass of Al = 0.00290 mol × 26.9815 g/mol = 0.0782 g.
Since the mass of the sample was 3.65 g, the percentage Al (by mass) is
0.0782 g
100%  2.1%
3.65 g
2.5
The mass of hydrogen peroxide solution = volume × density
= 100 mL × 1.01 g/mL = 101 g
The solution is 3.00% (be mass) hydrogen peroxide.
Therefore, the mass of hydrogen peroxide in the solution = 0.0300 × 101 g = 3.03 g.
the amount of H 2 O 2 decomposed 
3.03 g
 0.08908 mol
34.015 g/mol
This corresponds to the volume,
nRT 0.08908 mol  0.083145 bar mol1K 1  293 K

P
1.00 bar
 2.207 L .
V
The volume of O2 is one half the volume of H2O2,
V = (2.207 L) / 2 = 1.10 L
Note that all of the above questions make use of the ideal gas law, PV = nRT, simply rearranged,
as required, in order to solve for an unknown quantity.
25
3.
Electronic Structure
3.1 Light
The atomic structure of matter was first proposed because of observations of regularity in the
outcomes of chemical reactions. However, a detailed understanding of this structure – and
corroboration of ideas proposed to explain chemical reactions – was made possible by
experiments involving the interaction of matter with light. Thus, our journey into the structure of
matter begins with the properties of light. Light is an electromagnetic wave which carries energy.
The different colors of light result because these waves exist with different wavelength, and
corresponding frequency. The product of the wavelength, λ, and frequency, ν, equals the speed
of light, c.
c = 
3.1
To see this, consider “watching” a lightwave pass a reference point – as shown in Fig. 3.1. A
lightwave is an electric field wave, and orthogonal magnetic field wave, both orthogonal to the
direction in which the lightwave is traveling. There is a succession of peaks (maxima) and
troughs (minima), in the electric and magnetic fields, in the direction of lightwave propagation.
The frequency of the wave is the number of peaks passing the reference point, per unit time. The
SI unit of frequency is Hz, (1 Hz = 1 s‒1). The wavelength is the distance between peaks.
The peaks passing the reference point per unit time (i.e., the frequency) multiplied by the distance
between peaks (i.e., the wavelength) equals the distance traveled by the wave per unit time – i.e.,
the speed.
speed = wavelength × frequency
wavelength
direction of
propagation
Figure 3.1 Waves pass a reference point (the vertical dash on the left) with the same
speed, regardless of wavelength. Few waves pass per second for long wavelength
waves (top panel). These are low frequency waves. Many waves pass per second for
short wavelength waves (bottom panel). These are high frequency waves. The bottom
panel shows a wave with twice the frequency, and half the wavelength, of the wave
depicted in the top panel.
26
The speed of light in a vacuum, c = 2.9979×108 m s‒1, is the same for all wavelengths.
Consequently, wavelength and frequency have a simple reciprocal relationship. Specifically,

c

and

c

.
Example 3.1: Determine the range of frequencies that corresponds to visible light, given that the
range of wavelengths is 390. nm (violet) to 700. nm (red), where 1 nm (1 nanometer) = 10‒9 m.
Approach: Determine the frequencies that correspond to the given wavelengths using
Eq. 3.1. These are the frequency limits of the range of visible light.
Using Eq. 3.1, we have

c


2.9979 108 m s 1
 7.69 1014 s 1  769 THz
9
390 10 m
and
2.9979 108 m s1
 4.28 1014 s 1  428 THz .
 
9
700 10 m

c
Note that 1 THz (1 teraHertz) = 1012 Hz. Table 10.1 provides the most common SI unit
prefixes.
The range of visible frequencies is about 430 THZ (red) to 770 THz (violet).
Light with wavelengths (frequencies) outside the visible range includes infrared, microwave and
radio, at lower frequencies than red light, and ultraviolet, X-ray and gamma ray, at higher
frequencies – see Fig. 3.2.
wavelength, , in m
100
10-2
1
radio
TV
10-4
microwave
10-6
infrared
10-8
ultraviolet
10-10
X-ray
10-12
gamma ray
10-14
cosmic ray
visible
3x106
3x108
3x1010
3x1012
3x1014
3x1016
3x1018
3x1020
3x1022
frequency, , in Hz
red yellow green blue violet
Figure 3.2 The electromagnetic spectrum, showing wavelengths and corresponding
frequencies for each band of frequencies. The visible is almost exactly in the center of
the spectrum, on a logarithmic scale.
27
Light energy is quantized. The smallest increment of light energy is the energy of a single particle
of light, a photon. When light is absorbed by a material, the material gains the energy of the
photon.
Evidence for the particle character of light, together with its wave properties, comes from Planck’s
explanation of blackbody radiation – radiation in thermal equilibrium with a body with fixed
temperature. The classical wave theory of light predicts that the energy content of the radiation,
in each frequency band, increases without bound as frequency increases. Experimental
observations show that the energy content of blackbody radiation actually decreases
exponentially at high frequency. This failure of the accepted theory of Planck’s time is called the
ultraviolet catastrophe.
By positing that light consists of particles – in particular, photons with energy
Ephoton  h
3.2
,
where h = 6.62606957×10‒34 J s is called Planck’s constant – Planck showed that the energy
content of the radiation decreases exponentially at high frequency. In fact, the distribution of
energy derived by Planck agrees with experimental observations at all frequencies. The
increasing energy of individual high frequency photons prevents energy from accumulating in
these modes, thereby averting the ultraviolet catastrophe.
T = 600 K
500
400
300
200
T = 300 K
100
0
20
40
60
80

100
120
140
160
180
200
Figure 3.3 Blackbody radiation. The distribution of power emitted, per unit frequency
and unit surface area, is shown for blackbodies at temperature 300 K and 600 K, as
labeled. The horizontal axis gives frequency, in units of THz = 1012 s‒1. The units for the
vertical axis are pJ m‒2 = 10‒12 J m‒2.
Other evidence for the particle character of light (and that of matter) comes from Einstein’s
explanation of the photoelectric effect.
3.2 Photoelectric Effect
When light of sufficiently high frequency shines on a metal, electrons are ejected from the surface
of the metal. The kinetic energy of the ejected electrons equals the photon energy minus the
energy needed to eject the electron, which follows from energy conservation;
1
Ephoton  h  mu 2  metal
2
28
,
3.3
where m and u are the electron mass and velocity, respectively, and
metal is the work function of
the metal. The work function is the depth of the energy well occupied by an electron on the
surface of the metal – it is specific to each type of metal. Alkali metals have the smallest work
functions. This means it takes less energy to eject an electron from alkali metals. Ordered
according to decreasing work function, lithium > sodium > potassium > rubidium > cesium.
ammeter
measures current
0
1 2 34
5
an electron, e-
a photon, Ephoton = h 
variable
voltage
kinetic energy of
ejected electrons
Figure 3.4. The photoelectric effect. Two electrodes are separated by a vacuum. A
photon of sufficiently high frequency (and hence energy) is absorbed by an electron on
the surface of the metal of the top electrode. The electron is ejected and flies across to
the bottom electrode. Such events are recorded via current measurement using an
ammeter. The measured current is proportional to the intensity of the light (the number of
photons), except that if the frequency of the light is too low there is no current. The
variable voltage control lets us turn off the current by applying an opposing voltage. The
voltage at which the current stops determines the kinetic energy of the ejected electrons.
cesium
sodium
507 551
iron
1086
frequency, / THz
Figure 3.5 The kinetic energy of ejected electrons measured in the photoelectric
experiment for iron, sodium and cesium. Each metal has a characteristic threshold
frequency,  thres  metal / h . Beyond that frequency, electrons are ejected with kinetic
energy in accord with Eq. 3.3. Light with frequency below the threshold value does not
eject electrons – no photoelectric effect is seen in this case.
29
Example 3.2: The work function for calcium is 4.64×10‒19 J (or 279 kJ mol‒1 – this is the energy
needed to eject 1 mol of electrons).
(a) What is the threshold frequency above which calcium will exhibit the photoelectric effect?
(b) What is the kinetic energy of an electron ejected from calcium by a photon of wavelength,
300. nm?
Approach: (a) The threshold frequency corresponds to a photon with just enough
energy to eject the electron. Set the work function equal to the photon energy, then solve
for frequency. (b) Determine the photon energy using Eqs. 3.1 and 3.2. Substitute this
energy, and the work function for calcium, into Eq. 3.3 and solve for the kinetic energy of
the ejected electron.
(a) The threshold photon energy equals the work function,
h thres   calcium  4.64  1019 J
Therefore,
 thres 
calcium
h
4.64 1019 J

 700 THz
6.626 1034 J s
(b) The frequency of the incident light is

c


2.9979 108 m s 1
 999.3 THz
300 109 m
The kinetic energy of the ejected electron is
1
mu 2  h  calcium
2
 6.626  1034 J s  999  1012 s 1  4.64  1019 J
Ekin 
 1.98  1019 J .
3.3 Quantum Mechanics and the Hydrogen Atom
We have seen that electrons can absorb the energy of one6 photon of light, and be ejected from
the surface of a metal. In general, a material can absorb a photon of light, resulting in the
excitation of the material to a higher energy level – the increase in energy equals the photon
energy. The energy levels of a material are quantized – i.e., they are a discrete set of energy
values, as opposed to the continuously varying energy of classical mechanics. The observed
energy levels in materials can be computed using quantum mechanics. The agreement of such
computations with experimental measurements provides unprecedented evidence for a physical
theory. Quantum mechanics explains the microscopic world wherein particles (e.g., electrons)
have wave properties, and waves (e.g., light) have particle properties. In doing so, it also
explains the macroscopic world, as macroscopic properties are just the manifestation of the
properties of very many (interacting) microscopic systems (atoms).
6
Only at very high light intensity can an electron absorb the energy of two (or more) photons
simultaneously.
30
The measurement of the spectra of atoms in the gas phase provided early evidence for quantum
mechanics. A spectrum is a graph of absorption (or emission) intensity versus the frequency of
incident light.
Because the energy levels of an atom are discrete, only certain frequencies of light can be
absorbed (or emitted) – specifically, only photons with energies corresponding to the difference
between two energy levels of the atom.
The simplest set of energy levels is that of the hydrogen atom. This is because the hydrogen
atom has only one electron. Quantum mechanics predicts energy levels given by the following
formula
En  
hcRH
, n  1, 2,3,
n2
3.4
where RH = 1.0974×107 m‒1 is called the Rydberg constant, and n is called the principal
quantum number. The difference between two of these energies determines a transition energy
equal to the energy of the photon emitted or absorbed. The associated transition frequency is the
frequency of the photons emitted or absorbed – it is given by dividing the transition energy by h.
The set of transition frequencies, and corresponding wavelengths, predicted using Eq. 3.4
matches the observed spectrum for hydrogen atoms. Specifically, since the energy difference
must match the photon energy,
1 
 1
E  En '  En   hcRH  2  2   h , n '  n .
 n' n 
3.5
The wavelength associated with this transition is
1
  1
1 
    RH  2  2   .

  n n ' 
c
This equation, found empirically, is the origin of the Rydberg constant.
Example 3.3: Transitions of hydrogen.
(a) Determine the frequency of the emission lines resulting when an excited state with n’ = 3
drops to the (i) n = 2 , or (ii) n = 1 level. Are either of these emissions in the visible range?
(b) Of all the absorption frequencies associated with transitions between all levels up to n = 4,
which transitions have the largest and the smallest frequencies?
Approach: (a) Use Eq. 3.5, divided by h, to determine frequency,  . Note that
frequency is always positive. Always arrange the n and n’ values to get a positive result.
The sign of the difference in energies determines whether it is an emission or absorption
process. (b) Because the spacings between successive energy levels decrease rapidly
with increasing n, it is typically the smaller of n and n’ (call it n) that determines the size of
the transition frequency. This is true for all n’, when n is 1, 2 or 3. The largest transition
frequencies correspond to the smallest n value. For two transitions with the same n
value, larger n’ produces a larger the transition frequency. Note that a calculator is not
required to answer this question.
31
(a) From Eq. 3.5,
1 
 1
 2
2
 n n' 
  cRH 
 1 1
 2.9979  108 m s 1  1.0974 107 m 1   2  2 
2 3 
 457 THz
for n = 2. For n = 1, we have   2924 THz . The first of these absorption lines is in the
visible range. It corresponds to violet-blue light. In general, all transitions with n equal to
2 produce an absorption or emission line in the visible range. This series of lines is
called the Balmer series. If n is 1, then the line is the UV range. This series of lines is
called the Lyman series. Other named series are the Paschen (n = 3) and the Brackett
(n = 4) series, both in the infrared range.
(b) The largest transition frequency corresponds to the n = 1 to n’ = 4 absorption
transition - or the reverse, the n’ = 4 to n = 1 emission transition. The lowest frequency
corresponds to n = 3 to n’ = 4 absorption or corresponding emission transition.
3.4 Orbitals of Hydrogen
Quantum mechanics shows that the energy levels given in Eq. 3.4 do not determine unique
states of the hydrogen atom. In general, there is more than one state per energy level. There
are additional quantum numbers associated with each unique state of hydrogen – i.e., in addition
to principal quantum number, n. The additional quantum numbers result because of the spherical
symmetry of the hydrogen atom, and because electrons have an additional property called spin.
The electron in hydrogen has an orbital angular momentum associated with its orbital motion
about the proton nucleus. The angular momentum has a magnitude, with associated quantum
number, ℓ = 0, 1, 2, …, n ‒1 , and an orientation, with associated quantum number, mℓ = ‒ℓ,
‒ℓ +1, …, ℓ ‒1, ℓ. Quantum number ℓ is called the angular momentum quantum number, while mℓ
is the magnetic quantum number (its existence is detected by applying a magnetic field).
Specifying the (n, ℓ, mℓ) values determines an orbital of the hydrogen atom, which determines the
state of the electron with respect to its motion about the nucleus.
For n = 1, there is only one ℓ value, ℓ = 0. For ℓ = 0, there is only one mℓ value, mℓ = 0.
Therefore, the n = 1 energy level has only one orbital – the 1s orbital.
The n = 2 energy level has four orbitals specified by the quantum numbers, (n, ℓ, mℓ) = (2, 0, 0)
(the 2s orbital), (2, 1, ‒1), (2, 1, 0) and (2, 1, 1) (the three 2p orbitals).
Orbitals are labeled by their principal and angular momentum quantum numbers, n and ℓ. The ℓ
value is specified by a letter (s for ℓ = 0, p for ℓ = 1, d for ℓ = 2, f for ℓ = 3, g for ℓ = 4, …).
In addition to the orbital quantum numbers already introduced, the electron has a spin quantum
number, ms = ½ or ‒½, distinguishing the orientation of the spin angular momentum – pointing
up or down, respectively.
There are two states associated with the lowest energy level: (n, ℓ, mℓ, ms) = (1, 0, 0, ½) and (1,
0, 0, ‒½). Thus, the ground state of hydrogen is actually two states differing only in the
orientation of electron spin. All other states of hydrogen are excited states. If excited states are
32
produced, they lose energy by emitting photons until the ground state is reached. Thus, in the
absence of radiation exciting the hydrogen atoms, they will be in their ground state.
In general, because of spin, there are two states for every orbital. Hydrogen orbitals can be
arranged according to energy and angular momentum as shown in Fig. 3.6.
Energy

E  0
5
E5   hcRH / 52
4
E4   hcRH / 4 2
3
E3   hcRH / 32
2
E2   hcRH / 2 2
1
d orbitals
p orbitals
4s
4p
4d
3s
3p
3d
2s
2p
4f
1s
E1  hcRH
s orbitals
Figure 3.6. The energy levels of hydrogen, and associated orbitals. The orbitals are
arranged with energy on the vertical axis and angular momentum increasing (from one
block of orbitals to the next) on the horizontal axis. The inset shows the corresponding
arrangement of the (n, ℓ) labels. Each label corresponds to a block of 2 ℓ + 1 orbitals.
For the hydrogen atom, the orbital energy depends only on the principal quantum number.
Therefore, the number of orbitals with energy En is
1 + 3 + 5 + … + 2 n ‒ 1 = n2.
Table 3.1 shows the numbers of orbitals and states for n = 1, 2, 3 and 4.
Table 3.1. Numbers of orbitals and states of hydrogen
n
ℓ
1
0
2
0
1
0
1
2
0
1
2
3
3
4
Number of Orbitals
total
1
1
1
3
1
3
5
1
3
5
7
4
9
16
33
Number of States
total
2
2
2
6
2
6
10
2
6
10
14
8
18
32
Example 3.4: Spin and orbital quantum numbers
(a) Which of the following sets of quantum numbers, (n, ℓ, mℓ, ms), corresponds to an actual state
of a hydrogen atom?
(i) (1, 0, 1, ½)
(ii) (2, 1, ‒1, ‒½)
(iii) (2, 2, 0, ‒½)
(iv) (3, 1, ‒1, 0)
(v) (3, ‒2, ‒2, ½)
(b) How many orbitals are there with energy up to and including E4? How many electron states
are there with energy up to this value?
Approach: (a) For these quantum numbers to represent an actual state of the hydrogen
atom, they must satisfy the restrictions:
0 < n
0 ≤ ℓ ≤ n–1
–ℓ ≤ mℓ ≤ ℓ
ms = ±½
(b) There are n2 orbitals with energy En. We must sum n2 from n = 0 to 4. There are
two electron states (different spin states) for every orbital.
(a) Only (2, 1, ‒1, ‒½) corresponds to an actual state of hydrogen.
(1, 0, 1, ½) violates mℓ ≤ ℓ.
(2, 2, 0, ‒½) violates ℓ ≤ n – 1
(3, 1, ‒1, 0) violates ms = ±½
(3, ‒2, ‒2, ½) violates 0 ≤ ℓ
(b) There are 1 + 4 + 9 + 16 = 30 orbitals with energy up to and including E4. There
are 2×30 = 60 electron states up to this energy.
We have introduced orbitals as states of an electron in a hydrogen atom. According to quantum
mechanics, the states of an electron are wavefunctions. They are oscillatory functions of position
about the nucleus of the atom. They can take on positive or negative values – in fact, they can
even be complex valued (i.e., with a 1 term). Wavefunctions can be added or subtracted to
form new wavefunctions. Thus, both constructive and destructive interference is possible. When
an electron is in a state described by a wavefunction, it does not have a well defined position. If
the position of the electron is measured, it might be found (almost) anywhere. However, some
positions are much more likely than others. The square of the wavefunction at each position
gives the likelihood of finding the electron in that location. Where the magnitude of the
wavefunction is largest is where the electron is likely to be.
The orbitals of hydrogen are wavefunctions concentrated about the nucleus. Orbitals with higher
energy - i.e., larger n - extend further from the nucleus, and exhibit more oscillations. The shape
of the orbital is determined by the angular momentum – i.e., the ℓ quantum number. It is
customary to plot closed surfaces such that the likelihood of finding the electron within the surface
is 90%, for example. The shape of such a surface is the shape of the orbital. While the energy
depends on the principal quantum number, n, the shape of the orbital depends on angular
momentum quantum number, ℓ.
For ℓ = 0, orbitals are spherically shaped. All directions away from the nucleus are equivalent.
Only the distance from the nucleus is important. The 1s orbital is like a spherical ball. The 2s
orbital has a single oscillation in the radial direction. It consists of a small inner sphere
surrounded by a large outer spherical layer. Between these radial lobes is a nodal sphere where
the wavefunction is zero. The 3s orbital has two nodal spheres. The ns orbital has n – 1 nodal
spheres.
34
For ℓ = 1, orbitals have a single oscillation upon moving around the nucleus, in a circle with fixed
distance from the nucleus. This gives rise to a barbell shape with two lobes – positive and
negative – pointing in opposite directions along an axis (see Fig. 3.6). These lobes are separated
by a nodal plane where the wavefunction is zero. There are three np orbitals. The mℓ = 0 orbital
is called npz orbital. It is concentrated about the z axis. The xy plane is the corresponding nodal
plane. For mℓ = ±1, the orbitals are complex valued. However, since these orbitals have
identical properties – except in the presence of a magnetic field – it is admissible to form + and –
combinations that are real. These are npx and npy. They are exactly like npz, except that they
are aligned with the x and y axes, respectively. Like s orbitals, p orbital have nodal spheres. The
np orbital has n – 2 nodal spheres.
For ℓ = 2, orbitals have two oscillations upon moving around the nucleus, giving rise to a double
barbell or cloverleaf pattern with four lobes – alternately positive and negative. Each d orbital has
two nodal planes where the wavefunction is zero. There are five (2ℓ+1) nd orbitals. The mℓ = 0
orbital is called nd z 2 . It is concentrated about the z axis, with a doughnut-shaped lobe in the xy
nd xz and nd yz orbitals, while the mℓ = ±2
plane. The mℓ = ±1 orbitals are combined to make
orbitals are combined to make nd xy and
nd x2  y2 orbitals.
The orbital shapes are shown in Fig. 3.7.
z
z
z
x
d x2  y 2
y
+
z
+
-
x
py
-
d xy
+
+
x
-
y
+
y
d z2
x
+
z
y
z
y
z
+
-
x
+
y
-
x
y
-
+
d yz
pz
z
x
+
z
s
+
+
-
px
x
d xz
-
+
+
-
x
y
y
Figure 3.7. The shapes of orbitals.
Table 3.2. The shapes of orbitals
Type (ℓ)
s (ℓ = 0)
p (ℓ = 1)
Shape
sphere
barbell
d (ℓ = 2)
mℓ ≠ 0
mℓ = 0
35
double barbell
d xy , d x2  y2 , d xz , d yz
barbell with doughnut about its axis
d z2
3.5 The Periodic Table
The energy levels of hydrogen are given by Eq. 3.4. This equation also applies to He+, Li2+, …,
since these species have only one electron. However, since the charge on the nucleus of these
species is different (+Ze, rather than +e, the proton charge, as it is for hydrogen), Eq. 3.4 is
modified as follows:
En  
Z 2 hcRH
, n  1, 2,3,
n2
3.6
For example, the energy levels of He+ are four times deeper than those of H because the nuclear
charge is doubled – the Coulomb potential is doubled and the electron is brought closer to the
nucleus.
For atoms and ions with more than one electron – i.e., almost all atoms and ions – the atomic
states and associated energies are calculated on computers. This is because of electron
repulsion. If the electrons did not interact with each other, then each electron in a many electron
atom would be in a hydrogen-like orbital – an orbital of hydrogen, shrunken into the nucleus, and
with lower energy because of the +Ze nuclear charge. Electron repulsion modifies this picture.
However, much insight arises from the approximation wherein electrons are treated without
electron repulsion, and then the positive repulsion energy is added to the negative energies
obtained by ignoring electron repulsion. In this approximation, the electrons are treated
independently – each electron occupies a hydrogen-like orbital. Electron repulsion is added as a
correction to the total energy.
Using the approximation described above, the ground states of successive atoms corresponds to
the successive addition of all the electrons to hydrogen-like orbitals, two per orbital to account for
spin, starting with the lowest energy orbital – the 1s orbital – and going up in energy as the
orbitals become filled. Orbitals fill in accord with the Pauli exclusion principle: two electrons with
the same spin cannot occupy the same orbital. This is a fundamental property of fermions, and
electrons are fermions.
The electron configuration is the set of occupied orbitals. The order in which the orbitals are
occupied is determined by their relative energy. In the hydrogen atom, 2s and 2p orbitals have
the same energy. However, when there is electron repulsion, these orbitals do not have the
same energy. If there is an electron in the 1s orbital, then an electron in 2s or 2p is screened
from the nucleus – the repulsion from the 1s electron cancels some of the attraction to the
nucleus. However, an electron in the 2p orbital experiences greater screening. The 2s orbital
has an inner spherical lobe that is close to the nucleus and is not as well screened – this reduces
the repulsion energy. The 2p orbital has a larger positive repulsion energy, and consequently, a
higher energy than 2s when there is more than one electron. The electron configurations of H,
He and Li are as follows (see Fig. 3.8):
H:
He:
Li:
1s1
1s2
1s22s1
36
Energy
2s
2p
2p
2s
2p
1s
2s
H ground state
1s
He ground state
1s
Li ground state
Figure 3.8. Ground state electron configurations of H, He and Li, expressed on a vertical
energy scale with arrows depicting electrons occupying orbitals. The arrow direction
specifies the spin orientation. Note that 2s is lower than 2p for He and Li.
The electron configurations of Be, B and C are as follows (see Fig. 3.9):
Be:
B:
C:
1s22s2
1s22s22p1
1s22s22p2
37
Energy
2s
2p
2p
2s
2p
2s
1s
Be ground state
1s
B ground state
1s
C ground state
Figure 3.9. Ground state electron configurations of Be, B and C. Note that the two 2p
electrons in carbon occupy different orbitals, with the same spin.
In the case of carbon, there are two 2p electrons. Repulsion between these electrons is
minimized if the electrons are in different orbitals (this is possible because there are three orbitals
with the same energy), and if they have the same spin. Electrons in different orbitals will be
further apart, on average, than electrons in the same orbital. Electrons with the same spin simply
avoid each other and thereby minimize repulsion. The Pauli exclusion principle is just an
example of this avoidance. Two identical fermions intrinsically avoid each other. In general,
electrons in the same subshell – i.e., same n and ℓ values – spread out into different orbitals, as
much as possible, with the same spin, as much as possible. This is called Hund’s rule.
The electron configurations of all the atoms in their ground state can be determined with the
above picture. This is called the Aufbau principle (Aufbau is German for “building up”). The order
in which the orbitals are filled, determined by the orbital energies in ground state atoms, is given
by following the arrows in Fig. 3.10 – starting with the top left arrow then proceeding down to the
right. This gives the following orbital filling order:
1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s
Since, except for the s orbitals, these labels correspond to more than one orbital with the same
energy, these sets of orbitals are called subshells. A full set of subshells for given principal
quantum number, n, is called a shell.
38
1s
2s
2p
3s
3p
3d
4s
4p
4d
5s
5p
5d ...
6s
6p
...
7s
...
4f
Figure 3.10. The Aufbau principle orbital filling order for atomic ground state electron
configurations.
Example 3.5: Ground state electron configurations of atoms
(a) Determine ground state electron configurations for the following atoms. Use the noble gas
shorthand notation (except in case of Ne). The electron configuration of Ar determines the core
of fourth row elements. The electron configuration of fourth row elements is then written as [Ar]
followed by the valence configuration.
(i) N
(ii) Ne
(iii) P
(iv) Ca
(v) V
(b) To which element does the following atomic ground state electron configuration correspond?
1s22s22p63s23p64s23d7
Approach: (a) The number of electrons equals the atomic number. Fill up the orbitals
using the Aufbau principle, or use the periodic table which is arranged in terms of
subshell blocks to jump to the answer (or at least check your answer). (b) Count the
electrons. Since it is an atom (i.e., not an ion), this is just the atomic number.
(a)
N:
Ne:
P:
Ca:
V:
1s22s22p3
1s22s22p6
[Ne] 3s23p3
[Ar] 4s2
[Ar] 4s23d3
(b) There are 27 electrons in this atom. Therefore, it is cobalt (Z = 27).
Chromium and copper are the two exceptions to the orbital filling order of Fig. 3.10, that occur in
the first 40 elements. The electron configurations of chromium and copper are as follows ( noble
gas shorthand notation to the right):
Cr:
Cu:
1s22s22p63s23p64s13d5
1s22s22p63s23p64s13d10
[Ar] 4s13d5
[Ar] 4s13d10
There is a special stability associated with filled (copper, 3d10) and half-filled (chromium, 3d5)
subshells. This is a subtle effect explained in upper level chemistry texts.
39
The patterns in the electron configuration of the atoms give rise to the block structure of the
periodic stable. The periodic table is arranged in blocks according to the subshell of the highest
energy electrons. Elements in the same column, group, have the same number of valence
electrons – the number of electrons in the outer shell. Because of the orbital filling order, the
shells (for many electron atoms) are reshuffled to coincide with the rows in the block structure of
the periodic table shown in Fig. 3.11.
s block f block d block p block
1s
2s
3s
4s
5s
6s
7s
4f
5f
2p
3p
4p
5p
6p
7p
3d
4d
5d
6d
Figure 3.11. The periodic table block structure. The 1s block – i.e., H and He – is
typically split so that He appears on the far right above the other noble gases.
Elements in the same group have similar physical and chemical properties, as it is the highest
energy electrons in the atom that primarily determine its properties.
Problems
3.1 Magnesium exhibits a photoelectric effect.
(a) When photons of wavelength, 200. nm, are incident on magnesium, electrons are ejected
with velocity, 942 km s‒1. What is the work function of magnesium in kJ mol‒1? The
mass of an electron is 9.109×10‒31 kg.
(b) What is the longest wavelength of light that can eject electrons from magnesium?
3.2 An excited state of hydrogen emits photons with various wavelengths. One of the emission
lines is at 486 nm.
(a) What are the principal quantum numbers of the excited state, and the state to which it
decays, when it emits 486 nm photons?
(b) How many of the other emission lines arising from this excited state have wavelength(s)
longer (or shorter) than 486 nm?
3.3 List the (ℓ, mℓ) quantum numbers associated with the n = 3 shell. Group the quantum
numbers in labeled subshells.
3.4 Electron configurations that do not follow the Aufbau principle are excited states. Label each
of the following electron configurations as the ground or excited state of an element.
(a) 1s22s22p63s13p3
(c) 1s22s22p64s1
(b) 1s22s22p63s23p64s1
(d) 1s22s22p63s23p63d1
40
Solutions
3.1 (a) The kinetic energy of the ejected electron is
2
1 2 9.109 1031 kg
mu 
  942 103 m s 1 
2
2
19
 4.041 10 J
Ekin 
The incident photon energy is
6.626 1034 J s  2.998 108 m s 1
Ephoton  h 

200 109 m

 9.932 1019 J
hc
The work function is the difference between these energies:
magnesium  Ephoton  Ekin  9.932 1019  4.041 1019 J
 5.89 1019 J
The work function per mole is just this number × Avogadro’s number:
 magnesium  N A magnesium
 6.022 1023 mol1  5.89 1019 J
 355 kJ mol1
(b) The photon, with the longest wavelength that can eject electrons from magnesium, has
energy equal to the work function of magnesium (not the work function per mole).
Therefore,
Ephoton  5.89 1019 J  h 
and so
hc

hc
6.626 1034 J s  2.998 108 s1

5.89 1019 J
5.89 1019 J
 337 nm

3.2 (a) The photon energy equals the energy difference between the excited state and final
state, with principal quantum numbers, n’ and n, respectively.
1 
hc
 1
E  hcRH  2  2   h 

 n n' 
or
1 
1
1
 1

 0.1875
 2  2
1
9
7
 n n '  RH  1.0974 10 m  486 10 m
41
First, this tells us that n cannot be more than 2. This is because the left side of this equation
must be less than 1/n2 (the second term is negative) and 1/32 = 1/9 = 0.11 which is less than
0.187. There are two cases to consider: n = 1 and n = 2:
If we set n = 1, we can solve for n’:
1
1
 0.1875
n '2
or
n '  1.11
This cannot be correct. The initial excited state principal quantum number must be an integer
greater than 1.
If we set n = 2, the result is
1
1
 2  0.1875
2
2 n'
or
n'  4
Thus, the 486 nm transition occurs upon transition from n’ = 4 to n = 2.
Note that, as seen in Example 3.3, emission in the visible range occurs only for final n, n = 2.
Once we have n = 2, we can solve for n′, as above.
(b) The n’ = 4 excited state can also decay to n = 1 or n = 3. The transition to n = 1 produces
a larger energy photon – a shorter wavelength – while the transition to to n = 1 produces a
smaller energy photon – a longer wavelength. Thus, there is one emission line with shorter
wavelength, and one with longer wavelength.
3.3 The (ℓ, mℓ) quantum numbers associated with the n = 3 shell are listed (and grouped
according to subshell label) as follows:
(0,0)
3s
(1, ‒1) (1, 0) (1, 1)
3p
(2, ‒2) (2, ‒1) (2, 0) (2, 1) (2, 2)
3d
3.4 (a) 1s22s22p63s13p3
Here, there are electrons in 3p with an unfilled lower energy orbital – the 3s orbital.
Therefore, this is an excited state electron configuration. It is an excited state of Si – there
are 14 electrons in the neutral atom.
(b) 1s22s22p63s23p64s1
This is a ground state electron configuration – the ground state of K (Z = 19).
(c) 1s22s22p64s1
Here, there is an electron in 4s with an entire unfilled shell – the 3 shell. Therefore, this is an
excited state electron configuration. It is an excited state of Na (Z = 11).
(d) 1s22s22p63s23p63d1
42
Here, there are electrons in 3d with an unfilled lower energy orbital – the 4s orbital.
Therefore, this is an excited state electron configuration. It is an excited state of K (Z = 19).
43
4.
Periodic Properties
4.1 Atomic/Ionic Radius
The size of an atom is determined by its shell structure. Atoms with the same number of valence
electrons – i.e., atoms in the same group – increase in size with increasing number of shells.
Size of an atom is characterized by its atomic radius. Thus, atomic radius increases going down
a group. For example, with respect to atomic radius,
Li < Na < K < Rb < Cs.
Moving across a row of the periodic table, atoms have the same number of shells. They differ in
the number of valence electrons, and in the charge on the nucleus. The valence electrons are
screened by inner shell electrons. The number of inner shell electrons is the same for all atoms
in a row. Since nuclear charge increases across a row, with the number of screening electrons
fixed, the effective nuclear charge felt by the valence electrons increases across the row.
The effective nuclear charge is approximated by the core charge, the charge of the nucleus plus
the (negative) charge on the inner shell electrons. This approximation neglects the small
screening of valence electrons by other valence electrons, and treats the screening by inner shell
electrons as perfect – i.e., each inner shell electrons cancels +e from the nuclear charge.
The core charge increases by 1 (in units of proton charge) for each successive element in the
row. The increasing core charge pulls the valence electrons closer to the nucleus making the
atom smaller. Thus, atomic radius decreases across a row – from left to right. For example, with
respect to atomic radius,
Li > Be > B > C > N > O > F > Ne.
The trends in atomic radius are summarized in Fig. 4.1.
Atomic radius increases down and to the left
Figure 4.1. Periodic table trend in atomic radius
There are two additional rules regarding ionic radius, which characterizes the size of an ion.
(1) The radius of the ions of an element – i.e., nuclear charge is fixed – increases with increasing
number of electrons. Thus,
… < X3+ < X2+ < X+ < X < X‒ < X2‒ < X3‒ < …
(2) The radius of ions, or atoms, with the same number of electrons - isoelectronic species –
increases with decreasing nuclear charge. For example,
Ca2+ < K+ < Ar < Cl‒ < S2‒.
44
There are exceptions to the trends in atomic radius depicted in Fig. 4.1, among heavier elements
– especially among d and f block elements. Our concern is only with the general trends in atomic
and ionic radius, as described above.
Example 4.1: Order according to radius
(a) Order the following elements according to increasing atomic radius:
Cs
F
N
He
Ga
As
(b) Order the following ions/atoms according to increasing radius:
(i) O2‒
(ii) Se2‒
Mg2+
Ne
Na+
F‒
Br‒
Kr+
Br
Rb2+
Approach: (a) Look for the element closest to the upper right corner of the periodic
table, then find the find the next such element and so on. (b) Look for species with the
same number of electrons with different nuclear charge, or the same nuclear charge with
different numbers of electrons.
(a) He is the highest and rightmost element. The He atoms are the smallest. The next
such element is F, then N, As, Ga and Cs – i.e.,
He < F < N < As < Ga < Cs.
(b) (i) These are isoelectronic species. The smallest radius occurs for the highest
atomic number. Therefore,
Mg2+ < Na+ < Ne < F‒ < O2‒
(ii) Br‒ and Se2‒ are isoelectronic. Therefore,
Br‒ < Se2‒.
Br, Kr+ and Rb2+ are also isoelectronic. Therefore,
Rb2+
< Kr+ < Br.
Since, Br < Br‒, we have
Rb2+ < Kr+ < Br < Br‒ < Se2‒.
45
Table 4.1. Atomic Radii of Main Group Elements (in pm = 10‒12 m)
H
32
Li
130
Na
160
K
200
Rb
215
Cs
238
Be
99
Mg
140
Ca
174
Sr
190
Ba
206
B
84
Al
124
Ga
123
In
142
Tl
144
C
75
Si
114
Ge
120
Sn
140
Pb
145
N
71
P
109
As
120
Sb
140
Bi
150
O
64
S
104
Se
118
Te
137
Po
142
He
37
Ne
62
Ar
101
Kr
116
Xe
136
Rn
146
F
60
Cl
100
Br
117
I
136
At
148
Table 4.1 shows the atomic radii of main group elements. These values are such the the sum of
two of these atomic radii corresponds to the length of a single bond between the atoms. Values
for the three lightest noble gas atoms are based on computation, as these elements do not form
bonds. Bolded elements satisfy the general trend described above. The others do not – the first
three noble gases and three of the period six elements. From CRC Handbook of Chemistry and
Physics, 93rd Ed..
Table 4.2. Ionic Radii of Main Group Elements (in pm)
Li+
76
Na+
102
K+
138
Rb+
152
Be2+
45
Mg2+
72
Ca2+
100
Sr2+
118
Al3+
54
Ga3+
62
In3+
80
N3‒
171
P3‒
212
Sn4+
69
Sb3+
76
O2‒
140
S2‒
184
Se2‒
198
Te2‒
221
F‒
133
Cl‒
181
Br‒
196
I‒
220
Table 4.2 shows the ionic radii of main group elements. These values are such that the distance
between ions in an ionic solid is the sum of the associated ionic radii. Elements not shown do not
form ionic solids. These values correspond to ionic solids where the number of nearest
neighbors of each ion is six. From CRC Handbook of Chemistry and Physics, 93rd Ed.
4.2 Ionization Energy
The first ionization energy, I1, is the energy needed to eject an electron from an atom in the gas
phase. It gives the binding strength, to the atom, of the highest energy electron(s). It is the
change in energy for the process,
X(g) → X+(g) + e‒.
The second ionization energy, I2, of X is the ionization energy of the cation, X+ - i.e., the energy
change in the process,
X+(g) → X2+(g) + e‒.
Successive ionization energies are similarly defined.
First ionization energy increases, moving from left to right across the periodic table, as the
highest energy electron (the electron ejected) is in the same shell for these elements – while the
46
effective nuclear charge increases from left to right. A larger effective nuclear charge pulls the
valence electrons closer to the nucleus, holding them more tightly, resulting in larger ionization
energy.
First ionization energy decreases moving down a group. This is because the electrons ejected
upon ionization of each successive element come from successive shells. With each successive
outer shell, the valence electrons become more distant from the nucleus, and are less tightly
held. For example, with respect to ionization energy,
Li > Na > K > Rb > Cs.
The general trends in ionization energy are summarized in Fig. 4.2.
First ionization energy increases up and to the right
Figure 4.2. Periodic table trend in first ionization energy – the general trend without filled
and half-filled subshell effects.
first ionization energy (kJ/mol)
Scanning the periodic table from lightest to heaviest element, ionization energy decreases except
for large drops when a new valence shell is encountered – this produces the decreasing
downward trend – and small drops when a new subshell or the second half of a subshell is
encountered. The latter effects are contrary to the general trend depicted in Fig. 4.2. Thus, first
ionization energy actually increases in a zig-zag pattern, with local peaks in ionization energy at
elements with a filled (s, p, d, …) or half-filled (p, d, …) valence subshells. The second row first
ionization energies are shown in Fig. 4.3.
2000
Ne
1000
500
F
N
1500
Be
B
Li
3
O
C
4
5
6
7
8 9
10
atomic number
Figure 4.3. First ionization energies of the second row elements – showing filled and
half-filled subshell effects.
47
The filled subshell effect, wherein the first ionization energy drops from Be to B, results because
the electron ejected from Be is in the 2s subshell whereas the electron ejected from B is in the
higher energy 2p subshell.
In the case of N, the electron is ejected from a half-filled 2p subshell, 2p3. The half-filled subshell
has a special stability. Each 2p electron occupies a different orbital, all with the same spin. In
this configuration, the electrons avoid each other – reducing repulsion – as much as is possible.
Ionizing N consequently requires a higher energy. In contrast, ionizing O achieves the stable 2p3
configuration starting from 2p4. This makes the ionization of O more favorable, resulting in a
lower first ionization energy. The drop in ionization energy from N to O (or from P to S, and As to
Se) is called the half-filled subshell effect. There is no half-filled subshell effect in the 5th and 6th
periods, however. The 5p and 6p subshells are very large, and the special stability of the halffilled subshells is diminshed for these subshells.
The same increasing zigzag pattern is seen for the third row elements, Na, Mg, Al, Si, P, S, Cl
and Ar, with local peaks at Mg and P.
Table 4.3. First Ionization Energies of Main Group Elements (in kJ mol‒1)
H
1312.1
Li
520.2
Na
495.9
K
418.8
Rb
403.0
Cs
375.7
Be
899.5
Mg
737.8
Ca
589.8
Sr
549.5
Ba
502.9
B
800.6
Al
577.5
Ga
578.9
In
558.3
Tl
589.4
C
1086.5
Si
786.5
Ge
762.2
Sn
708.6
Pb
715.6
N
1402.3
P
1011.8
As
944.5
Sb
830.6
Bi
703.0
O
1313.9
S
999.6
Se
941.0
Te
869.3
Po
811.8
F
1681.1
Cl
1251.2
Br
1139.9
I
1008.4
At
He
2372.3
Ne
2080.7
Ar
1520.6
Kr
1350.8
Xe
1170.4
Rn
1037.1
Table 4.3 shows the first ionization energies of main group elements (excecpt At). The filled and
half-filled subshell effects occur in periods 2, 3 and 4. In the 5th and 6th periods, these effects do
not occur – ionization energy increases continuously across the period, in these cases. The filled
and half-filled subshell effects are less important for larger atoms with much larger valence shells
(electron repulsion within the shell is reduced). From CRC Handbook of Chemistry and Physics,
93rd Ed..
Example 4.2: Order according to ionization energy
(a) Order the following elements according to increasing first ionization energy:
Cs
F
N
He
Ga
As
(b) Which of the following orderings, according to increasing first ionization energy, is correct?
(i) Mg < Si < P < Ar
(ii) Li < N < O < F
(iii) Ga < Ca < Se < As
(iv) Mg < Al < S < P
48
Approach: Look for the element closest to the lower left corner of the periodic table.
This element has the lowest ionization energy. Then find the find the next such element,
and so on. Look out for the filled and half-filled subshell effects. The general trend is
observed except that group 2 and 13 elements are reversed in the order, as are group 15
and 16 elements.
(a) Cs has the lowest first ionization energy in the group. Next is Ga, then As, N, F and
He. The filled and half-filled subshell effects do not arise in this series of elements.
Therefore, according to increasing first ionization energy,
Cs < Ga < As < N < F < He.
(b) (i) This is correct – this ordering is in accord with the general trend.
(ii) N < O is incorrect. O < N because of the half-filled subshell effect.
(iii) This is correct. Here, we see the filled subshell effect, Ga < Ca, and the half-filled
suhshell effect, Se < As.
(iv) S < P is correct – the half-filled subshell effect. However, Mg < Al is not correct
because of the filled subshell effect.
Successive Ionization Energies
The shell and subshell structure of an atom is evident if one considers the sequence of
successive ionization energies – first, second, and so on. Each successive ionization process
takes more energy – there are successively fewer electrons and correspondingly less screening
of the nucleus. Thus, the sequence of successive ionization energies, I1, I2, … , is strictly
increasing. When the next electron comes from a lower energy shell than the previous electron,
there is a big jump in ionization energy. For example, in the case of aluminum, the first five
ionization energies have the sequence:
Ionization Energy (In) in kJ mol‒1
1
2
3
4
5
578
1817
2745
11577
14842
The ionization energies increase with a big jump between I3, I4. Al3+(g) → Al4+(g) + e‒ requires
much more energy than Al2+(g) → Al3+(g) + e‒ because the electron in Al3+ comes from the
second shell, whereas the electron in Al2+ comes from the third shell. Electrons in the second
shell are much more tightly held – they experience a much larger core charge.
Example 4.3:
An atom exhibits the following pattern of successive ionization energies:
I1 < I2 << I3 < I4 < I5 < I6
Which element would this most likely be, among Ne, Li, B or Be?
Approach: Look for an element with two valence electrons. Removing three electrons
requires taking the third electron from a core shell of the original atom.
The element is in group 2. Therefore, it is Be.
49
4.3 Electron Affinity
Electron affinity, A1, is the energy change when an gas phase atom gains an electron.
X(g) + e‒ → X‒(g).
The electron affinity is usually negative, as the process of an atom accepting an electron is
generally favorable. However, there are some positive electron affinities – e.g., He, Ne, Be and
N. These atoms will form singly charged anions in the gas phase, but the anions are not stable –
they lose the electron and return to the neutral atom state. Since the electron affinity is otherwise
negative, it is customary to compare electron affinities in terms of absolute value. For example,
we would say that the magnitude of the electron affinity of fluorine (A1 = ‒328 kJ mol‒1) is
greater than that of oxygen (A1 = ‒141 kJ mol‒1). A larger magnitude of electron affinity
corresponds to a more favorable singly charged anion.
The general trend for the magnitude of electron affinity is that it increases up and to the right just
like ionization energy. However, there are many exceptions – primarily because there are very
strong filled and half-filled subshell effects. Group 2 and 15 elements have much lower
magnitude of electron affinity than the general trend predicts because of filled and half-filled
subshell effects, respectively.
Magnitude of electron affinity generally increases up and to the right
Figure 4.4. Periodic table trend in magnitude of electron affinity – the general trend
without filled and half-filled subshell effects. Note that the noble gas atoms are excluded
from this trend as the anions are always unstable.
Example 4.4:
Order the following elements according to increasing magnitude of electron affinity.:
Rb
F
O
Sn
Ge
As
Approach: Look for the element closest to the lower left corner of the periodic table.
This element has the lowest magnitude of electron affinity. Then find the find the next
such element, and so on. There are no filled and half-filled subshell effects in this series.
Rb is the smallest magnitude electron affinity. Next is Sn, then Ge, As, O and F. In
summary,
Rb < Sn < Ge < As < O < F
50
Table 4.4. Electron Affinities of Main Group Elements (in kJ mol‒1)
H
‒72.8
Li
‒59.6
Na
‒52.9
K
‒48.4
Rb
‒46.9
Cs
‒45.5
Be
+
Mg
+
Ca
‒2.4
Sr
‒4.6
Ba
‒14.0
B
‒27.0
Al
‒41.8
Ga
‒41.5
In
‒29
Tl
‒36.4
C
‒121.8
Si
‒134.1
Ge
‒118.9
Sn
‒107.3
Pb
‒35.1
N
+
P
‒72.0
As
‒77.6
Sb
‒100.9
Bi
‒90.9
O
‒141.0
S
‒200.4
Se
‒195.0
Te
‒190.2
Po
‒183.3
F
‒328.2
Cl
‒348.6
Br
‒321.9
I
‒295.2
At
‒270.2
He
+
Ne
+
Ar
+
Kr
+
Xe
+
Rn
+
Table 4.4 shows the (first) electron affinities of main group elements. Note the half-filled subshell
effects occur only in periods 2, 3 and 4. In the 5th and 6th periods, the magnitude of electron
affinity increases across the period, in these cases. The half-filled subshell effect is less
important for larger atoms with much larger valence shells (electron repulsion within the subshell
is reduced). From CRC Handbook of Chemistry and Physics, 93rd Ed..
4.4 Chemical Properties
Ionization and electron affinity are properties of non-interacting atoms – separated atoms in the
gas phase. In nature, atoms are constantly interacting, chemically and otherwise. There are
periodic table trends in chemical properties, as these properties depend on the number of
valence electrons and the effective nuclear charge.
The oxides of third, fourth, … row elements exhibit a chemical trend. For example, the oxides of
potassium and calcium are basic, whereas the oxide of aluminum is amphoteric (can behave as
an acid or base), and the oxides of silicon, phosphorus, sulfur and chlorine are increasingly
acidic. An acidic or basic oxide produces an acidic or basic solution when dissolved in water.
Metal oxides are basic (group 1 and 2 oxides – except beryllium oxide) or amphoteric (e.g.,
beryllium, aluminum, zinc and iron oxides). Metals – bottom left of the periodic table – do not
hold their electrons as tightly as oxygen. For groups 1 and 2 (except Be), the metal loses its
valence electron(s) to form a cation, while oxygen forms the 2‒ anion. These oxides (e.g., Na2O
and BaO) are ionic. They dissolve in water forming O2‒(aq) – a very strong base – and
subsequently, OH‒(aq). In other metal oxides, the metal oxygen bond has covalent character –
electrons are shared. However, as oxygen is generally much more electronegative, the electrons
are unequally shared being shifted more towards oxygen than the metal.
Electronegativity quantifies the pull of an atom on the electrons in its covalent bonds. The
periodic table trend in electronegativity is the same as that of ionization energy and the
magnitude of electron affinity, as depicted in Fig. 4.5.
51
Electronegativity generally increases up and to the right
Figure 4.5. Periodic table trend in electronegativity. Electrons in a bond are held more
closely by the more electronegative atom.
The difference in electronegativity between bonded atoms indicates the polarization of the bond –
i.e., the shift of the bonding electrons towards the more electronegative atom. When the
difference is large (the case of a low electronegativity metal – left and bottom of periodic table –
and a high electronegativity nonmetal – top right), the bonding between the atoms is ionic. The
electrons are entirely transferred from the metal to the nonmetal.
Table 4.5. Electronegativities of the Main Group and Elements and Transition Metals
H
2.20
Li
0.98
Na
0.93
K
0.82
Rb
0.82
Cs
0.79
Be
1.57
Mg
1.31
Ca
1.00
Sr
0.95
Ba
0.89
Sc
1.36
Y
1.22
La
1.10
Ti
1.54
Zr
1.33
Hf
1.3
V
1.63
Nb
1.6
Ta
1.5
Cr
1.66
Mo
2.16
W
1.7
Mn
1.55
Tc
2.10
Re
1.9
Fe
1.83
Ru
2.2
Os
2.2
Co
1.88
Rh
2.28
Ir
2.2
Ni
1.91
Pd
2.20
Pt
2.2
Cu
1.90
Ag
1.93
Au
2.4
Zn
1.65
Cd
1.69
Hg
1.9
B
2.04
Al
1.61
Ga
1.81
In
1.78
Tl
1.8
C
2.55
Si
1.90
Ge
2.01
Sn
1.96
Pb
1.8
N
3.04
P
2.19
As
2.18
Sb
2.05
Bi
1.9
Table 4.5 shows the Pauling electronegativities of main group elements and transition metals.
The general trend is followed in the main group except for the increase in electronegativity
moving down groups 13 and 14 – from Al to Ga and Si to Ge. Specifically, electronegativity
otherwise increases moving up or to the right in the periodic table. From CRC Handbook of
Chemistry and Physics, 93rd Ed..
Electrons are more equally shared in nonmetal oxides. Oxygen still has the greater pull (except
in the case of F2O), but the pull of the nonmetal can make the oxide acidic. Consider the
reactions of some nonmetal oxides with water to form an acid:
CO2 + H2O → H2CO3
N2O5 + H2O → 2 HNO3
P4O10 + 6 H2O → 4 H3PO4
SO3 + H2O → H2SO4
Cl2O7 + H2O → 2 HClO4
carbonic acid
nitric acid
phosphoric acid
sulfuric acid
perchloric acid
The electrons of the chlorine atom in HClO4 feel a strong pull from four oxygen atoms. This
makes the opposing pull of chlorine even stronger. This pull of electrons – slightly back toward
chlorine – stabilizes the ClO4‒ formed when HClO4 donates an H+. This makes HClO4 a strong
52
O
3.44
S
2.58
Se
2.55
Te
2.1
Po
2.0
F
3.98
Cl
3.16
Br
2.96
I
2.66
At
2.2
acid. H2SO4 and HNO3 are similarly strong acids. H2CO3 and H3PO4 are weak acids because C
and P are not as electronegative as N, S and Cl.
The above acids are Bronsted-Lowry acids which can donate an H+ as shown below. The
strength of the acid is determined by the stability of the anion that results once the acid donates
an H+.
H2CO3 → HCO3‒ + H+
HNO3 → NO3‒ + H+
H3PO4 → H2PO4‒ + H+
H2SO4 → HSO4‒ + H+
HClO4 → ClO4‒ + H+
hydrogen carbonate
nitrate
dihydrogen phosphate
hydrogen sulfate
perchlorate
When fully stripped of H+ ions, the anions are called oxoanions. There is a mnemonic to help
students remember these oxoanions:
Nick the Camel had Clams for Supper at the Phoenix
CO32‒
ClO4‒
SO42‒
PO43‒
NO3‒
nitrate carbonate perchlorate sulfate
phosphate
The number of vowels in each capitalized word gives the magnitude of the negative charge on
the oxoanion associated with the indicated element - the underlined symbol. The number of
consonants gives the maximum number of O atoms in an oxoanion of the indicated element.
53
Problems
4.1 Which of the following orderings, according to increasing atomic radius, is/are correct?
(a) Al < Sr < Rb
(b) Li < B < Na
(c) Ge < Ga < In
4.2 Which of the following orderings, according to increasing ionization energy, are is/correct?
(a) Rb < Si < N < F
(b) K < Ga < Ca < Ge
(c) K < Si < N < O
4.3 Which of the following orderings, according to increasing magnitude of electron affinity, is/are
correct?
(a) Rb < Si < C < F
(b) K < Ga < Cl < S
4.4 Which of the following oxides of the each pair of oxides is more likely to be more acidic?
(a) RbO or CO2
(b) BaO or Cl2O7
(c) SO3 or SrO
54
Solutions
4.1
(a) Al < Sr < Rb is correct. It follows the trends depicted in Fig. 4.1.
(b) Li < B < Na is incorrect. Li atoms are bigger than B atoms.
(c) Ge < Ga < In is correct.
4.2
(a) Rb < Si < N < F is correct. It follows the general trends depicted in Fig. 4.2.
(b) K < Ga < Ca < Ge is correct. Note the filled subshell effect, Ga < Ca.
(c) K < Si < N < O is incorrect. O < N because of the half-filled subshell effect.
4.3
(a) Rb < Si < C < F is correct. It follows the trends depicted in Fig. 4.4.
(b) K < Ga < Cl < S is incorrect. Cl > S according to the general trend, which is followed in
this case.
4.4
(a) CO2 is more acidic. C is in group 14, whereas Rb is in group 1 – i.e., C is further to the right.
(b) Cl2O7 is more acidic. It is a nonmetal oxide.
(c) SO3 is more acidic. It is a nonmetal oxide.
55
5.
Bonding
5.1 Types of Bonding
The physical and chemical properties of a substance are largely determined by the nature of the
bonding between atoms of the substance. Different types of bonding are possible.
Ionic
When a metal reacts with a non-metal, the product is generally an ionic material. Non-metal
atoms take electrons from metal atoms to form anions. Metal atoms lose electrons and become
cations. An ionic material is an alternating lattice of cations and anions held together by the
strong electrostatic attraction between oppositely charged ions. The bonding in such a material is
called ionic. For example, NaCl(s) is an ionic material consisting of Na+ and Cl− ions. Ionic
materials are typically hard, brittle solids with high melting points.
Covalent
Non-metal atoms form covalent bonds, wherein they share valence electrons with a neighboring
non-metal atom. A covalent bond can be polar (as it is in case of H-Cl), wherein the bonding
electrons are closer to the more electronegative atom. A covalent bond can also be completely
non-polar (as it is in the case Cl-Cl, or any other homonuclear diatomic). An ionic bond is, in a
sense, the extreme of a polarized covalent bond. Covalent bonding frequently extends only over
a limited number of atoms. This defines a molecule. The associated material is called molecular.
In the case of a solid or liquid molecular material, the molecules are held together by
intermolecular forces. These forces are weaker, and such materials generally have lower melting
and boiling points than other types of materials, such as metals and covalent networks. Many
materials consisting of small molecules are gases or liquids at room temperature. CO2(g), H2O(l)
and sucrose,C12H24O12(s), are examples of molecular materials.
Covalent bonding can also extend over an array of vast numbers of atoms. In such cases, the
“molecule” is a vast network of atoms connected by covalent bonds. These materials are called
network covalent solids. They are generally very hard, and have high melting points. For
example, diamond is a 3-dimensional network of carbon atoms.
Metallic
In metallic solids, metal atoms share their valence electrons over an array of vast numbers of
atoms. The result is a lattice of metal ions held together with a “sea of electrons”. This type of
bonding is called metallic, and is responsible for the unique properties of metals familiar in
everyday life. For example, metals are typically solid, often with high melting points. They are
also shiny, and mostly silver-colored. They conduct electricity and are somewhat malleable.
Example 5.1: What type of bonding is present in the following substances? (states are shown for
each substance at 25C).
(a) KF(s)
(f) Na(s)
(b) Cu(s)
(g) CaBr2(s)
(c) LiNO3(s)
(h) C2H6(g)
(d) SF6(g)
(i) MgCO3(s)
(e) SiO2(s)
(j) Br2(l)
Approach: Look for metals alone (metallic bonding),metal-nonmetal compounds (ionic
bonding), and nonmetal compounds (covalent bonding).
(a) Ionic bonding. KF(s), potassium fluoride, is a compound formed from a metal
(potassium) and a non-metal (fluorine). It is ionic, consisting of K+ and F− ions. The
formula, KF, tells us that there is a one to one ratio of K+ to F− ions to balance the
charge.
- 56 -
(b) Metallic bonding. Cu(s) is a metal. The chemical formula simply says that the
material consists only of copper atoms.
(c) Ionic and covalent bonding. LiNO3(s) is an ionic compound composed of Li+ and
NO3− (nitrate) ions. The nitrate ion is a polyatomic ion. It consists of covalently
bonded N and O atoms, with a net charge of −1.
(d) Covalent bonding. SF6(g) is a gas consisting of distinct SF6 molecules. Each SF6
molecule has an S atom covalently bonded to six F atoms; both elements are nonmetals.
(e) Covalent bonding. Quartz, SiO2(s), is a network covalent solid. The chemical
formula for a network covalent solid tells us the relative amounts of the different types
of atoms – here, Si and O. Because it is a continuous network it can be represented
as (SiO2)n.
(f) Metallic bonding. Na(s) is a metal.
(g) Ionic bonding. CaBr2(s) is an ionic compound composed of Ca2+ and Br− ions.
Because the calcium has a +2 charge, whereas bromide is only −1, there are two
bromide ions for every calcium ion.
(h) Covalent bonding. Ethane, C2H6(g), is a gas of C2H6 molecules. The chemical
formula for molecular compounds tells us the actual number of each type of atom in
one molecule – it is not just the relative amounts of atoms. Here, there are two C
atoms and 6 H atoms in every ethane molecule.
(i) Ionic and covalent bonding. MgCO3(s) is an ionic compound composed of Mg2+ and
CO32− (carbonate) ions. The carbonate ion is a polyatomic ion. It consists of
covalently bonded C and O atoms, with a net charge of −2.
(j) Covalent bonding. Br2(l) is a liquid consisting distinct of Br2 molecules. Covalent
bonding holds each Br2 molecule together.
5.2 Lewis Structures
The electronic structure of a set of covalently bonded atoms – i.e., a molecule or a network solid
– is conveniently described in terms of a Lewis structure. In a Lewis structure, the electrons of a
molecule are placed on atoms as lone pairs (or single electrons in the case of free radicals), or
bonding pairs. A bond between atoms can have one, two or three pairs of electrons making it a
single, double or triple bond, respectively.
To construct a Lewis structure, start with a skeletal structure – you need to know how the atoms
are linked together by covalent bonds. The only general rules in formulating a reasonable
skeletal structure are that hydrogen makes only one bond, and is therefore always terminal (a
non-central atom), fluorine is similarly terminal, and – if there is a choice – the central atom is
generally the least electronegative atom.
A Lewis structure depicts only the valence electrons of the atoms involved. Core electrons are
held tightly to each nucleus, and do not participate in bonding. Their effect is only to screen the
nuclei. As such, the core charge is used to represent the effective nuclear charge. This
corresponds to perfect screening by the inner shell electrons. The core charge is just the group A
number (e.g., group 17 is also known as group 7A – it has 7 valence electrons), and it equals the
number of electrons that an atom brings to the molecule.
The following steps lead to a Lewis structure:
(1) Add the electrons from all atoms – i.e., sum the group A numbers. Add an extra electron for
each minus net charge – in case of a polyatomic anion. Subtract an electron for each plus net
charge – in case of a polyatomic cation.
(2) Use the electrons, in pairs, to form the skeletal structure of the molecule.
- 57 -
(3) Use the remaining electrons as lone pairs on atoms – starting with the most electronegative
peripheral atoms – until each atom has an octet. An octet corresponds to four electron pairs in
the valence shell of an atom, bonding or non-bonding (lone) pairs.
(4) If there is an atom remaining without an octet, move lone pairs from a bonded atom into the
associated bond, so that a single bond becomes double, or a double bond becomes triple.
Exceptions to this rule are often seen when the atom in question is Be, B or Al (e.g., BeH2, BF3,
AlCl3 – in the gas phase).
The Lewis structure of F2O is constructed with the first three steps – see Fig. 5.1.
(2)
(1)
F
6 + 2 x 7 = 20 electrons
O
F
Skeletal structure - two bonds use
4 electrons.
(3)
F
There are 16 electrons left.
O
F
Lewis structure - every atom has
an octet. 4 + 2 x 6 = 16 electrons
are used.
There are no electrons left.
Figure 5.1. Finding the Lewis structure of F2O.
The Lewis structure of CO2 requires step (4) – see Fig. 5.2.
- 58 -
(2)
(1)
4 + 2 x 6 = 16 electrons
O
C
O
Skeletal structure - two bonds use
4 electrons.
There are 12 electrons left.
(3)
O
C
(4)
O
The remaining 12 electrons are
used to make octets on the oxygen
atoms - the more electronegative
atom.
O
C
O
A lone pair (2 nonbonding electrons) on the right
oxygen becomes a second bond from that
oxygen to carbon.
The carbon atom now has 6 valence electrons.
There are no electrons left. But
this is not the Lewis structure
because the carbon atom has only
4 electrons in its valence shell.
O
C
O
The above process is repeated - this time for the
left oxygen - to spread the electrons evenly, as
suggested by the symmetric structure.
The carbon atom now has an octet. This is the
Lewis structure of CO2.
Figure 5.2. Finding the Lewis structure of CO2.
The octet is a stable valence electron configuration – it is the noble gas valence configuration.
However, if there are an odd number of electrons, it is not possible to have an octet for every
atom. In such cases, the least electronegative atom has less than eight electrons in its valence
shell. For example, NO2 has 17 valence electrons. The less electronegative nitrogen ends up
with 7 electrons in its valence shell – see Fig. 5.3.
O
N
O
Figure 5.3. The Lewis structure of NO2. The nitrogen has a single unpaired nonbonding electron.
Example 5.2: Draw Lewis structures for the following compounds.
(a) O2
(b) N2
(c) CS2
Approach: Follow the steps listed above. Once you have a structure, it applies to other
species with the same number of valence electrons, but different elements.
- 59 -
(a)
(2)
(1)
O
O
2 x 6 = 12 electrons
Skeletal structure - one bonds uses
2 electrons.
There are 10 electrons left.
(3)
(4)
O
O
O
O
There are not enough electrons for
an octet on both oxygen atoms.
The Lewis structure of O2. Now,
both atoms have an octet.
Remove a lone pair from the right
oxygen atom, and make a bonding
pair.
This structure represents a
limitation of Lewis structures.
Oxygen is actually a diradical. It is
better described as having a triple
bond and an "antibond", giving a
net bond order of 2.
(b)
(2)
(1)
N
N
2 x 5 = 10 electrons
Skeletal structure - one bonds uses
2 electrons.
There are 8 electrons left.
(3)
N
(4)
N
N
N
(c)
S
C
S
Metals have lower electronegativity and bring few valence electrons to the structure. When
metals appear in a Lewis structure it is not uncommon for the metal atom to have only 4 or 6
electrons in its valence shell. Beryllium and tin are metals at the edge of the metal region of the
periodic table. They form covalent bonds with nonmetals, but they do not always achieve an
octet. For example, BeF2 and SnCl2 exist as molecules in the gas phase with the structures
shown in Fig. 5.4.
- 60 -
F
Be
Cl
F
Cl
Sn
Beryllium has only four (shared)
electrons in its valence shell.
Tin has only six electrons in its valence shell.
Figure 5.4. Lewis structures of BeF2 and SnCl2.
Boron similarly has less than an octet in some Lewis structures - see BF3 in Fig. 5.5.
F
B
F
F
Figure 5.5. Lewis structure of BF3. Boron has only six electrons in its valence shell.
The vacancy in the valence shell of boron in BF3 provides a source of reactivity. BF3 is a Lewis
acid. For example, it combines with F‒ to form BF4‒ shown in Fig. 5.6.
F
B
F
F
F
Figure 5.6. Lewis structure of BF4‒. Boron has an octet. Note that the negative charge
of the BF4‒ anion is attributed to the boron atom as a formal charge.
Formal Charge
The charge distribution within a structure is modeled via formal charges. Formal charge on an
atom is core charge (group A number), minus the number of non-bonding electrons and half of
the bonding electrons. The idea is to split bonding electrons - attributing half to each of the
bonding atoms. The formula for formal charge can be expressed as
formal charge = core charge – (number of non-bonding electrons) – (number of bonds),
because each bond has two electrons. Formal charge is useful in cases when there is more than
possible Lewis structure. The first principle is that formal charges should be made as small as
possible – minimization of charge. For example, in the correct structure of CO2 (shown in Fig. 5.2
and on the left in Fig. 5.7 below), each atom has zero formal charge. Secondly, if there is a
choice, negative formal charges should go to the more electronegative atom, and positive formal
charges go to the less electronegative atom.
The alternative structure of CO2, with a single and triple bond, rather than two double bonds, is
rejected because it places plus and minus formal charges on the oxygen atoms. The structure
on the left in Fig. 5.7 has zero formal charges – it is the charge minimized structure.
- 61 -
C
O
O
O
Carbon:
Carbon:
4
- ( 0x2
core charge
+ 4x1 )
= 0
4
0 lone pairs + 4 bonds (1 electron each)
Oxygen:
6
O
C
- ( 0x2
core charge
+ 4x1 )
= 0
0 lone pairs + 4 bonds
Oxygen on left:
- ( 2x2
core charge
+ 2x1 )
= 0
6
2 lone pairs + 2 bond (1 electron each)
- ( 3x2
core charge
+ 1x1 )
= -1
3 lone pairs + 1 bond
Oxygen on right:
6
- ( 1x2
core charge
+ 3x1 )
= +1
1 lone pair + 3 bonds
Figure 5.7. Finding the formal charges in CO2. The best structure – the structure on the
left – is charge minimized.
Figure 5.8 shows the preferred structures for SCN‒ and OCN‒. The SCN‒ structure with the
minus formal charge on N is better than the structure with the minus formal charge on S, because
nitrogen is more electronegative than S. In OCN‒, the minus formal charge is better placed on O,
because oxygen is more electronegative than nitrogen.
S
C
O
C
C
N
N
O
better than
S
C
better than
O
C
better than
C
N
N
O
Figure 5.8. Lewis structures of thiocyanate, SCN‒, cyanate, OCN‒ and CO, showing
formal charges. The minus formal charge is on the more electronegative element in the
first two structures. In carbon monoxide, the formal charges are not in accord with
electronegativity, as it is otherwise impossible for the carbon to have an octet. For
carbon, the octet rule is more important than formal charge considerations.
Elements in the third row and beyond can expand their valence shell to beyond an octet. This is
possible because, while the octet is a stable configuration, it only represents filled s and p
subshells. Atoms beyond the second row also have a d subshell of orbitals available which can
accommodate additional electrons.
Expansion of the valence shell occurs when the Lewis structure obtained from the above four
rules produces a structure which has a separation of formal charge. In particular, it occurs when
an atom from the third row, or beyond, with a positive formal charge is bonded to an atom with a
negative formal charge.
- 62 -
Following the above four rules applied to the structure of the hydrogen sulfate anion leads to the
structure shown in Fig. 5.9, with formal charges included.
The formal charges in HSO4-.
Terminal oxygen:
H
6
O
S
O
( 3x2
core charge
+
1x1 )
=
-1
3 lone pairs + 1 bond (1 electron)
Hydroxyl oxygen:
2
O
O
-
6
-
( 2x2
core charge
+
2x1 )
=
0
=
+2
2 lone pairs + 2 bonds
Sulfur:
6
-
( 0x2
+
4x1 )
Figure 5.9. Finding formal charges in HSO4‒.
The above structure indicates a charge separation in the ion. Since formal charge attributes
bonding electrons equally to bonded atoms, it does not account for the difference in
electronegativities of bonded atoms. Thus, formal charge separation is distinct from the
polarization expected in the S-O bonds because of the difference in electronegativies. As we
have already seen, formal charge separation is minimized in the best Lewis structures.
It is possible to reduce the formal charge separation in the above structure by converting a lone
pair on one of the terminal oxygen atoms into an additional bond to sulfur. The formal charge on
the oxygen atom increases by one, as the bonding pair counts as one electron toward the formal
charge tally, whereas the original lone pair counted as two. At the same time, the formal charge
on sulfur decreases by one since there is an additional bond that was not there before. The net
effect is the transfer of formal charge from oxygen to sulfur, reducing the charge separation
indicated in the structure. After another such formal charge transfer, charge separation is
minimized. The resulting structure has a minus one formal charge on one atom – the remaining
oxygen with three lone pairs.
H
O
S
O
O
O
Figure 5.10. The charged-minimized Lewis structure of HSO4‒.
Charge minimization is possible here because sulfur is a third row element. Sulfur can expand its
valence shell to accept the two extra bonding pairs. In contrast, a second row element, such as
nitrogen, cannot expand its valence shell to accommodate more than eight electrons. Therefore,
the formal charge separation cannot be eliminated in the Lewis structure of nitrate, shown in Fig.
5.11.
- 63 -
O
N
O
O
Figure 5.11. Lewis structure of NO3‒. showing formal charges. No further formal charge
minimization is possible because nitrogen is in the second row.
The last rule for constructing Lewis structures is given as follows:
(5) Minimize formal charges, to the extent possible, by allowing atoms beyond the second row to
exceed eight electrons in the valence shell – via transfer of lone pairs from adjacent atoms into
bonding pairs with the element expanding its valence shell.
The Lewis structures drawn above for hydrogen sulfate and nitrate suggest an asymmetry in the
electron distribution among the terminal oxygen atoms. In nitrate, two of the terminal oxygen
atoms have a ‒1 formal charge and are singly bonded to nitrogen, while the third is doubly
bonded with no formal charge. This asymmetry is not consistent with measured properties – e.g.,
the N-O bond distances are exactly equal. This limitation of the Lewis structure is addressed by
appealing to the superposition principle of quantum mechanics. Quantum mechanics admits
superposition states – states constructed by adding or subtracting other states. In particular,
when there are states identical, due to symmetry, the lowest energy state is the sum of the
identical states. Incorporating these ideas into Lewis structures, we say that the electronic
structure of nitrate is described as a resonance hybrid – a superposition of the three equivalent
Lewis structures that can be constructed according the rules given above. The resonance
structures of nitrate are shown in Fig. 5.12.
O
N
O
O
O
N
O
O
N
O
O
O
Figure 5.12. NO3‒ has three equivalent Lewis structures because of the equivalence of
the three oxygen atoms. We say that the structure of nitrate is the resonance hybrid of
these three equivalent resonance structures.
Hydrogen sulfate similarly has three resonance structures shown in Fig. 5.13.
- 64 -
H
H
O
H
O
O
O
S
S
S
O
O
O
O
O
O
O
O
Figure 5.13. The three equivalent resonance structures of HSO4‒. Note that the hydroxy
oxygen is not equivalent to the other three. It is singly bonded in all three of structures.
In the case of HSO4‒, it is important to recognize that oxygen appears in two different
environments – there is a hydroxyl oxygen atom and three terminal oxygen atoms. It is the
terminal oxygen atoms that are equivalent, giving rise to the three equivalent resonance
structures.
Resonance structures provide a more flexible description of bond order than simple Lewis
structures admit - i.e., beyond just single, double and triple bonds. In nitrate, there are three
equivalent N-O bonds. Each equivalent Lewis structure depicts these bonds as a double bond
and two single bonds. The resonance hybrid model treats these bonds as the average of a
double bond and two single bonds.
The bond order is the average bond order given by the sum, over resonance structures, of the
bond order in each structure, divided by the number of resonance structures.
Equivalently, the average bond order is given by the sum, over equivalent bonds, of the bond
order of each equivalent bond, divided by the number of equivalent bonds.
The latter formula for average bond order is easier to compute, as it does not require drawing the
resonance structures. In some cases, it is much easier to compute – e.g., the sulfate ion has six
resonance structures.
Formal charge on an atom in a resonance hybrid is the average formal charge. Average formal
charge on an atom equals the sum, over resonance structures, of the formal charges on that
atom, divided by the number of resonance structures.
Equivalently, the average formal charge is the sum, over equivalent atoms, of formal charges
divided the number of equivalent atoms.
Example 5.3: What are the average formal charges and bond orders in the resonance hybrid
structures of (a) nitrate and (b) hydrogen sulfate?
Approach: Apply the above formulas. Average over equivalent atoms to get average
formal charge, and average over equivalent bonds to get average bond order.
(a) In the case of nitrate, the average formal charge on the three equivalent oxygen
atoms is
( (‒1) + (‒1) + 0 )/3 = ‒2/3.
The average formal charge on nitrogen is +1. It is the same on all resonance structures.
The average N-O bond order is
( 1 + 1 + 2 )/3 = 4/3.
- 65 -
(b) In the case of hydrogen sulfate, the average formal charge on the three equivalent
terminal oxygen atoms is
( (‒1) + 0 + 0 )/3 = ‒1/3.
The average formal charge on the sulfur atom, and hydroxyl oxygen atom, is zero.
The average bond order of the three equivalent S-O bonds is
( 1 + 2 + 2 )/3 = 5/3.
The order of the bond between S and the hydroxyl oxygen atom is one.
Example 5.4: Draw charge minimized Lewis structures, including the formal charges, for the
following species. How many resonance structures are there? What are the average formal
charges and bond orders?
(a) HPO42‒
(H is bonded to O)
(b) XeOF4
(Xe is the central atom)
Approach: Follow the steps listed above, including step (5).
(a)
Steps (1), (2) and (3) - step (4) does not apply.
H
Step (5) - applies only once. .
H
O
O
P
P
O
O
O
O
O
O
There are three resonance structures because the doubly bonded oxygen could be any
of the three equivalent terminal oxygen atoms. The average formal charge on the three
equivalent oxygen atoms is ‒2/3. The average P-O bond order – to a terminal oxygen –
is 4/3.
(b)
Steps (1), (2) and (3) - step (4) does not apply.
Step (5) - applies only once. .
O
O
F
F
Xe
F
F
F
Xe
F
F
F
There are no additional resonance structures. Therefore, the bond orders and formal
charges are simply as indicated in the structure. All formal charges equal zero, the Xe-F
bonds are single bonds, and the Xe-O bond is double.
- 66 -
Bond order is manifest in the bond length and bond energy (the energy needed to break a bond).
A double bond is stronger than a single bond between the same two atoms. The double bond is
shorter, and has a larger bond energy. With respect to bond length, we have the following
ordering:
triple < double < single.
With respect to bond energy, the order is reversed:
single < double < triple.
The fractional bond orders seen in resonance hybrids fit into these patterns. Thus, the 5/3 order
S-O bonds in hydrogen sulfate are shorter, and have higher bond energy, than the single S-O
bond (S-O-H). However, these bonds are longer, and with lower bond energy, than the three S-O
double bonds in SO3.
5.3 Molecular Shape and Polarity
Once a Lewis structure is known, it is possible to predict the shape of a molecule or ion.
Molecular shape and bond polarity are important determining factors for the physical and
chemical properties of a substance.
The valence shell electron pair repulsion (VSEPR) theory1 makes the connection between
electronic structure and molecular shape. The idea is simply that electron pairs in the valence
shell of an atom repel each other and are maximally separated in the atoms of a stable molecule.
This means that the arrangement of electron pairs – more generally, electron domains that
consist of single electrons, electron pairs or multiple pairs of bonding electrons – in the valence
shell of every atom has the greatest symmetry possible.
The shape of the molecule, about an atom A, is determined by the number of electron domains,
and the number of those that are bonding. For each bonding domain, there is an atom X. X is
any atom bonded to A. The shape of the molecule is the arrangement of the X atoms about atom
A. This arrangement is determined by the VSPER class, expressed as AXmEn, where m is the
number of bonded atoms and n is the number of non-bonding electron domains (usually lone
pairs).
If there are two electron domains about an atom, A, they are arranged on opposite sides of the
valence shell of A, as shown in Fig. 5.14.
A
Figure 5.14. Two electron domains – linear geometry.
This is a linear (about atom A) electron domain geometry. If both of the domains are bonding, we
have the AX2 VSEPR class (n = 0). These molecules or ions have a linear shape. The X-A-X
bond angle equals 180°.
For example, BeF2 has two bonding electron domains about beryllium, and is consequently
linear. Carbon dioxide is also linear. It has two double bond electron domains – each with four
electrons.
1
Ronald Gillespie and Ronald Nyholm created VSPER theory in 1957.
- 67 -
The carbon and the nitrogen in cyanide, CN‒, also have two electron domains – both have a lone
pair and a triple bond. For each atom, one of the domains is non-bonding. The VSEPR class is
AXE. It is a diatomic ion, and is therefore is linear - all diatomics are linear.
N
C
If there are three electron domains about A, they are arranged in the trigonal planar pattern
shown in Fig. 5.15.
A
Figure 5.15. Three electron domains – trigonal planar geometry.
Boron in BF3 and nitrogen in nitrate are examples of the trigonal planar electron domain geometry
with three bonding domains – i.e., the AX3 VSEPR class. Such molecules (ions) have a trigonal
planar shape.
Cl
Al
Cl
Cl
Figure 5.16. Lewis structure of AlCl3 showing the trigonal planar shape.
The bond angles are all 120°, the ideal bond angle, in symmetrical AX3 molecules and ions such
as BF3, AlCl3 and NO3‒.
Tin in SnCl2 and nitrogen in NO2 are examples of the AX2E VSEPR class. In the case of NO2,
one of the non-bonding domains is a single electron. Nitrogen in nitrous acid, HNO2, is another
example of the AX2E VSEPR class – see Fig. 5.17. Molecules with the AX2E VSEPR class have
a bent shape.
H
O
N
O
Figure 5.17. Lewis structure of HNO2 showing the bent shape. This structure is bent at
both the nitrogen atom and at the hydroxyl oxygen atom.
The attraction of a bonding electron domain to the associated X atom reduces the spreading of
the bonding domain within the valence shell of the A atom. Thus, nonbonding domains exhibit
more such spreading, making the bond angle in an AX2E molecule or ion less than its ideal value
– i.e., it is less than 120°.
AXE2 are diatomic and therefore linear.
- 68 -
If there are four electron domains about A, they are arranged in the tetrahedral pattern shown in
Fig. 5.18. The domain indicated with the thickest line comes forward out of the plane, while the
thinnest line depict a domain extending behind the plane. All six bond angles are 109.5°.
A
Figure 5.18. Four electron domains – tetrahedral geometry.
Boron in BF4‒, sulfur in HSO4‒, and carbon in CCl4 are examples of the tetrahedral electron
domain geometry with four bonding domains – the AX4 VSEPR class. Such molecules (ions)
have a tetrahedral shape.
Cl
CCH33
CH
HCl
3C
Cl
Cl
Cl
Figure 5.19. Lewis structure of CCl4 showing the tetrahedral shape using wedge bonds.
The solid wedge bond extends forward, while the dashed wedge bond extends backward.
Ammonia, NH3, and sulfite, SO32‒, are examples of the AX3E VSEPR class – see Fig. 5.20. Such
molecules have a trigonal pyramidal shape.
CH
NCH33
H3H
C
Cl
H
ammonia
CH
CH
SCH333
H
H3O
C
O
Cl
Cl
O
sulfite - there are three equivalent
resonance structures
Figure 5.20. Lewis structures of NH3 and SO32‒ showing the trigonal pyramidal shape.
AX2E2 molecules have a bent shape. Water, F2O and the hydroxyl group in hydrogen sulfate (the
shape about the H-O oxygen) are examples of the AX2E2 bent shape. The AX2E2 and AX2E bent
shapes differ in the ideal bond angle – 109.5° and 120°, respectively. Actual bond angles are
less than these values because non-bonding electron domains require more space in the valence
shell of A.
If there are five electron domains about A, they are arranged in the trigonal bipyramidal geometry
shown in Fig. 5.21.
- 69 -
A
Figure 5.21. Five electron domains – trigonal bipyramidal geometry.
The trigonal bipyramidal geometry is different from the others in that the positions of the five
electron domains are not all equivalent - there are two distinct position subsets – the two axial
(vertical) positions are distinct from the three equatorial positions. The latter determine a trigonal
planar arrangement in the plane perpendicular to the vertical axis. The ideal bond angles within
the equatorial plane are 120°, while those between axial and equatorial positions are 90°.
Antimony pentafluoride has five bonding electron domains – it is AX5. Such molecules have a
trigonal bipyramidal shape.
F
F
H3C
CH
Sb
CH33
F
F
Cl
F
Figure 5.22. Lewis structure of SbF5 showing the trigonal bipyramidal shape.
The AX4E VSEPR class has the seesaw shape - for example, SF4 shown in Fig. 5.23.
F
F
H3C
CH
SCH33
F
Cl
F
Figure 5.23. Lewis structure of SF4 showing the seesaw shape.
Note that the lone pair goes in an equatorial position because the lone pair requires more space
in the valence shell of A than the bonding domains. In this configuration, the lone pair is about
90° (a little less, since the bond angle is non-ideal in this case) from two bonding domains. If the
lone pair were in an axial position, it would be about 90° from three bonding domains. There is
more repulsion in the latter case. For the same reason, the bond angles are less than ideal, as
shown in figure (i.e., <90° and <120°).
If there are two non-bonding electron domains – i.e., the AX3E2 VSEPR class – then the second
non-bonding electron domain also goes in an equatorial position and the molecule is T-shaped.
BrCl3, shown in Fig. 5.24, is T-shaped. Again, the bond angles are less than ideal (<90°).
- 70 -
Cl
Br
Cl
Cl
Figure 5.24. Lewis structure of T-shaped BrCl3.
AX2E3 molecules or ions, such as I3‒, are linear.
I
I
I
Figure 5.25. Lewis structure of I3‒ showing the linear shape of the AX2E3 class.
If there are six electron domains about A, they are arranged in the octahedral geometry shown in
Fig. 5.26.
A
Figure 5.26. Six electron domains – octahedral geometry.
Sulfur hexafluoride, shown in Fig. 5.27, has six bonding electron domains – it is AX6. Such
molecules have an octahedral shape.
F
F
F
CH3
H3C
S
Cl
F
FCl
F
Figure 5.27. Lewis structure of SF6 showing the octahedral shape.
The six positions of the octahedral electron domain geometry are all equivalent – there is no
axial-equatorial distinction in the octahedral geometry. The ideal bond angles are all 90° in this
geometry.
Iodine pentafluoride, IF5, (shown in Fig. 5.28) has the AX5E VSEPR class – it is square pyramidal.
The bond angles are less than ideal (<90°).
- 71 -
F
F
F
CH3
H3C
I
F
Cl
Cl
F
Figure 5.28. Lewis structure of IF5 showing the square pyramidal shape.
Xenon tetrafluoride, XeF4, (shown in Fig. 5.29) has the AX4E2 VSEPR class – it is square planar.
The bond angles are ideal (90°) in this case because repulsion from the two lone pairs is
balanced.
F
F
CH 3
H 3C
Xe
F
Cl
FCl
Figure 5.29. Lewis structure of XeF4 showing the square planar shape.
Note that the two lone pairs on Xe are on opposite sides because they require more space in the
valence shell. Lone pairs 90° from each other corresponds to a higher energy structure that
would relax to the above square planar structure. For the same reason, the oxygen in XeOF4 is
opposite the lone pair – see Example 5.4. The double bond requires more space than the single
bonds.
Xenon trifluoride anion, XeF3‒, (shown in Fig. 5.30) has the AX3E3 VSEPR class – it is T-shaped,
like the AX3E2 class. The bond angles are less than ideal (<90°).
F
Xe
H3C
F
CH3
F
Figure 5.30. Lewis structure of XeF3‒, an AX3E3 class T- shaped molecule.
Molecular Polarity
The polarity of a molecule is determined by its shape, and the polarity of its bonds. For example,
water is polar because it is an asymmetric molecule with polar covalent bonds. The H-O bonds
are polar because oxygen is more electronegative than hydrogen. Thus, the bonding electrons
are shifted toward the O atom. The O atom is slightly negatively charged (), while the H atoms
are slightly positively charged (+), where  indicates a partial charge. Each water molecule has
a net permanent dipole.
In water, the two H-O bond dipole moments do not cancel. This is because two lone pairs of
electrons on the O atom push the H-O bonds to one side of the molecule.
- 72 -

H

H
O

Figure 5.31 Lewis structure of water, showing the bent shape, partial charges, and the
net dipole moment vector (on the left).
In SiCl4 (shown in Fig. 5.32), the bonds are also polar, with bond dipoles pointing from Si to Cl.
However, the molecule is symmetrical – it has a center of symmetry – and the bond dipoles
cancel. Consequently, SiCl4 is nonpolar.
Cl
CH3 Cl
Si
CH
3
HCl
3C
Cl
Cl
Figure 5.32. Lewis structure of SiCl4: The molecule has polar bonds. However, it is a
symmetric molecule (it has a tetrahedral shape). Because of the symmetry of the
molecule, all of the bond dipoles cancel out, and it has no permanent dipole (it is a nonpolar molecule).
In general, molecules with a center of symmetry (AXn, AX2E3 and AX4E2 molecules – with only
one type of X atom) are non-polar. Molecules with polar bonds, and without a center of symmetry
(all other classes, and/or more than one X atom), are polar.
Example 5.5: For the following molecules, determine the VSEPR class and shape. Is it a polar
or non-polar molecule?
(a) AlCl3
(b) HCCl3
(c) SO2
(d) IF3
(e) SeF6
Approach: Find Lewis structures to determine the number of lone pairs about the
central atom. Assign the VSEPR class which determines the shape. Assess bond
polarity and molecular symmetry to determine whether the molecule is polar or non-polar.
(a)
Cl
Al
Cl
Cl
Aluminum trichloride is AX3, and therefore has trigonal planar shape. The Al-Cl bonds
are polar, but the molecule is symmetric about the aluminum atom and the bond dipoles
cancel. AlCl3 molecules (i.e., in the gas phase) are non-polar.
- 73 -
(b)
H
CH
CCH33
Cl
H 3C
Cl
Cl
Trichloromethane is AX4, and is therefore tetrahedral. This shape has a center of
symmetry. However, the C-H bond dipole is different (both in magnitude and direction)
from the C-Cl bond dipoles, and the dipoles do not cancel. Trichloromethane
(chloroform) is polar. To distinguish this from symmetric AX4, we introduce the notation,
AX3Y, for CCl3H.
(c)
O
O
S
Sulfur dioxide is AX2E, and has the bent shape. Since the S-O bonds are polar, and the
molecule is asymmetric, SO2 is polar.
(d)
F
I
F
F
Iodine trifluoride is AX3E2. It is T-shaped, asymmetric and polar.
(e)
F
F
F
CH3
H 3C
Se
Cl
F
FCl
F
Selenium hexafluoride is AX6. It is octahedral, symmetric and non-polar.
5.4 Aqueous Solutions
The different types of materials described in Sec. 5.1 have distinguishing properties when those
materials are dissolved in a solvent (often, but not necessarily, water). For example, ionic
materials (salts) dissolve in water to form electrolyte solutions – though some salts have
extremely small solubility. An electrolyte solution conducts electricity. This is because the ions of
the ionic material move around freely in the water solution. Solvated ions can carry a current. In
- 74 -
contrast, an ionic solid does not conduct electricity as the ions are fixed in lattice positions and
cannot move to carry current.
Salts dissolve in water because the ions can be solvated by water molecules (or, “hydrated”).
Solvation is an example of intermolecular forces – forces of attraction that exist between
molecules or between molecules and ions. In this case, we have ions and water molecules.
Solvation is strong here because water molecules are polar and are able to align themselves to
interact favorably with either a positive or negative ion – as shown in Fig. 5.33. This is an
example of an ion-dipole force.

H
O


H
O

H


O
H

H

H


H

H

H

O
 
O
H
H

+
Na
O

H


O
H

H 


H
H

H
O
H


O
H

Cl
O 
-

H

H
O 
 H
H

O
 H
Figure 5.33 Solvation of sodium and chloride ions in aqueous solution. The δ+ and δ
are the partial charges on H and O atoms associated with the polarity of the OH bond.
Molecular compounds can also dissolve in water. This is again because of intermolecular forces.
Compounds with the highest solubilities are those consisting of molecules with net dipole
moments – i.e., polar molecules like water. When molecular compounds dissolve in water, in the
absence of reaction with water (e.g., acids and bases excluded), the result is a non-electrolyte
solution. Such solutions do not conduct electricity. When the molecular compound is an acid or
a base, it reacts when dissolved in water (see Chapter 6) producing H3O+ or OH− ions,
respectively. These ions move freely in solution, and will carry a current. The solutions of acids
or bases are electrolyte solutions. Strong acids and bases are strong electrolytes, while weak
acids and bases are weak electrolytes (they produce solutions with small ion concentration).
Soluble salts are considered strong electrolytes.
Dissolution of compounds in other molecular solvents is also important. Many organic solvents
(e.g., CCl4) are non-polar. These solvents best dissolve other non-polar materials. This is
because a polar molecule has stronger intermolecular forces when it is surrounded by other polar
molecules. The polar compound will not readily dissolve in a non-polar solvent, as it would be
solvated by weaker intermolecular forces than are already present in the compound. In
summary, “like dissolves like”.
- 75 -
Example 5.6: Consider the following Lewis structures:
O
H
C
H
Lewis structure and shape of formaldehyde, CH2O
O
H
N
H
H
H
H
H
H
Lewis structure and shape of ammonia NH3
H
H
Lewis structure of ethanol, C2H5OH,
showing shape around the O atom –
and the C atoms (these atoms appear
only as vertices from which four bonds
emerge)
Which of the following are electrolyte solutions?
(a)
(b)
(c)
(d)
(e)
(f)
potassium chlorate solution: KClO3(aq)
formaldehyde solution (pH = 7): CH2O(aq)
ethanoic acid solution (pH = 3): CH3COOH(aq)
sodium bromide: NaBr(aq)
ammonia solution (pH = 10): NH3(aq)
ethanol solution (pH = 7): C2H5OH(aq)
(a) Electrolyte solution. KClO3(aq) consists of K+ and ClO3− ions in aqueous solution. These
ions move freely and therefore can conduct electricity.
(b) Non-electrolyte solution. Formaldehyde solution contains CH2O molecules solvated by
water molecules. Formaldehyde is neither an acid nor a base in water. The information
that pH = 7 tells us that the solution is neutral. The H atoms in CH2O are bonded to
carbon. They are not acidic. Though the O atom has lone electron pairs, it does not
accept H+ from water. Formaldehyde remains neutral in solution. While it moves freely,
it cannot conduct electricity.
(c) Electrolyte solution. CH3COOH is a weak acid. It donates an H+ to H2O to form H3O+.
The H3O+ and CH3COO− ions move freely in solution and can conduct electricity.
CH3COOH(aq) + H2O(l)  CH3COO(aq) + H3O+(aq)
Because ethanoic acid is a weak acid (see Chapter 6), the equilibrium is shifted to the
left. It produces a dilute solution of ions, even when the acid concentration is high. We
are told that pH = 3 for the ethanoic acid solution in question. This corresponds to [H3O+]
= 10‒3 mol L1 (= [CH3COO]). Ethanoic acid is a weak electrolyte. Its solutions only
weakly conduct electricity.
(d) Electrolyte solution. NaBr(aq) consists of Na+ and Br− ions in aqueous solution. The
hydrated Na+ and Br− ions can conduct electricity.
(e) Electrolyte solution. NH3 is a weak base. It accepts an H+ from H2O to leaving OH− and
NH4+ ions. The OH− and NH4+ ions move freely in solution and can conduct electricity.
NH3(aq) + H2O(l)  NH4+(aq) + OH−(aq)
- 76 -
Because ammonia is a weak base, the equilibrium is shifted to the left, and ammonia is
only a weak electrolyte. pH = 10 corresponds to [OH−] = 10‒3 mol L1 (= [NH4+]).
Ammonia solutions only weakly conduct electricity.
(f) Non-electrolyte solution. A molecular solution, and ethanol is neither an acid nor a base
in water (pH = 7).
Example 5.7: Which of the following pairs of compounds form solutions?
(a)
(b)
(c)
(d)
(e)
(f)
potassium iodide and water
methane, CH4, and water
methane, CH4, and carbon tetrachloride, CCl4
formaldehyde, CH2O, and water
formaldehyde, CH2O, and carbon tetrachloride, CCl4
ethanol, C2H5OH, and water
(a) Solution. KI is ionic. It dissolves in water to form solvated K+ and I− ions.
(b) No solution. CH4 is symmetrical and non-polar. It has very little solubility in a polar
solvent such as water.
(c) Solution. CCl4 is symmetrical and non-polar. It dissolves non-polar compounds like
methane.
(d) Solution. CH2O is asymmetrical. It has a dipole moment arising mainly from the polar CO bond (a double bond in formaldehyde). It dissolves in polar solvents like water.
(e) No solution. Polar formaldehyde does not dissolve in non-polar carbon tetrachloride.
(f) Solution. Ethanol is polar (primarily because of the O-H bond) and dissolves in polar
water.
Problems:
5.1 Identify the types of bonding present in the following materials. Label the material as ionic,
molecular, network covalent, or metallic.
(a) Cl2(g)
(b) LiOH(s)
(c) H2SO4(l)
(d) Fe(s)
(e) CaCO3(s)
(f) C(diamond)
(g) brass – an alloy of mostly copper, zinc and (sometimes) Sn.
5.2 Draw charge minimized Lewis structures, including formal charges, for the following species.
How many resonance structures are there – if there is more than one? In such cases, what
are the average formal charges and bond orders? What are the VSEPR classes and
shapes?
(a) ClO4‒
(d) Methanoate, HCO2‒ (HCOO‒)
(b) SO42‒
(e) O3
(c) AsO3F2‒
(f) Azide, N3‒
5.3 What are the VSEPR classes and shapes of the following species? In cases of a neutral
molecule, is the molecule polar or non-polar?
(a) SnCl4
(d) Phosgene, COCl2
(b) SeF3‒
(e) SeF4
(c) XeF5+
(f) XeO3
- 77 -
5.4 Consider the C-O bonds in the following species:
CO2
CO32‒
HCOO‒
CO
(a) Order the bonds according to increasing bond energy.
(b) Order the bonds according to increasing bond length.
5.5 Which of the pairs of material form solutions?
(a)
(b)
(c)
(d)
(e)
ethane, C2H6, and butane, C4H10.
ethanol, C2H5OH, and water
carbon dioxide and carbon tetrachloride, CCl4
lithium bromide, LiBr, and water
sodium nitrate, NaNO3, and carbon tetrachloride, CCl4
5.6 Electrolytes can be strong or weak. Ionic solids are strong electrolytes (as long as they are
soluble). Acids and bases are also electrolytes. In Chapter 6, we see that (Bronsted-Lowry)
acids are molecules that react with water to form an anion (the conjugate base) and
hydronium ion, H3O+. The ions formed can conduct electricity. However, most acids are
weak which means they only dissociate to a small extent. Consequently, the ion
concentration in solution is small, and the electrical conductivity is also small. Weak acids
are therefore weak electrolytes. Weak bases are similarly weak electrolytes. Ionic materials
and strong acids and bases (see Chapter 6 for a list of strong acids and bases) are strong
electrolytes.
Label each of the following materials as a strong or weak electrolyte, or as a non-electrolyte.
(a) nitric acid, HNO3
(b) lithium hydroxide, LiOH
(c) ethanol, C2H5OH
(d) potassium iodide, KI
(e) ammonia, NH3
(f) hypochlorous acid, HOCl
(g) sodium hypochlorite, NaOCl
- 78 -
Solutions:
5.1
(a) Cl2(g). The bonding is covalent. This is a molecular gas. Chlorine gas consists of
pairs of chlorine atoms covalently bonded to form molecules.
(b) LiOH(s). Ionic and covalent bonding are present. This is an ionic material. Lithium
hydroxide consists of Li+ and OH− ions arranged in a lattice. Hydroxide, OH−, is a
diatomic anion with an oxygen atom covalently bonded to a hydrogen atom.
(c) H2SO4(l). The bonding is covalent within the sulfuric acid molecules. Since the
material is a liquid, there are also intermolecular forces holding the molecules together.
These are weaker forces. At room temperature, the sulfuric acid molecules have enough
thermal energy to easily move around one another. Consequently, sulfuric acid is a liquid
at room temperature.
(d) Fe(s). The bonding is metallic. The material is a metal. Iron atoms share valence
electrons over vast arrays of nuclei arranged in a dense lattice to optimize the sharing.
(e) CaCO3(s). Ionic and covalent bonding are present. This is an ionic material.
Calcium carbonate consists of Ca2+ and CO32− ions arranged in a lattice. Carbonate,
CO32−, is a polyatomic anion with three oxygen atoms covalently bonded to a carbon
atom.
(f) C(diamond). Only covalent bonding is present. Diamond is a network covalent solid.
It consists of a network (lattice) of carbon atoms held together by covalent bonds
between neighboring atoms.
(g) brass – an alloy of mostly copper, zinc and (sometimes) Sn. The bonding is metallic.
The material is a metal. It consists of vast arrays of copper, zinc and (sometimes) Sn
nuclei with valence electrons shared as a “sea” over the network.
5.2
(a)
O
CH
CH3333
CH
CH
Cl
HH3O
C
3C
O
Cl
OCl
There are four resonance structures – the single bond could be any one of the four Cl-O
bonds. The average formal charge on O is ‒1/4. The average bond order is 7/4.
The ion is AX4, and consequently tetrahedral.
- 79 -
(b)
O
CH
CH3333
CH
CH
S
HH3CC
O
3
O
Cl
OCl
There are six resonance structures. There are four bonds, all equivalent. Any two of
them could be the double bonds. The average formal charge on O is ‒1/2. The average
bond order is 3/2.
The ion is AX4, and consequently tetrahedral.
(c)
F
CH3
CH
As
3
O
H3O
C
Cl
O
There are three resonance structures – the double bond could be any one of the three
As-O bonds. The average formal charge on oxygen is ‒2/3. The average bond order is
4/3.
The ion is AX4, and consequently tetrahedral.
(d)
O
H
C
O
There are two resonance structures – either one of the C-O bonds could be the double
bond. The average formal charge on O is ‒1/2. The average bond order is 3/2.
The ion is AX3, and consequently trigonal planar.
(e)
O
O
O
There are two resonance structures (the single and double bond can be swapped). The
two terminal O atoms are equivalent in the resonance hybrid. Therefore, the middle O
atom has its own average formal change, +1 (it is the same in both structures). The
- 80 -
terminal O atoms have an average formal charge of ‒1/2. The average bond order is 3/2.
Ozone, O3, is isoelectronic (with respect to valence electrons) with SO2.
The molecule is AX2E, and consequently bent.
(f)
N
N
N
There are no additional resonance structures. The average formal charges are as
indicated: +1 on the middle atom, and ‒1 on the terminal atoms. The average bond order
is 2 for both bonds. Azide, N3‒, is isoelectronic (with respect to valence electrons) with
CO2.
This ion is AX2, and consequently linear.
5.3
(a)
Cl
Sn
SnH2 Cl
H
Cl
3C
Cl
Cl
This molecule is AX4, and consequently tetrahedral. It is non-polar because it has a
center of symmetry.
(b)
F
Se
F
F
This ion is AX3E2, and consequently T-shaped.
(c)
F
F
F
CH3
H 3C
Xe
Cl
F
F
Cl
This ion is AX5E, and consequently square pyramidal.
- 81 -
(d)
O
C
Cl
Cl
There are no additional resonance structures.
The molecule is AX3, and consequently trigonal planar. It has three bond dipoles all
pointing away from carbon. Since one is a C-O bond dipole, while the others are C-Cl
bond dipoles, they do not cancel perfectly – so COCl2 is polar, but with a relatively small
net dipole.
(e)
F
F
H3C
CH
Se
CH33
F
Cl
F
This molecule is AX4E, and consequently seesaw shaped. It is polar because it is
asymmetric, and has polar bonds.
(f)
CH
CH3333
CH
Xe
CH
HH3O
C
3C
O
Cl
OCl
This molecule is AX3E, and consequently trigonal pyramidal. It is polar because it is
asymmetric and has polar bonds.
Here are the Lewis structures:
H
O
C
C
O
C
O
O
O
O
C
O
O
5.4
two resonance structures
three resonance structures
average bond order = 3/2
average bond order = 4/3
(a) The average C-O bond order for these four structures are 2, 3/2, 4/3 and 3,
respectively. Bond energy increases with increasing bond order. Therefore, according to
increasing bond energy, we have
CO32‒ < HCOO‒ < CO2 < CO .
- 82 -
(b) Bond length decreases with increasing bond order. Therefore, according to
increasing bond length, we have
CO < CO2 < HCOO‒ < CO32‒ .
5.5
(a) ethane, C2H6, and butane, C4H10. These materials form a solution. They are both
non-polar molecules.
(b) ethanol, C2H5OH, and water. They form a solution. They are both polar molecules.
(c) carbon dioxide and carbon tetrachloride, CCl4. They form a solution, as they are both
non-polar molecules.
(d) lithium bromide, LiBr, and water. These form a solution, as LiBr is an ionic material
that dissolves in water and splits into ions that are solvated by the polar water molecules.
(e) sodium nitrate, NaNO3, and carbon tetrachloride, CCl4. These do not form a solution,
as carbon tetrachloride is non-polar molecular liquid and the ionic NaNO3 requires a polar
solvent to dissolve it and solvate its ions.
5.6
(a) nitric acid, HNO3. Nitric acid is a strong electrolyte. It is a strong acid - it fully
dissociates in water.
(b) lithium hydroxide, LiOH. Lithium hydroxide is a strong electrolyte. It is an ionic
material. It is also a strong base. However, since hydroxide is already ionic, reaction
with water is not needed to produce an electrolyte solution. In the case of hydroxide, the
products of reaction with water are water and hydroxide – there is no effect on the
composition of the solution.
(c) ethanol, C2H5OH. Ethanol is a non-electrolyte. It dissolves in water as neutral,
solvated ethanol molecules. It is not an acid, though it has an H bonded to an O as do
many acids.2
(d) potassium iodide, KI. Potassium iodide is a strong electrolyte – another ionic
material.
(e) ammonia, NH3. Ammonia is a weak electrolyte. It dissolves in water as neutral,
solvated NH3 molecules. However, as ammonia is a weak base, some of these
molecules react with water to form ammonium, NH4+, and hydroxide, OH−. These ions
conduct electricity, giving an electrolyte solution. However, the reaction occurs only to a
small extent. It is a weak electrolyte solution.
(f) hypochlorous acid, HOCl. Hypochlorous acid is a weak electrolyte. It dissolves in
water as neutral HOCl molecules. However, as HOCl is a weak acid, some of the HOCl
molecules react with water and form hypochlorite ion, OCl−, and hydronium ion, H3O+.
These ions conduct electricity, gving an electrolyte solution. However, the reaction
occurs only to a small extent. It is a weak electrolyte solution.
2
Ethanol is like water with one of the H atoms replaced by an ethyl group, C2H5 (think of H-O-H
vs. H-O-C2H5). The carbon atom of the ethyl group bonded to the O atom is not very
electronegative. It cannot pull electrons away from the O atom. In contrast, oxo acids (like HOCl)
have electrons pulled away from O atoms bonded to H. This makes the O atoms pull electrons
away from the H making donation of H+ occur more readily.
- 83 -
(g) sodium hypochlorite, NaOCl. Sodium hypochlorite is a strong electrolyte. It is an
ionic material – it splits into solvated ions in water. Now, OCl− is a weak base. As such,
to a small extent, it reacts with water to form HOCl and OH−. This makes the solution
slightly basic. However, it simply replaces some (a small amount) of the OCl− with OH−.
This has no impact on the conductivity – it is still a strong electrolyte solution.
- 84 -
6.
Types of Reactions
6.1 Solubility and Precipitation Reactions
Solubility
Ionic compounds (salts) can dissolve in water and form a solution of solvated cations and anions.
For every salt, there is a maximum concentration that can be achieved. The maximum
concentration of a salt is its solubility. Sometimes the solubility is very low. Such salts are said to
be insoluble. If an insoluble salt is added to water only a tiny (sometimes imperceptible) portion
of it dissolves before the solution becomes saturated – i.e., the concentration equals the
solubility. While salt solubilities range over all possible values, it is customary to define a series
of levels of solubility. In the simplest such system, salts with solubility below 0.01 mol L‒1 are
insoluble, while those with solubility above this value are soluble. Soluble salts with solubility
below 0.1 mol L‒1 are only slightly soluble.
Whether a salt is soluble or insoluble depends on the strength of the bonding in the ionic solid
compared with the strength of aqueous solvation of the ions. For common salts, there is a set of
solubility rules. Rules 1 to 4 concern mostly soluble salts, while rules 5 to 8 concern mostly
insoluble salts.
1. Alkali metal (Li+, Na+, K+, Rb+) and ammonium (NH4+) salts are soluble.
2. Nitrate (NO3‒), chlorate (ClO3‒), perchlorate (ClO4‒), hydrogen carbonate (HCO3‒) and
ethanoate (CH3COO‒) salts are soluble.
3. Sulfate (SO42‒) salts are soluble, except for the calcium (Ca2+), strontium (Sr2+) and barium
(Ba2+) (i.e., alkaline earth metals other than magnesium) salts which are insoluble.
4. Chloride (Cl‒), bromide (Br‒) and iodide (I‒) salts are soluble, except for the silver (Ag+), lead
(Pb2+) and mercury I (Hg22+) salts which are insoluble (see rule 5 below).
5. Silver (Ag+), lead (Pb2+) and mercury I (Hg22+) salts are insoluble, except for the salts already
covered by rules 2 and 3 above.
6. Sulfide (S2‒) salts are insoluble, except for the alkali metal, ammonium (covered by rule 1
above), and alkaline earth salts (Mg2+, Ca2+, Sr2+ and Ba2+) which are soluble.
7. Oxide (O2‒) and hydroxide (OH‒) salts are insoluble, except for the alkali metal, ammonium
(see rule 1 above), calcium (Ca2+), strontium (Sr2+) and barium (Ba2+) salts which are soluble –
though CaO and Ca(OH)2 are only slightly soluble.
8. Carbonate (CO32‒) and phosphate (PO43‒) salts are insoluble, except for the alkali metal and
ammonium salts (covered by rule 1 above).
Table 6.1 summarizes the above solubility rules in a grid of ionic compounds with a row for each
class of anions and a column for each class of cations.
- 85 -
Table 6.1. Solubility of Some Common Salts
NO3‒, ClO3‒, ClO4‒,
HCO3‒, CH3COO‒
Li+, Na+, K+, Rb+, Cs+
NH4+
Mg2+
Ca2+, Sr2+, Ba2+
Ag+, Pb2+, Hg22+
other
soluble
soluble
soluble
soluble
soluble
soluble
soluble
soluble
soluble
soluble
soluble
soluble
soluble
insoluble
insoluble
insoluble
soluble
soluble
soluble
insoluble
soluble
insoluble
insoluble
insoluble
insoluble
soluble
soluble
insoluble
insoluble
insoluble
SO42‒
‒
‒ ‒
Cl , Br , I
S2‒
O2‒, OH‒
CO32‒, PO43‒
Example 6.1: Which of the following salts are soluble?
(a) NaCl
(g) SrSO4
(b) AgCl
(c) Ba(OH)2
(h) Hg2(CH3COO)2
(d) Pb(NO3)2
(e) CaCO3
(f) PbSO4
Approach: First, identify the cation and ion. Apply the following steps, stopping once
solubility or insolubility is established. (1) If the cation is an alkali metal or ammonium, or if
the anion is nitrate, chlorate, perchlorate or ethanoate, the salt is soluble. (2) If the anion is
sulfate, the salt is soluble unless the cation is an alkaline earth metal other than magnesium.
(3) If the anion is a halide other than fluoride, the salt is soluble unless the cation is silver,
lead or mercury I. (4) If the anion is sulfide, the salt is insoluble unless the cation is an
alkaline earth metal. (5) If the anion is oxide or hydroxide, the salt is insoluble unless the
cation is an alkaline earth metal other than magnesium. (6) If the anion is carbonate or
phosphate, the salt is insoluble.
(a) Sodium salts are soluble. Therefore, NaCl is soluble.
(b) While most chlorides are normally soluble, silver chloride is not. AgCl is insoluble.
(c) While most hydroxides are insoluble, barium hydroxide is not. Ba(OH)2 is soluble.
(d) Nitrates are soluble. Therefore, Pb(NO3)2 is soluble.
(e) Carbonates are generally insoluble. Since the cation (calcium here) is not an alkali metal
or ammonium, CaCO3 is insoluble.
(f) Sulfates are generally soluble. Since the cation (lead here) is not calcium, strontium or
barium, PbSO4 is soluble.
(g) SrSO4 is an insoluble sulfate.
(h) Ethanoates are soluble. Therefore, Hg2(CH3COO)2 is soluble.
Precipitation Reactions
By mixing solutions, it is possible to obtain a solution with concentration above the solubility of a
salt. When this happens, the salt precipitates out of solution. For example, since sodium chloride
(NaCl) and silver nitrate (AgNO3) are both soluble salts, it is possible to prepare separate
aqueous solutions of these salts with significant concentration. These solutions can then be
mixed to produce a supersaturated solution with silver chloride concentration far in excess of the
solubility of silver chloride – silver chloride is an insoluble salt (see Table 6.1). Consequently,
solid silver chloride precipitates from solution. The balanced chemical equation for this
precipitation reaction is given by
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq) .
Here, AgNO3(aq) is shorthand for Ag+(aq) + NO3‒(aq), and similarly for the other aqueous salts.
Thus, the above equation can be written in ionic form as
- 86 -
Ag+(aq) + NO3‒(aq) + Na+(aq) + Cl‒(aq) → AgCl(s) + Na+(aq) + NO3‒(aq) .
When the reaction is written in ionic form, it becomes clear that the sodium and nitrate ions do not
participate – they are spectator ions. Sodium and nitrate ions can be cancelled from the equation
to give the net ionic equation which includes only the ions that participate in the precipitation
reaction. Specifically, the net ionic equation here is
Ag+(aq) + Cl‒(aq) → AgCl(s) .
Example 6.2: Write net ionic equations for the following precipitation reactions.
(a)
(b)
(c)
(d)
(e)
MgBr2(aq) + PbSO4(aq) → PbBr2(s) + MgSO4(aq)
SrBr2(aq) + (NH4)2SO4(aq) → 2 NH4Br(aq) + SrSO4(s)
Mg(ClO3)2(aq) + 2 KOH(aq) → 2 KClO3(aq) + Mg(OH)2(s)
3 Hg2(NO3)2(aq) + 2 Li3PO4(aq) → 6 LiNO3(aq) + (Hg2)3(PO4)2(s)
BaCl2(aq) + Ag2SO4(aq) → BaSO4(s) + 2 AgCl(s)
Approach: Identify the ions present in the ionic solutions. Look for the ions that form the
precipitate – i.e., the ions in the solid product. Leave the other ions out of the net ionic
equation.
(a) Pb2+(aq) + Br‒(aq) → PbBr2(s)
(b) Sr2+(aq) + SO42‒(aq) → SrSO4(s)
(c) Mg2+(aq) + 2 OH‒(aq) → Mg(OH)2(s)
(d) 3 Hg22+(aq) + 2 PO43‒(aq) → (Hg2)3(PO4)2(s)
(e) This reaction separates into two independent precipitation reactions:
Ba2+(aq) + SO42‒(aq) → BaSO4(s)
Ag+(aq) + Cl‒(aq) → AgCl(s)
The common factor of 2 is dropped from the latter reaction.
Example 6.3: Which of the following sets of solutions produce precipitation reactions when
mixed? If there is a precipitation reaction, what is the precipitate?
(a)
(b)
(c)
(d)
(e)
MgBr2(aq) + Rb2S(aq)
Ba(CH3COO)2(aq) + Na2SO4(aq)
Cu(ClO3)2(aq) + 2 NaOH(aq)
LiNO3(aq) + (NH4)3PO4(aq)
Na2CO3(aq) + ZnClO4(aq)
Approach: Identify the ions present in the ionic solutions. Match the cation from one
solution with the anion from the other – and vice versa. If either pair of ions forms an
insoluble salt, then there is a precipitation reaction.
(a) MgS and RbBr are both soluble salts. Therefore, there is no precipitation reaction.
(b) NaCH3COO is soluble, while BaSO4 is insoluble. Therefore, BaSO4(s) precipitates.
(c) NaClO3 is soluble, while Cu(OH)2 is insoluble. Cu2+ is in the “other” column of Table 6.1.
Cu(OH)2(s) is the precipitate.
- 87 -
(d) (Li)3PO4 and NH4NO3 are both soluble salts. Therefore, there is no precipitation reaction.
Since all lithium and nitrate salts are soluble, adding LiNO3(aq) to any other solution will not
produce a precipitation reaction.
(e) ZnCO3(s) is the precipitate. Zn2+ is in the “other” column of Table 6.1.
Precipitation reactions can be used to identify the ions present in a solution. For example, we
could distinguish three solutions, one containing magnesium, one containing calcium and one
containing silver, using two test solutions, Na2SO4(aq) and NaCl(aq). Na2SO4(aq) produces a
precipitate with calcium, but not magnesium or silver. NaCl(aq) produces a precipitate with silver,
but not magnesium or calcium.
6.2 Brønsted-Lowry Theory of Acid-Base Reactions
According to the Brønsted-Lowry definition of acids and bases, an acid reacts with a base by
exchanging an H+. For example,
HBr + NH3 → Br− + NH4+
Here, HBr is the H+ donor – the acid – while NH3 is the H+ acceptor – the base. The H+ is able to
leave the acid because the H-Br bond is polarized – the bonding electrons are shifted towards the
bromine atom. The Lewis definition of acids and bases explains that NH3 is able to accept the H+
because there is a lone pair of electrons on the nitrogen atom. These electrons are shared to
become the fourth N-H bond in NH4+.
Example 6.4: Identify the acid and base in the following reactions:
(a) NH2− + HF → NH3 + F−
(b) HCl + HCO3− → Cl− + H2CO3
(c) NH4+ + PO43− → NH3 + HPO42−
Approach: Look for the reactant that gained an H+, and that which lost an H+.
(a) HF loses an H+. It is the H+ donor – i.e., it is the acid. NH2− accepts the H+. It is the
base. Note that an acid always has an H bonded to an electronegative atom (N, O, F, S, Cl,
Br or I – often it is O). A base must have a lone pair of electrons to accept the H+. The lone
pair is usually on an N atom (often seen for uncharged basic molecules) or an O atom
(mostly seen for anionic bases).
(b) HCl is the acid – it loses an H+. HCO3− is the base – it accepts the H+.
(c) NH4+ is the acid – it loses an H+. PO43− is the base – it accepts the H+.
Acids and bases form acidic and basic aqueous solutions. Acids and bases can be molecules or
ions.
Salts form aqueous solutions by splitting into positive and negative ions that are solvated by
water. For example,
NaOH(s) → Na+(aq) + OH−(aq) .
Molecular compounds form aqueous solutions by becoming solvated by water. For example,
ethanoic acid (vinegar) dissolves in water, as indicated in the following reaction:
CH3COOH(l)
H2O
→ CH3COOH(aq) .
- 88 -
Acids and bases react with water in aqueous solution. Acids react by donating H+ to water. For
example,
HCl(aq) + H2O(l) → Cl−(aq) + H3O+(aq) .
In this reaction, H2O acts as a base. It accepts an H+.
Bases react by accepting an H+ from water. For example,
NH2−(aq) + H2O(l) → NH3(aq) + OH−(aq)
Here, H2O acts as an acid. It donates an H+. Some substances, such as water, can behave as
either acids or bases depending on what they react with. They are called amphoteric.
6.3 Strong and Weak Acids and Bases
The chemical reactions shown above have arrows to the right indicating the reaction proceeds to
completion. In reality, no reaction proceeds to completion; there is always a reverse reaction that
returns products back to reactants. However, we frequently encounter reactions that almost go to
completion. In such cases, the rate of the reverse reaction is very much smaller than the rate of
forward reaction. When equilibrium is obtained – i.e., when the rate of reverse reaction equals
the rate of forward reaction – there is very little reactant left. This amount is often negligible.
However, there are also many reactions where the rate of reverse reaction is not negligible; in
these cases there can be a significant amount of reactant present at equilibrium.
Many acids and bases react with water only to a limited extent – for example, ammonia:
NH3(aq) + H2O(l)
 NH4+(aq) + OH−(aq)
The limited degree of reaction is indicated here by the double arrow (showing forward and
reverse reactions). In fact, in an aqueous solution of ammonia there is only a small amount of
NH4+(aq). Most of the ammonia is unreacted. This is indicated by the asymmetry of the double
arrow. For this reason, NH3 is said to be a weak base. In contrast, the reaction of NH2− with
water shown above goes essentially to completion. NH2− is a strong base.
Acids are similarly labeled as strong or weak depending on whether the reaction with water goes
to completion. For example, ethanoic acid is a weak acid (we use a double arrow which shows
the preference for reactants):
CH3COOH(aq) + H2O(l)
 CH3COO−(aq) + H3O+(aq)
Nitric acid is strong acid (we use a single arrow which shows an overwhelming preference for
products):
HNO3(aq) + H2O(l) → NO3−(aq) + H3O+(aq)
The reverse of an acid-base reaction is itself an acid-base reaction. The acid and base of the
reverse reaction are the products of the forward reaction. When the acid of the forward reaction
loses an H+, the result is a base – the conjugate base. An H+ is accepted by the conjugate base
in the reverse reaction. Similarly, when the base of the forward reaction accepts an H+, the result
is an acid – the conjugate acid. The conjugate acid-base donates an H+ in the reverse reaction.
For example,
H2SO4(aq) + H2O(l)
acid
base
→
HSO4−(aq)
conjugate base of H2SO4
- 89 -
+
H3O+(aq)
conjugate acid of H2O
This reaction goes essentially to completion because H2SO4 is a stronger acid than the conjugate
acid of water, H3O+. The equilibrium is always shifted away from the stronger acid. Above, this
corresponds to the forward reaction. This is why we call H2SO4 a strong acid. Hydronium, H3O+,
is the prototype strong acid. Any acid stronger than H3O+ is also a strong acid.
In contrast,
HOCl(aq) + H2O(l)
acid
base

OCl−(aq)
conjugate base of HOCl
+
H3O+(aq)
conjugate acid of H2O
The acid, HOCl, is weaker than H3O+. The equilibrium is shifted towards reactants because an
acid-base equilibrium always shifts away from the stronger acid. Hypochlorous acid, HOCl, is
called a weak acid because it is weaker than H3O+.
Now consider some reactions of bases. When NaOH dissolves in water, Na+(aq) and OH−(aq)
result. Hydroxide, OH−, is the prototype strong base. Any alkali metal (group 1) or alkaline earth
(group 2) metal hydroxide produces OH−(aq) when dissolved in water (since MgO and Mg(OH)2
are insoluble, only very small concentrations are achieved). As such, these compounds are
considered strong bases.
Amides are also strong bases. For example, NaNH2 dissolves in water, producing Na+(aq) and
NH2−(aq). The amide ion, NH2−, is called a strong base because it is stronger than OH−.
NH2−(aq) + H2O(l)
base
acid
→
OH−(aq)
conjugate base of H2O
NH3(aq)
+
conjugate acid of NH2−
Here, the reaction essentially goes to completion. The equilibrium is (strongly) shifted toward
OH−, away from NH2−. Acid-base equilibria are always shifted away from the stronger base.
Thus, NH2− is a stronger base than OH−. Any base stronger than OH−, is also a strong base.
The two rules for the shift in any acid-base equilibrium, (1) the equilibrium always shifts away
from the stronger base, and (2) the equilibrium always shifts away from the stronger acid, are
equivalent. The stronger acid and stronger base are always on the same side of the reaction.
Ammonia, NH3, is called a weak base. In the reaction of ammonia and water, shown above, the
equilibrium is shifted away from OH−, towards NH3. Thus, NH3 is a weaker base than OH−. Any
base weaker than OH− is called a weak base.
Sodium hypochlorite, NaOCl, dissolves in water to produce aqueous hypochlorite ions, OCl−(aq).
Hypochlorite is a weak base which reacts with water as follows:
OCl−(aq) + H2O(l)
base
acid

HOCl(aq)
conjugate acid of OCl−
- 90 -
+
OH−(aq)
conjugate base of H2O
Table 6.2. Strong acids and some weak acids
Acid(s)
Conjugate
Base(s)
Strong
Acids
HCl, HBr, HI
HNO3
H2SO4
HClO3, HClO4
Cl−, Br−, I−
NO3−
HSO4−
ClO3−, ClO4−
The conjugate bases are
very weak bases
Weak
Acids
HF
H2S
H2CO3, HCO3−
HNO2
H3PO4, H2PO4−
HSO4−
H2SO3, HSO3−
HOCl, HClO2
CH3COOH
NH4+
F−
HS−
HCO3−, CO32−
NO2−
H2PO4−, HPO42−
SO42−
HSO3−, SO32−
OCl−, ClO2−
CH3COO−
NH3
The conjugate bases are
weak bases
Table 6.3. Strong bases and some weak bases
Strong
Bases
Weak
Bases
Base(s)
Conjugate Acid(s)
O2−, oxide ion
(e.g., Li2O, Na2O, CaO, SrO)
OH−
OH−, hydroxide ion
(e.g., LiOH, NaOH, KOH,
Ca(OH)2, Sr(OH)2, Ba(OH)2)
H2O
NH2−, amide ion (e.g., NaNH2)
H−, hydride ion (e.g., LiH)
S2−
NH3
H2
HS−
NH3, ammonia
CH3NH2, methylamine
HS−
F−
2−
CO3 , HCO3−
HPO42−, H2PO4−
OCl−, ClO2−
CH3COO−
NH4+, ammonium
CH3NH3+, methylammonium
H2S
HF
HCO3−, H2CO3
H2PO4−, H3PO4
HOCl, HClO2
CH3COOH
The conjugate acids are
very weak acids
The conjugate acids are
weak acids
Weak acids are conjugate to weak bases, whereas strong acids are conjugate to very weak
bases and strong bases are conjugate to very weak acids. These relationships are depicted in
Table 6.4.
- 91 -
Table 6.4. Acid and base strength of conjugate pairs
Acid
Conjugate Acid
Strong
Weak
Very Weak
Acid
Strength ↑
↔
↔
↔
↔
↔
Conjugate Base
Base
Very Weak
Weak
Strong
Base
Strength ↓
Example 6.5: Identify the conjugate acid and base pairs within the following set of species:
Cl−, OCl−, HSO3−, O2−, NH4+, NH3, OH−, H2O, HCl, HOCl, H2SO3, H3O+, NH2−, SO32−
Approach: Look for species differing by adding or subtracting a single H+.
Conjugate pairs are indicated here with double arrows (acid on the left and conjugate base
on the right):
HCl ↔ Cl−
H3O+ ↔ H2O ↔ OH− ↔ O2−
HOCl ↔ OCl−
NH4+ ↔ NH3 ↔ NH2−
H2SO3 ↔ HSO3− ↔ SO32−
Example 6.6: Given that
NH2−(aq) + H2O(l) → NH3(aq) + OH−(aq)
NH3(aq) + H2O(l)  NH4+(aq) + OH−(aq)
NH4+(aq) + OCl−(aq)  HOCl(aq) + NH3(aq)
(a) Order the bases (the species acting as a base in either the forward or reverse of one of these
reactions) according to increasing strength as a base (from weakest to strongest).
(b) Order the acids according to increasing strength.
Approach: (a) Each equilibrium shifts away from the stronger base. This tells us which base
is stronger in each reaction. Combining results from the three reactions allows the four bases
to be ordered according to strength. (b) Each equilibrium also shifts away from the stronger
acid. Therefore, proceed as in (a), except consider the acids. Alternatively, since the acids
are conjugate to the bases in part (a), the order according to acid strength is the reverse of
the order of the conjugate bases according to base strength.
(a) In the first reaction, NH2− is the reactant base. This strongly product-favored reaction
indicates that NH2− is a stronger base than OH−. The shift to the left of the second
equilibrium indicates that OH− is a stronger base than NH3. The shift to the left of the third
equilibrium indicates that NH3 is a stronger base than OCl−. Therefore. in order of increasing
strength as a base, we have
OCl− < NH3 < OH− < NH2− .
(b) The acids are just the conjugate acids of the bases listed. The order of acid strength is
the reverse of the above order (with conjugate acids replacing the bases); i.e., according to
strength as an acid,
NH3 < H2O < NH4+ < HOCl .
- 92 -
6.4 Types of Acids and Bases
Binary Acids
The simplest class of acids is the binary acids consisting of H bonded to an electronegative atom.
The most important of these are hydrosulfuric acid (H2S).and the hydrohalic acids: hydrofluoric
acid (HF), hydrochloric acid (HCl), hydrobromic acid (HBr) and hydroiodic acid (HI).
Oxoacids and Oxoanions
Many common acids are oxoacids. These are acids with a central atom bonded to one or more
oxygen atoms, with one or more of those oxygen atoms in turn bonded to a hydrogen atom.3 The
bases conjugate to the oxoacids are oxoanions. Some oxoacids can donate more than one H+ they are polyprotic. The common set of oxoacids and associated oxoanions are shown in Table
6.3.
Table 6.4. Oxoacids and oxoanions
Oxoacid
1st Conjugate Base
2nd Conjugate Base
H2CO3
carbonic acid
HCO3−
hydrogen carbonate
(bicarbonate)
HNO2
nitrous acid
NO2−
nitrite
HNO3
nitric acid
NO3−
nitrate
H3PO3
phosphorous acid
H2PO3−
dihydrogen phosphite
HPO32−
hydrogen phosphite
not acidic – see footnote
H3PO4
phosphoric acid
H2PO4−
dihydrogen phosphate
HPO42−
hydrogen phosphate
H2SO3
sulfurous acid
HSO3−
hydrogen sulfite
SO32−
sulfite
H2SO4
sulfuric acid
HSO4−
hydrogen sulfate
SO42−
sulfate
HClO
hypochlorous acid
OCl−
hypochlorite
HClO2
chlorous acid
ClO2−
chlorite
HClO3
chloric acid
ClO3
chlorate
HClO4
perchloric acid
ClO4
perchlorate
3rd Conjugate Base
CO32−
carbonate
PO43−
phosphate
−
−
3
There is one exception to this characterization of oxoacids: phosphorus acid (H3PO3), wherein a
(non-acidic) H is bonded to the central atom – phosphorus in this case..
- 93 -
Note that some elements as central atoms (e.g., N, S, P and Cl) have more than one oxoacid.
When there are two such species they are labeled ...ous acid and ...ic acid (e.g., nitrous and nitric
acids). The oxoacid with more O atoms (for which the central atom has the highest oxidation
number) is labeled ...ic acid, while that with fewer O atoms is labeled ...ous acid. The
corresponding oxoanions are labeled ...ite and ...ate. In the case of chlorine, there are four
oxoacids. The middle two (with respect to number of O atoms) acids are chlorous and chloric
acid, respectively. The acid with one fewer O than chlorous acid is called hypochlorous acid.
The acid with one more O than chloric acid is called perchloric acid. The conjugate base to
hypochlorous acid is hypochlorite, while the conjugate base to perchloric acid is perchlorate.
Other common oxoanions include the powerful oxidizing agents, dichromate, Cr2O72−, and
permanganate, MnO4−.
Organic Acids and Bases
There are many organic acids. The simplest are the alkanoic (carboxylic) acids derived from the
alkanes. The alkanes have the chemical formula, CnH2n+2. For example, for n = 1 and 2 we have
methane, CH4, and ethane, C2H6, respectively. The alkanoic acids are obtained by replacing 3 H
atoms attached to one of the end carbon atoms with a doubly bonded O and an OH group (the O
bonds to the carbon atom. The result is a compound with formula, CnH2n-1OOH (or RCOOH).
The simplest of these compounds are tabulated in Table 6.4. All carboxylic acids have structures
of the form,
O
R
O
H .
They are weak acids.
Simple amines are also derived from alkanes. We obtain a primary amine from an alkane by
replacing an H with an NH2, to get a compound with chemical formula, CnH2n+1NH2, The amines
derived in this way are weak bases similar to ammonia. Some of them are tabulated in Table 6.4.
Primary amines have structures of the form,
CH
NCH33
H3H
C
Cl
H
R
.
Table 6.5. Some simple organic acids and bases
Alkane
Alkanoic acid
Conjugate Base Alkanoate
Amine
CH4
methane
HCOOH
methanoic acid
(also known as formic acid)
HCOO−
methanoate
(also known as formate)
CH3NH2
methylamine
C2H6
ethane
CH3COOH
ethanoic acid
(also known as acetic acid)
CH3COO−
ethanoate
(also known as acetate)
CH3CH2NH2
ethylamine
C3H8
propane
CH3CH2COOH
propanoic acid
CH3CH2COO−
propanoate
CH3CH2CH2NH2
Propylamine
- 94 -
C4H10
butane
CH3CH2CH2COOH
butanoic acid
CH3CH2CH2COO−
butanoate
CH3CH2CH2CH2NH2
Butylamine
C5H12
pentane
CH3CH2CH2CH2COOH
pentanoic acid
CH3CH2CH2CH2COO−
pentanoate
CH3CH2CH2CH2CH2NH2
Pentylamine
C6H14
hexane
CH3CH2CH2CH2CH2COOH
hexanoic acid
CH3CH2CH2CH2CH2COO−
hexanoate
CH3CH2CH2CH2CH2CH2NH2
Hexylamine
Oxalic acid, H2C2O4 is another related, commonly encountered, organic acid. It is a dicarboxylic
acid. The associated 2nd conjugate base is oxalate, C2O42−. The structures of these two species
are given as follows:
H
.
O
O
O
O
-
H
O
O
O
O
-
oxalate
oxalic acid
6.5 Oxidation-Reduction Reactions
An oxidation-reduction (redox) reaction is an electron transfer reaction. As electrons are the
lightest particles present in ordinary matter, and they are a key ingredient of everything, this is the
most important class of chemical reactions. Electron transfer reactions are also an important
source of energy – in our bodies and to power everyday devices. For example, combustion of a
fuel is a redox reaction. Redox reactions also provide the energy delivered by a battery – see
Chapter 10. Also, cellular respiration involves a series of redox reactions that power our bodies
and every other living thing.
The simplest redox reactions are the reactions of metals and non-metals wherein the metal loses
one electron or more to form a positively charged cation, and the non-metal gains the electron(s)
to form a negatively charged anion. For example, magnesium reacts with chlorine to form
magnesium chloride:
Mg(s) + Cl2(g) → MgCl2(s) .
Magnesium chloride is an ionic solid – it consists of Mg2+ and Cl‒ ions in an alternating lattice. In
solid magnesium, electrons are shared equally by every magnesium atom – they are neutral
atoms. Thus, magnesium atoms lose two electrons in the above reaction, while chlorine atoms
gain one electron. We say that magnesium is oxidized, while chlorine is reduced. The reaction
can be split into half reactions that show the electrons lost by magnesium, and the electrons
gained by chlorine.
Mg → Mg2+ + 2 e‒
Cl2 + 2 e‒ → 2 Cl‒
lost electrons = oxidation
gained electrons = reduction
LEO
GER
The redox reaction is the sum of the oxidation and reduction half reactions, written with the same
number of electrons. Every electron lost in the oxidation half reaction is gained in the reduction
half reaction.
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In the above example, magnesium gets oxidized, while chlorine gets reduced. The oxidation of
magnesium requires the reduction of chlorine – the lost electrons must go somewhere. For this
reason, we say that chlorine is the oxidizing agent. It is the substance that oxidizes magnesium –
i.e., it accepts the lost electrons. Magnesium is the reducing agent. It is the source of the
electrons gained by chlorine – magnesium reduces chlorine.
In general, the reducing agent donates electrons, while the oxidizing agent accepts the electrons.
In this way, redox reactions are analogous to acid-base reactions – with electrons transferred
instead of H+ ions. In a redox reaction, a reducing agent, R, donates electrons to an oxidizing
agent, O. The oxidized form of R is the oxidizing agent (O′) in the reverse reaction, while the
reduced form of O is the reducing agent (R′) in the reverse reaction; i.e.,
R + O → O′ + R′ .
redox reaction
In the above reaction, R = Mg(s), O = Cl2(g), O′ = Mg2+ and R′ = Cl‒, with the latter two ions
existing together in the ionic solid, MgCl2(s). In the reaction of magnesium and chlorine the
products are strongly favored. This is typical for redox reactions. Redox reactions are shifted
away from the stronger reducing agent and stronger oxidizing agent. Thus, we can say that
Mg(s) is a stronger reducing agent than Cl‒, and Cl2(g) is a stronger oxidizing agent than Mg2+.
Oxidation Numbers
When covalent compounds are formed, it is not as easy to recognize a redox reaction. For
example, the combustion of carbon (e.g., in the form of coal) involves breaking CC and OO
bonds, and forming CO bonds:
C(s) + O2(g) → CO2(g) .
The carbon and oxygen atoms in the reactants are neutral because they equally share the
electrons. However, the electrons of the carbon oxygen bond are closer to oxygen, as oxygen is
more electronegative – i.e., the CO bond is polarized. In this reaction, electrons are effectively
transfered from carbon to oxygen. Oxidation numbers account for this electron transfer.
Assigning oxidation numbers is similar to assigning formal charge (see Chapter 5). When
assigning formal charges, electrons in a bond are split evenly between the two atoms bonded.
When assigning oxidation number, electrons in a bond are attributed entirely to the more
electronegative atom. Specifically, the oxidation number of an atom is given by
oxidation number = core charge – (number of non-bonding electrons)
– (number of bonding electrons to less electronegative element(s))
If an atom is bonded to another atom of the same type (same element), the two atoms have the
same electronegativity. In this case, the bonding electrons are shared equally between the
bonded atoms, and oxidation number is the same as formal charge.
In CO2, the four bonding electrons in each CO bond are attributed entirely to the oxygen atom.
Consequently,
the oxidation number of O = 6 – 4 – 4 = –2 , for each O atom
and
the oxidation number of C = 4 – 0 – 0 = +4 , for each C atom.
It is often not necessary to know the bonding within a compound to assign oxidation numbers to
atoms. There is a set of rules consistent with the above definition of oxidation number. First, if
the compound is ionic, break it into cation(s) and anion(s), and treat each species separately.
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1. The sum of all the oxidation numbers in a species equals its charge – zero for a neutral
molecule.
2. Atoms in elemental form (e.g., Br2(l), Fe(s), O2(g)) have oxidation number zero.
3. Fluorine (the most electronegative element) has oxidation number, –1, in all of its compounds.
4. Oxygen (the next most electronegative element) has oxidation number, –2, in its compounds –
except in peroxides (A-O-O-B compounds such as H2O2) where it has oxidation number, –1, and
in F2O where it has oxidation number, +2.
5. Chlorine, bromine and iodine have oxidation number –1, unless bonded to a more
electronegative element.
6. Group 16 elements, such as sulfur and selenium, have oxidation number –2, unless bonded to
a more electronegative element. Group 15 elements, such as nitrogen and phosphorus, have
oxidation number –3, unless bonded to a more electronegative element.
7. Hydrogen has oxidation number, +1, when bonded to non–metals (e.g., H2O and NH3). There
is one exception to this rule. Phosphorus has a lower electronegativity (2.19) than hydrogen
(2.20). Consequently, hydrogen has oxidation number, –1, in PH3. Hydrogen has oxidation
number –1 in main group metal (groups I and II, Al, Ga, In, Tl, Sn, Pb and Bi) and semimetal (B,
Si, Ge, As, Sb, Te and Po) hydrides. Some transition metals (e.g., Au) have higher
electronegativity than hydrogen, and hydrogen has oxidation +1 in the associated compound.
Example 6.7: Assign oxidation numbers for the atoms of the following species.
(a)
(b)
(c)
(d)
(e)
SO2(g)
H2SO3(aq)
HSO3–(aq)
H2SO4(aq)
CH3COOH(aq)
ethanoic acid
Approach: Use the above list of rules (using Lewis structures will also work).
(a) Both oxygen atoms have oxidation number, ON(O) = –2, and the sum of the oxidation
numbers must be zero. Therefore,
ON(S) + 2 ON(O) = 0
ON(S) + 2×(–2) = 0
or
ON(S) = +4 .
The oxidation number of sulfur is +4. We call this is the +IV state of sulfur.
(b) Hydrogen has oxidation number, ON(H) = +1. Also, ON(O) = –2 and
2 ON(H) + ON(S) + 3 ON(O) = 0
or
ON(S) = +4 .
In sulfurous acid, sulfur is in the +IV oxidation state – the same as sulfur in SO2. The
hydration reaction which produces sulfurous acid from sulfur dioxide and water is not a redox
reaction. It does not change the oxidation number of sulfur, oxygen or hydrogen.
(c) Here,
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ON(H) + ON(S) + 3 ON(O) = –1
or
ON(S) = +4 .
Again, sulfur is in the +IV oxidation state. The acid-base reaction, wherein sulfurous acid
donates a proton to form hydrogen sulfite, is not a redox reaction. Sulfur is in the +IV
oxidation in both of these species.
(d) Here,
2 ON(H) + ON(S) + 4 ON(O) = 0
or
ON(S) = +6 .
The oxidation number of sulfur in sulfuric acid is +6. This is the +VI oxidation state of sulfur
(the highest oxidation state of sulfur – it equals the core charge). A reaction producing
sulfuric acid from sulfurous acid is a redox reaction, since the oxidation number of sulfur
changes in such a reaction. Such a reaction requires an appropriate oxidizing agent.
Oxidation number and formal charge are opposing means of attributing charge to atoms within
compounds. Formal charge treats bonding electrons as equally shared between the bonded
atoms, while oxidation number treats bonding electrons as only belonging to the more
electronegative atom. The truth is somewhere in-between. Nevertheless, formal charge is the
right tool when choosing a best Lewis structure, and oxidation number is the right tool for
identifying electron transfer (i.e., redox) reactions. In the reaction of carbon and oxygen, the
oxidation number of carbon changes from 0 to +4. An increase in oxidation number corresponds
to oxidation. The oxidation number of oxygen changes from 0 to –2. A decrease in oxidation
number corresponds to reduction.
In general, a reaction is a redox reaction if there is a change in the oxidation number of some or
all of the atoms. Atoms whose oxidation number increases are oxidized, while atoms whose
oxidation number decreases are reduced.
Example 6.8: Which of the following reactions are redox reactions? How many electrons are
transfered in each redox reaction? Identify the oxidizing and reducing agents.
(a)
(b)
(c)
(d)
(e)
(f)
2 Na(s) + Cl2(g) → 2 NaCl(s)
2 Na(s) + 2 H2O(l) → 2 NaOH(aq) + H2(g)
HCl(g) + NH3(g) → NH4Cl(s)
OCl−(aq) + HSO3−(aq) → Cl−(aq) + HSO4−(aq)
HCl(g) + LiH(s) → LiCl(s) + H2(g)
H2SO4(aq) + 2 Br−(aq) + 4 H+(aq) → Br2(aq) + SO2(g) + 2 H2O(l)
Note that it is convenient to use H+(aq) as a shorthand for hydronium – otherwise, an additional
H2O is required on the other side of the reaction.
Approach: Assign oxidation numbers to the atoms in reactants and products. Changes in
oxidation numbers signal a redox reaction. The species with increasing oxidation number(s)
is the reducing agent (it gets oxidized). The species with decreasing oxidation number(s) is
the oxidizing agent (it gets reduced). The matching changes in oxidation number(s) (positive
for the reducing agent and negative for the oxidizing agent) correspond to the number of
electrons transfered.
(a)
Oxidation numbers:
2 Na(s) + Cl2(g) → 2 NaCl(s)
0
0
+1 −1
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Since the oxidation numbers of sodium and chlorine both change, this is a redox reaction.
Sodium loses electrons (it gets oxidized) – it is the reducing agent. Chlorine gains electrons
(it gets reduced) – it is the oxidizing agent. Since two sodium atoms each lose one electron,
two electrons are transfered.
Because electrons gained equals electrons lost, the number of electrons transfered is
determined by considered either electrons lost or electrons gained.
(b)
Oxidation numbers:
2 Na(s) + 2 H2O(l) → 2 Na+(aq) + 2 OH−(aq) + H2(g)
0
+1 −2
+1
−1 +1
0
The oxidation numbers of sodium and two of the hydrogen atoms change. Therefore, this is
a redox reaction. Sodium loses electrons (it gets oxidized) – it is the reducing agent. Two of
the hydrogen atoms in two water molecules gain electrons (H2O gets reduced) – water is the
oxidizing agent. Since two sodium atoms each lose one electron, two electrons are
transfered.
(c)
HCl(g) + NH3(g) → NH4Cl(s)
Oxidation numbers: +1 −1
−3 +1
−3 +1 −1
The oxidation numbers of every atom is the same of both side of the reaction. This is not a
redox reaction. It is an acid-base reaction.
(d)
OCl−(aq) + HSO3−(aq) → Cl−(aq) + HSO4−(aq)
Oxidation numbers: −2 +1
+1 +4 −2
−1
+1 +6 −2
The oxidation numbers of sulfur and chlorine atoms change. Therefore, this is a redox
reaction. Sulfur in hydrogen sulfite loses electrons (HSO3− gets oxidized) – it is the reducing
agent. Chlorine in hypochlorite gains electrons (OCl− gets reduced) – water is the oxidizing
agent. Since each sulfur atom loses two electrons, and there is only one sulfur atom in the
balanced reaction, two electrons are transfered.
(e)
HCl(g) + LiH(s) → LiCl(s) + H2(g)
Oxidation numbers: +1 −1
+1 −1
+1 −1
0
The oxidation numbers of two hydrogen atoms change – one in hydrogen chloride, and one
in lithium hydride. Therefore, this is a redox reaction. The hydride ion loses one electron (H−
gets oxidized) – hydride is the reducing agent. Hydrogen in hydrogen chloride gains one
electron (HCl gets reduced) – hydrogen chloride is the oxidizing agent. Since each hydride
ion loses one electron, one electron is transfered.
The above reaction of hydrogen chloride and lithium hydride is also a proton transfer reaction.
As such, it is both an acid-base and a redox reaction.
(f)
H2SO4(aq) + 2 Br−(aq) + 4 H+(aq) → Br2(aq) + SO2(g) + 2 H2O(l)
Oxidation numbers: +1 +6 −2
−1
+1
0
+4 −2
+1 −2
The oxidation numbers of sulfur and bromine atoms change. Therefore, this is a redox
reaction. Each bromide ion loses one electron (Br− gets oxidized) – bromide is the reducing
agent. Sulfur in sulfuric acid gains two electrons (H2SO4 gets reduced) – sulfuric acid is the
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oxidizing agent. Since each sulfur atom gains two electrons, and there is only one sulfur
atom in the balanced reaction, two electrons are transfered.
Note that, because the reaction of sulfuric acid and bromide to give sulfur dioxide and
bromine does not have a balanced number of oxygen and hydrogen atoms, the balanced
reaction involves the solvent (H2O) and hydronium. The latter is only present in significant
abundance in an acidic solution. This redox reaction is balanced for an acidic aqueous
solution – as is appropriate for a reaction involving the strong acid, H2SO4.
Table 6.6. Some oxidizing and reducing agents
Oxidizing Agent
product
(corresponding reducing
agent in reverse reaction)
Reducing Agent
product
(corresponding oxidizing
agent in reverse reaction)
O2
O2−
(or compound with
ON(O) = ‒2)
H2
H+
(or compound with
ON(H) = +1)
H2O2
H2O
Li, Na, K, Rb, Cs
Li+, Na+, K+, Rb+, Cs+
F2, Cl2, Br2 and I2
F−, Cl−, Br− and I−
Mg, Ca, Sr, Ba, Al
Mg2+, Ca2+, Sr2+, Ba2+, Al3+
H2SO4, HNO3
HClO3, HClO4, HBrO3
SO2, NO
−
−
Cl , Br
C, S, P
e.g., CO, CO2, SO2, SO3,
P4O6, P4O10
Cr2O72− (in acid)
CrO42− (in base)
3+
Cr (in acid)
Cr(OH)3 (in base)
hydrocarbons: CnHm
e.g., CO, CO2
HOCl (in acid)
−
OCl (in base)
Cl−
H2S, HS−
e.g., SO2
MnO4−
Mn2+ (in acid)
MnO2 (in base)
Fe, Zn
2+
2+
Fe , Zn
Ag+, Fe3+
Ag, Fe2+
Fe2+
Fe3+
Sn4+
Sn2+
I−
I2 or IO3−
H+
with active metal
H2
SO32−, NO2−
SO42−, NO3−
PbO2
Pb2+ (in acid)
alcohols such as
CnH2n+1OH
aldehyde or carboxylic acid,
CnH2nO or CnH2n-1OOH
Table 6.4 shows some common oxidizing and reducing agents. For example, the halogens are
powerful oxidizing agents. They are electronegative elements that form halides. In such
reactions, the oxidation number of each halogen atom decreases from 0 to ‒1. Other oxidizing
agents have elements with high oxidation number (e.g., ON(Cr) = +6 in Cr2O72−) for which stable
low oxidation states exist (e.g., ON(Cr) = +3 in Cr3+).
Oxidizing agents include an element whose oxidation number decreases. Identification of such
an element is aided by first ruling out any element in its lowest oxidation state. For example, in
sulfuric acid, the oxygen atoms are in the ‒2 oxidation state. This is the lowest oxidation state of
oxygen. The oxygen atoms cannot be further reduced. Therefore, when sulfuric acid acts as an
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oxidizing agent, it is the sulfur or hydrogen atom that gets reduced. In sulfuric acid, sulfur is in the
+VI oxidation state. It can gain two electrons to give the stable +IV oxidation state. If the
reducing agent is sufficiently strong, the hydrogen atoms of sulfuric acid can also be reduced.
Hydrogen in sulfuric acid is in the +I oxidation state. It can gain an electron to form the zero
oxidation state of hydrogen gas.
Similarly, reducing agents include an element whose oxidation number increases. In this case,
any element in its highest oxidation state can be ruled out. Thus, sulfuric acid cannot act as a
reducing agent under anything but extraordinary conditions. The hydrogen atom and sulfur atom
are already in their highest oxidation state. Only the oxygen atom can be oxidized. However, the
‒2 oxidation state of oxygen is very stable in sulfuric acid. Extraordinary oxidizing conditions are
required to make sulfuric acid act as a reducing agent. Such a reaction produces a species such
as persulfuric acid (H2SO5) which has an O-O bond.
The strength of oxidizing agents varies. The halogens are ordered according to oxidizing
strength with strength decreasing down the group. Fluorine is the strongest oxidizing agent.
Chlorine is next strongest, and so on. Specifically, according to oxidizing strength,
F2 > Cl2 > Br2 > I2 .
Since fluorine is a stronger oxidizing agent than chlorine, fluorine can oxidize chloride,
F2(g) + 2 Cl−(aq) → 2 F−(aq) + Cl2(g) .
Similarly, chlorine can oxidize bromide,
Cl2(g) + 2 Br−(aq) → 2 Cl−(aq) + Br2(g) .
These reactions proceed away from the stronger oxidizing agent. They also proceed away from
the stronger reducing agent. Thus, Cl− is a stronger reducing agent than F−. Br− is a stronger reducing
agent than Cl−, and so on. Specifically, according to strength as reducing agent,
I− > Br− > Cl− > F− .
The strongest oxidizing agent becomes the weakest reducing agent after it has accepted
electrons. Similarly, the strongest reducing agent becomes the weakest oxidizing agent upon
accepting electrons.
Concentrated sulfuric acid is a powerful oxidizing agent. It can oxidize bromide, but not chloride.
It is therefore a stronger oxidizing agent than bromine, but weaker than chlorine. The above
ordering of oxidizing agents can thus be extended:
F2 > Cl2 > H2SO4(conc) > Br2 > I2 .
6.6 Balancing Redox Reactions
Once you have identified the oxidizing and reducing agents, and their associated products, the
overall reaction can be balanced by matching the electrons lost by the reducing agent with the
electrons gained by the oxidizing agent. For example, when concentrated nitric acid solution
reacts with copper, the unbalanced half reactions are
and
HNO3(aq) + 3 e− → NO(g)
Cu(s) → Cu2+(aq) + 2 e− .
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To balance these half reactions, with respect to electrons, we must multiply the first half reaction by 2 and
the second by 3. In the balanced redox reaction, six electrons are transfered. Adding the scaled half
reactions gives
2 HNO3(aq) + 3 Cu(s) → 2 NO(g) + 3 Cu2+(aq) .
Here, electrons and atoms are balanced, except for hydrogen and oxygen. This reaction involves
solvent (in this case water) participation. This situation was encountered above – specifically, the
balanced reaction of sulfuric acid and bromide in aqueous solution involved H2O and H+. The
participation of solvent in aqueous reactions is indicated by an imbalance of hydrogen and
oxygen atoms in the otherwise balanced reaction (electrons and atoms other than hydrogen and
oxygen balanced) of oxidizing and reducing agent.
Balancing reactions in aqueous solution typically involves hydronium or hydroxide. As such, we
distinguish between acidic and basic solutions. In acidic solutions, only hydronium is readily
available, in addition to water. In basic solutions, hydroxide is the only other readily available
species. Therefore, in acidic solution, the reaction is balanced using H2O and H+, while in basic
solution, the reaction is balanced using H2O and OH−.
Acid Solution
The following steps balance a reaction in acid:
(1) Identify the oxygen deficient side of the reaction. Add one H2O to that side for each missing
oxygen atom.
(2) Identify the hydrogen deficient side of the reaction. Add one H+ to that side for each missing
hydrogen atom.
Base Solution
To balance a reaction in base, there are two strategies. The following steps provide one strategy:
(1) Identify the oxygen deficient side of the reaction. Add one OH− to that side for each missing
oxygen atom.
(2) Identify the hydrogen deficient side of the reaction. Add one H2O to that side, and one OH− to
the other side, for each missing hydrogen atom.
Alternatively, you can balance the reaction as though the solution were acidic – i.e. using H2O
and H+, and then add an equal amount of OH− to both sides to neutralize H+ – i.e. replace H+ +
OH− by H2O. In the last step, you include only the net water on one side of the reaction.
The above rules can be applied to the separate half reactions, as well as the overall redox
reaction. This means we can write balanced half reactions (with electrons appearing as reactants
or products), then combine them with electrons cancelling to produce the overall balanced
reaction (additional cancelation of H+, OH− and H2O may also be required). However, if the goal
is simply to find the overall balanced reaction, it is faster to balance the half reactions only with
respect to atoms other than hydrogen and oxygen, combine them (cancelling electrons), then
balance with H2O and H+ or OH− (as described above).
Example 6.9: Balance the following unbalanced redox reactions. Acidic or basic conditions are
indicated where applicable. Identify the oxidizing and reducing agents.
(a) Fe3+(aq) + Mn(s) → Fe2+(aq) + Mn2+(aq)
(b) Cr2O72−(aq) + CH3OH(aq) → Cr3+(aq) + HCOOH(aq)
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(in acid)
(c)
(d)
(e)
(f)
MnO4−(aq) + CH3CH2OH(aq) → Mn2+(aq) + CH3COOH(aq) (in acid)
(in acid)
H2O2(aq) + Al(s) → H2O(l) + Al3+(aq)
MnO4−(aq) + I−(aq) → MnO2(s) + I2(s)
(in base)
OCl−(aq) + Br−(aq) → Cl−(aq) + BrO3−(aq) (in base)
Approach: Assign oxidation numbers. Identify oxidizing and reducing agents, and
associated products. Separate the half reactions, balance them with respect to elements
other than hydrogen and oxygen, then insert the gained or lost electrons. Combine the half
reactions (scaled, if necessary) so that the electrons cancel. Balance hydrogen and oxygen
using the protocol appropriate to the given conditions, acidic or basic.
(a)
Fe3+(aq) + Mn(s) → Fe2+(aq) + Mn2+(aq)
Oxidation numbers: +3
0
+2
+2
Iron III is the oxidizing agent, while elemental manganese is the reducing agent. The half
reactions are
and
Mn(s) → Mn2+(aq) + 2 e−
oxidation
Fe3+(aq) + e− → Fe2+(aq) .
reduction
The half reactions are both balanced with respect to atoms. The balanced overall reaction is
given by combining these half reactions, with the second half reaction scaled by two to
balance the electrons.
2 Fe3+(aq) + Mn(s) → 2 Fe2+(aq) + Mn2+(aq) .
(b)
Cr2O72−(aq)
Oxidation numbers: +6 −2
+
CH3OH(aq)
→
−2 +1 −2 +1
Cr3+(aq) + HCOOH(aq)
+3
+1 +2 −2
+1
Dichromate is the oxidizing agent, while methanol is the reducing agent. The unbalanced
(with respect to H and O) half reactions are
and
CH3OH(aq) → HCOOH(aq) + 4 e−
oxidation
Cr2O72−(aq) + 6 e− → 2 Cr3+(aq) .
reduction
These reactions are balanced with respect to atoms other than hydrogen and oxygen. The
second reaction requires a coefficient of two for chromium III, to balance the chromium
atoms. To balance electrons, combine the oxidation half reaction scaled by three with the
reduction half reaction scaled by two, to get
2 Cr2O72−(aq) + 3 CH3OH(aq)
→
4 Cr3+(aq) + 3 HCOOH(aq) .
It remains to balance hydrogen and oxygen. There are 17 O atoms on the left, and 6 O
atoms on the right. Therefore, we add 11 H2O(l) to the right to balance oxygen. The result is
2 Cr2O72−(aq) + 3 CH3OH(aq)
→
4 Cr3+(aq) + 3 HCOOH(aq) + 11 H2O(l) .
There are now 12 H atoms on the left, and 28 H atoms on the right. Therefore, we add 16
H+(aq) to the left to balance hydrogen. The result is
2 Cr2O72−(aq) + 3 CH3OH(aq) + 16 H+(aq) →
- 103 -
4 Cr3+(aq) + 3 HCOOH(aq) + 11 H2O(l) .
(c)
MnO4−(aq)
Oxidation numbers: +7 −2
+
CH3CH2OH(aq)
→
Mn2+(aq) + CH3COOH(aq)
−3 +1 −1 +1 −2 +1
+2
−3 +1 +3 −2 +1
Here, oxidation numbers for atoms in ethanol and ethanoic acid are assigned using
structures. This gives different values for the two carbon atoms in both compounds: −3 and
−1 for carbon in ethanol, and −3 and +3 for carbon in ethanoic acid.
If the listed rules for assigning oxidation number are applied, the average oxidation number of
carbon in these compounds is determined: −2 for ethanol, and 0 for ethanoic acid. Either
approach – assigning oxidation numbers to individual atoms, or average oxidation number for
each type of atom – gives the same number electrons transfered. With individual atom
assignments, the carbon atom bonded to oxygen in ethanol loses four electrons. Using
average oxidation numbers, we say that the two carbon atoms of ethanol each lose two
electrons. In either case, four electrons are lost.
Permanganate is the oxidizing agent in the above reaction, while ethanol is the reducing
agent. The unbalanced half reactions are
and
CH3CH2OH(aq) → CH3COOH(aq) + 4 e−
oxidation
MnO4−(aq) + 5 e− → Mn2+(aq) .
reduction
These reactions are balanced with respect to atoms other than hydrogen and oxygen. To
balance electrons, combine the oxidation half reaction scaled by five with the reduction half
reaction scaled by four, to get
4 MnO4−(aq) + 5 CH3CH2OH(aq)
→
4 Mn2+(aq) + 5 CH3COOH(aq) .
It remains to balance hydrogen and oxygen. There are 21 O atoms on the left, and 10 O
atoms on the right. Therefore, we add 11 H2O(l) to the right to balance oxygen. The result is
4 MnO4−(aq) + 5 CH3CH2OH(aq)
4 Mn2+(aq) + 5 CH3COOH(aq) + 11 H2O(l) .
→
There are now 30 H atoms on the left, and 42 H atoms on the right. Therefore, we add 12
H+(aq) to the left to balance hydrogen. The result is
4 MnO4−(aq) + 5 CH3CH2OH(aq) + 12 H+(aq) →
(d)
4 Mn2+(aq) + 5 CH3COOH(aq) + 11 H2O(l)
H2O2(aq) + Al(s) → H2O(l) + Al3+(aq)
Oxidation numbers: +1 −1
0
+1 −2
+3
The unbalanced half reactions (the first is actually balanced) are
and
Al(s) → Al3+(aq) + 3 e−
oxidation
H2O2(aq) + 2 e− → 2 H2O(l) .
reduction
Here, aluminum is the reducing agent, while hydrogen peroxide is the oxdizing agent. In this
case, it is necessary to balance the O atoms in the reduction half reaction because it is the
oxygen whose oxidation number decreases. To balance the electrons, combine the half
reactions, with the first half reaction scaled by two and the second scaled by three. The
result is
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2 Al(s) + 3 H2O2(aq) → 2 Al3+(aq) + 6 H2O(l) .
This reaction is balanced with respect to all atoms except hydrogen. There are 6 H atoms on
the left, and 12 on the right. To balance the hydrogen, add 6 H+(aq) to the left to get
2 Al(s) + 3 H2O2(aq) + 6 H+(aq) → 2 Al3+(aq) + 6 H2O(l) .
(e)
MnO4−(aq) + I−(aq) → MnO2(s) + I2(s)
Oxidation numbers: +7 −2
−1
+4 −2
0
The unbalanced half reactions are
and
2 I−(aq) → I2(s) + 2 e−
oxidation
MnO4−(aq) + 3 e− → MnO2(s) .
reduction
In the oxidation half reaction, a coefficient of 2 was introduced to balance the I atoms. To
balance the electrons, combine the half reactions with the first scaled by three and the
second scaled by two to get
2 MnO4−(aq) + 6 I−(aq) → 2 MnO2(s) + 3 I2(s) .
Since the solution is basic, oxygen is balanced by added hydroxide. There are 8 O atoms on
the left, and 4 on the right. Therefore, add 4 OH−(aq) to the right.
2 MnO4−(aq) + 6 I−(aq) → 2 MnO2(s) + 3 I2(s) + 4 OH−(aq) .
There are 4 H atoms on the right, and none on the left. To balance hydrogen, add 4 H2O(l) to
the left and 4 OH−(aq) to the right to get
2 MnO4−(aq) + 6 I−(aq) + 4 H2O(l) → 2 MnO2(s) + 3 I2(s) + 8 OH−(aq) .
(f)
OCl−(aq) + Br−(aq) → Cl−(aq) + BrO3−(aq)
Oxidation numbers: −2 +1
−1
−1
+5 −2
The unbalanced half reactions are
and
Br−(aq) → BrO3−(aq) + 6 e−
oxidation
OCl−(aq) + 2 e− → Cl−(aq) .
reduction
To balance the electrons, combine the half reactions with the second scaled by three to get
3 OCl−(aq) + Br−(aq) → 3 Cl−(aq) + BrO3−(aq)
There are 3 O atoms on the left, and 3 on the right. Also, there no H atoms on either side.
Therefore, the reaction is balanced as written.
Disproportionation Reactions
Sometimes the oxiding and reducing agent are the same species – i.e. it reacts with itself. Such
reactions are called disproportionation reactions. Hydrogen peroxide is an example of a species
that reacts in this way. Hydrogen peroxide spontaneously decomposes into water and oygen
according to the decomposition reaction (shown here in aqueous solution),
- 105 -
2 H2O2(aq) → 2 H2O(l) + O2(g) .
Oxidation numbers:
+1 −1
+1 −2
0
Here, two oxygen atoms from two hydrogen peroxide molecules each lose one electron and form
oxygen gas, while the other two oxygen atoms each gain one electron and become the oxygen
atoms in two water molecules.
Hydrogen peroxide solution is used as a disinfectant. It is a powerful oxidizer that kills bacteria.
Hydrogen peroxide can be used, in spite of its spontaneous decomposition reaction, because the
decomposition reaction is very slow. However, the decomposition reaction does give hydrogen
peroxide a limited shelf life. Old bottles of hydrogen peroxide solution stored at room temperature
are mostly just water.
The decomposition of hydrogen peroxide is considerably faster in the presence of a catalyst.
Manganese dioxide, and enzymes in blood, are catalysts that speed up this reaction.
Another example of a disproportionation reaction of chlorine in aqueous solution,
Cl2(aq) + H2O(l)  HCl(aq) + HOCl(aq) .
Oxidation numbers:
0
+1 −1
+1 −1
+1 −2 +1
This reaction only occurs to a limited extent – it is an equilibrium. The extent of this reaction
depends on the pH of the solution. It is shifted toward reactants in acidic or neutral solutions, and
towards products in basic solutions – Cl−(aq) and OCl−(aq) are the products under basic
conditions.
6.7 Visible Reactions
Some reactions result in a visible change – i.e., we can “see” the reaction taking place. For
example, in a precipitation reaction, tiny solid particles emerge from solution to form a suspension
of particles within the liquid. We observe the solution turn cloudy. The solution might appear
white, like milk. This happens when the solid is white. For colored solids, the result is a colored
solid suspension that looks like some type of “milkshake”. The solid precipitate separates from
the solution over time - the solid particles generally collect at the bottom of the suspension, under
the force of gravity. A centrifuge quickens this process.
Visible changes are also evident in reactions that form a gaseous product. Acid-base reactions
sometimes result in gas formation. For example, when HCl(aq) reacts with NaHCO3(aq),
H2CO3(aq) is formed. Carbonic acid is subject to decomposition into water and carbon dioxide.
Carbon dioxide emerges from solution as bubbles of gas. Overall,
HCl(aq) + NaHCO3(aq) → NaCl(aq) + H2O(l) + CO2(g) .
Redox reactions can also produce gases. Metals which are stronger reducing agents than H2(g)
react with acids to produce H2(g). For example,
and
Zn(s) + 2 H+(aq) → Zn2+(aq) + H2(g)
Fe(s) + 2 H+(aq) → Fe2+(aq) + H2(g) .
Both acid-base and redox reactions can produce color changes. Some colored compounds
change color or become colorless when they are protonated (i.e., they accept a proton and form
the conjugate acid) or deprotonated (they donate a proton and form the conjugate base). Such
- 106 -
compounds are called acid-base indicators. They distinguish between acidic and basic solutions.
Similarly, some colored compounds change color or become colorless when they are oxidized or
reduced. These compounds are called redox indicators. They distinguish between oxidizing and
reducing solutions.
The roadside breathalyzer test uses the redox reaction of dichromate and ethanol to measure the
amount of ethanol in a driver’s breath.
2 Cr2O72−(aq) + 3 CH3CH2OH(aq) + 16 H+(aq) →
4 Cr3+(aq) + 3 CH3COOH(aq) + 11 H2O(l) .
Dichromate solution is orange. When it reacts with ethanol to form chromium III and ethanoic
acid, the chromium III produced is green (due to the presence of sulfate in the solution used –
otherwise it is violet-blue). The color change is quantified by a colorimeter, and calibrated to yield
the ethanol concentration in the driver’s blood. Here, we see that dichromate is a redox indicator
that can detect the reducing environment of an aqueous ethanol solution. In the breathalyzer, a
solution of potassium dichromate, sulfuric acid and silver nitrate is used. Sulfuric acid is included
because H+(aq) is a reactant in the balanced reaction – increasing the H+(aq) concentration
“pushes” the reaction forward. Silver I acts as a catalyst – i.e., it increases the speed of reaction.
Permanganate is also a redox indicator sensitive to a reducing environment. Permanganate is a
strong oxidizing agent which forms a deep purple aqueous solution. If the solution is acidic,
permanganate produces manganese II which appears pale pink. If the solution is basic, the
product is manganese dioxide, a dark brown solid that precipitates from solution. The powdered
solid appears black, while the precipitate appears brown.
Reactions with a visible change provide means of testing an unknown substance or solution.
Adding specific substances to samples of an unknown and watching for visible reactions helps to
identify the substance.
Example 6.10: Predict the type of reaction(s) (precipitation, acid-base or redox) for the following
reactants. Which reactions produce a visible change?
(a)
(b)
(c)
(d)
(e)
(f)
(g)
LiOH(aq) + HCl(aq)
LiHCO3(aq) + HNO3(aq)
Cr2O72−(aq) + CH3OH(aq)
H2O2(aq) (in the presence of MnO2(s))
H2SO4(aq) + Al(s)
MnO4−(aq) + H2S(aq) (acidic solution)
LiOH(aq) + Ca(HCO3)2(aq)
Approach: (1) Look for acids and bases. There must be both an acid and base for an acidbase reaction to occur. (2) Look for oxidizing and reducing agents. Both must be present in
a redox reaction. (3) Look for cations and anions that form an insoluble solid, and result in a
precipitation reaction. Precipitation reactions produce a visible change. (4) Look for gases
formed, or color changes (if you know the colors) that also produce a visible change.
(a) LiOH(aq) + HCl(aq)
or
Li+(aq) + OH‒(aq) + H+(aq) + Cl‒(aq)
Here, we have aqueous hydroxide and hydronium ions. They neutralize in an acid-base
reaction. The overall reaction is
LiOH(aq) + HCl(aq) → LiCl(aq) + H2O(l) .
This reaction produces no visible change.
- 107 -
If an acid-base indicator is added to one of the solutions which is then employed as a limiting
reactant – i.e., the other solution is more abundant – a color change results from this reaction.
In this case, there is a visible change.
(b) LiHCO3(aq) + HNO3(aq)
Li+(aq) + HCO3‒(aq) + H+(aq) + NO3‒(aq)
or
This is another acid-base reaction. Here, the base is hydrogen carbonate. The products are
LiNO3(aq) and H2CO3(aq). The carbonic acid formed in this reaction subsequently
decomposes into carbon dioxide and water. The overall reaction is
LiHCO3(aq) + HNO3(aq) → LiNO3(aq) + CO2(g) + H2O(l) .
The carbon dioxide gas formed bubbles out of solution. This is visible change.
(c) Cr2O72−(aq) + CH3OH(aq)
This is a redox reaction. Dichromate is a strong oxidizing agent and methanol is a reducing
agent. This is just the breathalyzer reaction, except that the alcohol is methanol instead of
ethanol. Dichromate changes color when it gets reduced – it changes from orange to blue (or
green, if sulfate is present). This is a visible change.
(d) H2O2(aq) (in the presence of MnO2(s))
Hydrogen peroxide decomposes in the presence of MnO2(s). It undergoes decomposition,
and produces water and oxygen. Oxygen gas bubbles out of solution, giving a visible
change. This reaction is exothermic – i.e., it produces heat. The heat produced by the
reaction accelerates the reaction. It can be explosive.
(e) H2SO4(aq) + Al(s)
This is a redox reaction. Aluminum is a relatively active metal – i.e., it is a strong reducing
agent. The hydronium in the sulfuric acid solution is a sufficiently strong oxidizing agent to
oxidize the aluminum. The product of the reduction of hydronium is hydrogen gas, which
bubbles out of solution. Again, we have a visible change.
(f) MnO4−(aq) + H2S(aq) (acidic solution)
This is another redox reaction. Permanganate is a strong oxidizing agent which easily
oxidizes hydrogen sulfide with sulfur in its lowest oxidation state (‒II). Since permanganate is
deep purple and the product of its reduction (manganese II) is pale pink, there is a visible
change.
(g) LiOH(aq) + Ca(HCO3)2(aq)
or
Li+(aq) + OH‒(aq) + Ca2+(aq) + HCO3‒(aq)
This is an acid-base reaction. Here, there is a strong base, OH‒(aq), that reacts with the
weak acid, HCO3‒(aq). We know that HCO3‒(aq) is a weak acid, because it is conjugate to a
weak base, CO32‒(aq).
None of the oxoanions in Table 6.3 are strong bases – they are either weak or very weak
bases. The conjugate acids are correspondingly weak or strong acids. Knowing this will help
you identify acid-base reactions.
Problems:
6.1 Which of the following sets of solutions produce precipitation reactions when mixed? If there
is a precipitation reaction, what is the precipitate?
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(a)
(b)
(c)
(d)
(e)
(f)
(g)
Ba(OH)2(aq) + MgSO4(aq)
NaHCO3(aq) + Hg2SO4(aq)
RbOH(aq) + SrCl2(aq)
Pb(CH3COO)2(aq) + MgI2(aq)
Li2S(aq) + Ca(NO3)2(aq)
Li2S(aq) + Cu(NO3)2(aq)
K2CO3(aq) + AgClO3(aq)
6.2 Which of the following reactions are acid-base reactions? For those that are acid-base
reactions, identify the acid and the base.
(a)
(b)
(c)
(d)
(e)
2 Li(s) + Cl2(g) → 2 LiCl(s)
LiOH(aq) + HBr(g) → LiBr(aq) + H2O(l)
KOCl(aq) + HClO3(aq) → KClO3(aq) + HOCl(aq)
CaCO3(s) → CaO(s) + CO2(g)
CH3COOH(g) + NH3(g) → NH4CH3COO(s)
6.3 Identify each of the following as a strong/weak acid/base, or neither acid nor base. What is
the difference between a strong acid and a weak acid?
(a) HF
(b) CH4
(c) KOH
(d) CH3NH2
(e) HClO4
6.4 Identify the conjugate acid-base pairs present in the following reactions:
(a) H3PO4(aq) + H2O(l)  H2PO4−(aq) + H3O+(aq)
(b) CH3NH2(aq) + H2O(l)  CH3NH3+(aq) + OH−(aq)
(c) HClO4(aq) + H2O(l) → ClO4−(aq) + H3O+(aq)
6.5 Some acids are cations, while some bases are anions. When we dissolve a salt in water, we
get an acidic solution if the cation is an acid (strong or weak, but not very weak), and a basic
solution if the anion is a base (strong or weak, but not very weak). If the salt is composed of
an acid and a base, the pH of the resulting solution depends on the relative strength of the
cation as acid and anion as base.
Which of the following salts will produce an acidic/basic solution when dissolved in water?
(a) LiF
(b) NH4Cl
(c) KBr
(d) CH3NH3Cl
(e) NaOCl
6.6 Write out the chemical equilibria for the reactions of the acidic/basic ions (in question 6.5)
with water. Indicate whether the equilibrium is shifted to the right or shifted to the left – i.e.,
use arrow size to depict the shift in equilibrium.
6.7 Identify each of the following reactions as a precipitation, acid-base or redox reaction. For
acid-base reactions, identify the acid and base. For redox reactions, identify the oxidizing
and reducing agents and specify the number of electrons transfered.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
C3H10(g) + 11/2 O2(g) → 3 CO2(g) + 5 H2O(l)
Na2SO4(aq) + SrBr2(aq) → SrSO4(s) + 2 NaBr(aq)
NH4NO3(s) → N2O(g) + 2 H2O(l)
OCl−(aq) + HSO4−(aq) → HOCl(aq) + SO42−(aq)
H2O(l) + NaH(s) → NaOH(aq) + H2(g)
3 S(s) + 2 KClO3(s) → 3 SO2(g) + 2 KCl(s)
3 BrO−(aq) → BrO3−(aq) + 2 Br−(aq)
6.8 Balance the following unbalanced redox reactions. Acidic or basic conditions are indicated
where applicable. Identify the oxidizing and reducing agents.
- 109 -
(a)
(b)
(c)
(d)
(e)
(f)
(g)
Co2+(aq) + Cr(s) → Co(s) + Cr3+(aq)
Hg2+(aq) + Ag(s) → Hg22+(aq) + Ag+(aq)
ClO4−(aq) + Sn2+(aq) → Cl−(aq) + Sn4+(aq)
Cl−(aq) + MnO2(s) → Mn2+(aq) + HOCl(aq)
PbO2(s) + SO42-(aq) + Pb(s) → PbSO4(s)
MnO4−(aq) + Mn2O3(s) → MnO2(s)
O2(g) + Ba(s) → OH−(aq) + Ba2+(aq)
(in acid)
(in acid)
(in acid)
(in base)
(in base)
6.9 Which of the following substances react? For those that do, specify the type of reaction, and
whether it is visible.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
LiClO3(aq) + BaCl2(aq)
LiOCl(aq) + NH4Cl(aq)
NaClO3(aq) + BaI2(aq)
K2CO3(aq) + HClO3(aq)
HBr(aq) + Fe(s)
SrBr2(aq) + Hg2(NO3)2(aq)
K2CrO7(aq) + NaHS(aq)
- 110 -
Solutions:
6.1
Ba2+(aq) + 2 OH‒(aq) + Mg2+(aq) + SO42‒(aq)
(a) Ba(OH)2(aq) + MgSO4(aq) or
When the above two solutions are mixed, Ba2+ and SO42‒ encounter one another and
form a solid precipitate, BaSO4(s). This is a precipitation reaction.
Na+(aq) + HCO3‒(aq) + Hg22+(aq) + SO42‒(aq)
(b) NaHCO3(aq) + Hg2SO4(aq) or
Since sodium and are hydrogen carbonate salts are soluble, there is no precipitation
reaction.
(c) RbOH(aq) + SrCl2(aq)
Rb+(aq) + OH‒(aq) + Mg2+(aq) + SO42‒(aq)
or
Since magnesium hydroxide is insoluble, this is a precipitation reaction. The precipitate
is Mg(OH)2(s).
(d) Pb(CH3COO)2(aq) + MgI2(aq)
or
Pb2+(aq) + 2 CH3COO‒(aq) + Mg2+(aq) + 2 I‒(aq)
Here, PbI2(s) is the precipitate. This is a precipitation reaction.
(e) Li2S(aq) + Ca(NO3)2(aq)
or
2 Li+(aq) + S2‒(aq) + Ca2+(aq) + 2 NO3‒(aq)
In reality, since S2‒ is a strong base, sulfide is only present as the conjugate acid, HS‒, in
aqueous solution. In any case, calcium sulfide is soluble and there is no precipitation
reaction.
(f) Li2S(aq) + Cu(NO3)2(aq)
or
2 Li+(aq) + S2‒(aq) + Cu2+(aq) + 2 NO3‒(aq)
In this case, copper II sulfide, CuS(s), precipitates from solution – the formation of this
solid competes favorably with the formation of HS‒ from S2‒, in aqueous solution. This is
a precipitation reaction.
(g) K2CO3(aq) + AgClO3(aq)
or
2 K+(aq) + CO32‒(aq) + Ag+(aq) + ClO3‒(aq)
In this case, silver carbonate, Ag2CO3(s), precipitates from solution. This is a
precipitation reaction.
6.2
(a) 2 Li(s) + Cl2(g) → 2 LiCl(s)
Not an acid-base reaction. There is no H+ transfer. This is an electron transfer reaction,
a redox reaction.
(b) LiOH(aq) + HBr(g) → LiBr(aq) + H2O(l)
An acid-base reaction. LiOH(aq) is shorthand for Li+(aq) + OH−(aq). In this reaction, HBr
donates an H+ to OH−. Therefore, HBr is the acid and OH− is the base.
(c) KOCl(aq) + HClO3(aq) → KClO3(aq) + HOCl(aq)
Another acid-base reaction. KOCl(aq) is shorthand for K+(aq) + OCl−(aq). Since HClO3
is a strong acid, HClO3(aq) is shorthand for H3O+(aq) + ClO3−(aq). Thus, in this reaction,
it is really H3O+ acting as an acid and donating an H+ to OCl−. However, as written, the
reaction shows HClO3 donating an H+ to OCl−. Here, HClO3 is the acid and OCl− is the
base.
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(d) CaCO3(s) → CaO(s) + CO2(g)
Not an acid-base reaction. It is a gas-forming decomposition reaction.
(e) CH3COOH(g) + NH3(g) → NH4CH3COO(s)
An acid-base reaction. CH3COOH donates an H+ to NH3. Therefore, CH3COOH is the
acid and NH3 is the base.
6.3
(a) HF is a weak acid.
(b) CH4 is neither acid not base. Carbon is not sufficiently electronegative to stabilize
the minus charge on the conjugate base, CH3−. Also, CH4 cannot act as a base because
it has no lone pairs of electrons.
(c) KOH is a strong base.
(d) CH3NH2 is a weak base.
(e) HClO4 is a strong acid.
A strong acid fully dissociates in water – the equilibrium wherein the acid donates an H+
to H2O is strongly shifted to the right. A weak acid is mostly undissociated in water - the
equilibrium wherein the acid donates an H+ to H2O is shifted to the left.
6.4
(a) H3PO4(aq) + H2O(l) 
H2PO4−(aq) + H3O+(aq)
Here, H2PO4− is the base conjugate to the acid, H3PO4, while H3O+ is the acid conjugate
to the base, H2O.
(b) CH3NH2(aq) + H2O(l) 
CH3NH3+(aq) + OH−(aq)
Here, OH− is the base conjugate to the acid, H2O, while CH3NH3+ is the acid conjugate to
the base, CH3NH2.
(c) HClO4(aq) + H2O(l) → ClO4−(aq) + H3O+(aq)
Here, ClO4− is the base conjugate to the acid, HClO4, while H3O+ is the acid conjugate to
the base, H2O.
6.5
(a) LiF will produce a basic solution. LiF is an ionic material that dissolves in water to
form Li+(aq) and F−(aq). Li+ is NOT an acid (nor a base). In fact, all group 1 cations are
neutral in water, as they do not react with water. However, F− is the weak base
conjugate to the weak acid, HF. It will react with water via
F−(aq) + H2O(l) 
HF(aq) + OH−(aq)
to sufficient extent to make the solution noticeably basic.
(b) NH4Cl will produce an acidic solution. NH4Cl is an ionic material that dissolves in
water to form NH4+(aq) and Cl−(aq). Cl− is a very weak base - it is conjugate to a strong
acid. It does not react to noticeable extent with water – i.e., we can simple say that it is
not a base. However, NH4+ is the weak acid conjugate to the weak acid, NH3. It will
react with water via
- 112 -
NH4+(aq) + H2O(l) 
NH3(aq) + H3O+(aq)
to sufficient extent to make the solution noticeably acidic.
(c) KBr will produce a neutral solution. Neither K+ (a group 1 cation), not Br−, reacts with
water to a noticeable extent. Br− is the very weak base conjugate to the strong acid, HBr.
(d) CH3NH3Cl will produce an acidic solution. This is an ionic material that dissolves in
water to form CH3NH3+(aq) and Cl−(aq). Cl− does not react with water – see part (b)
above. CH3NH3+ is the weak acid conjugate to the weak acid, CH3NH2. It will react with
water via
CH3NH3+(aq) + H2O(l) 
CH3NH2(aq) + H3O+(aq)
to sufficient extent to make the solution noticeably acidic.
(e) NaOCl will produce a basic solution. NaOCl dissolves in water to form Na+(aq) and
OCl−(aq). Na+ (a group 1 cation) does not react with water. However, OCl− is the weak
base conjugate to the weak acid, HOCl. It will react with water via
OCl−(aq) + H2O(l) 
HOCl(aq) + OH−(aq)
to sufficient extent to make the solution noticeably basic.
6.6
The requested chemical equilibria are shown in the solution to question 6.5. The shifts in
the equilibria are indicated by the size of the forward and reverse arrows. They are all
reactant-favored. In each case, a weak acid or weak base reacts with water. All such
reactions are reactant-favored.
6.7
(a) C3H10(g) + 11/2 O2(g) → 3 CO2(g) + 5 H2O(l)
The oxidation number of carbon changes from ‒10/3 (the average value for the three
carbon atoms in propane, C3H10) to +4 (the value in carbon dioxide). Propane is oxidized
in this reaction. At the same time, the oxidation number of oxygen changes from 0 to ‒2.
Oxygen is reduced. This is a redox reaction. Oxygen is the oxidizing agent, while
propane is the reducing agent.
In this reaction, 22 electrons are transfered from carbon atoms to oxygen atoms. This
number is determined by considering carbon or oxygen atoms. Specifically, 3 carbon
atoms each lose 4‒(‒10/3) = 22/3 electrons, while 11 oxygen atoms each gain 2
electrons.
(b) Na2SO4(aq) + SrBr2(aq) → SrSO4(s) + 2 NaBr(aq)
This is a precipitation reaction – an insoluble solid product is obtained by mixing aqueous
solutions.
(c) NH4NO3(s) → N2O(g) + 2 H2O(l)
The oxidation number of nitrogen changes from ‒3 to +1 (NH4+ to N2O), and from +5 to
+1 (NO3− to N2O). This is a redox reaction. Four electrons are transfered. Ammonium is
the reducing agent, while nitrate is the oxidizing agent.
- 113 -
(d) OCl−(aq) + HSO4−(aq) → HOCl(aq) + SO42−(aq)
This is an acid-base reaction. Hydrogen sulfate, the acid, donates an H+ to hypochlorite,
the base.
(e) H2O(l) + NaH(s) → NaOH(aq) + H2(g)
Here, water acts as an acid, donating an H+ to hydride, a strong base. It is thus an acidbase reaction. It is also a redox reaction, as the oxidation number of hydrogen varies
from +1 to 0 (H2O to H2), and from −1 to 0 (H− to H2). One electron is transfered in the
reaction. Water is the oxidizing agent, while hydride is the reducing agent.
(f) 3 S(s) + 2 KClO3(s) → 3 SO2(g) + 2 KCl(s)
The oxidation number of sulfur changes from 0 to +4, while the oxidation number of
chlorine changes from +5 to −1. This is a redox reaction. Twelve electrons are
transfered. Three sulfur atoms each lose four electrons, while two chlorine atoms each
gain six electrons. Sulfur is the reducing agent, while chlorate is the oxidizing agent.
(g) 3 BrO−(aq) → BrO3−(aq) + 2 Br−(aq)
The oxidation number of bromine changes from +1 to +5 (BrO− to BrO3−), and from +1 to
−1 (BrO− to Br−). This is a redox reaction. In particular, it is a disproportionation reaction,
since the hypobromite reacts with itself. Four electrons are transfered. One bromine
atom loses four electrons, while two bromine atoms each gain two electrons.
Hypobromate is both oxidizing and reducing agent.
6.8
(a)
Co2+(aq) + Cr(s) → Co(s) + Cr3+(aq)
Oxidation numbers:
+2
0
0
+3
The half reactions are
and
Cr(s) → Cr3+(aq) + 3 e−
oxidation
Co2+(aq) + 2 e− → Co(s) .
reduction
To balance the electrons, combine the half reactions with the first scaled by two and the
second scaled by three to get
3 Co2+(aq) + 2 Cr(s) → 3 Co(s) + 2 Cr3+(aq) .
There are no H or O atoms. The reaction is balanced as written.
(b)
Hg2+(aq) + Ag(s) → Hg22+(aq) + Ag+(aq)
Oxidation numbers:
+2
0
+1
+1
The half reactions are
and
Ag(s) → Ag+(aq) + e−
oxidation
2 Hg2+(aq) + 2 e− → Hg22+(aq) .
reduction
Mercury is balanced in the reduction half reaction by scaling Hg2+(aq) by two.
- 114 -
To balance the electrons, combine the half reactions with the first scaled by two to get
2 Hg2+(aq) + 2 Ag(s) → Hg22+(aq) + 2 Ag+(aq) .
There are no H or O atoms. The reaction is balanced as written.
(c)
ClO4−(aq) + Sn2+(aq) → Cl−(aq) + Sn4+(aq)
Oxidation numbers:
+7 −2
+2
−1
+4
The half reactions are
and
Sn2+(aq) → Sn4+(aq) + 2 e−
oxidation
ClO4−(aq) + 8 e− → Cl−(aq) .
reduction
To balance the electrons, combine the half reactions with the first scaled by four to get
ClO4−(aq) + 4 Sn2+(aq) → Cl−(aq) + 4 Sn4+(aq) .
There are four O atoms on the left, and none on the right. In acid, oxygen is balanced by
adding water. Here, add 4 H2O(l) to the right to get
ClO4−(aq) + 4 Sn2+(aq) → Cl−(aq) + 4 Sn4+(aq) + 4 H2O(l) .
Now, there are eight H atoms on the right, and none on the left. Add 8 H+(aq) to the left
to get
ClO4−(aq) + 4 Sn2+(aq) + 8 H+(aq) → Cl−(aq) + 4 Sn4+(aq) + 4 H2O(l) .
(d)
Cl−(aq) + MnO2(s) → Mn2+(aq) + HOCl(aq)
Oxidation numbers:
−1
+4 −2
+2
+1 −2 +2
The half reactions are
and
Cl−(aq) → HOCl(aq) + 2 e−
oxidation
MnO2(s) + 2 e− → Mn2+(aq) .
reduction
To balance the electrons, combine the half reactions to get
Cl−(aq) + MnO2(s) → HOCl(aq) + Mn2+(aq) .
There are two O atoms on the left, and one on the right. Add H2O(l) to the right to get
Cl−(aq) + MnO2(s) → HOCl(aq) + Mn2+(aq) + H2O(l) .
Now, there are three H atoms on the right, and none on the left. Add 3 H+(aq) to the left
to get
Cl−(aq) + MnO2(s) + 3 H+(aq) → HOCl(aq) + Mn2+(aq) + H2O(l) .
- 115 -
(e)
PbO2(s) + SO42-(aq) + Pb(s) → PbSO4(s)
Oxidation numbers:
+4 −2
+6 −2
0
+2 +6 −2
The incompletely balanced half reactions are
and
Pb(s) + SO42−(aq) → PbSO4(s) + 2 e−
oxidation
PbO2(s) + SO42−(aq) + 2 e− → PbSO4(s) .
reduction
Aqueous sulfate is added to the left of both reactions to balance the sulfate in the lead
sulfate product.
To balance the electrons, combine the half reactions to get
PbO2(s) + Pb(s) + 2 SO42−(aq) → 2 PbSO4(s) .
There are ten O atoms on the left, and eight on the right. Add 2 H2O(l) to the right to get
PbO2(s) + Pb(s) + 2 SO42−(aq) → 2 PbSO4(s) + 2 H2O(l) .
Now, there are four H atoms on the right, and none on the left. Add 4 H+(aq) to the left to
get
PbO2(s) + Pb(s) + 2 SO42−(aq) + 4 H+(aq) → 2 PbSO4(s) + 2 H2O(l) .
(f)
MnO4−(aq) + Mn2O3(s) → MnO2(s)
Oxidation numbers:
+7 −2
+3 −2
+4 −2
The incompletely balanced half reactions are
and
Mn2O3(s) → 2 MnO2(s) + 2 e−
oxidation
MnO4−(aq) + 3 e− → MnO2(s) .
reduction
The MnO2(s) in the oxidation half reaction is scaled by two to balance the manganese
atoms.
To balance the electrons, combine the first half reaction scaled by three with the second
half reaction scaled by two to get
2 MnO4− + 3 Mn2O3(s) → 8 MnO2(s) .
There are 17 O atoms on the left, and 16 on the right. Since this reaction is in base,
oxygen is balanced by adding OH−(aq) to the right to get
2 MnO4− + 3 Mn2O3(s) → 8 MnO2(s) + OH−(aq) .
Now, there is one H atoms on the right, and none on the left. Add H2O(l) to the left and
OH−(aq) to the left to get
2 MnO4− + 3 Mn2O3(s) + H2O(l) → 8 MnO2(s) + 2 OH−(aq) .
- 116 -
(g)
O2(g) + Ba(s) → OH−(aq) + Ba2+(aq)
Oxidation numbers:
0
0
−2 +1
+2
The incompletely balanced half reactions are
and
Ba(s) → Ba2+(aq) + 2 e−
oxidation
O2(g) + 4 e− → 2 OH−(aq) .
reduction
The OH−(aq) in the reduction half reaction is scaled by two to balance the oxygen atoms.
We need to balance oxygen atoms here because it is oxygen that gets reduced.
To balance the electrons, combine the first half reaction scaled by two with the second
half reaction to get
O2(g) + 2 Ba(s) → 2 OH−(aq) + 2 Ba2+(aq) .
There are two O atoms on both sides. Oxygen is already balanced. There are two H
atoms on the right, and none on the left. Add 2 H2O(l) to the left and 2 OH−(aq) to the
right to get
O2(g) + 2 Ba(s) + 2 H2O(l) → 4 OH−(aq) + 2 Ba2+(aq) .
6.9
(a) LiClO3(aq) + BaCl2(aq)
No reaction and, consequently, no visible change. Lithium and chlorate salts are soluble.
Chlorate is an oxidizing agent (chlorine is in a high oxidation state – specifically, +5).
However, there is no appropriate reducing agent. Barium is in its highest oxidation state.
Chloride is a candidate for a reducing agent. However, chlorate is not a sufficiently
strong oxidizing agent to oxidize chloride.
(b) LiOCl(aq) + NH4Cl(aq)
An acid-base reaction. Weak base, hypochorite, accepts an H+ from weak acid,
ammonium. Unless there is an acid-base indicator added to one of the solutions, there is
no visible change.
(c) NaClO3(aq) + BaI2(aq)
A redox reaction. Chlorate is a sufficiently strong oxidizing agent to oxidize iodide.
Iodine is the product of iodide oxidation. It is a purple solid. A purple solid coming out of
solution is a visible change. However, until the iodide reactant is consumed, any iodine
produced reacts with the remaining iodide to produce the triiodide ion. Specifically,
I2(aq) + I−(aq) → I3−(aq) .
However, as triiodide solution is brown, the net reaction remains visible.
(d) K2CO3(aq) + HClO3(aq)
This is an acid-base reaction. HClO3 is a strong acid that produces an H3O+ solution.
H3O+ donates an H+ to CO32− to form HCO3−. HCO3− then accepts another H+ to form
H2CO3. The carbonic acid produced decomposes into water and carbon dioxide. Carbon
dioxide gas bubbles out of solution, and the reaction is visible.
- 117 -
(e) HBr(aq) + Fe(s)
Iron is a sufficiently active metal to react with any acid (i.e., any source of H+) to produce
iron cations (specifically, Fe2+) and hydrogen gas. This is a visible redox reaction.
(f) SrBr2(aq) + Hg2(NO3)2
This is a precipitation reaction. Hg2Br2 is insoluble. It is a visible reaction.
(g) K2CrO7(aq) + NaHS(aq)
This is a redox reaction. Dichromate (chromium in the +6 oxidation state) is a strong
oxidizing agent and hydrogen sulfide (sulfur in the −2 oxidation state) is a reducing agent.
Because dichromate changes color (orange to blue-green) upon reduction, this is a
visible reaction.
- 118 -
7.
Chemical Equilibrium
7.1 Dynamic Equilibrium
We often speak of systems at equilibrium. This is a state where the properties of the system –
including chemical composition – do not change with time. As such, it seems as though all
chemical reactions have ceased at equilibrium. But this is not so - chemical reactions continue at
equilibrium. For every forward reaction, there is a reverse reaction. The rate of a forward
reaction generally depends on the concentrations of reactants, while the rate of a reverse
reaction depends on the concentrations of products (partial pressures for gas reactions). When
these rates are unequal, net reaction proceeds, changing the concentrations of reactants and
products such that the rates approach each other. Chemical equilibrium occurs when the rate of
forward reaction equals the rate of reverse reaction, and there is no further net reaction even
though the forward and reverse reactions continue.
All reactions are equilibrium reactions. However, for some reactions, products are strongly
favored. In such a case, the rate of forward reaction exceeds the reverse reaction until almost no
reactant remains – the reaction essentially goes to completion. When the reactants are strongly
favored, no reaction is observed. As seen in Section 7.3, there are cases between these
extremes, reactions where neither reactant nor product concentration is negligible at equilibrium.
These reactions are characterized by an equilibrium constant, which is neither extremely big nor
extremely small.
7.2 Equilibrium Constant
At equilibrium, the concentrations of dissolved species, and partial pressures of gases, stop
changing. Sets of experiments show that the final equilibrium concentrations of reactants and
products (consider reactions of aqueous species, for example) depend on the initial
concentrations – i.e., the concentrations at the start of the experiment. However, there is a
combination of concentrations that is the same for every experiment performed at the same
temperature. The value of this combination of concentrations is called the equilibrium constant.
Careful studies show that the equilibrium constant is actually a combination of reactant and
product activities, a. For a species A in solution, at low concentration, the activity is
approximately equal to the concentration of the species, [A] in mol L1. For a gas, the activity is
approximately the gas partial pressure, pA in bar. For pure (or almost pure) liquids and solids, the
activity is approximately 1. We take these approximations as exact, and henceforth adopt the
following definition the activity, aA, of species A:
for A in solution
[A]

aA   PA for A in the gas phase
 1 for pure liquid or solid A

7.1
The units of [A] and PA are mol L−1 and bar, respectively. However, these units are dropped from
activity; activity is dimensionless because an activity is a concentration (or a pressure) divided by
a standard concentration (or pressure) of 1. The standard conditions for chemical reactions (as
defined in thermochemical studies – see Chapter 8) are 1 mol L−1 concentration of dissolved
species and 1 bar partial pressure for gases. Thus, the activity of an aqueous species is really
[A] / (1 mol L−1). It is the concentration relative to its value under standard conditions, 1 mol L−1.
Further, the activities of pure liquids and solids are taken to be 1, so these species do not appear
in equilibrium constant expressions (see below).
- 119 -
The equilibrium constant, K, for the reaction,
cA A + cB B  cY Y + cZ Z
is given by
aYCY aZCZ
K  CA CB
aA aB
7.2
where aA, aB , aY and aZ are the activities of A, B, Y and Z under equilibrium conditions, and cA,
cB, cY and cZ are the stoichiometric coefficients of A, B, Y and Z, respectively. Moreover, any
combination of aA, aB , aY and aZ satisfying this equation constitutes equilibrium conditions. The
equilibrium constant depends only on temperature.
For more general reactions, there are additional
aXCX factors in the numerator, if there are
additional products; and in the denominator, if there are additional reactants. Examples are
provided as follows:
Reaction
Equilibrium constant
2
PBrCl
PBr2 PCl2
Br2(g) + Cl2(g)  2 BrCl(g)
K
H2(g) + I2(s) 
PHI2
K
PH 2
2 HI(g)
N2(g) + 3 H2(g) 
CaO(s) + H2O(l) 
HClO(aq) + H2O(l) 
NaF(s) 
K
2 NH3(g)
PN 2 PH32
K  [Ca 2+ ][OH ]2
Ca2+(aq) + 2 OH−(aq)
ClO−(aq) + H3O+(aq)
Na+(aq) + F−(aq)
2
PNH
3
K  Ka 
[H3O+ ][ClO  ]
[H + ][ClO  ]
or
[HClO]
[HClO]
K  K sp  [Na + ][F ]
The last two examples show named equilibrium constants, Ka and Ksp, the acid dissociation
constant and the solubility product, respectively.
Example 7.1: Write down the equilibrium constant, K, for each of the following reactions.
(a)
(b)
(c)
(d)
(e)
CaCO3(s)  CaO(s) + CO2(g)
NH4HS(s)  NH3(g) + H2S(g)
NO(g) + O3(g)  NO2(g) + O2(g)
Cl2(aq) + H2O(l)  H+(aq) + Cl−(aq) + HClO(aq)
Cl2(g) + 2 Fe2+(aq)  2 Cl−(aq) + 2 Fe3+(aq)
- 120 -
Approach: Look for gases or aqueous species. The equilibrium constant is constructed
from the partial pressures of gases and the concentrations of aqueous species, raised to the
associated stoichiometric coefficient in the balanced chemical reaction. Products appear in
the numerator, while reactants appear in the denominator.
(a)
K  PCO2
The equilibrium constant is just the partial pressure of carbon dioxide. K here is the vapour
pressure of solid carbon dioxide. In an enclosed space, CO2(s) will sublimate until the partial
pressure of CO2(g) equals K and the solid and gas are in equilibrium.
(b) K  PNH3 PH 2S
NH4HS(s) does not appear in the equilibrium constant because it is a pure solid. All solids
and liquids you will encounter in chemical reactions are pure (water, in aqueous reactions, is
treated as though it were pure).
(c) K 
PNO 2 PO 2
PNO PO3
[H  ][Cl  ][HClO]
(d) K 
[Cl2 ]
H2O(l) does not appear. It is treated as a pure liquid. The concentration of pure water, [H2O],
is about 56 mol L−1. Changes in this value caused by dissolved aqueous species are small,
and neglected.
(e)
K
[Cl ]2 [Fe3+ ]2
PCl2 [Fe2+ ]2
This reaction has a gas and three aqueous species. In this case, the equilibrium constant is
constructed from a partial pressure and three concentrations. Note the stoichiometric
coefficients in the exponents.
An equilibrium constant is evaluated by preparing an equilibrium mixture of reactants and
products, then measuring the resulting concentrations and partial pressures. The value of K
obtained does not depend on how the mixture was prepared. It depends only on temperature.
Thus, temperature is generally noted when an equilibrium constant is given.
Once an equilibrium constant is known, an unknown concentration or partial pressure can be
determined in an equilibrium mixture – if the other concentrations and partial pressures are
known. For example, the equilibrium constant for the combustion of sulfur dioxide at 350°C is
K = 5.60×104.
2 SO2(g) + O2(g) 
2 SO3(g)
If the partial pressures of O2 and SO3 are both 0.500 bar, then the partial pressure of SO2 is given
by substituting all known quantities into the equilibrium constant equation and solving for PSO2 .
Specifically,
- 121 -
K
2
PSO
3
2
PSO
P
2 O2

0.500 2
 5.60  10 4
2
PSO
0.500
2
from which we get
2
PSO

2
0.5002
 8.929 106
4
5.60 10  0.500
and
PSO2  2.99 103 bar
Example 7.2: What is the concentration of OH− in an equilibrium solution with [CO32−] = [HCO3−]
= 1.0 mol L−1, given the following equilibrium constant?
CO32−(aq) + H2O(l) 
HCO3−(aq) + OH−(aq)
K = 2.1×10−4
Approach: Write down the equilibrium constant expression. Substitute the known
concentrations. Solve for the unknown concentration.
K
[HCO3 ][OH  ] 1.0  [OH  ]

 [OH  ]  2.1  104
2
[CO3 ]
1.0
Therefore, [OH−] = 2.1×10−4 mol L−1.
Suppose we have a table of equilibrium constants for a set of reactions that does not include a
reaction of interest. If the reaction of interest can be expressed as a combination of the tabulated
reactions, then the equilibrium constant can be expressed as a product or quotient of the
tabulated equilibrium constants.
The simplest case is when the reaction of interest is the reverse of a tabulated reaction.
Considering the above definition of the equilibrium constant in terms of activities, one can deduce
K reverse 
1
K forward
.
This follows because the reactants of the reverse reaction are the products of the forward
reaction, and vice versa. If the reaction of interest is 2× a tabulated reaction, then the equilibrium
constant of interest is the square of the tabulated equilibrium constant. This follows because all
the stoichiometric coefficients are doubled if a reaction is doubled. To see this, write down
expressions for K1 and K2 – the equilibrium constants for the following reactions - and see that K2
= K12.
Rxn 1 A  B + C
Rxn 2 2A  2 B + 2 C
K1
K2
In general, the equilibrium constant of interest is a product of tabulated equilibrium constants,
each raised to a power - the coefficient of the tabulated reaction in the combination required to
reproduce the reaction of interest.
K  K1n1 K 2n2 K 3n3  ,
where
- 122 -
7.3
Reaction of interest   ni  ith tabulated reaction  ,
i
ni is negative for tabulated reactions that need to be reversed.
7.3 The Reaction Quotient
The above expressions for the equilibrium constant only hold under equilibrium conditions – i.e.,
the concentrations and partial pressures are for an equilibrium reaction mixture. When a reaction
mixture is not at equilibrium, combining the concentrations and partial pressures as they are in
the equilibrium constant gives the reaction quotient, Q. The reaction quotient can be computed
under equilibrium or non-equilibrium conditions. If Q = K, then the system is at equilibrium.
Otherwise, Q ≠ K and there is net forward or reverse reaction.
If Q < K, there is net forward reaction. Net forward reaction depletes reactants, reducing the
denominator of Q. At the same time, product concentrations/partial pressures increase,
increasing the numerator of Q. The net effect is an increase in Q. Net forward reaction continues
until Q = K.
If Q > K, there is net reverse reaction. Net reverse reaction increases reactant
concentrations/partial pressures, increasing the denominator of Q. At the same time, product is
depleted, decreasing the numerator of Q. The net effect is a decrease in Q. Net reverse reaction
continues until Q = K.
Example 7.3: Consider each of the following reaction mixtures. Is the mixture at equilibrium, or
is there net forward or reverse reaction?
(a) Solid calcium carbonate and calcium oxide at 700°C are in a vessel with 0.1 bar partial
pressure carbon dioxide. At 700°C, K = 0.056 for the decomposition of CaCO3(s) into CaO(s)
and CO2(g).
(b)
Consider the following equilibrium:
Cu2+(aq) + 4 NH3(aq) 
Cu(NH3)42+(aq)
K = 6.7×1012 at 25°C
Suppose [Cu2+] = [Cu(NH3)42+] = 1.0 mol L−1, while [NH3] = 5.0×10−4 mol L−1 at 25°C.
(c)
Consider the following equilibrium:
SnO2(s) + 2 CO(g) 
Sn(s) + 2 CO2(g)
K = 11 at some T
Suppose the partial pressure of CO2 is 3 bar, while that of CO is 1 bar at the required T.
Approach: Compute Q and compare it with K.
(a) Q = PCO 2 = 0.1 > 0.056. Therefore, there is net reverse reaction. At this partial
pressure, CO2 in the gas phase combines with available CaO(s) to form CaCO3(s).
(b) Here,
[Cu  NH 3 4 ]
2
Q
2
4
[Cu ][NH 3 ]


1.0
1.0 5.0  10
Again, Q > K and there is net reverse reaction.
- 123 -
4

4
 1.6  1013  K
(c) Here,
Q
2
PCO
2
2
PCO

32
 9  11
1
In this case, Q < K and there is net forward reaction.
7.4 Shift in Equilibrium – le Châtelier's Principle
Non-equilibrium conditions can result, starting with an equilibrium mixture, by adding or removing
a reactant or a product. For example, if we have an equilibrium solution with the reaction
CO32−(aq) + H2O(l) 
HCO3−(aq) + OH−(aq)
K = 2.1×10−4
and we dissolve a small amount of NaOH(s) into the solution, a non-equilibrium initial state
results. Increasing the concentration of hydroxide makes Q larger (a factor in the numerator is
increased). Since Q is now larger than K, there is net reverse reaction – product concentrations
decrease, while the reactant concentration increases. Net reaction continues until Q = K is
restored. When equilibrium is re-established, the hydroxide concentration is smaller than it was
immediately after the added NaOH(s) dissolves, but larger than it was before the NaOH(s) was
added.
We can also add a small amount of concentrated sulfuric acid (so as not to significantly change
the total volume) to the solution. The added strong acid reacts completely with OH− present, and
reduces the hydroxide concentration. As such, it removes a product from the chemical reaction.
This has the effect of making Q smaller. Net forward reaction results until equilibrium is reestablished.
Whether we add or remove product, the net reaction acts to reduce the change imposed on the
system – the addition or removal of some hydroxide. The hydroxide concentration moves back
towards (but does not reach1) its value before the change was imposed. This is a general
principle called le Châtelier's principle:
When a change is imposed on an equilibrium system, the change is followed by net
reaction that reduces the change.
If a reactant is added – in the above example, some Na2CO3(s) can be dissolved in the solution le Châtelier's principle tells us that there will be net forward reaction consuming some of the
added CO32−. This is consistent with reaction quotient considerations. Note that the principle
only applies to changes in activity. Changes in the amount of a pure solid or liquid do not affect
an equilibrium. Thus, changing the amount of CaCO3(s) or CaO(s) in an equilibrium mixture with
CO2(g) does not affect the partial pressure of the gas.
Le Châtelier's principle also applies to the effect of changing temperature. The equilibrium
constant generally depends upon temperature. Thermodynamics shows that K increases or
decreases with temperature in accord with the sign of ∆H (change in enthalpy – see more in
Chapter 6). Specifically, K increases with T for endothermic reactions (∆H > 0), and decreases
with T for exothermic reactions (∆H < 0). If K increases with T, as it does for endothermic
reactions, products will be favored at higher T. Thus, increasing T for an endothermic reaction
causes net forward reaction until the new (product favored) equilibrium is established. This is
consistent with le Châtelier's principle in the following sense: Increasing T makes more heat
"available". A net forward endothermic reaction consumes some of this additional heat. In
1
In the case of an equilibrium with just one activity – for example, the decomposition of CaCO3(s)
– if the activity is changed, it will return to original value. Otherwise, shifts in other activities will
prevent a changed activity from returning to its original value.
- 124 -
general, the endothermic direction of reaction (the reverse reaction, in the case of an exothermic
reaction) is favored with increasing temperature. Thus, reactants are favored at high T in the
case of an exothermic reaction.
Example 7.4: Predict the direction each of the following equilibriums shifts when the indicated
changes are imposed.
NO(g) + O3(g) 
(a)
(b)
(c)
(d)
NO2(g) + O2(g)
∆H° = −200.8 kJ mol−1
The O3 partial pressure is increased
Some NO2 is removed from the reaction mixture
The reaction mixture is transferred to a vessel with twice the volume
Temperature is increased
NH4Cl(s)
H
HO
H
H 2 OC
l
H
HH
+
2HCl
HO
+
H
OH
H
2HO
+
H
∆H° = 176.2 kJ mol−1
NH3(g) + HCl(g)
(e) The HCl partial pressure is increased
(f) Some solid NH4Cl is added to the mixture
(g) Temperature is decreased
Approach: Is a reactant or product activity changed – i.e., a partial pressure or
concentration? If so, apply le Châtelier's principle. If temperature changes, note the sign of
the enthalpy of reaction.
(a) O3 is a reactant. Increasing its partial pressure results in net forward reaction – to
consume some of the added O3.
(b) NO2 is a product. Removing it from the reaction mixture results in net forward reaction.
This is a common means of improving product yield. Removing the product from a
reaction mixture drives the reaction in the forward direction.
(c) Doubling the volume of the vessel reduces all partial pressures to 1/2 their initial values.
However, this does not change the reaction quotient because there are two partial
pressure factors in both the numerator and denominator of Q and they have the same
exponents. Decreasing all partial pressures by the same factor thus has no affect on the
reaction quotient. Therefore, the system is still at equilibrium. This is consistent with le
Châtelier's principle, which says that the reaction will shift to the side with more moles of
gas so as to counter the reduction in total pressure caused by the increase in volume.
However, neither forward nor reverse reaction increases the moles of gas. Therefore,
the mixture remains at equilibrium.
(d) The reaction is exothermic. Increasing temperature favors reactants (the endothermic
direction of reaction) – i.e., there is net reverse reaction.
(e) HCl is a product. Increasing its partial pressure results in net reverse reaction.
(f) NH4Cl(s) is a pure solid. Adding it or removing some of it from the reaction mixture has
no effect on the equilibrium as long as some of the solid remains. No net reaction results.
(g) The reaction is endothermic. Decreasing temperature favors reactant (the exothermic
direction of reaction) – i.e., there is net reverse reaction.
The equilibrium constant allows quantitative determination of final equilibrium conditions. Such
determinations are carried out by constructing an ICE table to relate equilibrium activities to the
extent of reaction. This begins with the balanced chemical reaction. Below that are three rows.
The first row shows the initial (I), non-equilibrium concentrations/partial pressures of reactants
- 125 -
and products. The second row expresses the changes (C) in these concentrations (that occur
upon achieving equilibrium) in terms of the unknown extent of reaction. This row accounts for the
stoichiometric coefficients in the reaction. The third row is the sum of the first two rows. It gives
the concentrations/partial pressures at equilibrium (E). For example, consider the reaction of
water vapor and carbon monoxide, both initially at 1.00 bar partial pressure. K = 0.63 for this
reaction at a certain T.
H2O(g)
Initial
Change
Equilibrium
+
CO(g)



H2(g)
+
CO2(g)
1.00
1.00
0
0
−x
−x
+x
+x
1.00 − x
1.00 − x
x
x
Here, x is the extent of reaction, appearing as a change in partial pressure. If one of the species
had a stoichiometric coefficient greater than 1, it would multiply x in the associated cell. Having
expressed the equilibrium concentrations in terms of the unknown x, we construct the equilibrium
constant equation in terms of x.
 PH PCO 
xx
K   2 2 

 0.63
 PH O PCO 
(1.00
x
)(1.00
x
)


 2
eqm
from which we get
2


x

  0.63
 (1.00  x) 
x
 0.63  0.794
(1.00  x)
x  0.794 (1.00  x)
x
0.794
 0.443
1.794
With x = 0.443, we can get the equilibrium partial pressures – the bottom row in the ICE table.
Specifically,
PH2O = PCO = 1.00 − x = 0.56 bar
and
PH2 = PCO2 = x = 0.44 bar .
The above equation for x is easily solved because the right side of the first line is a perfect square
(resulting from the reaction stoichiometry and because we started with equal partial pressures of
H2O and CO). For a simple equilibrium, a linear equation might result directly. However, in
general, a polynomial equation is obtained. For many important reactions this is a quadratic
equation that can be solved approximately, or exactly using the formula for the solution of a
quadratic equation – see below.
- 126 -
Example 7.5: Consider the gas phase equilibrium of hydrogen and iodine,
H2(g) + I2(g) 
2 HI(g)
K = 33 at some temperature, T
(a) Suppose a vessel is filled to 0.100 bar of both H2 and I2. What is the partial pressure of HI
when equilibrium is established at T?
(b) Suppose the vessel is filled to 1.00 bar of H2 and 0.100 bar of I2. What is the partial pressure
of HI when equilibrium is established at T?
Approach: Construct an ICE table. Substitute into the equilibrium constant equation and
solve for the unknown extent of reaction. Evaluate the equilibrium partial pressures in the
third row of the ICE table.
(a)
+
H2
Initial
Change
Equilibrium



I2
2 HI
0.100
0.100
0
−x
−x
+2 x
0.100 – x
0.100 − x
2x
Substitute the equilibrium partial pressures into the equilibrium constant equation.
PHI2
K 
PH2 PI2

 2x
2
(0.100  x)(0.100  x)
 33
This equation simplifies.
2


2x

 = 33
 (0.100  x) 
2x
= 33  5.7 4
(0.100  x)
2 x  5.7 4 (0.100  x)
0.57 4
 0.0742
7.7 4
Thus, the final equilibrium partial pressure of HI is
x
PHI = 2 x = 2 × 0.0742 = 0.148 bar .
- 127 -
(b)
+
H2



2 HI
1.00
0.100
0
−x
−x
+2 x
1.00 – x
0.100 − x
2x
Initial
Change
Equilibrium
I2
Substitute the equilibrium partial pressures into the equilibrium constant equation.
K=
PHI2
(2 x)2
=
= 33
PH2 PI2 (1.00  x)(0.100  x)
This equation requires solution of a quadratic equation.
4 x2
= 33
(0.100  1.10 x  x 2 )
4 x 2 = 33(0.100  1.10 x  x 2 )
29 x 2  36.3 x  3.3  0
ax 2  bx  c  0
The solution of this quadratic equation is
x

b  b 2  4ac
2a
36.3  36.32  4  29  3.3
2  29
 1.15

0.0987
We get two solutions for x. However, only one of them is admissible. The choice of x = 1.15
yields negative equilibrium partial pressures for H2 and I2. It is an inadmissible solution.
Thus,
x = 0.0987
is the correct solution, and
PHI = 2 x = 2 × 0.0987 = 0.197 bar
- 128 -
7.5 Solubility
When the equilibrium corresponds to the dissolution of an ionic solid (a salt), the equilibrium
constant is called a solubility product, Ksp. The Ksp value determines the concentration of the ions
in equilibrium with the solid. If excess salt is added to pure water, the resulting cation and anion
concentrations are related by the stoichiometric coefficients in the dissolution process.
For example, the equilibrium constant of the reaction,
Ba3(PO4)2(s)  3 Ba2+(aq) + 2 PO43(aq)
is Ksp for Ba3(PO4)2(s). The overall concentration of the salt is called the molar solubility. Thus,
the molar solubility of Ba3(PO4)2(s) is 1/3 the barium concentration, or 1/2 the phosphate
concentration at equilibrium, as per the stoichiometry of the Ksp expression.
Example 7.6: Determine the molar solubilities of the following salts at 25°C. (Ksp values are for
25°C.)
(a) BaCO3 Ksp = 2.6 × 10−9
(b) CaF2 Ksp = 3.45 × 10−11
Approach: Construct an ICE table for the addition of excess solid salt to pure water. Solve
for the extent of reaction and evaluate the equilibrium concentrations.
(a)
BaCO3(s)


Ba2+(aq)
+
CO32−(aq)
Initial
0
0
Change
+x
+x
Equilibrium
x
x
Substitute the equilibrium ion concentrations into the solubility product equation.
K sp = [Ba 2+ ][CO32 ] = x 2 = 2.6  109
which gives
x = 2.6 109  5.1105
At equilibrium,
[Ba2+] = [CO32−] = 5.1 × 10−5 mol L−1
In this case, the molar solubility is just the barium or carbonate concentration, i.e.,
5.1 × 10−5 mol L−1.
- 129 -
(b)
CaF2(s)
Ca2+(aq)


2 F−(aq)
+
Initial
0
0
Change
+x
+2 x
Equilibrium
x
2x
Substitute the equilibrium ion concentrations into the solubility product equation.
K sp = [Ca 2+ ][F ]2 = x  2 x  = 3.45  1011
2
which gives
1/3
 3.45  1011 
x=

4


 2.05  104
At equilibrium,
[Ca2+] = x = 2.05 × 10−4 mol L−1
[F−] = 2 x = 4.10 × 10−4 mol L−1
In this case, the molar solubility is just the calcium concentration or 1/2 the fluoride
concentration, i.e., 2.05 × 10−4 mol L−1.
In general, if a salt, MiXj, dissolves in water to form i solvated metal cations and j anions, then its
molar solubility (the extent of dissolution at equilibrium) satisfies the equilibrium condition,
K sp   ix 
i
 jx 
j
 ii j j xi j
or
K 
x   i spj 
i j 
7.4
1
i j
.
The Common Ion Effect
If we attempt to dissolve a salt in a solution that already contains a solvated ion in common with
an ion in the salt, less of the salt will dissolve. This is called the common ion effect. For example,
consider dissolving CaCO3(s) in a saturated solution of CaF2(s). Ksp = 3.36 × 10−9 for calcium
carbonate at 25°C. Construct the ICE table for this process.
CaCO3(s)


Initial
Change
Equilibrium
- 130 -
Ca2+(aq)
+
CO32−(aq)
2.05 × 10−4
0
+x
+x
2.05 × 10−4 + x
x
Note that the initial concentration of calcium is not zero – it is the calcium concentration of a
solution saturated in calcium fluoride.
Substitute the equilibrium ion concentrations into the solubility product equation.
K sp = [Ca 2+ ][CO32 ] =  2.05104  x  x = 3.36  109
which gives
x 2  2.05  104 x  3.36  109  0
Using the quadratic formula,
x

b  b 2  4ac
2a
2.05  104 
 2.05  10 
4 2
 4  3.36  109
2
 1.526  105

4
 2.203  10
The negative solution produces a negative carbonate concentration at equilibrium and is
therefore inadmissible.
At equilibrium,
[Ca2+] = 2.05 × 10−4 + 1.53 × 10−5 mol L−1 = 2.20 × 10−4 mol L−1
[CO32−] = 1.53 × 10−5 mol L−1
If the initial calcium concentration had been zero, Ksp equation would have given
K sp = [Ca 2+ ][CO32 ] = x 2 = 3.36  10 9
x  3.36  109  5.80  10 5
In the absence of an initial calcium ion concentration, the extent of reaction (dissolution of calcium
carbonate) is larger; 5.80 × 10−5 versus 1.53 × 10−5. If the original solution contained neither
calcium nor carbonate ions (e.g., a potassium bromide solution), then the dissolution of calcium
carbonate would be unaffected.1
Example 7.7: What is the extent of dissolution of AgCl(s) (Ksp = 1.8 × 10−10) in the following
solutions when equilibrium is established? What are the equilibrium ion concentrations?
(a) 1.0 × 10−3 mol L−1 KBr(aq)
(b) 1.0 × 10−3 mol L−1 KCl(aq)
1
Actually, the presence of ions already in solution does have an affect on the dissolution of
CaCO3. This is because concentrations are not good approximations for ions in solutions. We
must include a correction factor for each ion – the activity coefficient. The correction factor is
different if there are already ions in solution, however we will not take this into account in Chem
1A03/1E03. In any case, a common ion has a much greater effect on solubility, and we can
account for this effect.
- 131 -
Approach: Construct ICE tables. Look for common ions. Substitute equilibrium
concentrations into the solubility product and solve for the extent of reaction (extent of
dissolution in this case).
In the first case, there are no initial common ions. In the second case, there are chloride ions
already present.
(a) Initially, there are potassium and bromide ions present in solution, but no lead or chloride.
AgCl(s)


Initial
Change
Equilibrium
Ag+(aq)
Cl−(aq)
+
0
0
+x
+x
x
x
Substitute the equilibrium ion concentrations into the solubility product equation.
K sp = [Ag + ][Cl  ] = x 2 = 1.8  10 10
which gives the extent of dissolution,
x = 1.8  10 10 
1/2
 1.34 105 mol L1
At equilibrium,
[Ag+] = [Cl−] = x = 1.3 × 10−5 mol L−1 .
(b) Initially, there is 1.0 × 10−3 mol L−1 Cl−(aq) present in solution.
AgCl(s)


Initial
Change
Equilibrium
Ag+(aq)
+
Cl−(aq)
0
1.0 × 10−3
+x
+x
x
1.0 × 10−3 + x
Substitute the equilibrium ion concentrations into the solubility product equation.
K sp = [Ag + ][Cl ] = 1.0 103  x  x = 1.8  1010
which gives
x 2  1.0  10 3 x  1.8  1010  0
Using the quadratic formula,
- 132 -
x

b  b 2  4ac
2a
1.0  10 3 
1.0  10 
3 2
 4  1.8  1010
2
7
 1.80  10

3
 1.00  10
The negative solution produces a negative silver concentration at equilibrium and is therefore
inadmissible. The extent of dissolution is therefore
x = 1.8 × 10−7 mol L−1 .
At equilibrium,
[Ag+] = 1.8 × 10−7 mol L−1
[Cl−] = 1.0 × 10−3 + 1.80 × 10−7 mol L−1 = 1.0 × 10−3 mol L−1
In this example, the extent of dissolution is reduced almost 100-fold when 1.0 × 10−3 mol L−1
is already present in solution. Note that the chloride ion concentration changes a negligible
amount.
7.6 Acid-base Equilibrium
Autoionization of Water
We consider acid-base equilibria in water. As such, it is important to start with the acid-base
reaction of water with itself – the autoionization of water
2 H2O(l) 
H3O+(aq) + OH−(aq) .
Kw = 1.00 × 10−14
This process occurs naturally in all aqueous solutions, and in pure water. In pure water, the
equilibrium concentrations of hydronium and hydroxide are determined using an ICE table.
2 H2O(l)


H3O+(aq)
Initial
Change
Equilibrium
+
OH−(aq)
0
0
+x
+x
x
x
The activity of H2O(l) is 1 and does not appear in the Kw expression; pure liquids and solids do
not appear in K expressions.2
K w  [H3O+ ][OH  ]  x 2  1.00 1014 .
2
Strictly speaking, the concentration of water is not constant in the autoionization reaction.
However, since the concentration of water in pure water is about 55.6 mol L−1, the change in
concentration upon autoionization is negligible – it decreases only by 2.0 × 10−7 mol L−1.
- 133 -
The solution to this equation is x = 1.00 × 10−7. This is the concentration of hydronium and
hydroxide in pure water, in mol L−1.
In aqueous solutions of acids and bases, the concentration of hydronium and hydroxide varies
dramatically. However, the autoionization reaction imposes the equilibrium condition,
K w  [H3O+ ][OH  ]  1.00 1014 ,
7.5
on the hydronium and hydroxide concentrations. If we know one of these concentrations, then
Eq. 7.5 determines the other. For example, in a 1.00 mol L−1 HCl solution, the concentration of
hydronium is 1.00 mol L−1, while the concentration of hydroxide is given by
Kw
1.00 1014
[OH ] 

 1.00 1014 mol L1 .
+
[H3O ]
1.00

Because hydronium and hydroxide concentrations vary over many orders of magnitude, and they
are typically very small, it is customary to specify pH and pOH values:
pH = ‒log10([H3O+])
and
pOH = ‒log10([OH‒]) .
7.6

[OH ]  pH  pOH .
7.7
Applying ‒log10 to Eq. 7.5 gives

pK w   log10 1.00  1014  14.00


  log10 [H 3O + ]  log10

We can easily determine pH from pOH, and vice versa. For example, a 5.00 × 10−3 mol L−1
solution of sodium hydroxide has pOH given by
pOH = ‒log10(5.00 × 10−3) = 2.30
and pH given by
pH = 14.00 ‒ 2.30 = 11.70 .
Acid Dissocation Equilibrium
When the equilibrium corresponds to the dissociation of an acid, HA(aq), into H+(aq) and A−(aq),
the equilibrium constant is called the acid dissociation constant and is denoted by Ka. Ka
determines the extent of dissociation of the acid and the resulting H+ concentration (and pH).
Note that H+ is actually transfered to water making H3O+(aq). H+(aq) is a convenient shorthand
for H3O+(aq), that eliminates the need to carry an extra water on the reactant side of the equation.
Strong acids dissociate completely and do not require an equilibrium analysis.
Consider the dissociation of ethanoic acid, a weak acid – showing the water and hydronium
explicitly.
CH3COOH(aq) + H2O(l) 
CH3COO−(aq) + H3O+(aq)
Ka = 1.8 × 10−5
The equilibrium concentrations for a 1.00 mol L−1 ethanoic acid solution are determined in the
usual way, starting with an ICE table.
- 134 -
CH3COOH(aq)
H2O(l)


CH3COO−(aq)
+
H3O+(aq)
1.00
0
0
−x
+x
+x
1.00 − x
x
x
Initial
Change
Equilibrium
+
Substitute the equilibrium concentrations into the equilibrium constant expression. Note that the
activity of H2O(l) is 1 and does not appear in the Ka expression; pure liquids and solids do not
appear in K expressions.3
Ka =
[H 3O + ][CH 3COO  ]
x2
=
= 1.8  105
[CH 3COOH]
1.00  x
This equation can be rearranged to quadratic form,
x 2  1.8  10 5 x  1.8  10 5  1.00  0
with solution,
x
1.8  105 
1.8 10 
5 2
 4  1.8  105  1.00
2
3
 4.23  10

3
 4.25  10
Only the positive solution is admissible. Therefore, the extent of dissociation is 4.2 × 10−3 and
the equilibrium concentrations are given by
[CH3COOH] = 1.00 − x = 0.996 mol L−1 = 1.00 mol L−1
[CH3COO−] = [H3O+] = x = 4.2 × 10−3 mol L−1
The extent of reaction is very small compared to the initial concentration of CH3COOH. In fact,
the concentration of CH3COOH does not change within the accuracy of the given data. This
suggests an approximation method for solving the equilibrium equation.
An approximate solution is provided if we neglect x (i.e., assume x is small and negligible) in the
denominator of the above expression for Ka. In the current example, 1.00 − x ≈ 1.00 is an
excellent approximation. This simplifies the equilibrium equation; specifically,
x2
= 1.8  105
1.00
x  1.8  105  1.00  4.24  103
3
For concentrated solutions, variation in the concentration of water can be non-negligible.
However, neglecting this variation makes an error smaller than the error made by using
concentration for the activity of solutes.
- 135 -
The approximation gives the same extent of reaction as the quadratic formula, to within the
accuracy of the given data.
In general, given an initial weak acid concentration, [HA]initial, and acid dissociation constant, Ka,
the extent of reaction is approximated by
x  K a [HA]initial .
Example 7.8: Use the approximation,
[HA]eqm = [HA]initial − x ≈ [HA]initial
to determine the pH of 2.00 mol L−1 solutions of
(a) HF
(b) HCN
(a) H2S
Ka = 6.3 × 10−4
Ka = 6.2 × 10−10
Ka1 = 8.9 × 10−8
Approach: Set up ICE table as above. Substitute equilibrium concentrations into the
expression for the equilibrium constant. Determine x using the approximation method.
(a)
HF(aq)
Initial
Change
Equilibrium
+
H2O(l)


F−(aq)
+
H3O+(aq)
2.00
0
0
−x
+x
+x
2.00 − x
x
x
Substitute the equilibrium concentrations into the equilibrium constant. expression.
Ka =
[H 3O + ][F ]
x2
=
= 6.3  10 4
[HF]
2.00  x
Upon invoking the approximation, 2.00 − x ≈ 2.00, this equation simplifies to
x2
= 6.3  104
2.00
x  6.3  104  2.00  3.55  102
Therefore,
[H3O+] = x = 3.55 × 10−2 mol L−1
pH = −log10([H3O+]) = −log10(3.55 × 10−2) = 1.45
(b) The calculation is the same as in part (a) except that Ka = 6.2 × 10−10.
The extent of reaction at equilibrium is given by
- 136 -
x2
= 6.2  1010
2.00
x  6.2  1010  2.00  3.52  105
Therefore,
[H3O+] = x = 3.52 × 10−5 mol L−1
pH = −log10([H3O+]) = −log10(3.52 × 10−5) = 4.45 .
(c) Here, we have
x  8.9  108  2.00  4.22  104
and
[H3O+] = x = 4.22 × 10−4 mol L−1
pH = −log10([H3O+]) = −log10(4.22 × 10−4) = 3.37 .
How do we know when the acid is sufficiently concentrated to permit use of the above
approximation? There is a rule based on the requirement that x be no more than [HA]initial/10.
Since x 
K a [HA]initial (the approximation is good enough for this purpose), the rule takes the
form,
K a [HA]initial 
or
[HA]initial
10
[HA]initial
 10
Ka
[HA]initial
 100
Ka
If the initial HA concentration is too small (<100 Ka), we must solve the quadratic equation. In
practice, it is best to always use the quadratic equation, and then use the approximation to check
your answer.
Example 7.9: What is the pH of a 1.00 × 10−3 mol L−1 solution of methanoic acid, HCOOH(aq)
(Ka = 1.8 × 10−4)? What is the percent dissociation of methanoic acid in this solution – i.e., the
percentage of the initial methanoic acid that dissociates?
Approach: Set up the ICE table. Substitute equilibrium concentrations into expression for
the equilibrium constant. Solve for the extent of reaction. Determine [H3O+] and pH.
In this case, the initial HA concentration is less than 6 times bigger than Ka. We must solve
the quadratic equation.
- 137 -
HCOOH(aq)
Initial
H2O(l)


HCOO−(aq)
+
H3O+(aq)
1.00 × 10−3
0
0
−x
+x
+x
1.00 × 10−3 − x
x
x
Change
Equilibrium
+
Substitute the equilibrium concentrations into the equilibrium constant.
Ka =
[H 3O + ][HCOO  ]
x2
=
= 1.8  10 4
3
[HCOOH]
1.00  10  x
This equation is rearranged to quadratic form,
x 2  1.8  10 4 x  1.8  10 4  1.00  10 3  0
with solution,
x
1.8  104 
1.8 10 
4 2
 4  1.8  104  1.00  103
2
 3.44  104

4
5.24  10
Only the positive solution is admissible. Therefore, the extent of dissociation is 3.4 × 10−4
and the equilibrium concentrations are given by
[HCOOH] = 1.00 × 10−3 − x = 1.00 × 10−3 − 3.4 × 10−4 = 6.6 × 10−4 mol L−1
[HCOO−] = [H3O+] = x = 3.4 × 10−4 mol L−1 .
pH = −log10([H3O+]) = −log10(3.4 × 10−4) = 3.47.
The percent dissociation of methanoic acid here is given by
100% × [HCOO−] / [HCOOH]initial = 3.4 × 10−4 / 1.00 × 10−3 = 34% .
For an initial concentration of methanoic acid of 1.00 mol L−1, the extent of reaction at equilibrium
is given using the approximate formula to be
x  K a [HA]initial
 1.8  10 4  1.00  0.013 .
In this case, the percent dissociation is given by
100% × [HCOO−] / [HCOOH]initial = 0.013 / 1.00 = 1.3% .
We see that a weak acid is mostly undissociated when present at high concentration. In Example
7.9, we saw that the same weak acid is 34% dissociated in a dilute solution. The percent
- 138 -
dissociation of methanoic acid increases to 50% for a 1.8 × 10−4 mol L−1 solution, and 99.5%
dissociated for a 1.00 × 10−6 mol L−1 solution.
Figure 7.1 shows the percent dissociation of methanoic acid (solid line) versus the total acid
concentration – i.e., dissociated plus undissociated form (this is just the initial acid concentration)
– with both a linear and logarithmic concentration scale. The percent dissociation of
corresponding hydrochloric acid (dashed line) solutions is also shown.
Figure 7.1. Percent dissociation of methanoic acid and hydrochloric acid for
concentrations from 10−7 to 1 mol L−1.
Figure 7.1 shows that a strong acid such as HCl is fully dissociated at all concentrations. In
contrast, a weak acid such as methanoic acid is mostly undissociated except in dilute solutions.
At sufficient dilution (below 10−6 mol L−1 for methanoic acid), even a weak acid becomes fully
dissociated.
At very low concentration (below 10−6 mol L−1), it is not sufficient to satisfy the acid dissociation
equilibrium condition. The autoionization of water must be taken into account. In principle, we
must satisfy two equilibrium conditions simultaneously. However, since we are dealing with very
dilute solutions, the acid will generally be fully dissociated. This assumption is good as long as
Ka > 100 [HA]initial.
For example, consider a 10−8 mol L−1 solution of hydrochloric acid (or any other monoprotic acid
with Ka > 10−6). In this case, we can take the acid to be fully dissociated resulting in a
concentration of hydronium of 10−8 mol L−1. However, this concentration is less than the
concentration of hydronium in pure water. To correct this result, we must account for the
- 139 -
autoionization of water. The 10−8 mol L−1 hydronium ion concentration provides the initial
concentration of hydronium in the associated ICE table.
In terms of the extent of the water autoionization reaction, Eq. 7.5 gives
10−14 = (x + 10−8) x = x2 + 10−8 x
or
0 = x2 + 10−8 x − 10−14,
which has the admissable root, x = 9.5 × 10−8. The equilibrium hydronium concentration is
[H3O+] = x + 10−8 = 1.05 × 10−7 mol L−1
and
pH = 6.98 .
This pH is slightly lower than that of a neutral solution, since it is an acid solution.
Base Association Equilibrium
Weak bases can be treated in exactly the same fashion as weak acids. The equilibrium constant
for the reaction of a base with water to form OH− is called Kb. For example,
NH3(aq) + H2O(l) 
NH4+(aq) + OH− (aq) .
Kb = 1.8 × 10−5
This is the base association reaction.
Example 7.10: What is the pH of a (a) 1.00 mol L−1 and (b) 1.00 × 10−3 mol L−1 solution of
ammonia, NH3(aq)?
Approach: Exactly like a weak acid calculation, except that B replaces HA, HB+ replaces A−
and OH− replaces H3O+.
(a)
NH4+(aq)
OH−(aq)
1.00
0
0
−x
+x
+x
1.00 − x
x
x
NH3(aq)
Initial
Change
Equilibrium
+
H2O(l)


Substitute the equilibrium concentrations into the equilibrium constant expression.
Kb =
[NH4+ ][OH ]
x2
=
= 1.8 105
[NH3 ]
1.00  x
The approximation, 1.00 − x ≈ 1.00 is good here. Therefore,
x 2 = 1.8 105 1.00
x  1.8 105  4.24  103
- 140 -
Therefore,
[OH−] = x = 4.24 × 10−3 mol L−1
pOH = −log10([OH−]) = −log10(4.24 × 10−3) = 2.37
pH = 14.00 − pOH = 14.00 − 2.37 = 11.63
(b)
NH3(aq)
Initial
H2O(l)


NH4+(aq)
+
OH−(aq)
1.00 × 10−3
0
0
−x
+x
+x
1.00 × 10−3 − x
x
x
Change
Equilibrium
+
Substitute the equilibrium concentrations into the equilibrium constant.
[NH 4 + ][OH  ]
x2
Kb =
=
= 1.8 105
3
[NH3 ]
1.00 10  x
The solution is dilute, 1.00 × 10−3 − x cannot be approximated. The equation is rearranged to
quadratic form,
x 2  1.8  10 5 x  1.8  10 5  1.00  10 3  0
with solution,
x
1.8  105 
1.8 10 
5 2
 4  1.8  105  1.00  103
2
4
 1.2  10
 5
4
1.43  10
The admissible solution (x = 1.25 × 10−4) gives
[OH−] = x = 1.25 × 10−4 mol L−1
pOH = −log10([OH−]) = −log10(1.25 × 10−4) = 3.90
pH = 14.00 − pOH = 14.00 − 3.90 = 10.10 .
Analogous to weak acids, weak bases are mostly unassociated except in dilute solutions –
specifically, concentrations lower than or comparable to Kb. In very dilute solutions
(concentration less than Kb/100), the base is almost completely present in protonated form (the
conjugate acid).
- 141 -
7.7 Strength of Acids and Bases
Acids are ordered according to acid dissociation constant, Ka. The strongest acids have large Ka,
while weak acids have small Ka. Bases are similarly ordered according to base association
constant, Kb. Acid and base equilibrium constants vary over many orders of magnitude. For this
reason, it is customary to introduce
pKa = ‒log10(Ka)
and
pKb = ‒log10(Kb) .
7.8
Because of the minus sign, the strongest acids have the lowest pKa values, while the weakest
acids have the highest pKa values. Similarly, a lower pKb corresponds to a stronger base.
The distinction between strong, weak and very weak acid is made quantitative by Ka (or pKa).
Hydronium, with pKa = 0, is the dividing line between strong acids (pKa < 0) and weak acids
(14 > pKa > 0). Water is the dividing line between weak acids and very weak acids (pKa >
14). Similarly, hydroxide separates weak (14 > pKb > 0) and strong (pKb < 0) bases, while
water separates weak and very weak bases (pKb > 14). Table 7.1 lists pKa and pKb values for
some acids and (conjugate) bases.
- 142 -
Table 7.1. Acid pKa and base pKb values
From CRC Handbook of Chemistry and Physics, 93rd Ed., except for the negative values which
are from M.B. Smith and J. March, March’s Advanced Organic Chemistry: Reactions,
mechanisms and Structure, 5th Ed. (Wiley-Interscience, New York, 2001).
Strong
Acids
Weak
Acids
Very
Weak
Acids
Acid
pKa
Base
pKb
HI
HClO4
HBr
HCl
H2SO4
HNO3
HClO3
H3O+
Cl3CCOOH
HIO3
Cl2CHCOOH
H2SO3
HClO2
HSO4‒
H3PO4
ClCH2COOH
BrCH2COOH
HF
HNO2
HCOOH
CH3COOH
CH3CH2COOH
NH3OH+
H2CO3
H2S
HSO3‒
H2PO4‒
HOCl
HOBr
HCN
NH4+
HCO3‒
+
CH3NH3
H2O
CH3OH
HCCH
H2
NH3
CH3CH3
‒10
‒10
‒9
‒7
‒3
‒1.4
‒1
0.00
0.66
0.78
1.35
1.85
1.94
1.99
2.16
2.87
2.90
3.20
3.25
3.75
4.76
4.87
5.94
6.35
7.05
7.2
7.21
7.40
8.55
9.21
9.25
10.33
10.66
14.00
15.5
25
35
38
50
I
ClO4‒
Br‒
Cl‒
HSO4‒
NO3‒
ClO3‒
H2O
Cl3CCOO‒
IO3‒
Cl2CHCOO‒
HSO3‒
ClO2‒
SO42‒
H2PO4‒
ClCH2COO‒
BrCH2COO‒
F‒
NO2‒
HCOO‒
CH3COO‒
CH3CH2COO‒
NH2OH
HCO3‒
HS‒
SO32‒
HPO42‒
OCl‒
OBr‒
CN‒
NH3
CO32‒
CH3NH2
OH‒
CH3O‒
HCC‒
H‒
NH2‒
CH3CH2‒
24
24
23
21
17
15.4
15
14.00
13.34
13.22
12.65
12.15
12.06
12.01
11.84
11.13
11.10
10.80
10.75
10.25
9.24
9.13
8.06
7.65
6.95
6.8
6.79
6.60
5.45
4.79
4.75
3.67
3.34
0.00
‒1.5
‒11
‒21
‒24
‒36
‒
Very Weak
Bases
Weak
Bases
Strong
Bases
The pKa of an acid and the pKb of its conjugate base are related. To see how they are related,
start with the expressions for Ka of an acid, HA, and Kb of its conjugate base, A‒.
Ka 
[H  ][A  ]
[HA]
and
Kb 
The product of these two expressions simplifies:
- 143 -
[OH  ][HA]
[A  ]
K a Kb 
[H  ][A  ] [OH  ][HA]
[A ]
[HA]
 [H ][OH  ]  K w  1.00 1014 .
The product, Ka Kb, is the same (equal to Kw) for any acid and its conjugate base. Thus, strong
acids with large Ka are conjugate to bases with small Kb, and vice versa. Applying ‒log10 to the
above equation gives
or
‒log10 (Ka Kb) = ‒log10(Kw) = 14.00
pKa + pKb = 14.00 .
7.9
The relative strength of an acid depends on the relative stability of the conjugate base. Consider
a neutral acid. When it loses an H+, an anion is formed which polarizes surrounding water
molecules. This polarization induces the return of the donated H+, and limits the stability of the
anion. A more stable anion results if its polarizing effect is reduced. Thus, the strongest neutral
acids are conjugate to the most stable anions. Properties affecting anion stability are listed as
follows:
(1) Anion stability increases with the size of the atom carrying the charge. Spreading out the
minus charge reduces the polarizing effect on the surrounding water molecules.
(2) Anion stability increases with the extent to which the charge is spread out due to resonance
structures.
(3) Anion stability increases with the electronegativity of the atom carrying the negative charge.
Electronegative atoms better accomodate a negative charge.
(4) Anion stability increases with the electronegativity of neighboring atoms. More
electronegative neighboring atoms pull electrons toward themselves, through bonds, reducing the
electron density at the atom(s) carrying the charge. This is called the inductive effect.
Strength of Binary Acids
Binary acids are of the form, HX. If we consider a series of binary acids with X atoms belonging
to the same group, rule 1 tells us that the acid with the largest X atom is the strongest.
Specifically, the strength of HX acids increases as X varies down a group. For example, the
hydrohalic acids are ordered according to strength as follows:
HI > HBr > HCl > HF .
pKa
‒10
‒9
‒7
3.2
According to stability, the conjugate base anions are correspondingly ordered:
I‒ > Br‒ > Cl‒ > F‒ .
As electronegativity decreases down the group, rule 3 supports the reverse ordering of these
acids. However, the size effect is the most important here. The large iodide ion (the conjugate
base of HI) only weakly polarizes the surrounding water molecules. Iodide is very stable. So, HI
is a very strong acid. In contrast, the small fluoride ion is sufficiently polarizing to make HF only a
weak acid – in spite of the large electronegativity of fluorine.
- 144 -
If we vary X across a period, the atoms are comparable in size. In this case, the electronegativity
of X determines the relative strength of the acids – rule 3. Thus, the strength of HX acids
increases as X goes from left to right across a period. For example,
CH4 < NH3 < H2O < HF .
pKa
60
38
14.00
3.2
According to stability, the anions are correspondingly ordered:
CH3‒ > NH2‒ > OH‒ > F‒ .
Carbon is not sufficiently electronegative to stabilize the negative charge on CH3‒. This methyl
carbanion is a very strong base. The conjugate acid, methane, is a very weak acid. Under
anything but most extreme conditions, hydrocarbons are not acids. Ammonia and water are also
very weak acids. Only HF is a weak acid in this series.
Strength of Oxoacids
In oxoacids, the acidic hydrogen is bonded to an oxygen atom. Variations in acid stength are
determined by rules 2 and 4. Resonance structures can spread the negative charge over a
number of atoms. Anion stability increases with the number of such atoms. It also increases with
the electronegativity of neighboring atoms – the inductive effect.
Consider the following ordering of acids:
H2O < HOBr < HOCl .
pKa
14
8.55
7.40
According to anion stability,
OH‒ < OBr‒ < OCl‒ .
This ordering of acids follows from the inductive effect (rule 4). In each of these molecules, the
acidic hydrogen is bonded to oxygen. The other atom bonded to oxygen determines the acid
strength. When the other atom is electronegative, as are bromine of chlorine, the pull of electron
density away from the oxygen atom gives the oxygen a larger effective electronegativity. This
stabilizes the negative charge on the conjugate base anion. Hypochlorous acid is a stronger acid
than hypobromous acid because chlorine is more electronegative than bromine.
Carboxylic acids are more acidic than alcohols because of the inductive effect and resonance
stabilization (rules 2 and 4).
CH3CH2OH < CH3COOH < Cl3CCOOH .
pKa
15.5
4.76
0.66
According to anion stability,
CH3
CH2
O
CCl3
CH3
<
C
O
C
<
O
O
- 145 -
O
In each of these anions, the minus charge is on an oxygen atom bonded to a carbon atom. In
ethanoate, this carbon is doubly bonded to an oxygen atom. The electronegative oxygen pulls
electron density away from the carbon atom. This increases the effective electronegativity of the
carbon atom, making it pull harder on the electron density of the oxygen with the minus charge.
The charge spreads out, through bonds, stabilizing the anion. Trichloroethanoate is even more
stable because of the electronegative chlorine atoms. The pull of the chlorine atoms further
spreads the charge through the bonds of the molecule. The inductive effect predicts the
observed order of acid strength. However, carboxylates also have resonance stabilization.
CH3
CH3
C
O
C
O
O
O
Ethanoate, like all carboxylates, has two resonance structures which spread the negative charge
equally over two oxygen atoms. This makes ethanoate less polarizing of its surroundings, and
much more stable than ethanoxide, the conjugate base of ethanol.
The relative strength of sulfurous and sulfuric acid also follows from the inductive effect and
resonance stabilization.
H2SO3 < H2SO4 .
pKa
1.85
‒3
The sulfur atom in sulfuric acid is bonded to four oxygen atoms. This gives it a larger effective
electronegativity than the sulfur atom in sulfurous acid which is bonded to only three oxygen
atoms. Also, hydrogen sulfate has three resonance structures, speading the charge over three
oxygen atoms, whereas hydrogen sulfite has only two:
H
H
O
H
O
O
O
S
S
S
O
O
O
O
O
S
H
O
O
O
O
O
O
S
H
O
O
O
For polyprotic acids – acids with more than one acidic hydrogen – the conjugate base is also an
acid which can donate an H+ to a base. The sequence of pKa values associated with the
successive deprotonation steps is generally labeled pKa1, pKa2, etc.. With each deprotonation
step, the conjugate base takes on an additional minus one charge. Since a doubly charged anion
- 146 -
is less stable than the singly charged conjugate acid, we must have pKa1 < pKa2. There is
generally a distinct gap between these values. Subsequent steps have even larger pKa values.
Strength of Bases
Many bases are anions – they are conjugate bases to a neutral acid. We have already seen that
the strength of an acid increases with stability of the conjugate base anion. Thus, the strength
(as a base) of the anion decreases with its stability. A stable anion has little propensity to accept
an H+ from the solution. A stable anion is therefore a very weak base.
Some bases are neutral. They accept an H+, and become positively charged. The strength of
the base is determined by the stability of the cation formed. For example, according to base
strength, we have
NH3 > H2O > HF .
pKb
4.75
14.00
much larger
This matches the ordering of the associated cations according to stability,
NH4+ > H3O+ > H2F+ .
The positive charge is more stable attached to a less electronegative element, nitrogen. H2F+ is
not sufficiently stable to ever be present in an aqueous solution.
The inductive effect also plays a role. Consider the following amines:
CH3NH2 > NH3 > NH2OH .
pKb
3.34
4.75
8.06
According to cation stability,
CH3NH3+ > NH4+ > NH3OH+ .
Here, the positively charged nitrogen atom can take methyl group electron density to stabilize the
positive charge, since nitrogen is more electronegative than carbon. In contrast, the more
electronegative oxygen atom takes electron density away from positively charged nitrogen, and
destabilizes the cation.
Example 7.11: Order the following sets of acids and bases according to increasing strength as
an acid or base, respectively:
(a)
(b)
(c)
(d)
H2O, H2S, H2Se, H2Te (as acids)
SiH4, PH3, H2S, HCl (as acids)
CH3COOH, Cl3CCOOH, BrCH2COOH, ClCH2COOH (as acids)
CH3CH2NH2, NH3, NHCl2 (as bases)
Approach: For neutral (or negatively charged) acids, assess the relative stability of the
conjugate base anion. The most stable anion is conjugate to the strongest acid. For
neutral bases, assess the relative stability of the conjugate acid cation. Look for inductive
effects that stabilize or destabilize the cation. The most stable cation is conjugate to the
strongest base.
(a) According to increasing acid strength,
- 147 -
H2O < H2S < H2Se < H2Te .
The larger atoms can better accommodate the negative charge of the conjugate base
anion.
(b) According to increasing acid strength,
SiH4 < PH3 < H2S < HCl .
The more electronegative atoms can better accommodate the negative charge of the
conjugate base anion. The atom X, in HX, varies across the third period. These atoms
are comparable in size.
(c) According to increasing acid strength,
CH3COOH < BrCH2COOH < ClCH2COOH < Cl3CCOOH .
The inductive effect is larger with one chlorine, than with one bromine atom – chlorine is
more electronegative. It is larger still if there are three chlorine atoms.
(d) According to increasing strength as bases,
NHCl2 < NH3 < CH3CH2NH2 .
In this case, the inductive effect of the electronegative chlorine atoms destabilizes the
conjugate acid cation by pulling electron density away from positively charged nitrogen.
Problems:
7.1 Write down expressions for the equilibrium constants of the following reactions, in terms of
equilibrium concentrations and/or partial pressures.
(a)
(b)
(c)
(d)
(e)
Cl2(g) + Br2(g)  2 BrCl(g)
Fe3+(aq) + Cu+(aq)  Fe2+(aq) + Cu2+(aq)
2 IO3−(aq) + 5 SO32−(aq) + 2 H+(aq)  I2(s) + 5 SO42−(aq) + H2O(l)
Ca3(PO4)2(s)  3 Ca2+(aq) + 2 PO42−(aq)
HCN(aq) + H2O(l)  CN−(aq) + H3O+(aq)
7.2 An equilibrium mixture of N2O4 and NO2, at 25°C, has a N2O4 partial pressure of 2.00 bar.
What is the partial pressure of NO2?
N2O4(g) 
2 NO2(g)
K = 0.323 at 25°C
7.3 Write down the equilibrium constants for the following reactions given the data table of
equilibrium constants below.
(a) Fe3+(aq) + Cu+(aq)  Fe2+(aq) + Cu2+(aq)
(b) 2 Fe2+(aq) + Cl2(aq)  2 Fe3+(aq) + 2 Cl−(aq)
(c) 2 NaF(s) + Mg2+(aq)  MgF2(s) + 2 Na+(aq)
- 148 -
DATA:
Reaction
Equilibrium Constant
1) 2 Fe3+(aq) + Cl2(aq) + 2 H2O(l)  2 Fe2+(aq) + 2 HOCl(aq) + 2 H+(aq)
2) 2 HOCl(aq) + 2 Cu+(aq) + 2 H+(aq)  Cl2(aq) + 2 H2O(l) + 2 Cu2+(aq)
3) 2 Cu+(aq) + Cl2(aq)  2 Cu2+(aq) + 2 Cl−(aq)
4) MgF2(s)  Mg2+(aq) + 2 F−(aq)
5) NaF(s)  Na+(aq) + F−(aq)
K1
K2
K3
K4
K5
7.4 Predict the direction each of the following equilibriums shifts when the indicated changes are
imposed.
N2O4(g) 
(a)
(b)
(c)
(d)
∆H° = 55.3 kJ mol−1
2 NO2(g)
The N2O4 partial pressure is increased
Some NO2 is removed from the reaction mixture
The reaction mixture is transferred to a vessel with twice the volume
Temperature is increased
MgF2(s)
 Mg2+(aq) + 2 F−(aq)
∆H° = −13.6 kJ mol−1
(e) Some KF is dissolved in the solution
(f) Some solid MgF2 is added to the mixture
(g) Temperature is decreased
7.5 Determine the equilibrium partial pressure of all gases in the following equilibrium, after a
vessel is prepared at 400°C with 10.0 bar partial pressure of both methane and bromine.
CH4(g) + Br2(g) 
CH3Br(g) + HBr(g)
K = 0.012 at 400°C
7.6 (a) What is the extent of dissolution of Ba3(PO4)2(s) (Ksp = 3.4 × 10−23), and the equilibrium
ion concentrations, in a 1.0 × 10−5 mol L−1 KBr(aq) solution?
(b) How would the extent of dissolution of Ba3(PO4)2(s) be different in a 1.0 × 10−5 mol L−1
K3PO4(aq) solution? (Just answer qualitatively – no calculations required.)
7.7 What is the pH of a (a) 1.00 mol L−1 and (b) 1.00 × 10−6 mol L−1 solution of hypochlorous
acid? Ka(HOCl) = 4.0 × 10−8. Determine the percent dissociation in both cases.
7.8 What is the pH of a (a) 1.00 mol L−1 and (b) 1.00 × 10−3 mol L−1 solution of ethylamine,
CH3CH2NH2? Kb(CH3CH2NH2) = 4.5 × 10−4. Determine the percent association in both
cases.
7.9 Order the following sets of acids and bases according to increasing strength as an acid or
base, respectively:
(a) AsH3, PH3, H2Te, H2Se, HI (as acids)
(b) H2SO3, H2SeO3, HClO3, HClO4 (as acids)
(c) CCl3CH2NH2, CH2CICH2NH2, CH2BrCH2NH2 (as bases)
- 149 -
Solutions:
2
PBrCl
7.1 (a) K 
PCl2 PBr2
(b)
K
[Fe2+ ][Cu 2  ]
[Fe3+ ][Cu  ]
[SO 24  ]5
(c) K 
[IO3 ]2 [SO32  ]5 [H + ]2
The I2(s) and H2O(l) do not appear in this expression. Iodine is a pure solid, and the liquid
water concentration varies very little except in very concentrated aqueous solutions.
(d)
K  K sp  [Ca 2+ ]3[PO4 2  ]2
Pure solid, Ca3(PO4)2(s), does not appear. This is the solubility product of calcium
phosphate.
(e)
K  Ka 
[H3O + ][CN  ]
[HCN]
Liquid water does not appear in the Ka expression.
7.2
K
2
PNO
2
PN2O4

2
PNO
2
2.00
 0.323
PNO2  2.00  0.323  0.804 bar
7.3 (a) Since
1/2 ×
1/2 ×
2 Fe3+(aq) + Cl2(aq) + 2 H2O(l) 
2 HOCl(aq) + 2 Cu+(aq) + 2 H+(aq)
Net exn:
Fe3+(aq) + Cu+(aq) 
2 Fe2+(aq) + 2 HOCl(aq) + 2 H+(aq)
 Cl2(aq) + 2 H2O(l) + 2 Cu2+(aq)
Fe2+(aq) + Cu2+(aq)
K  K11/ 2 K 21/ 2
(b) Since
−1×
−1×
1×
2 Fe3+(aq) + Cl2(aq) + 2 H2O(l)  2 Fe2+(aq) + 2 HOCl(aq) + 2 H+(aq)
2 HOCl(aq) + 2 Cu+(aq) + 2 H+(aq)  Cl2(aq) + 2 H2O(l) + 2 Cu2+(aq)
2 Cu+(aq) + Cl2(aq)  2 Cu2+(aq) + 2 Cl−(aq)
Net rxn: 2 Fe2+(aq) + Cl2(aq)
 2 Fe3+(aq) + 2 Cl−(aq)
- 150 -
K
K3
K1 K 2
(c) Since
−1×
2×
Net rxn:
MgF2(s)  Mg2+(aq) + 2 F−(aq)
NaF(s)  Na+(aq) + F−(aq)
2 NaF(s) + Mg2+(aq) 
MgF2(s) + 2 Na+(aq)
K
K52
K4
7.4 (a) N2O4 is the reactant. Increasing reactant results in net forward reaction – consuming
some of the added N2O4.
(b) Removing product results in net forward reaction.
(c) We start with an equilibrium mixture with
2
PNO
2 ,eqm 1
PN 2O4 ,eqm 1
K
Doubling the volume decreases both partial pressures to 1/2 their initial values. The reaction
quotient is now equal to
2
1

2
 PNO2 ,eqm 1 
1 PNO2 ,eqm 1 1
2


Q

 K.
1
2 PN2O4 ,eqm 1 2
PN2O4 ,eqm 1
2
Since Q < K, there is net forward reaction. This is predicted by le Châtelier's principle. The
reaction shifts to the side with greater moles of gas when volume is increased (decreasing
partial pressure). This increases the total pressure, countering the reduction caused by the
increase in volume.
(d) This reaction is endothermic. The equilibrium constant increases with increasing
temperature. Therefore, there is net forward reaction. In terms of le Châtelier's principle, the
reaction shifts so as to consume some of the additional heat available at higher T.
(e) Dissolving some KF in the solution increases the fluoride concentration. Fluoride is a
product. This causes net reverse reaction. Less of the MgF2(s) is dissolved when equilibrium
is re-established. This is the common ion effect.
(f) Adding pure solid MgF2 to the mixture has no effect on the equilibrium.
(g) This reaction is exothermic. The equilibrium constant increases with decreasing
temperature. Therefore, there is net forward reaction. In terms of le Châtelier's principle, the
reaction shifts, releasing heat, to counter the reduced availability of heat at lower T.
- 151 -
7.5
CH4(g)
Initial
Change
Equilibrium
+
Br2(g)
CH3Br(g)



+
HBr(g)
10.0
10.0
0
0
−x
−x
+x
+x
10.0 − x
10.0 − x
x
x
K 
PCH3Br PHBr
PCH4 PBr2

xx
 0.012
(10.0  x)(10.0  x)
from which we get
2


x

  0.012
 (10.0  x ) 
x
 0.012  0.110
(10.0  x )
x  0.110 (10.0  x)
x
1.10
 0.991
1.110
With x = 0.991, we can get the equilibrium partial pressures. Specifically,
PCH4 = PBr2 = 10.0 − x = 9.0 bar
and
PCH3Br = PHBr = x = 0.99 bar .
7.6 (a) Initially, there are potassium and bromide ions present in solution. However, there are
no barium or phosphate ions.
Ba3(PO4)2(s)


3 Ba2+(aq)
+
2 PO43−(aq)
0
0
Change
+3 x
+2 x
Equilibrium
3x
2x
Initial
Substitute the equilibrium ion concentrations into the solubility product equation.
K sp = [Ba 2+ ]3 [PO 43 ]2 =  3 x   2 x  = 3.4  1023
3
which gives the extent of dissolution,
- 152 -
2
1/5
 3.4  1023 
x=

 108 
 1.26  105 .
At equilibrium,
[Ba2+] = 3 x = 3.8 × 10−5 mol L−1
[PO43−] = 2 x = 2.5 × 10−5 mol L−1
(b) Initially, there is 1.0 × 10−5 mol L−1 PO43−(aq) present in solution. The phosphate already
present suppresses the dissolution of Ba3(PO4)2(s) – the common ion effect. The extent of
dissolution is smaller in this case.
7.7 (a) What is the pH of a (a) 1.00 mol L−1 and (b) 1.00 × 10−3 mol L−1 solution of hypochlorous
acid? Ka(HOCl) = 4.0 × 10−8
H3O+(aq)
OCl−(aq)
1.00
0
0
−x
+x
+x
1.00 − x
x
x
HOCl(aq)
Initial
Change
Equilibrium
+
H2O(l)


Substitute the equilibrium concentrations into the equilibrium constant expression.
Ka 
[H 3O + ][OCl ]
x2

 4.0  108
[HOCl]
1.00  x 
The approximation, 1.00 − x ≈ 1.00 is good here. Therefore,
x 2  4.0  108  1.00  4.0  108
x  4.0  108  2.0  104
and
[H3O+] = x = 2.0 × 10−4 mol L−1
pH = −log10([H3O+]) = −log10(2.0 × 10−4) = 3.70
The percent dissociation of hypochlorous acid is given by
100% × [OCl−] / [HOCl]initial = 2.0 × 10−4 / 1.00 = 0.020% .
- 153 -
(b)
H3O+(aq)
OCl−(aq)
1.00 × 10−6
0
0
−x
+x
+x
1.00 × 10−6 − x
x
x
HOCl(aq)
Initial
Change
Equilibrium
+
H2O(l)


Substitute the equilibrium concentrations into the equilibrium constant expression.
Ka 
[H 3O + ][OCl  ]
x2

 4.0  10 8
6
[HOCl]
1.00  10  x


The solution is dilute, we cannot neglect x in (1.00 × 10−6 − x). The equation is rearranged to
quadratic form,
x 2  4.0  10 8 x  4.0  10 8  1.00  10 6  0
with solution,
x
4.0  108 
 4.0 10 
8 2
 4  4.0  108  1.00  106
2
 1.8  107
 1
7
2.21  10
The admissible solution (x = 1.81 × 10−7) gives
[H3O+] = x = 1.81 × 10−7 mol L−1
pH = −log10([H3O+]) = −log10(1.81 × 10−7) = 6.74
In this case, the percent dissociation of hypochlorous acid is
100% × [OCl−] / [HOCl]initial = 1.81 × 10−7 / 1.00 × 10−6 = 18 % .
7.8 (a)
CH3CH2NH3+(aq)
OH−(aq)
1.00
0
0
−x
+x
+x
1.00 − x
x
x
CH3CH2NH2(aq)
Initial
Change
Equilibrium
+
H2O(l)


Substitute the equilibrium concentrations into the equilibrium expression.
- 154 -
Kb =
[CH3CH 2 NH3+ ][OH  ]
x2
=
= 4.5 104 .
[CH3CH 2 NH 2 ]
1.00  x
The approximation, 1.00 − x ≈ 1.00 is good here. Therefore,
x 2 = 4.5 104 1.00
x  4.5 104  2.12  102
Therefore,
[OH−] = x = 2.12 × 10−2 mol L−1
pOH = −log10([OH−]) = −log10(2.12 × 10−2) = 1.67
pH = 14.00 − pOH = 14.00 − 1.67 = 12.33
The percent association of ethanamine (i.e., the percentage of the initial ethanamine in
conjugate acid form) is
100% × [CH3CH2NH3+] / [CH3CH2NH2]initial = 2.12 × 10−2 / 1.00 = 2.1 % .
(b)
CH3CH2NH3+(aq)
OH−(aq)
1.00 × 10−3
0
0
−x
+x
+x
1.00 × 10−3 − x
x
x
CH3CH2NH2(aq)
Initial
Change
Equilibrium
+
H2O(l)


Substitute the equilibrium concentrations into the equilibrium constant expression.
Kb =
[CH3CH 2 NH3+ ][OH  ]
x2
=
= 4.5 104
3
[CH3CH 2 NH 2 ]
1.00 10  x
The solution is dilute, 1.00 × 10−3 − x cannot be approximated. The equation is rearranged to
quadratic form,
x 2  4.5  10 4 x  4.5  10 4  1.00  10 3  0
with solution,
x
4.5  104 
 4.5 10 
4 2
 4  4.5  104  1.00  103
2
4
 4.83  10

4
9.33  10
- 155 -
The admissible solution (x = 4.83 × 10−4) gives
[OH−] = x = 4.83 × 10−4 mol L−1
pOH = −log10([OH−]) = −log10(4.83 × 10−4) = 3.32
pH = 14.00 − pOH = 14.00 − 3.32 = 10.68 .
In this case, the percent association of ethanamine is
100% × [CH3CH2NH3+] / [CH3CH2NH2]initial = 4.83 × 10−4 / 1.00 × 10−3 = 48.3 % .
7.9 (a) According to increasing strength as acids,
PH3 < AsH3 < H2Se < H2Te < HI .
AsH3 is a stronger acid than PH3 because the larger arsenic atom better accommodates the
minus charge of AsH2‒. H2Se is stronger than AsH3, because selenium atoms have
comparable size to arsenic atoms, and are more electronegative. H2Te is stronger than H2Se
because of the larger size of the tellurium atoms. HI is stronger than H2Te because of the
higher electronegativity of iodine.
(b) According to increasing strength as acids,
H2SeO3 < H2SO3 < HClO3 < HClO4 .
H2SO3 is a stronger acid than H2SeO3 because sulfur is more electronegative than selenium.
The larger size of selenium is not relevant here because the minus charge of the anion is
attached to oxygen atoms. Selenium plays its role through the inductive effect. More
electronegative sulfur has a larger inductive effect. Since chlorine is even more
electronegative than sulfur, HClO3 is a stonger acid than H2SO3. Also, chlorine in HClO3 is in
the +V oxidation, whereas sulfur in H2SO3 is in the +IV oxidation state. HClO4 is stronger
than HClO3 because, in perchloric acid, chlorine is in the +VII oxidation state. A higher
oxidation state results in a larger effective electronegativity.
(c) According to increasing strength as bases,
CCl3CH2NH2 < CH2CICH2NH2 < CH2BrCH2NH2 .
CH2BrCH2NH2 is a stronger base than CH2CICH2NH2 because bromine is less
electronegative than chlorine. Here, a larger inductive effect destabilizes the conjugate acid
cation, weakening the strength of the base. CCl3CH2NH2 has an even larger inductive effect,
and is therefore an even weaker base.
- 156 -
8. Thermochemistry
Chemical reactions, such as 2 H2(g) + O2(g) → 2 H2O(l), and phase changes, such as
CO2(s) → CO2(g), are accompanied by changes in energy. For example, two moles of liquid
water has less energy than two moles of hydrogen and one mole of oxygen gases.
Consequently, the chemical reaction of hydrogen and oxygen releases energy – ultimately in the
form of heat. Chemical reactions are an important source of energy – e.g., heat in homes in
winter, mechanical energy in the engine of a car, electrical energy in a cellphone battery. In order
to predict the energy released by a given chemical reaction or physical process, chemists have
measured the heat released by many processes and developed an energy scale for a wide range
of substances. The closely related enthalpy scale is used to determine heat released under
standard conditions - see Sec. 8.5. To understand this scale and how to use it requires the
introduction of precise terminology.
8.1 System and Properties - Energy
We divide the universe into system(s) and surroundings (i.e., everything else). A system is an
amount of a substance or collection of substances with a volume. If the system is in equilibrium
(i.e., it is unchanging, or changing slowly), then it also has a temperature and pressure. It has
other properties also. Most notably, the system has an energy, the sum total of all potential and
kinetic energies of particles in the system. Enthalpy is another property – it is energy plus a small
extra term. An energy or enthalpy scale for substances can be constructed from a sufficiently
large set of measured energy or enthalpy changes. Changes in energy or enthalpy are
determined by measuring the flow of heat to or from the system.
8.2 Heat
Heat is the energy that flows from a hot system to a cold system, when they are in thermal
contact (this simply means they are touching, and not separated by thermal insulation). Heat
flows until the final temperatures of the systems are the same. We use the symbol q to denote
heat flow into a system, when there is one system of interest. When heat flows from the system,
q is negative. When there are two systems, we use subscripts to distinguish the properties.
When heat flows from system 2 to system 1 their q values are equal in magnitude but opposite in
sign;
q2 = −q1
q1 is positive because heat flowed into system 1, while q2 is negative.
Temperature measurements are used to determine the amount of heat flow. When heat flows
into a system, its temperature generally increases. The temperature change is proportional to the
heat transferred, and vice versa. Thus, a measured temperature change can be used to
determine the amount of heat transferred via
q = C ∆T
C is the heat capacity of the system. It is proportional to the amount of substance – the size of
the system. A system twice as big – but otherwise identical – has twice the heat capacity.
Consequently, heat capacities for substances are tabulated for substances per mole, or per gram.
Specific heat capacity,
s = C/m
is the heat capacity per gram of substance; i.e., m is the mass in grams. Specific heat capacities
are tabulated for various substances. The value for water is 4.18 J K−1 g−1 at 25°C and 1 bar
pressure. We will use this value for water under any conditions. This is an approximation that
- 157 -
greatly simplifies all our calculations. Copper has a much lower specific heat capacity, 0.385 J
K−1 g−1 at 25°C and 1 bar pressure (we will treat s as independent of temperature for all
substances). In terms of specific heat capacity, heat is related to temperature change according
to
q = ms ∆T .
8.1
Example 8.1: How much heat must flow into a 100. g sample of water to raise its temperature
from 20.00°C to 25.00°C? (swater = 4.18 J K−1 g−1)
Approach: Substitute given mass, specific heat capacity and temperature difference into the
above formula.
Note that
∆T = Tfinal − Tinitial = 25.00 °C − 20.00 °C = 5.00 °C or 5.00 K .
Differences in temperature, ∆T, expressed in °C, can also be written with the unit K. A 1 °C
difference is the same as 1 K difference.
The heat transferred is
q = ms ∆T
= 100. g × 4.18 J K−1 g−1 × 5.00 K
= 2092 J = 2.09 kJ .
This example demonstrates how a calorimeter works. A chemical reaction or phase transition
releases or absorbs heat within a calorimeter. The calorimeter has a well-defined heat capacity.
Sometimes you will be given the heat capacity, C. Other times you are given a mass of a
substance (usually it is water, or a dilute aqueous solution) and you use ms for the heat capacity.
Occasionally, you must add the heat capacities of the (empty) calorimeter and a reacting solution.
The temperature change inside the calorimeter is measured to determine how much heat was
released or absorbed by the reaction or phase transition.
Example 8.2: A 20.0 g sample of copper at 300.°C is placed within a thermally insulated vessel
containing 100. g of water at 20.0°C. What is the final temperature when thermal equilibrium is
re-established? (swater = 4.18 J K−1 g−1 and scopper = 0.385 J K−1 g−1)
Approach: Express the heat lost by copper, and the heat gained by water, in terms of the
respective changes in temperature. Equate these expressions and solve for the unknown
final temperature, Tf. Note that heat lost by copper is −qcopper.
The heat lost by copper is
−qcopper = −mcopper scopper ∆Tcopper
= −20.0 g × 0.385 J K−1 g−1 × (Tf − 300. °C)
= 7.70 J K−1 (300. °C − Tf)
The heat gained by water is
qwater = mwater swater ∆Twater
= 100. g × 4.18 J K−1 g−1 × (Tf − 20. °C)
= 418 J K−1 (Tf − 20. °C)
Heat lost by copper is gained by water. So
qwater = −qcopper
or
418 J K−1 (Tf − 20. °C) = 7.70 J K−1 (300. °C − Tf)
- 158 -
which rearranges to
(418 + 7.70) Tf = 7.70 × 300. + 418 × 20.
Tf = 10670/425.7 °C = 25.1 °C
The final temperature has units of °C because the initial temperatures of copper and water
were expressed in °C. Only when dealing with temperature differences do we use °C in a
thermodynamic formula. Otherwise, temperature must be expressed in Kelvin.
When there is a phase transition, heat can be absorbed or released without an accompanying
change in temperature. Melting, vaporization and sublimation absorb heat, while the reverse
processes – freezing, condensation and deposition – release heat while a constant temperature
is maintained.
8.2 Work
The energy of a system can be changed in another way. Work can be done on a system (or by a
system). Work is mechanical energy transferred when force applies over a distance. A system
exchanges work with its surroundings when the boundary between them moves, increasing or
decreasing the volume of the system. Consider the gas system depicted in Fig. 8.1. Here, the
piston applies a force to a boundary of the gas. The boundary moves down a distance, d. Work
is done on the gas, as its boundary moves and its volume decreases.
F
F
apply force on gas in
cylinder via piston
Compressed gas has
higher energy
d
Vi
Vf
Figure 8.1. Force applied on a gas over a distance decreases the volume of the gas while
increasing its energy.
The force on the piston is applied over the distance, d. The work done on the gas is
w = Fd
= PAd
= −P ∆V
8.2
where A is the area of the moving boundary and P is the pressure on that boundary (force per
unit area). The third line follows from A d = − ∆V ; i.e., the volume of the gas decreases by this
amount, – ∆V is negative.
Example 8.3: How much work is done on the air in a bicycle pump when the volume of air is
decreased by 50 mL under 5.0 bar pressure?
- 159 -
Approach: Substitute into the above formula for work.
w = −P ∆V
= −5.0 bar × 0.050 L = −0.25 L bar
= −100 J L−1 bar−1 × 0.25 L bar
= 25 J
The L bar energy units are converted to J by the conversion factor 100 J = 1 L bar. This is an
important step if you need to combine work and heat. Heat is determined in J.
The conversion factor between L bar and J follows from the conversion of L bar to SI units.
1 L = 103 mL = 103 × (10‒2 m)3 = 10‒3 m3
and
1 bar = 105 Pa = 105 J m‒3 .
A system can gain or lose energy via work. The system gains energy when work is done on it –
i.e., when it is compressed. The system loses energy when work is done by the system – it
expands, compressing the surroundings. The sign of the work associated with a process is
simply assessed by asking whether the volume of the system increased or decreased during the
process. Because gases generally occupy much larger volumes than solids and liquids, the
volume of a system is primarily determined by the number of moles of gas in the system.
Moreover, changes in volume are generally connected with changes in the number of moles of
gas. Thus, work is done by (i.e., w < 0) a system undergoing a reaction that increases the
number of moles of gas. For example, a system of calcium carbonate decomposing into calcium
oxide and carbon dioxide gas does work on the surroundings – i.e., work is done by the system.
The energy is lost to the surroundings.
CaCO3(s) → CaO(s) + CO2(g)
Example 8.4: For which of the following processes is work positive – i.e., work is done on the
system?
(a)
(b)
(c)
(d)
(e)
2 Na(s) + 2 H2O(l) → 2 NaOH(aq) + H2(g)
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)
2 NI3NH3(s) → 2 N2(g) + 3 I2(s) + 3 H2(g)
H2O(g) → H2O(l)
CO2(s) → CO2(g)
Approach: Count the number of moles of reactant and product gases. Work is positive if the
number of moles of gas decreases.
(a) The only gas here is a product. ∆ngas = nproduct gas – nreactant gas = 1. Work is done by the
system. w < 0
(b) ∆ngas = –3. Work is done on the system. w > 0
(c) ∆ngas = 5. Work is done by the system. w < 0
(d) ∆ngas = –1. Work is done on the system. w > 0
(e) ∆ngas = 1. Work is done by the system. w < 0
- 160 -
8.4 Energy Conservation – the First Law of Thermodynamics
We have seen that energy can be transferred, either as heat or work, between systems, but is
never created or destroyed. This is the first law of thermodynamics – energy is conserved. Thus,
if the energy of a system changes it is because heat or work has been transferred between the
system and surroundings. Specifically,
∆U = q + w
8.3
is the change in energy of the system expressed as the sum of heat flowing into the system and
work done on the system - negative values accounting for reverse transfers. The above equation
is the first law of thermodynamics in equation form.
Example 8.5: What is the change in energy for the following combinations of q and w?
(a) q = 50 J and w = 50 J
(b) q = 0 J and w = 100 J
(a) q = 100 J and w = 0 J
Approach: Add heat and work to get ∆U, the change in energy.
(a) ∆U = q + w = 50 J + 50 J = 100 J
(b) ∆U = q + w = 0 J + 100 J = 100 J
(c) ∆U = 100 J
Note that the three combinations of q and w listed above produce the same change in the energy.
The final energy depends only on the initial energy and the sum of heat and work transfers. In
general, every system has a thermodynamic state with well-defined values of all of system
properties – the state functions (they depend only on the state of the system). There are many
paths connecting two final states. Paths correspond to processes that generally involve heat or
work transfer. Heat and work are characteristics of processes – they are not state functions.
8.5 Enthalpy
Enthalpy is state function like energy (volume, pressure, temperature, etc). It is closely related to
energy and we sometimes speak of energy of enthalpy as though they were interchangeable.
Enthalpy, H, is just energy, U, plus an extra term:
H = U + PV
8.4
The extra term is the energy required to "make room" for the system in an environment with
constant pressure, P. Solids and liquids occupy small spaces and the associated P V term in the
enthalpy is generally negligible. Gases occupy much larger volumes. The P V term is small, but
not negligible, for gases.
Enthalpy is introduced because it is common to perform heat measurements under constant
pressure conditions. Calorimetry is used to measure heat. If the experiment takes place under
constant pressure conditions, the change in energy is not equal to the measured q. The first law
gives
∆U = q + w
= q − P ∆V
Under constant pressure conditions, the volume can change.
- 161 -
Bomb calorimeters are calorimeters with constant volume. Heat measurements made with bomb
calorimeters directly give ∆U (the work term is zero);
∆U = q + 0 = qV
8.5
where qV denotes heat for a constant volume process. However, bomb calorimeters require a
strong, rigid (steel) container to withstand large pressure differences resulting from chemical
reactions inside the calorimeter. Constant pressure calorimeters, known as coffee cup
calorimeters, are much safer – but do not give ∆U directly. This is not a problem, however.
Under constant pressure conditions,
P ∆V = ∆(P V)
Work is thus the change in a state function (pressure × volume), when pressure is constant. This
allows the first law to be rearranged to
∆U = q + w
∆U = q − P ∆V
∆U + ∆(P V) = q
∆(U + P V) = q
or
∆H = q = qp
8.6
where qp denotes heat for a constant pressure process, and H = U + P V as introduced above.
Here, we see that heat measured in a coffee cup calorimeter gives the change in enthalpy, a
state function. It is sufficient (and convenient) to construct an enthalpy scale by doing a series of
coffee cup calorimeter experiments. Each experiment measures the enthalpy difference between
the reactants and products for some reaction – i.e., the enthalpy of reaction.
Example 8.6: The neutralization of 100. mL 0.200 M NaOH solution with 100. mL 0.200 HCl
solution is performed in a coffee cup calorimeter. The temperature of the combined solution rises
from 20.00°C to 21.34°C. What is the enthalpy of this reaction in kJ mol−1? Assume the solution
has density 1.00 g mL−1 and specific heat capacity 4.18 J K−1 g−1 – i.e., the properties of pure
water at 20.00°C.
Approach: Determine the heat transferred from the temperature change and the heat
capacity (use mass × specific heat capacity here). Write a net balanced chemical reaction
and determined the number of moles reacted. Divide the heat by the number of moles to get
the enthalpy of reaction in kJ mol−1. The sign is negative because heat is released by the
reaction and the temperature increased.
To use
q = m s ∆T
we need the total mass of solution.
m = Vd
= (100. + 100. mL) × 1.00 g mL−1 = 200. g
Therefore,
q = 200. g × 4.18 J K−1 g−1 (21.34 − 20.00) K
= 1120 J = 1.12 kJ
The net balanced reaction is
H+(aq) + OH−(aq) → H2O(l)
- 162 -
The number of moles of H+(aq) is the volume of the HCl solution × its concentration – i.e.,
0.100 L × 0.200 mol L−1 = 0.0200 mol. The number of moles of OH− is the same. They both
react to completion. Therefore the number of moles reacted – the extent of reaction – is
0.0200 mol.
The enthalpy of reaction is
∆H = −1.12 kJ / 0.0200 mol
= −56.0 kJ mol−1
The sign reversal in the last step - recognizing that the reaction releases heat - can be confusing
for students. The sign reversal results because the heat that determines ∆H is the heat that must
be released to return the contents of the calorimeter to the initial temperature. The system – the
calorimeter and its contents – would lose heat in this imagined process, a process not actually
carried out since the calorimeter is thermally insulated. Nevertheless, the heat associated with
the imagined process is calculated exactly as above, except that the sign of ∆T is reversed. In
this process, 21.34°C is the initial temperature and 20.00°C is the final temperature. The heat
determined in this way gives the enthalpy of reaction at 20.00°C.
In the above example, the reaction released heat and ∆H is negative. This is an exothermic
reaction. Processes with positive ∆H absorb energy – they are called endothermic processes.
For example, the dissolution of ammonium nitrate in water is endothermic.
NH4NO3(s) → NH4NO3(aq)
∆H = 24.9 kJ mol−1
This process is endothermic because strong ionic bonds in the reactant were broken in order to
form weaker intermolecular forces (solvation of the separated ions) in solution. Other
endothermic processes include melting, vaporization, sublimation and bond breaking. Bonds
(covalent, ionic or intermolecular) are stronger in a solid, than in a liquid. In the gas phase
intermolecular forces are essentially absent.
Exothermic reactions are reactions that break weaker bonds to form stronger bonds.
Condensation (the reverse of vaporization) is exothermic because it forms intermolecular bonds
in the liquid without breaking any bonds.
Having measured a number of enthalpies of reaction, how do we go on to construct an enthalpy
scale? The key is that because enthalpy is a state function, we can get the enthalpy of a specific
reaction (the "target" reaction) by adding the enthalpies of reaction for a series of reactions as
long as they begin with the specified reactants and end with the specified products. This is called
Hess' Law.
The simplest application of Hess' Law is to combine a reaction with its reverse. In this case, the
final products are just the reactants. ∆H = 0 for the closed path. Therefore,
or
∆H = ∆Hforward + ∆Hreverse = 0
∆Hreverse = − ∆Hforward
If a reaction must be scaled (e.g., doubled) so the stoichiometric coefficients match what is
required to get the target reaction, then the enthalpy of the reaction is scaled by the same factor.
For example, burning twice as much gas produces twice as much heat. If multiple reactions are
combined to form a target reaction then their ∆H values are added together.
- 163 -
Example 8.7: Determine the enthalpy of the following reaction given the data below.
Fe(s) + S(s) → FeS(s)
∆Hn (kJ mol−1)
n
Reaction
1
2
3
4
2 Fe(s) + O2(g) → 2 FeO(s)
2 S(s) + 3 O2(g) → 2 SO3(g)
FeO(s) + SO3(g) → FeSO4(s)
FeS(s) + 2 O2(g) → FeSO4(s)
−544.0
−791.4
−260.7
−828.4
Approach: Manipulate and combine the given reactions as needed so they add up to the
target reaction; perform the same manipulations to the ∆H values for each reaction . The
target reaction ∆H is the combination of the given ∆H data after they have been manipulated.
Look for the target reaction reactants and products in the given reaction.
Reaction 1 has reactant 2 Fe(s) among reactants. Scale reaction 1 by 1/2 to match the Fe(s)
coefficient in the target reaction. Reaction 2 has 2 S(s) among reactants. Again, scale by
1/2. Reaction 4 has FeS(s) as a reactant, whereas FeS(s) is a product in the target reaction.
Reverse reaction 4. Reaction 3 is needed to cancel out compunds that do not participate in
the target reaction. Note that the ∆H values must be adjusted similarly.
1/2 × Rxn 1
1/2 × Rxn 2
−Rxn 4
Rxn 3
∆Hn (kJ mol−1)
Fe(s) + 1/2 O2(g) → FeO(s)
S(s) + 3/2 O2(g) → SO3(g)
FeSO4(s) → FeS(s) + 2 O2(g)
FeO(s) + SO3(g) → FeSO4(s)
Fe(s) + S(s) → FeS(s)
−272.0
−395.7
828.4
−260.7
−100.0
It is useful to have a more systematic approach to determining an enthalpy of reaction from
known data. It turns out that the enthalpy of any reaction can be determined from a table of all
formation reactions. The formation reaction of a compound is the reaction that forms the
compound from its elements in their standard state. The standard state of an element is its
stable form at 1.00 bar pressure and 25°C. Stable form refers to the stable allotrope (structure) of
the element and the stable phase. The standard state of most elements is a solid. The noble
gases, hydrogen, nitrogen, oxygen, fluorine and chlorine are gases, while bromine is a liquid.
The standard state of carbon is graphite, C(graphite), not diamond, C(diamond).
Any reaction, under standard conditions (1.00 bar pressure and 25°C), can be written as the sum
of the formation reactions of all products (weighted by the stoichiometric coefficients) plus the
sum of the reverse formation reactions of all reactant (again, with stoichiometric weightings).
Consequently, the enthalpy of reaction can correspondingly be expressed in terms of enthalpies
of formation:
H o 

cP H of [P] 
products, P

cR H of [R] .
reactants, R
The superscript “o” refers to standard conditions. The terms cP and cR are the stoichiometric
coefficients of products and reactants, respectively.
For example, the enthalpy of reaction of
Fe2O3(s) + 3 SO3(g) → Fe2(SO4)3(s)
is
- 164 -
8.7
H o  H of [Fe 2 (SO 4 )(s)]  H of [Fe 2 O 3 (s)]  3H of [SO 3 (g)]
Note that an element in the standard state has an enthalpy of formation of zero, ∆Hf o = 0. The
formation reaction in this case is a non-reaction.
The equation for enthalpy of reaction in terms of enthalpies of formation can be depicted
graphically.
"Unform" all reactants
into elements (i.e., sum
the reverse of reactant
formation reactions):
Elements in
standard states
∆H1° = −ΣR cR ∆Hf°[R]
"Form" all products from
elements (i.e., sum the
product formation
reactions):
∆H2° = ΣP cP ∆Hf°[P]
∆H° = ∆H1° + ∆H2°
Reactants
Products
Figure 8.2. Graphical depiction of the expression of enthalpy of reaction in terms of enthalpies of
formation – Equation 8.7 above.
Example 8.8: Determine the enthalpy of combustion of propane from the given enthalpies of
formation.
Compound
CO2(g)
H2O(l)
C3H8(g)
∆Hf° (kJ mol−1)
−393.5
−285.8
−103.8
Approach: Write down the balanced chemical reaction. Combine the enthalpies of
formation as described above – products minus reactants.
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)
H o  3H of [CO 2 (g)]  4H of [H 2O(l)]  H of [C3H8 (g)]
 3  393.5   4  285.8    103.8  kJ mol1  2219.9 kJ mol1
Note that the enthalpy of formation of oxygen is left out of the equation because it is zero.
A table of enthalpies of formation is not the only enthalpy we can construct. It is also convenient
determine enthalpies of reaction using bond enthalpies. A bond enthalpy is the enthalpy of a
bond dissociation reaction – a reaction breaking a type of bond. For example, the bond enthalpy
of the Cl-Cl bond, D(Cl-Cl), is ∆H for the dissociation of Cl2 gas into Cl atoms:
Cl2(g) → 2 Cl(g)
∆H = D(Cl-Cl)
- 165 -
Bond dissociation is an endothermic process. Bonds are broken – this requires energy – while no
bonds are formed. D is always positive.
For molecules with more than one bond, ∆H for the dissociation of the molecule into atoms is the
sum of all bond enthalpies – one for each bond in the molecule. For example,
CHCl3(g) → C(g) + H(g) + 3 Cl(g) ∆H = D(C-H) + 3 D(C-Cl)
The enthalpy of any gas phase reaction can be written as the sum of all reactant bond enthalpies
minus the sum of all product bond enthalpies:
H 


nr D[r] 
reactant bonds, r
np D[p]
8.8
product bonds, p
where nr is the number of occurrences of bond type r (e.g., C-Cl bond) in all reactant species.
Similarly, np is the number of occurrences of bond type p among products.
The enthalpy of reaction for
C2H6(g) + 3 Br2(g) → C2H3Br3(g) + 3 HBr(g)
H
H
H
H
H
H
+
3 Br
Br
Br
Br
H
H
H
+
3 H
Br
Br
is approximated by
H  D[C-C]  6 D[C-H]  3D[Br-Br]  D[C-C]  3D[C-H]  3D[C-Br]  3D[H-Br]
 3D[C-H]  3D[Br-Br]  3D[C-Br]  3D[H-Br] .
Note that some of the bonds are unchanged from reactant to product. The associated bond
enthalpies cancel out leaving only the broken and formed bonds.
The structures of reactants and products must be drawn in order to identify the bonds present.
Single, double and triple bonds are tabulated separately, as they have different bond enthalpies.
In general, enthalpies of reaction determined from bond enthalpies are only approximate. This is
because tabulated bond enthalpies are average bond enthalpies for each bond type – each
distinct occurrence of a bond has a slightly different dissociation enthalpy.
The use of bond enthalpies to estimate reaction enthalpies is restricted to gas phase reactions.
However, bond enthalpies can be used to determine enthalpies of reaction for reactions with
liquids and/or solids, if the bond enthalpy data is supplemented with enthalpies of vaporization
and/or sublimation. The target reaction is written as a gas phase reaction plus necessary phase
transitions, making use of Hess’ Law. For example, the combustion of propane at 25C can be
written as the sum of a gas phase reaction and the condensation of water:
C3H8(g) + 5 O2 → 3 CO2(g) + 4 H2O(g)
4× H2O(g) → H2O(l)
C3H8(g) + 5 O2 → 3 CO2(g) + 4 H2O(l)
Therefore, the enthalpy of combustion of propane is just
- 166 -
 H   H gas phase rxn  4  H vap [H 2 O]
where
 H gas phase rxn  2 D [C-C]  8 D [C-H]  5 D [O=O]  3  2 D [C=O]  4  2 D [O-H ] .
The equation for enthalpy of reaction in terms of bond enthalpies can be depicted graphically.
Break all reactant bonds
(i.e. sum the reactant
bond dissociations):
Atoms
∆H1 = Σr nr D[r]
Form all product bonds (i.e.
sum the reverse product bond
dissociations):
∆H2 = ‒Σp np D[p]
∆H = ∆H1 + ∆H2
Reactants
Products
Figure 8.3. Graphical depiction of the expression of enthalpy of reaction in terms of bond
enthalpies – Equation 8.8 above.
Example 8.9: Approximate the enthalpy of combustion of oxidation of SO2(g) to form SO3(g)
from the given bond enthalpies.
Bond
S=O
O=O
D (kJ mol−1)
515
495
Approach: Write down the balanced chemical reaction. Combine the enthalpies of
formation as described above – reactant bonds minus product bonds.
2 SO2(g) + O2 → 2 SO3(g)
H  2  2 D[S=O]  D[O=O]  2  3D[S=O]
 D[O=O]  2 D[S=O]
 495  2  515  kJ mol1  535 kJ mol1
8.6 Lattice Enthalpy and the Born-Haber Cycle
Bonding in an ionic solid results from the electrostatic attraction of cations to anions. Since
cations repel each other, as do anions, the electrostatic potential energy is the sum of negative
and positive contributions. The lattice arrangement of the ions in the solid gives a net negative
electrostatic potential energy. It corresponds to the minimum enthalpy configuration of the ions.
The net negative enthalpy is called the lattice enthalpy. For example, the lattice enthalpy of
NaCl(s) is the enthalpy change associated with
Na+(g) + Cl−(g) → NaCl(s) ∆H = ∆Hlatt[NaCl] .
- 167 -
Lattice enthalpy is the enthalpy of the solid relative to separated ions. Its units are kJ mol−1, i.e.,
kJ per mole of compound formula (NaCl in the above example).
Since it is not possible to actually carry out the process of bringing the ions together to form the
ionic solid, lattice enthalpy is determined by constucting a sequence of steps with measured
enthalpy changes that add up to the lattice enthalpy. To this end, we start with the formation
reaction of the ionic solid. The enthalpy of the formation reaction can generally be obtained using
calorimetry. In the case of sodium chloride, the formation reaction is
Na(s) + ½ Cl2(g) → NaCl(s)
∆H = ∆Hf[NaCl(s)] .
This reaction is the sum of the following steps:
Na(s) → Na(g)
Na(g) → Na+(g) + e−
½ Cl2(g) → Cl(g)
Cl(g) + e− → Cl−(g)
Na+(g) + Cl−(g) → NaCl(s)
Na(s) + ½ Cl2(g) → NaCl(s)
∆Hsub[Na]
I1[Na]
½ D[Cl-Cl]
A1[Cl]
∆Hlatt[NaCl]
∆Hf[NaCl(s)]
The sum of the enthalpy changes gives the enthalpy of formation of sodium chloride. Specifically,
∆Hf[NaCl(s)] = ∆Hsub[Na] + I1[Na] + ½ D[Cl-Cl] + A1[Cl] + ∆Hlatt[NaCl] ,
8.9
where (1) ∆Hsub[Na] is the enthalpy of sublimation of sodium, (2) I1[Na] is the first ionization
energy of sodium, (3) D[Cl-Cl] is the bond enthalpy of a chlorine-chlorine bond, and (4) A1[Cl] is
the first electron affinity of chlorine. Note that ∆Hsub[Na] is sometimes written as ∆Hatom[Na], the
enthalpy of atomization of sodium. ∆Hsub and ∆Hatom are the same for sodium, since sodium
sublimates to form an atomic gas.
The above equation for ∆Hf[NaCl(s)] can be rearranged to give ∆Hlatt[NaCl(s)] in terms of
measurable quantities:
∆Hlatt[NaCl(s)] = ∆Hf[NaCl(s)] − ∆Hsub[Na] − I1[Na] − ½ D[Cl-Cl] − A1[Cl] .
8.10
For sodium chloride, the lattice enthalpy is
∆Hlatt[NaCl(s)] = −411.2 − 107.5 – 495.8 − ½ × 242.9 – (−348.5) kJ mol−1
= −787.5 kJ mol−1 .
The two pathways – direct and indirect, through the alternative set of steps – of the formation
reaction of solid sodium chloride correspond to a cycle in the state space of sodium and chlorine
atoms. This cycle is called the Born-Haber cycle of sodium chloride.
The Born-Haber cycle of sodium chloride is depicted in Fig. 8.4.
- 168 -
Na+(g) + Cl(g)
H
Na+(g) + 1/2 Cl2(g)
1/2 D[Cl-Cl]
A1[Cl]
Na+(g) + Cl(g)
I1[Na]
Na(g) + 1/2 Cl2(g)
sub H[Na] Na(s) + 1/2 Cl (g)
2
latt H[NaCl]
f H[NaCl(s)]
NaCl(s)
Figure 8.4. The Born-Haber cycle for sodium chloride. The horizontal lines correspond to the
different states of sodium and chlorine placed on a vertical enthalpy scale. The vertical arrows
(drawn to scale) correspond to the various steps in the Born-Haber cycle.
Even though chlorine has the largest magnitude electron affinity among all elements, and the
ionization energy of sodium is among the smallest, the formation of separated sodium and
chloride ions from the atoms is endothermic. The electron affinity is not enough to compensate
for the ionization energy. It is the large drop in enthalpy when the ions form a lattice that makes
the formation of sodium chloride exothermic, and favorable.
Example 8.10: Determine the enthalpy of formation of zinc bromide from the following data (in
kJ mol−1):
∆Hlatt[ZnBr2]
∆Hsub[Zn]
I1[Zn]
I2[Zn]
∆Hvap[Br2]
D[Br-Br]
A1[Br]
−2674.3
130.4
906.3
1733.1
30.9
193.9
−324.5
- 169 -
Approach: Here, we are asked for the enthalpy of formation of zinc bromide – the lattice
enthalpy is given. Simply sum the steps in the Born-Haber cycle for zinc bromide – no
rearrangement is required. For doubly charged ions, there is a second ionization energy or
electron affinity. Also, look for any required phase changes. For example, the formation
reaction of zinc bromide starts with bromine in the liquid phase. There must be an additional
vaporization of bromine step, because the bond enthalpy of bromine corresponds to breaking
the bond in the gas phase.
The formation reaction of zinc bromide is
Zn(s) + Br2(l) → ZnBr2(s) ∆H = ∆Hf[ZnBr2(s)] .
This reaction is the sum of the following steps:
Zn(s) → Zn(g)
Zn(g) → Zn+(g) + e−
Zn+(g) → Zn2+(g) + e−
Br2(l) → Br2(g)
Br2(g) → 2 Br(g)
2 Br(g) + e− → 2 Br−(g)
Zn2+ (g) + 2 Br−(g) → ZnBr2(s)
Zn(s) + Br2(l) → ZnBr2(s)
∆Hsub[Zn]
I1[Zn]
I2[Zn]
∆Hvap[Br2]
D[Br-Br]
2 A1[Br]
∆Hlatt[ZnBr2]
∆Hf[ZnBr2(s)]
The sum of the enthalpy changes equals the enthalpy of formation of zinc bromide:
∆Hf[ZnBr2(s)] = ∆Hsub[Zn] + I1[Zn] + I2[Zn] + ∆Hvap[Br2] + D[Br-Br] + 2 A1[Br] + ∆Hlatt[ZnBr2]
= 130.4 + 906.3 + 1733.1 + 30.9 + 193.9 + 2×(−324.5) + (−2674.3) kJ mol−1
= −328.7 kJ mol−1 .
The magnitude of the lattice enthalpy of zinc bromide is much larger than that of sodium chloride:
2674.3 versus 787.5 kJ mol−1. This results primarily because of the +2 charge on the zinc ion
and the fact that one mole of ZnBr2 has two moles of bromide ions. The size of the charge is
significant because the electrostatic potential energy of a cation of charge, +Ze, a distance r from
an anion of charge, −Ye, (−e is the charge on an electron) is given by
Ze  Ye 
r
ZY
  Ke2
rZ  rY
EK
Coulomb's law
8.11
at closest approach.
If the ions are at closest approach (i.e., just touching), then the distance between them is the sum
of their radii; r  rZ  rY .
The magnitude of the negative electrostatic potential is proportional to the charge on both the
cation and anion. It is inversely proportional to the sum of the radii of the ions. Because ion
charges come only in multiples of e, larger ion charge has more effect on the electrostatic
potential than smaller ion radius. Thus, (1) the largest magnitude lattice enthalpies are
associated with the largest ion charges. For sets of solids with same ion charges, (2) the largest
magnitude lattice enthalpies occur for the smallest ions.
- 170 -
Example 8.11: Order the following sets of ionic solids according to the magnitude of lattice
enthalpy:
(a) LiF, LiCl, NaCl, MgCl2, CaCl2
(b) CaO, SrS, NaF, Al2O3, SrBr2
Approach: Group the solids according to the size of the charges. Solids with the largest
charges have the largest magnitude of lattice enthalpy. Within each group (same size of
charges), the solids with the smallest ions have the largest magnitude of lattice enthalpy.
(a) MgCl2, CaCl2 have a +2 cation and a −1 anion. These solids have the largest magnitude
lattice energies. Of these solids, MgCl2 has the smaller cation (anions are the same) and,
therefore, the larger magnitude lattice enthalpy. LiF, LiCl, NaCl have a +1 cation and a −1
anion. LiF has the smallest ions, and largest magnitude lattice enthalpy. LiCl is next,
followed by NaCl. Altogether, according to increasing magnitude lattice enthalpy,
NaCl < LiCl < LiF < CaCl2 < MgCl2 .
(b) Al2O3 has a +3 cation and a −2 anion. It has the largest magnitude lattice enthalpy. CaO
and SrS have a +2 cation and a −2 anion. They have the next largest magnitude lattice
enthalpies. Of these, CaO has the smaller cation and smaller anion, and hence the larger
magnitude lattice enthalpy. SrBr2 has a +2 cation and a −1 anion. It has the next largest
magnitude lattice enthalpy. NaF, with a +1 cation and a −1 anion, has the smallest
magnitude lattice enthalpy. Altogether, according to increasing magnitude lattice enthalpy,
NaF < SrBr2 < SrS < CaO < Al2O3 .
Problems:
8.1 How much heat is released when 100. g of ethanol cools from 50.0°C to 20.0°C? (sethanol =
2.44 J K−1 g−1)
8.2 How much heat is required to produce steam at 150.°C from 200. mL of liquid water at
20.°C? (s[H2O(l)] = 4.18 J K−1 g−1, s[H2O(g)] = 1.85 J K−1 g−1 and ∆Hvap°[H2O] = 44.0 kJ
mol−1)
8.3 When 8.16 g of NH4NO3(s) is dissolved in 100. mL of water, in a calorimeter, the temperature
is observed to drop from 19.5°C to 15.9°C. What is the enthalpy change for this dissolution
process?
8.4 The energy of a gas increases by 125 J, after its volume decreases by 2.00 L under 1.00 bar
pressure. How much heat was absorbed/released by the gas?
8.5 For which of the following processes is (i) q > 0 (i.e., endothermic), or (ii) w > 0 (i.e., work is
done on the system)?
(a)
(b)
(c)
(d)
(e)
Hg(l) → Hg(g)
Br2(g) → 2 Br(g)
H(g) + Cl(g) → HCl(g)
C(g) → C(s)
2 S(s) + 3 O2(g) → 2 SO3(g)
8.6 Determine the enthalpy of the following reaction from the given data:
N2(g) + 3 H2(g) → 2 NH3(g)
- 171 -
n
1
2
3
Reaction
∆Hn (kJ mol−1)
2 H2(g) + O2(g) → 2 H2O(g)
−483.6
N2(g) + O2(g) → 2 NO(g)
182.6
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) −260.7
8.7 Determine the enthalpy of the reaction below, known as the "mother of all bombs" reaction,
2 Al(s) + NH4NO3(s) + O2(g) → Al2O3(s) + N2(g) + 2 H2O(l)
from the given data. How much heat is released per gram of reactant (excluding oxygen – it
is supplied by the atmosphere? What is the final temperature of 1.0 L of water at 20.°C, if it is
heated by reaction of 10. g of this reactant?
Compound
NH4NO3(s)
Al2O3(s)
H2O(l)
∆Hf° (kJ mol−1)
−356.6
−1675.7
−285.86
8.8 Estimate the enthalpy of the oxidation of ethanol with hypochlorous acid using the given data.
CH3CH2OH(l) + 2 HOCl(g) → CH3COOH(l) + 2 HCl(g) + H2O(l)
∆Hvap°[CH3CH2OH] = 38.6 kJ mol−1
∆Hvap°[CH3COOH] = 23.7 kJ mol−1
∆Hvap°[H2O] = 44.0 kJ mol−1
Bond
C-C
C-H
C-O
C=O
O-H
Cl-O
Cl-H
D (kJ mol−1)
345
415
360
750
460
220
430
8.9 Determine the ionization energy of potassium from the following data (in kJ mol−1):
∆Hf[K2O]
∆Hlatt[K2O]
∆Hsub[K]
D[O=O]
A1[O]
A2[O]
−361.5
−2238
89.0
498.4
−141.0
752.7
8.10 Order the following sets of ionic solids according to increasing magnitude of lattice enthalpy.
(a) Na2O, K2S, Rb2S, Cs2Se, MgO
(b) SrCl2, CaF2, BaI2, Ga2O3, In2S3
- 172 -
Solutions:
8.1 Heat is related to temperature change of ethanol via
q = msethanol ∆T
= 100. g × 2.44 J K−1 g−1 × (20.0 °C − 50.0 °C)
= 100. g × 2.44 J K−1 g−1 × (−30.0 K)
= −7320 J = −7.32 kJ
This heat is negative indicating heat is released by the ethanol as it cools.
8.2 There are three processes to consider here: (1) heat the water from 20.°C to 100.°C; (2)
vaporize the H2O(l) to H2O(g) at 100.°C; (3) heat the H2O(g) from 100.°C to 150.°C.
First, get the mass of water from its volume and density:
m = d V = 200. mL × 1.00 g mL−1 = 200. g
Next, determine the heat needed to raise the liquid water temperature to 100. °C - the boiling
point of water.
q1 = mswater liq ∆T
= 200. g × 4.18 J K−1 g−1 × (100.0 °C − 20.0 °C)
= 66880 J = 66.88 kJ
Now consider the vaporization of the H2O(l) to H2O(g) at 100.°C. The heat of vaporization is
given as kJ mol−1. The number of moles of water is
n = m/M = 200. g / 18.015 g mol−1 = 11.1 mol
Vaporization of all the water requires
q2 = n ∆Hvap°[H2O] = 11.1 mol × 44.0 kJ mol−1 = 488.4 kJ
Finally, heating the water vapour from 100°C to 150°C requires
q3 = mswater vap ∆T
= 200. g × 1.85 J K−1 g−1 × (150.0 °C − 100.0 °C)
= 18500 J = 18.5 kJ
The total heat required is
q = q1 + q2 + q3
= 66.88 + 488.4 + 18.5 kJ
= 573.8 kJ = = 574 kJ
Note that the bulk of this heat is needed to vaporize the water.
8.3 We assume the solution has the same density and specific heat as pure water. The mass of
solution is the mass of water (volume x density) plus the mass of NH4NO3,
m = 100. mL × 1.00 g mL−1 + 8.16 g
The drop in temperature indicates an endothermic reaction. The heat absorbed by the
reaction (we include the sign reversal by reversing the final and initial temperatures) is given
by
- 173 -
qrxn = mswater (Tinitial – Tfinal)
= 108 g × 4.18 J K−1 g−1 × (19.5 °C – 15.9 °C)
= 1625 J = 1.625 kJ
To get the enthalpy of dissolution per mole of NH4NO3, we must divide this heat by the moles
of NH4NO3,
n = 8.16 g / 80.04 g mol−1 = 0.1019 mol
The enthalpy of dissolution is
∆H = qrxn / n
= 1.625 kJ / 0.1019 mol = 15.9 kJ
8.4 The change in energy of the gas is given by the first law of thermodynamics,
∆U = q + w = 125 J
The work done on the system is given by
w = −P ∆V
= 1.00 bar × (−2.00 L) = −2.00 L bar × (100 J L−1 bar−1) = 200 J
Note the insertion of the conversion factor in the last step.
The first equation can now be solved for q,
q = ∆U − w = 125 − 200 J = −75 J
Because this heat is negative, heat was released by the gas during the process.
8.5 (a) (i) Vaporization of mercury is endothermic (q > 0). The system requires input of heat to
break the bonds between mercury atoms in the liquid. (ii) Work is done by the system (w <
0) because the volume increases. The volume of the system increases because one mole of
gas is produced from one mole of liquid – gas occupies a much greater volume.
(b) (i) Atomization of bromine (breaking the diatomic molecule into atoms) is endothermic (q
> 0). Breaking a bond requires energy input. (ii) Work is done by the system (w < 0), as
volume increases.
(c) (i) Recombination of hydrogen and chlorine atoms is exothermic (q < 0). A bond is
formed, releasing energy. (ii) Work is done on the system (w > 0), as volume decreases.
(d) (i) Deposition of carbon atoms into solid carbon is exothermic (q < 0) – bonds are
formed. (ii) Work is done on the system (w > 0). Volume decreases.
(e) (i) In the combustion of sulfur, S-S and O=O bonds are broken, while S=O bonds are
formed. In this case, experience tells us that this is exothermic (q < 0). The burning of sulfur
(brimstone) releases energy. (ii) There are two moles of product gas and three moles of
reactant gas. Work in done on the system (w > 0).
8.6 Use Hess' law. Reactions 1 and 2 have H2 and N2 as reactants. Reaction 3 has ammonia as
a reactant. Reverse reaction 3 because ammonia is a product in the target reaction.
- 174 -
3
 2H 2 (g) + O 2 (g) 
 2H 2 O(g)
2
N 2 (g) + O 2 (g) 
 2NO(g)
1
 4 NO(g) + 6 H 2 O(g) 
 4 NH 3 (g) + 5O 2 (g)
2
N 2 (g) + 3H 2 (g) 
 2 NH3 (g)
3
 (483.6 kJ mol1 )
2
ΔH o = 182.6 kJ mol1
1
ΔH o =   (260.7 kJ mol1 )
2
o
ΔH =  412.5 kJ mol1
ΔH o =
8.7 Subtract reactant enthalpies of formation from those of the products (leave out elements in
their standard state – their contribution is zero):
H o  H of [Al2O3 (s)]  2H of [H 2O(l)]  H of [NH 4 NO3 (s)]
 1675.7  2  285.8    356.6  kJ mol1  1890.7 kJ mol1
The molar mass of reactants, not including oxygen, is 134.0 g mol−1. Per gram of explosive,
the heat released is
1890.7 kJ mol−1 / 134.0 g mol−1 = 14.11 kJ g−1
Note the sign reversal because we are asked for the heat released (‒q).
The heat released by 10. g of explosive is 141.1 kJ. This heat would increase the
temperature of 1.0 L of water by
T 
qwater
141.1 kJ
=
= 33.8 K
mwater swater
1000 g  4.18 J K 1g 1
The final temperature of 1.0 L of water initially at 20.°C is
= 20. + 33.8 °C = 54°C
8.8 First relate the reaction to a corresponding gas phase reaction by adding appropriate
vaporization and condensation steps. (If solids were present sublimation and/or deposition
processes would be included.) Specifically,
o
CH 3CH 2OH  g   2 HOCl  g   CH 3COOH  g   2 HCl  g   2 H 2 O  g 
ΔH o = ΔH gas
phase rxn
o
CH 3CH 2 OH  l  
 CH 3CH 2OH  g 
ΔH o = ΔH vap
[ethanol]= 38.6 kJ mol1
o
CH 3COOH  g  
 CH 3COOH  l 
ΔH o =  ΔH vap
[ethanoic acid]=  23.7 kJ mol1
o
o
2  H 2 O  g  
 H 2O  l 
ΔH =  2ΔH vap
[ water]=  88.0 kJ mol1
CH 3CH 2OH  l   2 HOCl  g   CH 3COOH  l   2 HCl  g   2 H 2 O  l 
o
1
ΔH o = ΔH gas
phase rxn  73.1 kJ mol
Subtract product bond enthalpies from those of the reactants to estimate the enthalpy of the
gas phase reaction. To identify all the bonds, rewrite the reaction using molecular structures:
- 175 -
H
O
H
H
H
H
H
H
+
2
O
Cl
H
H
2 H
O
H
O
H
Cl
+
+
2 H
O
H
H gas phase rxn  5 D[C-H]  D[C-O]  D[O-H]  2 D[O-H]  2 D[Cl-O]
 3D[C-H]  D[C-O]  D[C=O]  D[O-H]  4 D[O-H]  2 D[Cl-H]
 2 D[C-H]  2 D[Cl-O]  D[C=O]  2 D[O-H]  2 D[Cl-H]
 2  415  2  220  750  2  460  2  430 kJ mol1
 1260 kJ mol1
Note that there is much cancelation of bond energies. Reactant bonds unchanged in the
reaction (i.e., they also appear in the products) cancel out. The intermediate step, shown
above, includes only the bonds that are changing – either broken (+D) or formed (‒D).
The enthalpy of the oxidation of liquid ethanol to liquid ethanoic acid is therefore
∆H = ∆Hvap°[CH3CH2OH] −1260 kJ mol−1 − ∆Hvap°[CH3COOH] – 2 ∆Hvap°[H2O]
= 38.6 −1260 – 23.7 – 2 × 44.0 kJ mol−1
= −1333.1 kJ mol−1
Note that the terms of the above equation correspond to the following steps that, together,
achieve the target reaction: (1) vaporize the ethanol, (2) carry out the reaction in the gas
phase, (3) and (4) condense the ethanoic acid and water produced.
8.9 The enthalpy of formation of potassium oxide is the sum of the following steps:
2 K(s) → K(g)
2 K(g) → K+(g) + e−
½ O2(g) → O(g)
O(g) + e− → O−(g)
O−(g) + e− → O2−(g)
2 K+ (g) + O2−(g) → K2O(s)
2 K(s) + ½ O2(l) → K2O(s)
2 ∆Hsub[K]
2 I1[K]
½ D[O=O]
A1[O]
A2[O]
∆Hlatt[K2O]
∆Hf[K2O(s)]
The sum of the enthalpy changes equals the enthalpy of formation of potassium oxide:
∆Hf[K2O(s)] = 2 ∆Hsub[K] + 2 I1[K] + ½ D[O=O] + A1[O] + A2[O] + ∆Hlatt[K2O]
= 2 I1[K] + 2 × 89.0 + ½ × 498.4 + (−141.0) + 752.7 + (−2238) kJ mol−1
= −361.5 kJ mol−1 .
The ionization energy of potassium is the one unknown in this Born-Haber cycle. Solving for
I1[K] gives
I1[K] = ½ ( −361.5 – (−1199.1) ) kJ mol−1
= 418.8 kJ mol−1 .
- 176 -
8.10 (a) The solid consisting of 2+ cations and 2− anions, MgO, has the largest magnitude lattice
enthalpy. The other solids have 1+ cations and 2− anions. Of these, the smallest ions are
found in Na2O. K2S has the next smallest cation, and next smallest anion. Rb2S has the next
smallest cation, while Cs2Se has the largest cation and the largest anion. According to
increasing magnitude lattice enthalpy, the solids are ordered as follows:
Cs2Se < Rb2S < K2S < Na2O < MgO
(b) Here, there are three solids consisting of 2+ cations and 1− anions, and two solids
consisting of 3+ cations and 2− anions. The latter have the largest magnitude lattice
enthalpy. Of these, Ga2O3 has the smallest ions, and largest | ∆Hlatt |. In2S3 has the next
largest value. Of the remaining solids, CaF2 has the smallest ions and largest | ∆Hlatt |. SrCl2
is next. BaI2 has the smallest | ∆Hlatt |. Ordered according to increasing magnitude lattice
enthalpy, we have
BaI2 < SrCl2 < CaF2 < In2S3 < Ga2O3 .
- 177 -
9. Entropy and Gibbs Free Energy
9.1 Spontaneous Processes
For every balanced reaction that can be written, there is either net forward reaction, net reverse
reaction or the reactants and products are in equilibrium. If there is net forward reaction, the
reaction is said to be spontaneous. If there is net reverse reaction, it is the reverse reaction that
is spontaneous. In this case, the forward reaction is non-spontaneous. In either case, reactants
and products are initially not in equilibrium. Net reaction results, shifting the reaction mixture
toward equilibrium.
A spontaneous reaction need not be fast. Diamond decomposes spontaneously into graphite.
However, the reaction is so slow that it takes millions of years before one half of a diamond
converts to graphite. Consequently, no one worries about this process before buying a diamond.
Nevertheless, the conversion of diamond to graphite is spontaneous, and diamonds are not really
“forever”.
For some reactions, spontaneity is obvious. For example, sodium reacts vigorously with water:
Na(s) + H2O(l) → NaOH(aq) + ½ H2(g) .
This is a spontaneous exothermic reaction. Other spontaneous exothermic reactions include the
combustion of gasoline, natural gas, hydrogen or any other fuel, and the oxidation of iron (rusting
of steel),
Fe(s) + O2(g) → Fe2O3(s).
The latter reaction is slow – but, unfortunately, much faster than the decomposition of diamond.
There are also spontaneous endothermic reactions. For example, ammonium nitrate dissolves
spontaneously in water, though the dissolution process is endothermic. This spontaneous
process is used in cold packs, activated by breaking a seal that separates NH4NO3(s) from
H2O(l). Other endothermic spontaneous processes include melting of any solid above its melting
pointing, and boiling of any liquid above its boiling point. For example, ice melts spontaneously
above 0°C.
The first law of thermodynamics tells us that energy is conserved for all of these processes.
However, it does not distinguish between the spontaneous and non-spontaneous reaction
directions. It is the second law of thermodynamics that tells us whether a reaction is spontaneous
or non-spontaneous.
A clue to the origins of the second law of thermodynamics is provided by considering
spontaneous processes with no change in enthalpy (or nearly no change). For example, argon
and krypton gases mix spontaneously, as do liquid methanol and ethanol. For such a mixing
process, spontaneity is evidently just the natural tendency toward disorder. In fact, it turns out
that every spontaneous process is just a transition to greater disorder.
- 178 -
Ar
Ar
Kr
Kr
Ar
Kr
Ar
Kr
Kr
Ar
Ar
Ar
Ar
Kr
Kr
Kr
Kr
Ar
Ar
Ar
Kr
Kr
Kr
Ar
Kr
Kr
Ar
Kr
Ar
Kr
Ar
Kr
Ar
Ar
Ar
Kr
Kr
Kr
Ar
Ar
Ar
Ar
Ar
Kr
Ar
Kr
Kr
Kr
Figure 9.1. Spontaneous mixing of argon and krypton gas. In the top panel, argon is
restricted to the left box, while krypton is restricted to the right box. If the two sides are
connected via an opening in the box interface, the gases will mix. At equilibrium (bottom
panel), argon and krypton are both spread evenly across the two boxes. In practice,
there are fluctuations in the number of atoms of each type in each box. These
fluctuations would be significant if there were only 12 atoms of each type – as shown
here. However, the number of atoms of each type is typically on the order of the
Avagadro constant. Consequently, fluctuations in real gas mixtures are negligible, when
expressed as a fraction of the total number of atoms.
9.2 Entropy
The spontaneity of the mixing of argon and krypton gas, shown in Fig. 9.1, can be understood as
an increase in the freedom of argon and krypton atoms. Here, the argon and krypton have the
same initial temperature and pressure. In the top panel, argon and krypton are restricted to the
box on the left and right, respectively. When the boxes are connected, argon and krypton atoms
are subsequently free to explore both boxes (bottom panel).
The freedom of a system corresponds to the number of microstates available to the system.
Microstates are the states in a completely microscopic description of the system which specifies
the quantum state of every atom. A microstate of an argon or krypton atom, in the above gas
- 179 -
mixing example, is a small volume whose size depends on the speed of the atom – via the de
Broglie wavelength. Since temperature does not change in the process, average speed does not
change, and average atomic microstate volume is fixed throughout the process. Consequently, if
the left and right boxes have equal volume, then the number of microstates available to both
types of atoms doubles when the boxes are connected. If there are N argon atoms, the number
of microstates of the N atom system increases by the factor 2N. An extensive quantity is
obtained if we take the logarithm of the number of microstates. Ludwig Boltzmann introduced the
following formula for entropy based on such observations:
where
S = kB ln W,
9.1
kB = R/NA = 1.3806 × 10‒23 J K‒1
is called the Boltzmann constant. W is the number of microstates. The Boltzmann constant can
be thought of as the gas constant per particle – as opposed to per mole.
In Boltzmann`s time, entropy was already known from the study of heat engines. Boltzmann
provided the above microscopic definition of entropy which reproduces values obtained using
thermodynamic entropy formulas.
To better appreciate the Boltzmann formula, it is instructive to use it to evaluate the entropy
change associated with the above example of mixing gases. Let Win, Ar and Wfin, Ar be the initial
and final number of microstates of the NAr argon atoms. According to the Boltzmann formula, the
change in the entropy of argon upon mixing of gases is given by
SAr  kB ln Wfin, Ar  kB ln Win, Ar
 kB ln
Wfin, Ar
Win, Ar
 kB ln 2 NAr
 N Ar kB ln 2  nAr R ln 2 .
In the second line above, we use (1) the fact that the total number of microstates of NAr argon
atoms equals the number of microstates of one atom raised to the NAr th power, and (2) the ratio
of the final to initial number of one-atom microstates equals 2 in this example. The first of these
two points follows because the argon atoms move almost entirely independently. Argon and
krypton are ideal gases under ordinary conditions. In an ideal gas, particles have negligible
interactions. A microstate of an N particle ideal gas consists of specifying the microstate of each
particle. Thus, every combination of N one-particle microstates constitutes a microstate of the Nparticle system. This corresponds to the number of microstates of the N-particle system equal to
the N th power of the number of one-particle microstates.
As the boxes have equal size, temperature and pressure, the number of krypton atoms is the
same as the number of argon atoms. Also, the relative change in the volume is the same for
krypton and argon. Consequently, the change in entropy of krypton is the same as that for argon
and
S  S Ar  S Kr
 nAr R ln 2  nKr R ln 2
 2 mol  R ln 2  11.53 J K 1 ,
if there is one mole of both argon and krypton.
- 180 -
The change in entropy upon mixing of 1 mol each of argon and krypton is not particularly large
when expressed here in standard entropy units. However, because there are so many atoms,
this entropy change corresponds to a huge increase in the number of microstates available to the
full set of all atoms. In particular,
Wfin, ArWfin,Kr
 Wfin 
 2 NAr  NKr

 
 Win  mix Win, ArWin, Kr
 226.02210
23
.
To best appreciate the size of this ratio, consider the reverse of the mixing process wherein all
argon and krypton atoms end up back in the left and right box, respectively. In this case, the ratio
of final to initial number of microstates is just the reciprocal of the above ratio for the mixing
process – i.e.
24
 Wfin 
 21.204410


 Win  unmix
.
This is just the probability that the atoms are unmixed in the final state. For each atom, the
probability that it is in the correct box (i.e. as in the unmixed state) is ½. As there are
1.2044×1024 atoms, the probability of a completely unmixed state is just the above very small
ratio of numbers of microstates. This probability is absurdly small. There is a lottery wherein six
numbers are picked from 1 to 49. The probability of the above unmixing event is the same as the
probability of winning this lottery 2 billion times a day, every day for about 70 billion years – more
than 5 times the current lifetime of the universe. It is safe to say that such events do not occur.
Hence, the second law of thermodynamics simply says that the things that happen are those
such that the probability that they did not happen is absurdly small. According to Boltzmann’s
formula, Eq. 9.1, this law says that entropy must increase.
It is important to realize that when we say entropy increases for spontaneous processes, it is the
entropy of the universe. Many spontaneous processes impact the environment, as well as the
system. In general, there is a change in entropy, for both system and surroundings that
accompanies the process. Thus, the second law of thermodynamics takes the form,
∆Suniv = ∆Ssys + ∆Ssurr > 0
9.2
for all spontaneous processes. Here, ∆Ssurr is the change in entropy in the surroundings that
arises due to the process. It is not necessary to account for the vast number of other processes
going on at the same time, all over the rest of the universe. Only the direct impact of the process
on the surroundings is relevant.
- 181 -
(b)
(a)
Ssurr > 0
Suniv > 0
Ssurr < 0
Ssys > 0
Suniv > 0
Ssys > 0
(c)
Suniv > 0
Ssurr > 0
Ssys < 0
Figure 9.2. Three types of spontaneous processes: (a) ∆Ssys > 0 and ∆Ssurr > 0.
(b) ∆Ssys > 0, ∆Ssurr < 0 and ‒∆Ssys < ∆Ssurr. (c) ∆Ssys < 0, ∆Ssurr > 0 and
‒∆Ssurr < ∆Ssys. In each case, ∆Suniv > 0. When either the change in system or
surroundings is negative, the other positive change must exceed the magnitude of the
negative change in order to give a net positive change in the entropy of the universe.
Figure 9.2 depicts the three types of spontaneous processes. Panel (a) corresponds to a process
with an increase in system and surroundings entropy. All such processes are spontaneous.
Panel (b) corresponds to a process which increases system entropy at the expense of a smaller
in magnitude decrease in entropy of the surroundings. Panel (c) depicts the case of a decrease
in system entropy that is spontaneous because it is accompanied by a larger in magnitude
increase in entropy of the surroundings. These processes are spontaneous because the entropy
of the universe increases in each case. A non-spontaneous process corresponds to the reverse
of one of these processes.
Properties of Entropy
Like energy, entropy is extensive. Suppose we have two identical systems, each with entropy, S.
The combined entropy of the two systems is 2 S. Molar entropy, expressed in J K‒1 mol‒1, is an
intensive quantity which is characteristic of a substance with specific temperature and pressure.
Molar entropies of substances, under standard conditions (25°C and 1 bar pressure), are
tabulated along with enthalpies of formation.
Entropy depends on temperature and pressure. In thermodynamics, entropy is defined by the
following formula:
∆S = q / T ,
9.3
for reversible processes at constant temperature, T (also, for small reversible changes with small
changes in temperature). Here, q is the heat flow into the system. Since temperature is always
positive, positive q produces a positive change in entropy. Except at phase transition
temperatures (melting and boiling points), positive heat flow increases temperature. Therefore,
entropy increases with increasing temperature. It also increases when a solid melts or a liquid
boils – melting and boiling are endothermic (q = ∆H > 0).
- 182 -
Figure 9.3. The molar entropy of water at 1 bar pressure, as a function of temperature.
Entropy increases with increasing temperature, with positive discontinuities when the
solid melts and when the liquid boils. Determined using temperature dependent heat
capacity models found in S.I. Sandler, Chemical and Engineering Thermodynamics, 3rd
Ed. (John Wiley and Sons, New York, 1999).
Figure 9.3 shows the molar entropy for water at 1 bar pressure, from 0 to 500 K. Note that it is
possible to determine the molar entropy of water (or any substance), as opposed to just the value
(e.g. entropy of formation) relative to some convenient reference state. This is because there is a
reference state with exactly zero entropy. The third law of thermodynamics states that a pure,
perfect crystalline substance at zero Kelvin has zero entropy. This law follows from Boltzmann’s
formula (Eq. 9.1), since a pure, perfect crystal at absolute zero has just one microstate – the
quantum ground state of the substance.
The increase in the molar entropy of water, with increasing temperature, is in seen in Fig. 9.3.
The entropy of ice at zero Kelvin is not zero because it is not a perfect crystal. There is residual
entropy associated with disorder in the arrangement of hydrogen atoms within the crystal – due to
hydrogen bonding.
(1) When the solid is heated, intramolecular and intermolecular vibrations increase in magnitude.
Consequently, entropy of the solid increases with increasing temperature.
(2) When the solid melts, there is a discontinuous increase in entropy that reflects the greater
disorder in the liquid. In the liquid, molecules are able to flow past one another, and molecular
vibrations include longer range motions.
(3) Vibrations in the liquid also increase in magnitude with increasing temperature, and entropy
continues to increase with temperature. While the molar entropy curve always has positive slope,
because of the 1 / T factor in Eq. 9.3, the slope decreases with increasing temperature – i.e. the
curve bends slightly to the right.
(4) When the liquid boils, the molecules experience a large increase in entropy (disorder).
Molecules in the gas are free to explore the entire volume of the gas container, rather than sitting
in a pool at the bottom of the container as they do in the liquid phase. We see in Fig. 9.3 the
entropy of vaporization, as a large jump in molar entropy. Because of the large entropy of
- 183 -
vaporization, the molar entropy of gases is generally much larger than that of solids or liquids
consisting of molecules of comparable size.
(5) Molar entropy continues to increase with temperature, for the gas, as intramolecular
vibrations, translations and rotations become more energetic and consequently more disordered.
The heat required to melt 1 mol of a solid, at fixed pressure, is just the enthalpy of fusion.
Therefore, from Eq. 9.3, we get
∆Sfus = ∆Hfus / T
.
9.4
Similarly, the heat required to boil 1 mol of a liquid is the enthalpy of vaporization, and
∆Svap = ∆Hvap / T
.
9.5
Standard Molar Entropy
The standard molar entropy, S°, of a substance is the entropy of 1 mol of the substance at 25°C,
under standard conditions – specifically, 1 bar pressure and 1 mol L‒1 concentration for aqueous
species. These values are tabulated just like standard enthalpies of formation. For example,
S°[H2O(l)] = 70.0 J K‒1mol‒1 and S°[H2O(g)] = 188.7 J K‒1mol‒1. The entropy of the gas at 25°C is
much larger than that of the liquid at the same temperature.
There are some general trends in standard molar entropy values:
(1) Given two substances in the same phase, consisting of small molecules of the same
complexity (e.g. both diatomic), the substance with the larger molar mass has the greater molar
entropy. This is because of the uncertainty principle of quantum mechanics. One microstate of a
heavier molecule corresponds to a smaller volume in space than a microstate of a lighter
molecule. Consequently, there are more microstates available to a heavier molecule confined to
the same volume. For example, S°[H2(g)] = 130.6 J K‒1mol‒1 and
S°[O2(g)] = 205.0 J K‒1mol‒1 > S°[H2(g)]
.
(2) More complex molecules have larger standard molar entropy. For example,
S°[NO(g)] < S°[NO2(g)] < S°[N2O4(g)]
.
This is because more complex molecules have more types of motion. NO has 1 vibration and 2
rotations (rotation about the axis of a linear molecule does not count). NO2 has 3 vibrations (2
distinct stretches and one bend) and three rotations. N2O4 has 12 vibrations (5 stretches and 7
bends). Each of these motions stores thermal energy and contributes to the total number of
microstates, and the total molar entropy.
- 184 -
N
O
N
O
O
O
1 stretch vibration
2 rotations
N
N
O
O
O
N
O
12 vibrations:
3 vibrations:
2 stretches and 1 bend
3 rotations
5 stretches and 7 bends
3 rotations
Figure 9.4. The intramolecular motions of NO, NO2 and N2O4. In the top left, the stretch
vibration of NO and its two rotations are depicted. In the bottom left, the three vibrations
and 3 rotations of NO2 are depicted. On the right, the vibrations of N2O4 are depicted –
there are 5 distinct stretches (equal to the number of bonds) and 7 bends (this number is
not obvious from the motions shown in the figure). The molar entropies have the order,
NO < NO2 < N2O4.
Example 9.1: Order the following sets of substances according to increasing molar entropy.
(a) Ne(g), He(g), Xe(g), Ar(g) and Kr(g)
(b) CH3CH2OH(g), CH3OH(g), CH3OH(l), CH3OH(s) and CO2(g).
(c) CO2(g), CO2(s), CS2(g) and CS2(s).
Approach: Look for differences in phase (solid < liquid << gas), molecular complexity
(more complex means greater molar entropy), and mass for molecules with the same
complexity (larger mass means greater molar entropy).
(a) These substances are all monatomic gases (same phase and molecular complexity).
However, they differ in molar mass. Molar entropy increases with increasing molar mass.
Therefore, according to increasing molar entropy, we have
He(g) < Ne(g) < Ar(g) < Kr(g) < Xe(g) .
(b) First, note that the gases all have greater molar entropy than the liquid and solid.
Thus, the lowest molar entropy is for the solid, CH3OH(s). The corresponding liquid has
the next lowest molar entropy. Of the three gases, CO2 has the least complexity, while
CH3CH2OH has the greatest. Therefore, according to increasing molar entropy,
CH3OH(s) < CH3OH(l) < CO2(g) < CH3OH(g) < CH3CH2OH(g) .
(c) There are two solids and two gases. The solids have the lower molar entropies.
Since SO2 and CO2 are both triatomic, it is the mass of the molecules that determines the
order of molar entropies. Thus,
CO2(s) < CS2(s) < CO2(g) < CS2(g) .
- 185 -
Entropy of Reaction
Since entropy is a state function, standard molar entropies can be combined to determine the
standard entropy of reaction,
S o 

cP S o [P] 
products, P

cR S o [R]
,
9.6
reactants, R
where cP and cR are the stoichiometric coefficients of product, P, and reactant, R, respectively.
The entropy of reaction is the change in entropy upon converting reactants to products. The
standard entropy of reaction is the entropy of reaction under standard conditions.
Equation 9.6 is analogous to the formula for the enthalpy of reaction in terms of enthalpies of
formation (Eq. 8.7). Note that, while elements in their standard state have zero standard enthalpy
of formation, they do not have zero standard molar entropy. Do not forget to include elements in
their standard state when using Eq. 9.6.
Example 9.2: Determine the standard entropy of formation of liquid water, given the following
data:
Compound
H2(g)
O2(g)
H2O(l)
S° (J K−1 mol−1)
130.6
205.0
70.0
Approach: Write down the balanced formation reaction for liquid water, and then use
Eq. 9.6 to determine the entropy of reaction.
The formation reaction for liquid water is the reaction that forms one mole of liquid water:
H2(g) + ½ O2(g) → H2O(l) .
The associated standard entropy of reaction is given in terms of the tabulated absolute
entropies, using Eq. 9.6.
∆S° = S°[ H2O(l)] ‒ { S°[ H2(g)] + ½ S°[ O2(g)] }
= 70.0 ‒ ( 130.6 + ½ × 205.0 ) J K −1mol−1
= ‒163.1 J K −1mol−1
In the formation reaction of liquid water, 1 ½ moles of gas react to form 1 mole of liquid, ∆ngas =
‒1 ½. Since gases have much larger standard molar entropy than solids or liquids, the entropy
change is negative with a relatively large magnitude.
In general, if ∆ngas < 0, then ∆S is negative and large in magnitude – roughly in proportion to
∆ngas . If ∆ngas > 0, then ∆S is positive and large in magnitude. If ∆ngas = 0, the sign of ∆S
depends on ∆naq or ∆nliq , the changes (products ‒ reactants) in the number of moles of aqueous
and liquid components, respectively. In this case, the sign of ∆S matches that of ∆naq or ∆nliq (if
∆naq = 0). If ∆ngas , ∆naq and ∆nliq are all zero, ∆S is small in magnitude, and its sign cannot be
determined by simple inspection of the reaction. If ∆ngas = 0, and at least one of ∆naq and ∆nliq is
not zero, then ∆S is intermediate in magnitude.
- 186 -
Example 9.3: Label the following reactions according to whether ∆S is (i) positive and large in
magnitude, (ii) negative and large in magnitude, (iii) positive and intermediate in magnitude,
(iv) negative and intermediate in magnitude, or (v) small in magnitude. Order the reactions
according to increasing ∆S.
(a) NaCl(s) → Na+(aq) + Cl‒(aq)
(b) Na(s) + H2O(l) → Na+(aq) + OH‒(aq) + H2(g)
(c) C2H4(g) + H2(g) → C2H6(g)
(d) MgCl2(s) → Mg2+(aq) + 2 Cl‒(aq)
(e) NH4NO3(s) → N2(g) + 2 H2O(g) + ½ O2(g)
(f) H2(g) + Cl2(g) → 2 HCl(g)
(g) Mg2+(aq) + 2 OH‒(aq) → Mg(OH)2(s)
(h) H2O(l) + SO3(l) → H2SO4(l)
Approach: If ∆ngas is not equal to zero, it determines the sign of ∆S and that its
magnitude is large – in proportion to |∆ngas|. If ∆ngas equals zero, look for non-zero ∆naq
and then non-zero ∆nliq .
(a) No gases and ∆naq = +2. ∆S is positive and intermediate in magnitude.
(b) ∆ngas = +1 and ∆naq = +2. ∆S is positive and large in magnitude.
(c) ∆ngas = ‒1. ∆S is negative and large in magnitude.
(d) No gases and ∆naq = +3. ∆S is positive and intermediate in magnitude – larger than
reaction (a) though, as ∆naq = +2 in that case.
(e) ∆ngas = +3 ½ . ∆S is positive and large in magnitude – larger than reaction (b).
(f) ∆ngas = 0 and there are no aqueous or liquid reactants or products. ∆S is small in
magnitude.
(g) No gases and ∆naq = ‒3. ∆S is negative and intermediate in magnitude.
(h) No gases and ∆nliq = ‒1. ∆S is negative and intermediate in magnitude – smaller
than reaction (g) because aqueous components typically have more entropy than pure
liquid components.
The order of reactions according to increasing ∆S (from negative and large in magnitude
to positive and large in magnitude) is given by
c < g < h < f < a < d < b < e .
9.3 The Second Law of Thermodynamics
We return now to the second law of thermodynamics. It is the law that determines which
processes occur and which do not. It simply says that processes that occur are such that (Eq.
9.2)
- 187 -
∆Suniv = ∆Ssys + ∆Ssurr > 0 .
The contribution to the total change in entropy, ∆Ssurr, arises because the process that occurs in
the system is accompanied by heat transfer to or from the system – i.e., from or to the
surroundings. This heat transfer changes the entropy of the surroundings. Equation 9.3 (see
also 9.4 and 9.5) gives the relationship between heat transferred and change in entropy – at
constant temperature. Specifically,
∆Ssurr = qsurr / T
Since the heat transferred to the surroundings comes from the system,
qsurr = ‒qsys .
At constant pressure, the heat flow into the system equals the change in enthalpy of the system.
Therefore,
qsurr = ‒∆Hsys
and
∆Ssurr = ‒∆Hsys / T
9.7
Substituting this equation into Eq. 9.2, the criterion for spontaneity (i.e., the second law of
thermodynamics) is expressed in terms of changes in system properties only. Specifically,
∆Suniv = ∆Ssys ‒ ∆Hsys / T > 0
9.8
for spontaneous processes occurring at constant temperature and pressure.
This explains why so many spontaneous processes are exothermic. Exothermic reactions have
negative ∆Hsys , and consequently release heat into the surroundings. This increases the entropy
of the surroundings, making spontaneity of the reaction more likely – it also depends on ∆Ssys .
Exothermic processes are spontaneous, at least in part, because of the increased disorder that
results from the heat released to the surroundings.
Example 9.4: Determine ∆Suniv for the formation reaction of one mole of liquid water, at 25°C,
given
∆Sf°[ H2O(l)] = ‒163.1 J K −1mol−1
and
∆Hf°[ H2O(l)] = ‒285.8 kJ mol−1 .
Is the reaction of hydrogen and oxygen to form liquid water spontaneous at 25°C?
Approach: Determine ∆Ssys and ∆Ssurr, and then add them to get ∆Suniv . Use Eq. 9.7 to
get ∆Ssurr .
The entropy change of the system for the formation reaction of one mole of liquid water,
at 25°C, is just the standard entropy of formation of liquid water.
∆Ssys = ∆Sf°[ H2O(l)] = ‒163.1 J K −1mol−1 .
The entropy change in the surroundings that accompanies this process is given by Eq.
9.7, with
∆Hsys = ∆Hf°[ H2O(l)] = ‒285.8 kJ mol−1 ;
- 188 -
i.e.,
∆Ssurr = ‒ ∆Hf°[ H2O(l)] / T = 285.8 kJ mol−1 / 298.15 K
= 958.6 J K −1mol−1 .
Altogether,
∆Suniv = ∆Ssys + ∆Ssurr
= ‒163.1 + 958.6 J K −1mol−1 = 796 J K −1mol−1 .
Since ∆Suniv > 0, this process is spontaneous at 25°C and 1 bar pressure.
9.4 Gibbs Free Energy
Equation 9.8 expresses the criterion for spontaneity, for constant temperature and pressure
processes, in terms of changes to the system. Here, we drop the sys subscript on ∆Hsys and
∆Ssys, and consider only properties of the system. Multiplying Eq. 9.8 by T and reversing the sign
(which reverses the sense of the inequality), gives
∆H ‒ T ∆S < 0
or
∆ (H ‒ T S) < 0 ,
since T is constant. This inequality can be expressed in terms of a new state function, G, the
Gibbs free energy.
G = H ‒ TS
9.9
The criterion for spontaneity of constant T and P processes now takes the form,
∆G < 0 .
9.10
Since
∆G = ∆H ‒ T ∆S ,
9.11
for constant temperature processes, the criterion for spontaneity also takes the form
∆H < T ∆S .
There are four types of processes, distinguished by the signs of ∆H and ∆Ssys . These processes
are listing in Table 9.1, and identified as spontaneous or non-spontaneous. The third and fourth
types of process, in the list, are just the reverse of the second and first types of process,
respectively. For the second and third types of process, spontaneity depends on temperature.
Sign of ∆H
Sign of ∆S
‒ exothermic
+
‒
+ endothermic
+
‒
‒
‒
+
+
+
Sign of ∆G
Spontaneous?
at all temperatures
at low temperature
at high temperature
at low temperature
at high temperature
at all temperatures
spontaneous
spontaneous
non-spontaneous
non-spontaneous
spontaneous
non-spontaneous
Table 9.1. Spontaneity of exothermic and endothermic processes
- 189 -
In the case of exothermic processes with positive change in (system) entropy, both contributions
to ∆G in Eq. 9.11 are negative. Such processes are always spontaneous – i.e., spontaneous at
all temperatures. They include the combustion of ethane (or any larger hydrocarbon), and the
decomposition of hydrogen peroxide. The reverse of such processes, endothermic processes
with negative entropy change, are never spontaneous.
In the case of exothermic processes with negative entropy change, the positive ‒T ∆S term
cancels the negative ∆H term, if T = Tcross , where
Tcross = ∆H / ∆S
9.12
Below the crossover temperature, Tcross, the sign of ∆H gives the sign of ∆G , and the process is
spontaneous. Above the crossover temperature, the process is non-spontaneous. Examples of
processes of this type include rusting of iron (or any metal), freezing of any liquid, and
condensation of any gas. For example, a liquid freezes only below its freezing point – the melting
point of solid. The crossover temperature of the freezing process is the melting point.
Endothermic processes with positive change in entropy are the reverse of the processes just
described. These processes are spontaneous only at high temperatures – above Tcross. For
example, boiling of a liquid occurs only above its boiling point – the crossover temperature of the
boiling process.
Standard Gibbs Free Energy of Reaction
The standard Gibbs free energy of reaction – for any reaction – can be determined from tabulated
standard Gibbs free energies of formation. Specifically, analogous to Eqs. 8.7 (for enthalpy) and
9.6 (for entropy, except Eq. 9.6 is in terms of absolute entropies), we have
G o 

cP G of [P] 
products, P

cR G of [R] ,
9.13
reactants, R
where cP and cR are the stoichiometric coefficients of product, P, and reactant, R, respectively.
Equation 9.13 determines the Gibbs free energy of reaction at 25°C, and under standard
conditions – all gases at 1 bar partial pressure and solutes at 1 mol L‒1 concentration. At other
temperatures, or under non-standard conditions, the Gibbs free energy of reaction generally
varies. To determine ∆Go at other temperatures requires knowing ∆Ho and ∆So separately, and
then using Eq. 9.11. The enthalpy and entropy both typically vary only weakly with temperature.
A useful approximation is obtained if we neglect the temperature dependence of ∆Ho and ∆So.
For example, the crossover temperature (under standard conditions) is estimated using Eq. 9.12
with ∆Ho and ∆So obtained at another temperature (typically 25°C). With this approximation, ∆Go
varies linearly with temperature, and is easily estimated using Eq. 9.11.
- 190 -
Figure 9.5. The temperature dependence of Gibbs free energy in the case of an
endothermic reaction (∆H > 0) and ∆S > 0. In this example, ∆H = 70 kJ mol‒1 and ∆S =
200 J K‒1mol‒1 are independent of T. For real reactions, ∆H and ∆S vary with
temperature – though only weakly.
Example 9.5: Consider the decomposition of dinitrogen tetraoxide:
N2O4(g) → 2 NO2(g) .
Data at 25°C:
Compound
H of (kJ mol−1)
N2O4(g)
NO2(g)
11.1
33.2
G of (kJ mol−1)
99.8
51.3
(a) Determine the standard Gibbs free energy of this decomposition reaction at 25°C. Is it
spontaneous at 25°C?
(b) Determine the standard Gibbs free energy of the decomposition at 50°C. Is it spontaneous at
50°C?
(c) Estimate the crossover temperature for the decomposition of dinitrogen tetraoxide under
standard conditions (1 bar partial pressure of both gases).
Approach: (a) Use Eq. 9.13, since the given data are for 25°C. (b) Use the enthalpy
data to get the enthalpy of reaction (Eq. 8.7). From the enthalpy and Gibbs free energy
of reaction at 25°C (298.15 K), we get the entropy of reaction from Eq. 9.11. Equation
9.11 is applied again – this time to get the Gibbs free energy of reaction at 50°C (323.15
K). Here, we neglect the weak temperature dependence of enthalpy and entropy of
o
reaction. In both part (a) and (b), we note the sign of G . The decomposition is
spontaneous only if G  0 . (c) Use Eq. 9.12 to estimate the crossover
temperature.
o
- 191 -
(a) From Eq. 9.13 and the given data, we get
G o  2G of [NO2 (g)]  G of [N 2O4 (g)]
 2  51.3  99.8 kJ mol1
 2.8 kJ mol1
at 25°C. This Gibbs free energy of reaction is positive. Therefore, the decomposition of
dinitrogen tetraoxide is not spontaneous at 25°C.
(b) From Eq. 8.7, and the given data, we get
H o  2H of [NO2 (g)]  H of [N 2O4 (g)]
 2  33.2  11.1 kJ mol1
 55.3 kJ mol1 ,
also at 25°C. Combining this result with the result obtained in part (a), we get
H o  G o
S 
T
55.3  2.8 kJ mol1

1000 J kJ 1
298.15 K
o
 176.1 J K 1 mol1 ,
from Eq. 9.11. Now we can determine the Gibbs free energy of reaction at any
temperature – assuming enthalpy and entropy of reaction are temperature independent.
In particular, at 50°C,
G o  H o  T S o
 55.3 kJ mol1  323.15 K 176.1 J K 1 mol1 /1000 J kJ 1
 1.6 kJ mol1 .
Since this Gibbs free energy of reaction is negative, the reaction is spontaneous at 50°C.
(c) The crossover temperature, under standard conditions, is given by Eq. 9.12 which
here takes the form,
Tcross = ∆H °/ ∆S°
= 55.3 kJ mol−1 × 1000 J kJ−1 / 176.1 J K−1 mol−1
= 314. K ,
or 41°C. Note that below the crossover temperature (e.g., 25°C), the reaction is not
spontaneous. Above the crossover temperature (e.g., 50°C), the reaction is
spontaneous.
- 192 -
9.5 Gibbs Free Energy and Equilibrium
We have seen that reactions with ∆G < 0 are spontaneous, and therefore proceed in the forward
direction. Reactions with ∆G > 0 are non-spontaneous. However, the reverse of such a reaction
is spontaneous. Therefore, the reaction proceeds in the reverse direction.
We also know that all reactions are equilibria which proceed in the forward direction, starting with
only reactants, until equilibrium conditions are achieved. At equilibrium, forward reaction is
balanced by reverse reaction – there is no net reaction. In this case, neither the forward nor
reverse reaction is spontaneous. Equilibrium conditions correspond to
∆G = 0 . equilibrium conditions
Reactions of gases and solutions achieve equilibrium under conditions with both reactant and
product present.
In Example 9.5, above, standard conditions are equilibrium conditions at the crossover
temperature, 41°C. Below the crossover temperature, the reverse reaction is spontaneous under
standard conditions. In this case, there is net reverse reaction which continues until equilibrium
conditions – which are non-standard – are achieved. Above the crossover temperature, there is
net forward reaction, which again produces non-standard conditions at equilibrium. In this case
the partial pressure of reactant (dinitrogen tetraoxide) is less than 1 bar, while that of product
(nitrogen dioxide) is greater than 1 bar.
We see that the Gibbs free energy of reaction depends on the partial pressures of reactant and
product gases in the above example. In general, the Gibbs free energy of reaction depends on
the reaction quotient, Q. If Q < K, there is net forward reaction. This corresponds to ∆G < 0.
If Q < K, ∆G > 0 are there is net reverse reaction.
The dependence of ∆G on Q arises because the entropy of gases and solutes depend upon their
activities. For example, a gas with low pressure has more entropy (per mole of gas). A low
pressure gas has large volume (per mole) which corresponds to greater freedom or disorder –
i.e., higher entropy. From Eq. 9.9, this corresponds to a lower Gibbs free energy. Combining the
activity dependence of reactant and product to construct the Gibbs free energy of reaction gives
the following dependence of ∆G on Q:
∆G = ∆G° + R T ln Q .
9.14
To see how this equation provides the quantitative counterpart to the above qualitative
relationship between equilibrium and Gibbs free energy, consider the following observations:
Under equilibrium conditions, Q = K and ∆G = 0. Substituting these equations into Eq. 9.14
gives:
0 = ∆G° + R T ln K
or
∆G° = ‒R T ln K .
9.15
Since the logarithm of a number greater than one is positive, we see that an equilibrium constant
greater than one corresponds to a negative standard Gibbs free energy of reaction – i.e. the
reaction is spontaneous under standard conditions. Since Q =1 for standard conditions, we have
Q < K in this case, and there is net forward reaction – the forward reaction is spontaneous.
Similarly, if K is less than one, Q > K under standard conditions, and the reverse reaction is
spontaneous.
Substituting Eq. 9.15 into 9.14 gives
- 193 -
∆G = ‒R T ln K + R T ln Q ,
or
Q
G  R T ln   .
K
9.16
This equation makes clear that whenever Q < K, ∆G < 0, and the forward reaction is
spontaneous. This equation explains the origins of equilibria phenomena – as described in
Chapter 7. Ultimately, the shift toward equilibrium is the natural tendency toward greater disorder
in the universe.
Equation 9.15 can be inverted to give the equilibrium constant in terms of the Gibbs free energy
of reaction:
 G  
K  exp  
 .
 RT
9.17
Equation 9.16 can also be written in terms of log10. Specifically, since ln(10)=2.303,
G
Q
 log10   .
2.303R T
K
This equation is featured in Fig. 9.6 which shows logarithmic scales of reaction quotient values
alongside linear Gibbs free energy of reaction scales for a reactant favored reaction (K < 1) in the
top panel, and a product favored reaction in the bottom panel (K > 1).
K<1
Q<K
net forward rxn
Q>K
10-3
10-2
reactant favored equilibrium
net reverse rxn
Q
10-4
10-1
1
103
102
10
104
K
-1
-2
2
1
0
7
6
G
standard
2.303 R T
conditions
K>1
5
4
3
product favored equilibrium
Q<K
net forward rxn
Q>K
net reverse rxn
Q
10-3
10-4
10-2
10-1
1
103
102
10
104
K
-6
-5
G
-4
2.303 R T
-3
-2
-1
0
1
2
3
standard
conditions
Figure 9.6. Reaction quotient and Gibbs free energy of reaction for reactant favored (top
panel) and product favored (bottom panel) reactions.
- 194 -
Example 9.6: Consider the hydration of anhydrous calcium sulfate:
CaSO4(s) + 2 H2O(g) 
CaSO4·2H2O(s) ,
for which ∆H° = ‒104.9 kJ mol−1 and ∆S° = ‒290.2 J K−1 mol−1, at 25°C. Calcium sulfate acts as
a desiccant because of this reaction – specifically, it removes water vapor from air.
(a) Determine ∆G° for this reaction at 25°C. Is it spontaneous under standard conditions – i.e.
water vapor with 1 bar partial pressure?
(b) What is the equilibrium constant for hydration of calcium sulfate at 25°C?
(c) Determine ∆G, at 25°C, if the partial pressure of water is 3.2 kPa. A water partial pressure of
3.2 kPa corresponds to 100% relative humidity at 25°C – i.e., water vapor in equilibrium with
liquid water at that temperature. Is the hydration reaction spontaneous with this partial pressure
of water?
(d) At what partial pressure of water vapor is the hydration reaction at equilibrium, at 25°C?
Approach: (a) Use Eq. 9.11 to determine ∆G° from the given data. (b) Use Eq. 9.17 to
get the equilibrium constant from ∆G°. (c) Write the expression for the reaction quotient
for this reaction. Evaluate the reaction quotient for 3.2 kPa water partial pressure. Then
use Eq. 9.14 to determine ∆G, and note its sign. Hydration is spontaneous under these
conditions if ∆G is negative. (d) Set Q = K, and then solve for the partial pressure of
water.
(a) From Eq. 9.11, the Gibbs free energy of reaction at 25°C is given by
∆G = ∆H ‒ T ∆S = ‒104.9 kJ mol−1 ‒ 298.15 × (‒290.2 J K−1 mol−1) / (1000 J kJ−1)
= ‒18.38 kJ mol−1 .
(b) From Eq. 9.17, the equilibrium constant at 25°C is given by
 18.38 kJ mol1  1000 J kJ 1 
 G  

K  exp  
exp



1
1
 RT
 8.314 J K mol  298.15 K 
 18.38 kJ mol1  1000 J kJ 1 
3
 exp 
  1.66  10 .
1
1
 8.314 J K mol  298.15 K 
(c) The reaction quotient for this reaction is
Q 

1
PH22O
1
 3.2 kPa 


 100. kPa 
2
 9.7 7  102 under the given conditions.
Note the division of the partial pressure of H2O (in kPa) by the standard pressure, 1 bar =
100 kPa.
- 195 -
Using Eq. 9.14, we find the Gibbs free energy of reaction for this partial pressure of
water:
G  G o  RT ln Q
 18.38 kJ mol1 
8.314 J K 1mol1  298.15 K
ln  9.77 102 
1000 J kJ 1
 1.3 kJ mol1
Since this ∆G is negative, the hydration reaction is spontaneous under the given
conditions – i.e., anhydrous calcium sulfate will spontaneously hydrate when exposed to
water saturated air at 25°C.
(d) Setting Q = K gives
Q 
1
 K  1.66  103
2
PH2O
or
PH2O 
1
1.66 103 
1/ 2
 0.0245 bar
= 100.0 kPa bar 1  0.0245 bar = 2.45 kPa .
9.6 Coupled Reactions
We have seen that spontaneous processes increase the entropy of the universe. This
corresponds to processes with negative ∆G. Reactions with positive ∆G never occur by
themselves. However, it is possible for such reactions to occur, if they are coupled to other
reactions with negative ∆G, such that the total change in Gibbs free energy – the sum of the
positive and negative Gibbs free energies of reaction – is negative. When reactions are coupled,
a spontaneous reaction can drive an otherwise non-spontaneous reaction. This is how
electrolysis works – a spontaneous reaction (inside a battery) drives another reaction in its nonspontaneous direction. For example, electrolysis of water produces H2(g) and O2(g) according to
the non-spontaneous reaction,
∆Go = 237 kJ mol‒1
H2O(l) → H2(g) + ½ O2(g) .
There are many other examples of coupled reactions in the chemistry of life. For example,
metabolism of glucose is a spontaneous process:
C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l) .
∆Go = ‒2880 kJ mol‒1
This process occurs through a series of steps known as cellular respiration. In the overall
process, Gibbs free energy liberated by the oxidation of glucose is stored in ATP molecules
(adenosine triphosphate), as the spontaneous oxidation of glucose drives the otherwise nonspontaneous condensation reaction,
H3PO4(aq) + ADP(aq) → H2O(l) + ATP(aq) ,
shown with structures in Fig. 9.7.
- 196 -
∆Go = 30.5 kJ mol‒1
ADP
phosphoric acid
O
O
HO P OH
NH2
adenosine diphosphate
+
N
O
HO P O P O
OH
OH
O
OH
N
N
N
OH OH
ATP
NH2
adenosine triphosphate
O
H2O
+
HO P
OH
O
N
O
O P O P O
OH
O N
OH
N
N
OH OH
Figure 9.7. The condensation of phosphoric acid and ADP (adenosine diphosphate) to
form ATP (adenosine triphosphate). The reverse reaction is the hydrolysis of ATP.
Under conditions in the body, the Gibbs free energy of the above condensation reaction is closer
to 50 kJ mol‒1 – it depends on the concentration of Mg2+. Typically, about 30 ATP molecules are
produced in the net cellular respiration of one glucose molecule. The corresponding positive
Gibbs free energy change (about 30×50 kJ mol‒1) is more than accommodated by the negative
Gibbs free energy associated with oxidation of glucose.
Having stored Gibbs free energy in the form of ATP, the stored Gibbs free energy is released in
the reverse reaction – the hydrolysis of ATP to form ADP. This spontaneous reverse reaction can
be coupled to other non-spontaneous reactions, driving the otherwise non-spontaneous
production of other molecules with high Gibbs free energy. For example, the non-spontaneous
(by itself) condensation of glucose and fructose is driven by free energy derived from hydrolysis
of ATP.
C6H12O6(aq) +
C6H12O6(aq) → C12H23O11(aq) + H2O(l) . ∆G = +23 kJ mol‒1
glucose
fructose
sucrose
Problems
9.1 Order the following sets of substances according to increasing molar entropy.
(a) CH3Cl(g), CH2Cl2(g), CH4(g) and CH3F(g)
(b) SO2(g), CO2(g), SBr2(g) and CO2(s).
(c) SO2(g), SO3(g), H2SO3(g) and H2SO3(l).
9.2 Determine the standard entropy of vaporization of sulfur trioxide, given the following data:
∆S°[ 2 S(s) + 3 O2(g) → 2 SO3(l)] = ‒395.4 J K−1 mol−1
Compound
S(s)
O2(g)
SO3(g)
S° (J K−1 mol−1)
3.98
205.0
256.8
- 197 -
Here, S(s) is shorthand for 1/8 S8(rhombic), where S8(rhombic) is the standard state of sulfur.
9.3 Label the following reactions according to the sign and magnitude (small, intermediate or
large) of the entropy of reaction. Order the reactions according to increasing ∆S.
(a) Br2(l) → Br2(g)
(b) I2(s) → I2(g)
(c) 2 C2H4(g) + H2(g) → C4H10(g)
(d) 2 Mg(s) + O2(g) → MgO(s)
(e) Cl2(aq) + 2 H2O(l) → 2 H3O+(l) + Cl‒1(aq) + OCl‒1(aq)
9.4 In process 1, 5.00 kJ of heat is transfered to the surroundings from the system. At the same
time, the entropy of the system decreases by 20.0 J K−1. In process 2, 1.00 kJ of heat is
transfered to the system from the surroundings, while the entropy of the system increases by
10.0 J K−1. Both processes occur at a constant temperature, T.
(a) For what temperatures is process 1 spontaneous?
(b) For what temperatures is process 2 spontaneous?
(c) There is a temperature range over which both processes are spontaneous. What is this
range?
9.5 Consider the decomposition of Ag2O(s):
Ag2O(s) → 2 Ag(s) + ½ O2(g)
Standard data at 25°C:
Compound
Ag(s)
O2(g)
Ag2O(s)
 H of (kJ mol−1)
0
0
‒31.1
S o (J K−1mol−1)
42.6
205.15
121.3
(a) Determine the standard Gibbs free energy of this decomposition at 25°C. Standard
conditions here corresponds to 1 bar partial pressure of oxygen. Is the decomposition
spontaneous under these conditions?
(b) Determine the standard Gibbs free energy of the decomposition at 200°C. Is the
decomposition spontaneous at 200°C, under standard conditions?
(c) Estimate the crossover temperature (in °C) for the decomposition of silver oxide under 1
bar partial pressure of oxygen.
(d) What is the equilibrium constant for the above reaction at 25°C, and at 200°C?
(e) What is the partial pressure of oxygen in equilibrium with silver oxide at 25°C, and at
200°C?
- 198 -
9.6 Consider the redox reaction,
Cd(s) + 2 Tl+(aq) → Cd2+(aq) + 2 Tl(s)
Standard data at 25°C:
Species
Cd2+(aq)
Tl+(aq)
G of (kJ mol−1)
‒77.6
‒32.4
(a) Determine the standard Gibbs free energy of reaction at 25°C. Standard conditions
correspond to 1.00 mol L‒1 Tl+(aq) and Cd2+(aq). Is the reaction spontaneous under these
conditions?
(b) Determine the Gibbs free energy of reaction if [Tl+] = 0.100 mol L‒1 and [Cd2+] =
0.100 mol L‒1. Is the reaction spontaneous under these conditions?
(c) Determine the Gibbs free energy of reaction if [Tl+] = 1.00×10‒5 mol L‒1 and [Cd2+] =
1.00 mol L‒1. Is the reaction spontaneous under these conditions? What about the reverse
reaction?
(d) Find the maximum concentration of Tl+(aq) such that the reverse reaction is spontaneous,
when [Cd2+] = 1.00 mol L‒1.
- 199 -
Solutions
9.1 (a) According to increasing entropy, we have
CH4(g) < CH3F(g) < CH3Cl(g) < CH2Cl2(g) .
These molecules all have the same number of atoms, and they are all in the gas phase.
Each molecule in the sequence, starting with methane, is just the previous molecule with a
lighter atom swapped for a heavier atom.
(b) Here, the order is
CO2(s) < CO2(g) < SO2(g) < SBr2(g) .
The solid has the lowest entropy. After that the gases are ordered according to increasing
molecular weight as in part (a).
(c) Here,
H2SO3(l) < SO2(g) < SO3(g) < H2SO3(g) .
The liquid has the lowest entropy. After that, the gases are ordered according to increasing
molecular complexity – i.e., increasing number of atoms.
9.2 The standard entropy of reaction for
2 S(s) + 3 O2(g) → 2 SO3(l)
is expressed in terms of the standard molar entropies of the reactants and products – see Eq.
9.6:
S o  2S o [SO3 (l)]  2S o [S(s)]  3S o [O2 (g)]
 2S o [SO3 (l)]  2  3.98  3  205.0 J K 1mol1
 2S o [SO3 (l)]  622.96 J K 1mol1
 395.4 J K 1mol1 .
Therefore,
395.4  622.96
J K 1mol1
2
1
1
 113.78 J K mol .
S o [SO3 (l)] 
The standard entropy of vaporization of sulfur trioxide is just the standard molar entropy of
the gas minus that of the liquid – another application of Eq. 9.6. Thus,
S o  S o [SO3 (g)]  S o [SO3 (l)]
 256.8  113.78 J K 1mol1
 143.0 J K 1mol1 .
- 200 -
9.3 (a) Br2(l) → Br2(g)
This reaction has a large positive entropy change. ∆ngas = +1.
(b) I2(s) → I2(g)
This reaction has a large positive entropy change. ∆ngas = +1. We expect ∆S of this reaction
to be greater than that of reaction (a) because the initial phase is solid rather than liquid, and
because iodine is heavier than bromine.
(c) 2 C2H4(g) + H2(g) → C4H10(g)
This reaction has a large, in magnitude, negative entropy change. ∆ngas = ‒2.
(d) 2 Mg(s) + O2(g) → MgO(s)
This reaction has a large, in magnitude, negative entropy change. ∆ngas = ‒1. ∆S for this
reaction has a smaller magnitude than reaction (d).
(e) Cl2(aq) + 2 H2O(l) → H3O+(l) + Cl‒1(aq) + HOCl(aq)
This reaction has positive entropy change which is intermediate in magnitude. ∆naq = +1,
while ∆nliq = ‒1.
The reactions are ordered, according to increasing entropy of reaction, as follows:
c < d < e < a < b .
9.4 (a) In process 1, ∆Ssys = ‒20.0 J K−1, while ∆Ssurr = qsurr / T , where qsurr = 5.00 kJ. The
process is spontaneous if
∆Suniv = ∆Ssys + ∆Ssurr = ∆Ssys + qsurr / T > 0 ,
which follows if
qsurr / T > ‒∆Ssys
or
qsurr > ‒∆Ssys T .
Since ‒∆Ssys is positive (∆Ssys is negative), dividing this inequality by ‒∆Ssys does not reverse
the sense of the inequality (it remains greater than). Consequently,
qsurr / (‒∆Ssys) > T ;
i.e., the reaction is spontaneous only if temperature is below
qsurr / (‒∆Ssys) = 5.00 kJ / (20.0 J K−1) × 103 J kJ−1 = 250. K .
(b) In process 2, ∆Ssys = 10.0 J K−1, while ∆Ssurr = qsurr / T , where qsurr = ‒1.00 kJ. The
process is spontaneous if
qsurr > ‒∆Ssys T .
In this case, ‒∆Ssys is negative. The sense of the inequality is reversed (> becomes <) when
we divide by ‒∆Ssys . Specifically,
- 201 -
qsurr / (‒∆Ssys) < T ;
i.e., the reaction is spontaneous only if temperature is above
qsurr / (‒∆Ssys) = ‒1.00 kJ / (‒10.0 J K−1) × 103 J kJ−1 = 100. K .
(c) Both reactions are spontaneous between 100. and 250. K.
9.5 (a) Using the given data, the standard enthalpy and entropy of the decomposition reaction at
25°C are determined using Eq. 8.7 and 9.6, respectively. The standard Gibbs free energy of
reaction is given by Eq. 9.11. Since the decomposition reaction is just the reverse of the
formation reaction of Ag2O(s), the standard enthalpy of reaction is just
∆H° = ‒ ∆Hf°[Ag2O(s)]
= 31.1 kJ mol−1 .
The entropy of reaction is given using Eq. 9.6:
∆So = 2 S°[Ag(s)] + ½ S°[O2(g)] ‒ S°[Ag2O(s)]
= 66.48 + ½ × 205.15 ‒ 121.3 J K−1mol−1
= 66.48 J K−1mol−1 .
The standard Gibbs free energy of reaction at 25°C is now determined to be
∆Go = ∆Ho ‒ T ∆So = 31.1 kJ mol−1 ‒ 298.15 K × 66.48 J K−1mol−1 / (103 J kJ−1)
= 11.3 kJ mol−1 .
Since the Gibbs free energy of reaction is positive, the reaction is not spontaneous under
standard conditions (1 bar oxygen partial pressure), at 25°C.
(b) To determine the standard Gibbs free energy of the decomposition at 200°C, we again
use Eq. 9.11. We neglect the temperature dependence of ∆Ho and ∆So, and use the values
determined in part (a) at 25°C. The result is
∆Go = ∆Ho ‒ T ∆So = 31.1 kJ mol−1 ‒ 473.15 K × 66.48 J K−1mol−1 / (103 J kJ−1)
= ‒0.36 kJ mol−1 .
Silver oxide will decompose spontaneously at 200°C, under 1 bar partial pressure of oxygen,
as ∆Go is negative.
(c) The crossover temperature is the temperature at which ∆G = 0. Under standard
conditions, we have ∆Go = 0. Substituting Eq. 9.11 and solving for T gives the crossover
temperature (Eq. 9.12):
Tcross = ∆Ho / ∆So = (31.1 kJ mol−1) × (103 J kJ−1) / (66.48 J K−1mol−1) = 467.8 K .
This corresponds to 468 ‒ 273.15 °C = 195°C.
(d) The equilibrium constant is related to the Gibbs free energy under standard conditions –
see Eq. 9.17:
- 202 -
 G  
K  exp  

 RT
 11.3 kJ mol1  103 J kJ 1 
 exp  
 at 25C
1
1
 8.314 J K mol  298.15 K 
 0.01048
At 200°C, we have
 0.36 kJ mol1 103 J kJ 1 
K  exp  
 at 200C
1
1
8.314
J
K
mol
473.
K

15


 1.16 .
(e) For the decomposition of silver oxide, the reaction quotient is given by
Q  PO1/2
.
2
K
at equilibrium.
Thus,
PO2
 K2
1 bar
 0.01048 2  1.10  104
at 25C
or
PO2  1.10  104 bar ,
at 25°C. A partial pressure determined from an equilibrium constant is an activity – it is the
partial pressure relative to the standard pressure, 1 bar. Thus, the partial pressure is
determined in bar.
At 200°C, we have
PO2  1.16 2 bar at 25C
= 1.3 bar .
9.6 (a) The standard Gibbs free of reaction is given by
G o  G of [Cd 2 (aq)]  2G of [Tl (aq)]
 77.6   32.4  kJ mol1
 45.2 kJ mol1 .
Since this value is negative, we conclude that the reaction is spontaneous under standard
conditions.
(b) Under non-standard conditions, we use Eq. 9.14:
- 203 -
G  G o  RT ln Q
 [Cd 2 (aq)] 
 G o  RT ln  
2 
 [Tl (aq)] 
8.314 J K 1mol1  298.15 K  0.100 
1
ln 
 45.2 kJ mol 
2 
103 J kJ 1
 0.100 
 39.5 kJ mol1 .
The Gibbs free energy of reaction is still negative, but with smaller magnitude than under
standard conditions. The reaction is spontaneous if [Tl+] = [Cd2+] = 0.100 mol L‒1.
(c) If [Tl+] = 1.00×10‒5 mol L‒1 and [Cd2+] = 1.00 mol L‒1, then
 [Cd 2 (aq)] 
G  G o  RT ln  
2 
 [Tl (aq)] 
 45.2 kJ mol1 

8.314 J K 1mol1  298.15 K 
1.00
ln
3
1
 1.00 105
10 J kJ





2


 11.9 kJ mol1 .
Note that concentrations appear in the reaction quotient without units – they are activities defined
as the ratio of concentration to standard concentration (exactly 1 mol L‒1).
The Gibbs energy of reaction is positive under the above conditions. Thus, the forward reaction
is not spontaneous. However, the reverse reaction has a negative ∆G and is spontaneous under
these conditions.
(d) Set ∆G = 0, and solve for Q. This gives Q under equilibrium conditions – i.e., Q = K.
Solving for K in terms of ∆Go is just Eq. 9.17.
The maximum Tl+(aq) concentration for which the reverse reaction is spontaneous corresponds to
equilibrium conditions – at which point, the reverse reaction is no longer spontaneous.
Q
[Cd 2 (aq)]
[Tl (aq)]2
 G o 
 K  exp  

 RT 
 45.2 kJ mol1 103 J kJ 1 
 exp  

1
1
 8.314 J K mol  298.15 K 
 8.30 107 .
Therefore,
- 204 -
1/2
 [Cd 2 (aq)] 
[Tl (aq)]  

K



1/2
 1.00 

7 
 8.30 10 
 1.10 104 mol L1 (re-inserting units).
Here, we see that though the reverse reaction was spontaneous in part (c) (there, [Tl+] =
1.00×10‒5 mol L‒1), the reverse reaction does not proceed very far. It is only spontaneous until
[Tl+] = 1.10×10‒4 mol L‒1 is achieved, and there is no further net reverse reaction.
- 205 -
10. Electrochemistry
10.1 Electron Transfer Reactions
Oxidation-reduction (redox) reactions – the most important class of reactions – are typically
exothermic, and provide the basis for chemical sources of energy. This includes the burning of
fossil fuels and cellular respiration. Redox reactions are electron transfer reactions. Electrons
are transferred from the reducing agent to the oxidizing agent. When we balance redox
reactions, the electrons lost by the reducing agent must match the electrons gained by the
oxidizing agent. For example, Cu(NO3)2(aq) oxidizes Zn(s). The net ionic equation is
Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq) .
Here, two electrons are transferred from a zinc atom to a copper II ion to give a zinc ion and a
copper atom. The half reactions are written as
Cu2+(aq) + 2 e‒ → Cu(s) reduction
and
Zn(s) → Zn2+(aq) + 2 e‒ oxidation.
If granulated zinc is added to an aqueous copper II nitrate solution, electrons are transferred
directly from zinc atoms on the surface of zinc granules to copper ions in solution. Copper metal
forms on the surface of zinc granules, while the granules dissolve into solution replacing the
copper II ions with zinc ions. This process is depicted in Fig. 10.1.
2+
Cu
-
2e
2+
Cu
-
2e
2+
Cu
2+
Cu
Zn
Zn
Zn
Zn
2+
Cu
Cu2+
Zn Zn Zn Zn
Zn
Zn Zn
Zn
Zn
Zn
Zn
Zn
Zn
Zn
Zn2+
Zn2+
2+
Zn
Zn2+
Zn2+
Cu Cu
Zn
Cu Zn Zn
Zn
Zn2+
Cu
Cu
Zn
Cu
Zn
Zn Zn
Zn
Zn
Zn
Zn Zn Zn
Zn
Zn
Zn
Zn
Zn
Figure 10.1. A depiction of the oxidation of zinc by copper ions at the surface of a zinc
granule in an aqueous copper II nitrate solution – not to scale. Two Cu2+ ions, in the top
panel, are shown receiving two electrons (each) from two zinc atoms. After four more
such electron transfer reactions, the bottom panel results showing six copper atoms
deposited onto the zinc and the release of six zinc ions into the solution.
- 206 -
Enthalpy released by the above reaction is manifest as heat which results in products at higher
temperature than reactants. However, because the reaction involves the transfer of electrons, it
is possible to harness the energy released by the reaction (or any other redox reaction) as
electrical energy. This is achieved by constructing an electrochemical cell wherein oxidation and
reduction half reactions are physically separated.
10.2 Electrochemical Cells
In an electrochemical cell – also known as a Galvanic or Voltaic cell – electrons released by the
oxidation half reaction must travel through a wire (and do electrical work) before they are
consumed by the reduction half reaction. The half reactions take place at the surface of metallic
electrodes: (1) oxidation occurs at the anode (2) reduction occurs at the cathode. Electrons
are liberated by the oxidation half reaction at the anode. They travel through wires to the
cathode, and are consumed there by the reduction half reaction.
The zinc-copper electrochemical cell is depicted in Fig. 10.2. Zinc metal provides the anode, and
the reactant in the oxidation half reaction. Copper metal provides the cathode, and the product of
the reduction half reaction. In the electrochemical cell, as copper is produced, it plates onto the
copper cathode rather than forming at the surface of the zinc granules as it does in the direct
reaction.
1.10 V
Voltmeter
e-
esalt bridge
NaNO3(aq) in a porous gel
anode
NO3-
cathode
Na+
Zn2+
NO3-
Zn
Cu
NO3-
Cu2+
1 mol L-1 Zn(NO3)2(aq)
Zn(s)
1 mol L-1 Cu(NO3)2(aq)
Zn2+ + 2 e-
Cu2+ + 2 e-
Cu(s)
Figure 10.2. The zinc-copper electrochemical cell under standard conditions – 1 mol L‒1
concentration of Cu2+(aq) and Zn2+(aq). By convention, the anode is always drawn on
the left, with the cathode on the right.
The redox half reactions take place in the half cells on the left (oxidation at the anode) and on the
right (reduction at the cathode). The half cells are connected by a wire, between anode and
cathode, and a salt bridge – required to complete the circuit.
- 207 -
To understand the role of the salt bridge, consider a cell without the salt bridge. When the anode
and cathode are connected by a wire, electrons flow from anode to cathode – but only very
briefly. A negative charge quickly accumulates in the cathode half cell, while a corresponding
positive charge accumulates in the anode half cell. These charges give rise to an electric field
that opposes further flow of electrical current. Adding the salt bridge allows ions within the salt
bridge to carry current and maintain charge balance in the half cells as current flows around the
cell. The salt bridge consists of an aqueous ionic solution held firmly within the bridge by a
porous gel that prevents the salt bridge solution from mixing with the half cell solutions – though
ions do flow between these solutions, as indicated in the figure.
Cell Diagrams
The above zinc-copper cell is represented by the following cell diagram:
Zn(s) | Zn(NO3)2(aq) || Cu(NO3)2(aq) | Cu(s)
The anode is shown on the left, and cathode on the right, as in Fig. 10.2. The vertical lines
denote phase boundaries. Here, they separate the solid and aqueous reactants and products in
the two half-reactions – oxidation on the left and reduction on the right. The double vertical lines
denote the salt bridge.
Sometimes the cell diagram is written with spectator ions left out. For the zinc-copper cell, the
cell diagram corresponding to the net ionic cell equation is written as
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) .
Here, the reactant in the oxidation half reaction is the anode, while the product of the reduction
half reaction is the cathode. Sometimes there are no metals among the reactants and products in
a half reaction. In such cases, an inert metal such as platinum serves as anode and/or cathode,
as required. Such an electrode is called inactive because it does not participate in the redox
reaction, except as a medium for transport of electrons.
In the zinc-hydrogen cell, under standard conditions, the hydrogen electrode acts as cathode. In
the associated reduction half reaction,
2 H+(aq) + 2 e‒ → H2(g) ,
neither reactant nor product is a metal. Consequently, an inactive electrode is required – usually,
platinum provides the inactive electrode in the hydrogen half cell. The cell diagram for the zinchydrogen cell is written as
Zn(s) | Zn2+(aq) || H+(aq) | H2(g) | Pt(s) .
The inactive platinum cathode is written last – the cathode is always rightmost in a cell diagram.
In the hydrogen-copper cell diagram, the platinum electrode appears first because the hydrogen
electrode acts as the anode – for the cell under standard conditions. Specifically,
Pt(s) | H2(g) | H+(aq) || Cu2+(aq) | Cu(s)
is the hydrogen-copper cell diagram.
Another cell diagram featuring inactive electrodes is provided by the chlorine-permanganate cell:
Pt(s) | Cl‒(aq) | Cl2(g) || MnO4‒(aq), H+(aq), Mn2+(aq) | Pt(s) .
- 208 -
This cell features two inactive electrodes and three active species – MnO4‒, H+ and Mn2+ – all in
the same (aqueous) phase in the cathode half cell. H+ appears here because it is a reactant in
the reduction half reaction for MnO4‒.
Example 10.1: Write the cell diagrams for electrochemical cells based on the net ionic
equations:
(a) Sn2+(aq) + Fe(s) → Sn(s) + Fe2+(aq)
(b) 2 Ag+(aq) + H2(g) → 2 Ag(s) + 2 H+(aq)
(c) Cl2(g) + 2 Fe2+(aq) → 2 Cl‒(aq) + 2 Fe3+(aq)
Approach: Identify the species that are oxidized and reduced (look for changing
oxidation numbers), and then split the reaction into oxidation and reduction half reactions.
The oxidation half reaction takes place at the anode, while reduction takes place at the
cathode. If the oxidation half reaction includes a metal, the metal is the anode.
Otherwise, an inactive anode is required – use Pt(s). The anode appears first, on the left,
in the cell diagram. The reactants (remaining reactants, if the anode is a reactant) and
products in the oxidation half reaction appear next – with distinct phases separated by a
vertical line. A double vertical line follows, representing the salt bridge. The reactants
and products in the reduction half reaction come next – again, with vertical lines
separating distinct phases. If the reduction half reaction includes a metal, then it serves
as the cathode and comes last. Otherwise, an inactive electrode – Pt(s) – appears last,
on the right.
(a) Iron atoms lose electrons, while tin ions gain electrons. The half reactions are
Fe(s) → Fe2+(aq) + 2 e‒
and
oxidation
Sn2+(aq) + 2 e‒ → Sn(s) .
reduction
Metallic iron provides the reactant in the oxidation half reaction, and serves as anode.
Similarly, metallic tin is the product of the reduction half reaction, and provides the
cathode. Both half cells have two phases – metallic solid and aqueous. The cell diagram
is
Fe(s) | Fe2+(aq) || Sn2+(aq) | Sn(s) .
(b) Hydrogen molecules lose electrons, while silver ions gain electrons. The half
reactions are
and
H2(g) → 2 H+(aq) + 2 e‒
oxidation
Ag+(aq) + e‒ → Ag(s) .
reduction
An inactive anode is required – use Pt(s). Silver metal provides the cathode. The cell
diagram is
Pt(s) | H2(g) | H+(aq) || Ag+(aq) | Ag(s) .
(c) Iron II ions lose electrons to form iron III, while chlorine molecules gain electrons and
form chloride ions. The half reactions are
Fe2+(aq) → Fe3+(aq) + e‒
and
Cl2(g) + 2e‒ → 2 Cl‒(aq) .
- 209 -
oxidation
reduction
An inactive anode and cathode are required – use Pt(s). The cell diagram is
Pt(s) | Fe2+(aq), Fe3+(aq) || Cl2(g) | Cl‒(aq) | Pt(s) .
10.3 Standard Cell Potential
A voltmeter (as shown in Fig. 10.2) can measure the electric potential difference between the
anode and cathode. This is called the cell potential, Ecell. The unit of electric potential (i.e.,
voltage) is the volt: 1 V = 1 J C‒1 where C is the SI unit of charge – the Coulomb. The cell
potential measured for an electrochemical cell under standard conditions, at 25°C, is called the
standard cell potential, Ecell°. For the zinc-copper cell under standard conditions the cell potential
is 1.10 V. We write
Ecell°(Zn| Zn2+ || Cu2+| Cu) = 1.10 V .
Current flows through an electrochemical cell for the same reason that Cu2+ will oxidize Zn when
it comes in contact with a zinc granule – there is a chemical driving force. The sum of the half cell
reactions, the cell reaction, is the same reaction that takes place when reactants are in direct
contact. The half reactions take place remotely in a cell because the wire allows electrons to
move freely. Electrons liberated at the anode easily flow toward the cathode. Electrons within
the wire move in the same direction, resulting in a net transfer of electrons to the cathode where
they are consumed by the reduction half-reaction, at its surface.
In Chapter 9, we saw that a chemical driving force corresponds to a negative Gibbs free energy
of reaction. With an electrochemical cell, the chemical driving force of a redox reaction is
measured as a positive cell potential – the measured voltage. The electric potential difference
between anode and cathode is analogous to the gravitational potential energy difference for water
at the top and bottom of a waterfall. The potential energy of water at the top of the waterfall is
released as kinetic energy, as the water falls. Similarly, electric potential energy is released as
electrons flow from anode to cathode. The amount of energy released is proportional to the
amount of charge transferred. The cell potential equals the released energy divided by the
charge transferred.
In the case of 1 mol of Cu2+ reacting with 1 mol of Zn, 2 mol of e‒ are transferred. The charge on
two moles of electrons is
2 mol × (‒F) = 2 mol × (‒96485 C mol‒1) = ‒192970 C ,
where F is the Faraday constant. It equals the magnitude of charge per mole of electrons. The
magnitude of the charge on one mole of electrons is
1 mol × F = 1 mol × 96485 C mol‒1 = 96485 C .
In general, if n is the number of electrons transferred in a cell reaction, then ‒n F is the charge
transferred, per mole of cell reaction. The drop in electric potential energy of this charge is given
by
‒n F Ecell = ∆G .
10.1
It is equal to the Gibbs free energy of reaction. Gibbs free energy released by a cell reaction
equals the electrical energy delivered by the cell. If the cell runs under standard conditions (1 mol
L‒1 concentration of aqueous species and 1 bar partial pressure of gases) Eq. 10.1 takes the
form,
∆G° = ‒n F Ecell° .
- 210 -
10.2
Here we see that a positive cell potential results when the Gibbs free energy of reaction is
negative – i.e., when the redox reaction is spontaneous. If the cell potential were negative, the
Gibbs free energy of reaction would be positive and the reaction would run in the reverse
direction. In such cases, we generally swap the anode and cathode so that the reaction is written
in the spontaneous direction, and the cell has a positive potential.
Gibbs free energy is a state function. As such, the Gibbs free energy of the sum of reactions is
the sum of the associated Gibbs free energy of reactions – see, e.g., Eq. 9.13. Equation 9.13 is
important because it allows the Gibbs free energy of reaction for a very large set of reactions to
be stored in terms of a table of the Gibbs free energies of formation of all reactants and products.
Equations 10.1 and 10.2 tell us that the cell potential is just the change in Gibbs free energy per
unit of charge transferred. Consequently, one expects a similarly efficient means of tabulating
cell potentials.
Measuring the standard cell potentials for the copper-silver and zinc-silver cells,
and
Cu(s) | Cu(NO3)2(aq) || AgNO3(aq) | Ag(s)
Zn(s) | Zn(NO3)2(aq) || AgNO3(aq) | Ag(s) ,
gives
and
Ecell°(Cu|Ag) = 0.46 V
Ecell°(Zn|Ag) = 1.56 V .
We see that, just as adding the cell reactions for the zinc-copper and copper-silver cells gives the
cell reaction for the zinc-silver cell, adding the zinc-copper and copper-silver cell potentials gives
the cell potential for the zinc-silver cell. Specifically,
Ecell°(Zn| Zn2+ || Ag+ |Ag) = Ecell°(Zn| Zn2+ || Cu2+| Cu) + Ecell°(Cu| Cu2+ || Ag+ |Ag)
= 1.10 + 0.46 V = 1.56 V .
Oxidation and Reduction Half-Cell Potential
Every cell reaction is the sum of an oxidation half reaction and a reduction half reaction. Also,
since an oxidation half reaction is just the reverse of a reduction half reaction, every cell reaction
can be written as the difference of two reduction half reactions. Consequently, a large number of
cell potentials are encapsulated in a table of reduction half-cell potentials. Because of the
additive property of cell potentials described above, the cell potential of any cell constructed from
reduction half reactions in the table is just the difference of the associated reduction half-cell
potentials. For example, we write
Ecell°(Zn| Zn2+ || Ag+ |Ag) = Eox°(Zn| Zn2+) + Ered°(Ag+ |Ag)
= E°[Zn(s) → Zn2+(aq) + 2 e‒] + E°[Ag+(aq) + e‒ → Ag(s)]
= E°[Ag+(aq) + e‒ → Ag(s)] ‒ E°[ Zn2+(aq) + 2 e‒ → Zn(s)]
= Ered°(Ag+ |Ag) ‒ Ered°(Zn2+| Zn) ,
where Ered° is standard reduction half-cell potential, and Eox° is standard oxidation half-cell
potential. The oxidation half-cell potential (or just oxidation potential) is just minus the
corresponding reduction half-cell potential (or just reduction potential). Note that the cell potential
is always the difference of two reduction potentials without regard for the number of electrons
- 211 -
transferred in the two reduction half-reactions. This is because cell potential is energy released
per amount of charge transferred. Thus, E°[Ag+(aq) + e‒ → Ag(s)] and E°[Zn(s) → Zn2+(aq) +
2 e‒] are combined directly, even though the two reduction half-reactions are written with different
numbers of electrons.
Tabulating reduction potentials provides a compact way of storing the cell potentials. Every halfcell can be paired with any other half-cell to produce a large number of distinct cell potentials.
The only trouble with such a table of standard reduction potentials is that it is impossible to
measure a half-cell potential. Only cell potentials, corresponding to full redox reactions, can be
measured. However, because of the additive property of cell potentials, if we define the reduction
potential to be zero for some reference reduction half-cell, then all other reduction potentials
become measurable cell potential values.
The standard reference electrode is chosen to be the hydrogen electrode – the reduction half cell
with half reaction,
2 H+(aq) + 2 e‒ → H2(g) ,
under standard conditions (1 mol L‒1 H+ and 1 bar partial pressure of H2). Figure 10.3 depicts this
standard hydrogen electrode.
e-
1 atm partial
pressure H2(g)
Cl-
Pt
H2
H+
1 mol L-1 HCl(aq)
2 H+ + 2 e-
H2(g)
Figure 10.3. The standard hydrogen electrode is the standard reference electrode. The
associated standard reduction potential is defined to be zero.
By setting (this is a convention)
Ered°(H+| H2) = E°[2 H+(aq) + 2 e‒ → H2(g)] = 0 V ,
we can determine other reduction potentials. For example,
Ered°(Cu2+| Cu) = Ered°(Cu2+| Cu) ‒ Ered°(H+| H2) = Ecell°(H2| H+ || Cu2+| Cu) = 0.34 V .
- 212 -
Subtracting Ered°(H+|H2) = 0 V from any reduction potential shows that the reduction potential is
just the cell potential for a cell with the hydrogen electrode serving as the anode. Here, we see
that Ered°(Cu2+| Cu) is just the cell potential for the hydrogen-copper cell, Ecell°( H2| H+ || Cu2+| Cu).
Once a few reduction potentials are known, it is not necessary to construct cells using the
hydrogen electrode as the anode to get other reduction potentials. For example, since the
reduction potential for copper is known, the reduction potential for silver is determined from
Ecell°(Cu| Cu2+ || Ag+| Ag) = 0.46 V .
Specifically, since
Ecell°(Cu| Cu2+ || Ag+| Ag) = Ered°(Ag+| Ag) ‒ Ered°(Cu2+| Cu) ,
Ered°(Ag+| Ag) = Ecell°(Cu| Cu2+ || Ag+| Ag) + Ered°(Cu2+| Cu) = 0.46 + 0.34 V = 0.80 V .
Similarly, since
Ecell°(Zn| Zn2+ || Cu2+| Cu) = 1.10 V
= Ered°(Cu2+| Cu) ‒ Ered°(Zn2+| Zn) = 0.34 V ‒ Ered°(Zn) ,
we must have
Ered°(Zn2+| Zn) = 0.34 ‒ 1.10 V = ‒0.76 V .
In this way, a table of standard reduction potentials is constructed. The table is then used to
determined standard cell potentials connecting any two electrodes in the table, whenever such a
potential is required. In general,
Ecell° = Ered°(cat) + Eox°(an)
Ecell° = Ered°(cat) ‒ Ered°(an) ,
10.3
where Ered°(cat) is the standard reduction potential for the reduction half reaction occurring at the
cathode, Eox°(an) is the standard oxidation potential for the oxidation half reaction occurring at the
anode, and Ered°(an) = ‒Eox°(an) is the standard reduction potential for the reverse of the latter
half reaction. Table 10.1 provides a list of 48 reduction potentials that can be used in Eq. 10.3 to
determine 1128 distinct cell potentials.1
1
1128 distinct cell potentials are determined from 48 reduction potentials. This number excludes
using the same electrode for anode and cathode – such cells have zero standard cell potential –
and it counts only the cell running in its spontaneous direction – i.e., with positive cell potential.
- 213 -
Table 10.1. Standard reduction potentials at 25°C
Reduction half reaction
‒
‒
F2(g) + 2e → 2 F (aq)
O3(g) + 2 H+(aq) + 2e‒ → O2(g) + H2O(l)
Co3+(aq) + e‒ → Co2+(aq)
H2O2(aq) + 2 H+(aq) + 2e‒ → 2 H2O(l)
Ce4+(aq) + e‒ → Ce3+(aq)
+
PbO2(s) + 4 H (aq) + SO42‒(aq) + 2e‒ → PbSO4(s) + 2 H2O(l)
MnO4‒(aq) + 8 H+(aq) + 5e‒ → Mn2+(aq) + 4 H2O(l)
Au3+(aq) + 3e‒ → Au(s)
HOCl(aq) + H+(aq) + 2e‒ → Cl‒(aq) + H2O(l)
ClO3‒(aq) + 6 H+(aq) + 6e‒ → Cl‒(aq) + 3 H2O(l)
Cl2(g) + 2e‒ → 2 Cl‒(aq)
2‒
Cr2O7 (aq) + 14 H+(aq) + 6e‒ → 2 Cr3+(aq) + 7 H2O(l)
O2(g) + 4 H+(aq) + 4e‒ → 2 H2O(l)
MnO2(s) + 4 H+(aq) + 2e‒ → Mn2+(aq) + 2 H2O(l)
Br2(l) + 2e‒ → 2 Br‒(aq)
‒
NO3 (aq) + 4 H+(aq) + 3e‒ → NO(g) + 2 H2O(l)
2 Hg2+(aq) + 2e‒ → Hg22+(aq)
‒
OCl (aq) + H2O(l) + 2e‒ → Cl‒(aq) + 2 OH‒(aq)
Hg22+(aq) + 2e‒ → 2 Hg(l)
Ag+(aq) + e‒ → Ag(s)
Fe3+(aq) + e‒ → Fe2+(aq)
O2(g) + 2 H+(aq) + 2e‒ → H2O2(aq)
‒
MnO4 (aq) + 2 H2O(l) + 3e‒ → MnO2(s) + 4 OH‒(aq)
I2(s) + 2e‒ → 2 I‒(aq)
O2(g) + 2 H2O(l) + 4e‒ → 4 OH‒(aq)
Cu2+(aq) + 2e‒ → Cu(s)
AgCl(s) + e‒ → Ag(s) + Cl‒(aq)
2‒
SO4 (aq) + 4 H+(aq) + 2e‒ → SO2(g) + 2 H2O(l)
Cu2+(aq) + e‒ → Cu+(aq)
2 MnO2(s) + H2O(l) + 2e‒ → Mn2O3(s) + 2 OH‒(aq)
Sn4+(aq) + 2e‒ → Sn2+(aq)
2 H+(aq) + 2e‒ → H2(g)
Pb2+(aq) + 2e‒ → Pb(s)
Sn2+(aq) + 2e‒ → Sn(s)
Ni2+(aq) + 2e‒ → Ni(s)
Co2+(aq) + 2e‒ → Co(s)
PbSO4(s) + 2e‒ → Pb(s) + SO42‒(aq)
Cd2+(aq) + 2e‒ → Cd(s)
Fe2+(aq) + 2e‒ → Fe(s)
Cr3+(aq) + 3e‒ → Cr(s)
Zn2+(aq) + 2e‒ → Zn(s)
2 H2O(l) + 2e‒ → H2(g) + OH‒(aq)
Mn2+(aq) + 2e‒ → Mn(s)
ZnO(s) + H2O(l) + 2e‒ → Zn(s) + 2 OH‒(aq)
Al3+(aq) + 3e‒ → Al(s)
Mg2+(aq) + 2e‒ → Mg(s)
Na+(aq) + e‒ → Na(s)
Ca2+(aq) + 2e‒ → Ca(s)
Sr2+(aq) + 2e‒ → Sr(s)
Ba2+(aq) + 2e‒ → Ba(s)
K+(aq) + e‒ → K(s)
Li+(aq) + e‒ → Li(s)
This data comes from HCP.
- 214 -
Ered° (V)
+2.87
+2.08
+1.92
+1.78
+1.72
+1.69
+1.51
+1.50
+1.48
+1.45
+1.36
+1.36
+1.23
+1.22
+1.07
+0.96
+0.92
+0.81
+0.80
+0.80
+0.77
+0.70
+0.60
+0.54
+0.40
+0.34
+0.22
+0.20
+0.15
+0.15
+0.15
0
‒0.13
‒0.14
‒0.26
‒0.28
‒0.36
‒0.40
‒0.45
‒0.74
‒0.76
‒0.83
‒1.19
‒1.26
‒1.66
‒2.37
‒2.71
‒2.87
‒2.90
‒2.91
‒2.93
‒3.04
Example 10.2: Use Table 10.1 to determine the standard cell potential at 25°C for cells with the
following cell diagrams.
(a)
(b)
(c)
(d)
Zn(s) | Zn2+(aq) || Ni2+(aq) | Ni(s)
Sn(s) | Sn2+(aq) || Fe3+(aq), Fe2+(aq) | Pt(s)
Pt(s) | I‒(aq) | I2(s) || NO3‒(aq), H+(aq) | NO(g) | Pt(s)
Li(s) | Li+(aq) || F2(g) | F‒(aq) | Pt(s)
Approach: Identify the half reactions represented in the cell – oxidation on the left, and
reduction on the right. The species appearing in the half reaction are written in the cell
diagram. The standard cell potential is the standard reduction potential of the reduction
reaction occurring at the cathode minus the standard reduction potential of the reverse of
the oxidation occurring at the anode.
(a) The half reactions are
and
Zn(s) → Zn2+(aq) + 2 e‒
oxidation
Ni2+(aq) + 2 e‒ → Ni(s) .
reduction
The associated standard reduction potentials are ‒0.25 V for the reduction of Ni2+(aq)
and ‒0.76 V for the reduction of Zn2+(aq) (the reverse of the oxidation half reaction).
Therefore, from Eq. 10.3, the cell potential is
Ecell° = Ered°(Ni2+| Ni) ‒ Ered°(Zn2+| Zn)
= ‒0.26 ‒ (‒0.76) V = +0.50 V .
(b) The half reactions are
and
Sn(s) → Sn2+(aq) + 2 e‒
oxidation
Fe3+(aq) + e‒ → Fe2+(aq) .
reduction
The cell potential is
Ecell° = Ered°(Fe3+| Fe2+) ‒ Ered°(Sn2+| Sn)
= +0.77 ‒ (‒0.14) V = +0.91 V .
(c) The half reactions are
and
2 I‒(aq) → I2(s) + 2 e‒
NO3‒(aq) + 4 H+(aq) + 3 e‒ → NO(g) + 2 H2O(l) .
The cell potential is
Ecell° = Ered°(NO3‒| NO) ‒ Ered°(I2| I‒)
= +0.96 ‒ 0.54 V = +0.42 V .
(d) The half reactions are
- 215 -
oxidation
reduction
and
Li(s) → Li+(aq) + e‒
oxidation
F2(g) + 2 e‒ → 2 F‒(aq) .
reduction
The cell potential is
Ecell° = Ered°(F2| F‒) ‒ Ered°(Li+| Li)
= +2.87 ‒ (‒3.04) V = +5.91 V .
In each reduction half reaction, there is a reactant (frequently the only reactant) that gets
reduced. These are the oxidizing agents. The strongest oxidizing agent, fluorine gas, appears at
the top of the table. Descending through the table, each successive row gives the next strongest
oxidizing agent in the list. In Table 10.1, these are ozone, aqueous cobalt III, hydrogen peroxide,
and so on.
The reducing agents appear as products in the reduction half reactions. The strongest reducing
agent, lithium metal, appears at the bottom of the table. Ascending through the table, each
successive row gives the next strongest reducing agent in the list. In Table 10.1, these are
potassium, barium, strontium, and so on.
Here, we see that an oxidizing agent in a higher row (i.e., associated with a higher reduction
potential) can oxidize a reducing agent from a lower row in the table. Such a combination
produces a cell with a spontaneous cell reaction and positive cell potential.
Example 10.3. Which of the following cell diagrams have a spontaneous cell reaction, under
standard conditions? For those that do not, re-write the cell diagram to have a spontaneous cell
reaction. For each cell, identify the anode and cathode, and the oxidizing and reducing agents,
when the cell runs in the spontaneous direction.
(a) Al(s) | Al3+(aq) || Mn2+(aq) | Mn(s)
(b) Pb(s) | Pb2+(aq) || Ni2+(aq) | Ni(s)
(c) Pt | H2(g) | H+(aq) || Sn4+(aq), Sn2+(aq) | Pt
(d) Au(s) | Au3+(aq) || Cl2(g) | Cl‒(aq) | Pt
Approach: Determine the cell potential, as in Example 10.2. If it is positive, the cell is
set up to run in the spontaneous direction. If the cell potential is negative, the cell runs in
the reverse direction. The correct cell diagram is then given by wapping the anode and
cathode half cells.
(a) The reduction potential of Mn2+(aq) is above that of Al3+(aq); ‒1.19 V versus ‒1.66 V,
respectively. Therefore, the cell potential is positive, and the cell diagram is written in the
correct spontaneous direction. Aluminum is the anode, and manganese is the cathode.
Aluminum is the reducing agent, and manganese II is the oxidizing agent.
(b) The reduction potential of Pb2+(aq) is above that of Ni2+(aq); ‒0.13 V versus ‒0.26 V,
respectively. Therefore, the cell potential is negative, and the cell diagram is written in
the non-spontaneous direction. The cell diagram is re-written as
Ni(s) | Ni2+(aq) || Pb2+(aq) | Pb(s) ,
to make the cell potential positive. Nickel is the anode, and lead is the cathode. Nickel is
the reducing agent, and lead II is the oxidizing agent.
- 216 -
(c) The reduction potential of Sn4+(aq) is above that of H+(aq); +0.15 V versus 0 V,
respectively. Therefore, the cell potential is positive, and the cell diagram is written in the
correct spontaneous direction. The anode and cathode are both inactive, and consist of
platinum. Hydrogen is the reducing agent, and tin IV is the oxidizing agent.
(d) The reduction potential of Au3+(aq) is above that of Cl2(g); +1.50 V versus +1.36 V,
respectively. Therefore, the cell potential is negative, and the cell diagram is written in
the non-spontaneous direction. The cell diagram is re-written as
Pt | Cl‒(aq) | Cl2(g) || Au3+(aq) | Au(s) ,
to make the cell potential positive. Platinum is the anode, and gold is the cathode.
Chloride is the reducing agent, and gold III is the oxidizing agent.
Combining Reduction Potentials
Atoms can exist in more than two oxidation states, and can therefore undergo a sequence of
reduction steps. For example, iron exists as Fe3+(aq) which can be reduced to Fe2+(aq), which in
turn can be reduced to Fe(s). Under some conditions, Fe3+(aq) can be reduced directly to Fe(s).
Knowing the reduction potential of Fe3+(aq) to Fe2+(aq), and the reduction potential of Fe2+(aq) to
Fe(s), the reduction potential for Fe3+(aq) to Fe(s) is determined by Hess’ law. However, the
combined standard reduction potential, Ered°(Fe3+| Fe), is not just the sum of Ered°(Fe3+| Fe2+) and
Ered°(Fe2+| Fe).
Hess’ law applies to changes in Gibbs free energy. To use it to determine Ered°(Fe3+| Fe), we
recall that a reduction potential is just the cell potential of the cell with the specified reduction half
cell as cathode and the hydrogen electrode as anode. Thus, Eq. 10.2 gives
∆G°(H2| H+|| Fe3+| Fe) = ‒n F Ered°(Fe3+| Fe)
= ‒3 F Ered°(Fe3+| Fe),
as three electrons are transferred (i.e., n = 3) in the cell reaction.
Since the cell reaction of H2| H+|| Fe3+| Fe equals the sum of the cell reactions of
H2| H+|| Fe3+| Fe2+ and H2| H+|| Fe2+| Fe, Hess’ law gives
∆G°(H2| H+|| Fe3+| Fe) = ∆G°(H2| H+|| Fe3+| Fe2+) + ∆G°(H2| H+|| Fe2+| Fe)
or
‒3 F Ered°(Fe3+| Fe) = ‒1 F Ered°(Fe3+| Fe2+) + (‒2 F Ered°(Fe2+| Fe)),
from which we get
Ered°(Fe3+| Fe) = Ered°(Fe3+| Fe2+)/3 + 2 Ered°(Fe2+| Fe)/3
= +0.77/3 + 2(‒0.45)/3 = ‒0.04 V.
10.4 Cell Potential and Equilibrium
We have seen – in Eqs. 10.1 and 10.2 – that the measured potential of a cell is just the Gibbs
free energy released by the cell reaction per charge transferred. Since the Gibbs free energy of
the cell reaction is related to the associated equilibrium constant – see Eq. 9.15 and 9.17 – cell
potential is clearly related to the equilibrium constant. Combining Eqs. 10.2 and 9.15, gives
- 217 -
‒R T ln K = ‒n F Ecell° ,
from which we get
o
Ecell

RT
ln K ,
nF
10.4
or
o
 nFEcell

K  exp 
 .
 RT 
10.5
A positive cell potential corresponds to a negative Gibbs free energy of reaction and an
equilibrium constant greater than one – i.e., a favorable forward reaction.
Non-standard Conditions and the Nernst Equation
So far, we have considered cell potential only under standard conditions. Combining Eqs. 10.1
and 9.14 shows how cell potential varies under non-standard conditions. Specifically, we get
or
‒n F Ecell = ‒n F Ecell° + R T ln Q ,
o
Ecell  Ecell

RT
ln Q .
nF
10.6
This is called the Nernst equation.
It is convenient to write the Nernst equation specifically for T = 298.15 K (25°C), and in terms of
the base 10 logarithm:
o
Ecell  Ecell

0.0257 V
ln Q
n
and
o
Ecell  Ecell

0.0592 V
log10 Q .
n
The Nernst equation shows that non-standard conditions generally produce small, but nonnegligible, shifts in the cell potential. The shift is to larger cell potential if the reaction quotient is
less than one – in this case, ln Q < 0. A reaction quotient less than one corresponds to reactant
activities greater than one and/or product activities less than one, while a larger than standard cell
potential corresponds to a cell reaction more favorable than under standard conditions. The
Nernst equation is therefore in accord with le Châtelier’s principle – more reactant and/or less
product makes the forward reaction more favorable. Similarly, a reaction quotient greater than
one corresponds to a cell potential below the standard value. Reducing the activity of reactants
and/or increasing the activity of products makes the forward reaction less favorable – again, in
accord with le Châtelier’s principle.
Figure 10.4 shows the cell potential for a zinc-copper cell as a function of Q. Significant
deviations from the standard cell potential require either a very small or very large reaction
quotient.
- 218 -
Figure 10.4. The cell potential of the zinc-copper cell versus reaction quotient, Q =
[Zn2+]/[Cu2+].
Example 10.4. Determine the cell potentials, at 25°C, for the following cells using Table 10.1 and
the Nernst equation.
(a) Zn(s) | 0.100 mol L‒1 ZnCl2(aq) || 2.00 mol L‒1 CdCl2(aq) | Cd(s)
(b) Zn(s) | 2.00 mol L‒1 ZnCl2(aq) || 0.100 mol L‒1 CdCl2(aq) | Ni(s)
(c)
Pt(s) | 2.00 bar NO(g) | 0.100 mol L‒1 HNO3(aq) || 0.100 mol L‒1 KCl(aq) | 2.00 bar Cl2(g) | Pt(s)
(d)
Pt(s) | 0.100 bar NO(g) | 2.00 mol L‒1 HNO3(aq) || 2.00 mol L‒1 KCl(aq) | 0.100 bar Cl2(g) | Pt(s)
Approach: Identify the reduction and oxidation half reactions in Table 10.1 to determine
the standard cell potential as a difference of standard reduction potentials. Use the
balanced cell reaction to find the expression for the reaction quotient, and the number of
electrons transferred. Use Eq. 10.6 to determine the cell potential under the given nonstandard conditions.
(a) The half reactions are
and
Zn(s) → Zn2+(aq) + 2 e‒
oxidation
Cd2+(aq) + 2 e‒ → Cd(s) .
reduction
Under standard conditions, the cell potential is
Ecell° = Ered°(Cd2+| Cd) ‒ Ered°(Zn2+| Zn)
- 219 -
= ‒0.40 ‒ (‒0.76) V = +0.36 V .
Under the given conditions, the cell potential shifts to
o
Ecell  Ecell

0.0257 V
ln Q .
n
We need the reaction quotient, Q. The balanced cell reaction is (with n = 2 electrons
transferred)
Zn(s) + Cd2+(aq) → Zn2+(aq) + Cd(s) .
Therefore,
Q
[Zn 2+ ] 0.100

 0.0500
[Cd 2+ ] 2.00
and
Ecell  0.36 V 
0.0257 V
ln  0.0500 
2
 0.40 V .
Depleted product (Zn2+(aq)) and excess reactant (Cd2+(aq)), compared to standard
conditions, yield a larger than standard cell potential.
(b) This is the same cell as in part (a), except the non-standard conditions are different.
Here,
[Zn 2+ ] 2.00
Q

 20.0
[Cd 2+ ] 0.100
and
Ecell  0.36 V 
0.0257 V
ln  20.0 
2
 0.32 V .
Excess product (Zn2+(aq)) and depleted reactant (Cd2+(aq)), compared to standard
conditions, yield a lower than standard cell potential.
(c) The half reactions are
NO(g) + 2 H2O(l) → NO3‒(aq) + 4 H+(aq) + 3 e‒
oxidation
and
Cl2(g) + 2 e‒ → 2 Cl‒(aq) .
Under standard conditions, the cell potential is
Ecell° = Ered°(Cl2| Cl‒) ‒ Ered°( NO3‒| NO)
= +1.36 ‒ (+0.96) V = +0.40 V .
Under the given conditions, the cell potential shift to
- 220 -
reduction
o
Ecell  Ecell

0.0257 V
ln Q .
n
The balanced cell reaction is (with n = 6 electrons transferred)
3 Cl2(g) + 2 NO(g) + 4 H2O(l) → 6 Cl‒(aq) + 2 NO3‒(aq) + 8 H+(aq) .
Therefore,
Q

[Cl ]6 [NO3 ]2 [H + ]8
2
PCl3 2 PNO
0.1006  0.1002  0.1008
 3.125 1018
2.003  2.002
and
Ecell  0.40 V 

0.0257 V
ln 3.125 1018
6

 0.57 V .
Depleted products (Cl‒(aq), NO3‒(aq) and H+(aq)) and excess reactants (Cl2(g) and
NO(g)), compared to standard conditions, yield a larger than standard cell potential.
(d) This is the same cell as in part (c), except the non-standard conditions are different.
Here,
Q

[Cl ]6 [NO3 ]2 [H + ]8
2
PCl3 2 PNO
2.006  2.002  2.008
 6.554 109
0.1003  0.1002
and
Ecell  0.40 V 

0.0257 V
ln 6.554 109
6

 0.30 V .
Excess products (Cl‒(aq), NO3‒(aq) and H+(aq)) and depleted reactants (Cl2(g) and
NO(g)), compared to standard conditions, yield a lower than standard cell potential.
- 221 -
Measuring Concentrations with Electrochemical Cells
Because of the sensitivity of cell potential to the concentration of active aqueous species, and the
partial pressure of active gases, cell potential measurements can be used to determine unknown
concentrations in sample solutions, and unknown partial pressures of gases. For example, the
solubility of silver bromide can be determined by setting up a silver or bromine electrode and
using a saturated solution of silver bromide as the associated electrolyte solution – the source of
Ag+ or Br‒, respectively.
Example 10.5. Determine the molar solubility of silver bromide, at 25°C, given the following
electrochemical cell potential, and data from Table 10.1.
Ag(s) | AgBr(aq, sat) || Br2(l) | 0.100 mol L‒1 NaBr(aq) | Pt(s)
Ecell = 0.69 V at 25°C .
AgBr(aq, sat) is a saturated solution of silver bromide – it is prepared by dissolving silver bromide
in water until no more silver bromide dissolves – i.e., undissolved solid silver bromide remains in
anode half cell at equilibrium.
Approach: Balance the cell equation to determine the expression for Q and the value of
n. The expression for Q has one unknown factor – the concentration of Ag+ in the
saturated solution of AgBr. Determine the standard cell potential using data from Table
10.1. Use the Nernst equation to determine the reaction quotient from the measured
potential, and then solve for the unknown silver I concentration in the saturated solution –
this is the molar solubility of silver bromide.
Note that there is bromide in both the cathode and anode half cells. However, only the
bromide in the cathode half cell is active.
The half reactions are
Ag(s) → Ag+(aq) + e‒
oxidation
Br2(l) + 2 e‒ → 2 Br‒(aq) .
reduction
and
Under standard conditions, the cell potential is
Ecell° = Ered°(Br2| Br‒) ‒ Ered°(Ag+| Ag)
= 1.07 ‒ (0.80) V = 0.27 V .
Under the given conditions, the cell potential shifts to
o
Ecell  Ecell

0.0257 V
ln Q
n
or
0.69 V  0.27 V 
from which we get
0.0257 V
ln Q ,
2
 2   0.42  
15
Q  exp 
 =6.39 10 .
 0.0257 
The balanced cell reaction is (with n = 2 electrons transferred)
- 222 -
Br2(l) + 2 Ag(s) → 2 Br‒(aq) + 2 Ag+(aq) ,
where Br‒(aq) refers to bromide concentration in the cathode half cell – not the spectator
bromide concentration in the anode half cell. Therefore,
Q  [Br  ]2 [Ag  ]2
= 0.100 2 [Ag  ]2  6.39 1015 ,
or
[Ag  ] 
6.39 1015
=8.0 107 mol L1 .
2
0.100
The molar solubility of silver bromide is the concentration of Ag+ or Br‒ in the saturated
silver bromide solution in the anode half cell – it is 8.0×10‒7 mol L‒1.
10.5 Concentration Cells
It is possible to use the same type of electrode for both anode and cathode in the same cell. For
example, we could make the cell with the cell diagram,
Cu(s) | Cu(NO3)2(aq) || Cu(NO3)2(aq) | Cu(s) .
The associated net the ionic equation is
Cu2+(aq) → Cu2+(aq) ,
and the standard cell potential is
Ecell° = Ered°(Cu2+| Cu) ‒ Ered°(Cu2+| Cu) = 0 V .
However, because the reactant and product copper ions in the cell reaction are in different half
cells, it possible to get a non-zero cell potential by using different concentrations of Cu2+ on the
two sides of reaction. In this case, conditions are necessarily non-standard, and the Nernst
equation shows that the cell potential is not zero. In particular, if the reactant Cu2+ concentration
(the concentration at the cathode) is larger than the product Cu2+ concentration (the concentration
at the anode), then the cell will have positive cell potential. The cell reaction is written
Cu2+(aq, cathode) → Cu2+(aq, anode) ,
or
Cu2+(aq, conc) → Cu2+(aq, dil) ,
since the concentrated copper II solution is at the cathode and the dilute solution is at the anode.
The above cell reaction corresponds to the natural direction of flow of copper II ions – from
concentrated portions of a solution, toward dilute portions. In the electrochemical cell, copper II
ions are removed from a concentrated solution, while other copper II ions are added to a dilute
solution. The net effect is the same for ions flowing between concentrated and dilute solutions.
At 25°C, the cell potential is
- 223 -
0.0257 V
ln Q
n
0.0257 V  [Cu 2 , anode] 
ln 


2
2
 [Cu , cathode] 
o
Ecell  Ecell


10.7
0.0257 V  [Cu 2 , dil] 
ln 
0 .
2
2
 [Cu , conc] 
The potential is positive because the reaction quotient is less than one – the logarithm of the
reaction quotient is negative, canceling the minus sign in the above expression. If the
concentrated and dilute solutions were switched, the cell potential would be negative. The cell
reaction always proceeds from concentrated to dilute. The reverse process never happens – it is
non-spontaneous.
Example 10.6. Determine the cell potential for the copper II concentration cell described above
given [Cu2+, cathode] = 1.00 mol L‒1 and (a) [Cu2+, anode] = 0.100 mol L‒1, (b) [Cu2+, anode]
= 0.0100 mol L‒1, and (c) [Cu2+, anode] = 0.00100 mol L‒1.
Approach: Use the Nernst equation, as given above (or, as is convenient for this
problem, in terms of log10), to evaluate the cell potential.
(a)
Ecell  
 [Cu 2 , anode] 
0.0592 V
log10 

2
2
 [Cu , cathode] 
 [Cu 2 , dil] 
 0.0296 V  log10 

2
 [Cu , conc] 
 1.00 
= 0.0296 V  log10 
 = 0.0296 V.
 0.100 
(b)
 1.00 
Ecell  0.0296 V  log10 
 = 0.0592 V.
 0.0100 
(c)
 1.00 
Ecell  0.0296 V  log10 
 = 0.0888 V.
 0.00100 
The pH Meter
A pH meter measures the concentration of hydrogen ions in sample solutions. It works by using
a glass membrane electrode consisting of a glass tube containing an HCl solution, with accurately
determined concentration (e.g., 1.000 mol L‒1), and a silver electrode coated with a thin layer of
solid silver chloride. The solution also contains a tiny silver ion concentration so that the solution
is in equilibrium with the solid silver chloride. At the end of the tube is a bulb of very thin glass in
a gel state – a hydrated state of glass – with hydrogen ions within the glass. Hydrogen ions flow
across the glass, carrying current flow to or from the sample solution surrounding the bulb. A
different hydrogen ion concentration in the sample solution results in a potential difference across
- 224 -
the glass membrane that corresponds to a hydrogen ion concentration cell. The rest of the glass
membrane electrode is a silver-silver chloride half cell under standard (or accurately determined
non-standard) conditions. This glass membrane electrode is typically surrounded by another
glass tube containing the counter electrode – usually another silver-silver chloride half cell.
Figure 10.5 depicts a pH meter constructed from two silver-silver chloride half cells within
concentric glass tubes. The inner half cell is the glass membrane electrode described above.
The outer half cell is connected to the sample solution by a porous sintered glass membrane that
provides a salt bridge.
4
5 6 78
pH
9
voltmeter - with readout
calibrated to give pH of
the sample solution
Ag(s)
AgCl(s)
Ag(s)
porous sintered
glass membrane
- a salt bridge
AgCl(s)
Sample
Solution
unknown H+
concnetration
KCl(aq, sat)
thin glass
membrane in
gel state with
mobile H+ ions
1.000 mol L-1
HCl(aq)
Figure 10.5. A pH meter constructed from two silver-silver chloride electrodes with
accurately known chloride and hydrogen ion concentrations. The outer half cell is
connected to the sample solution through a sintered glass membrane that acts as a salt
bridge. The inner half cell is connected to the sample solution via a thin glass membrane
permeable to hydrogen ions. The change in hydrogen ion concentration across the glass
membrane constitutes a hydrogen ion concentration cell in series with the net chloride
concentration cell provided by the two silver-silver chloride half cells.
The cell diagram for the pH meter depicted in Fig. 10.5 is
Ag(s) | AgCl(s) | Cl‒(aq, saturated KCl solution) || Cl‒(aq, 1.000 mol L‒1) | AgCl(s) | Ag(s) ,
except that imbedded within this cell (after the salt bridge) is a hydrogen ion concentration cell,
H+(aq, sample solution) || H+(aq, 1.000 mol L‒1) .
The hydrogen ion concentration cell is in series with the silver-silver chloride cell, since current
must pass through it (i.e., across the glass membrane) in order to complete the circuit. Thus, the
- 225 -
potential of the hydrogen ion concentration adds to the potential of the silver-silver chloride cell.
The potential of these two cells in series is measured by a voltmeter connecting the two silver
electrodes. The output of the voltmeter is calibrated so that the voltage measured appears as the
pH of the sample solution.
To see how this works, we express the overall voltage using the Nernst equation applied to the
two cells in series. To do this, we need the cell reactions.
For the silver-silver chloride cell, the half reactions are
Ag(s) + Cl‒(aq, saturated KCl solution) → AgCl(aq) + e‒
oxidation
AgCl(aq) + e‒ → Ag(s) + Cl‒(aq, 1.000 mol L‒1) .
reduction
and
Therefore, n = 1 for this cell, and the cell reaction is
Cl‒(aq, saturated KCl solution) → Cl‒(aq, 1.000 mol L‒1) .
We see that the silver-silver chloride cell is itself a chloride concentration cell. However, the
chloride concentrations are fixed. This cell provides a reference potential that ensures a good
electrical connection around the circuit by driving a small current.
For the hydrogen concentration cell, the half reactions are
H2(g) → 2 H+(aq, sample solution) + 2 e‒
oxidation
2 H+(aq, 1.000 mol L‒1) + 2 e‒ → H2(g) .
reduction
and
Therefore, n = 2 for this cell, and the cell reaction is
2 H+(aq, 1.000 mol L‒1) → 2 H+(aq, sample solution) .
The factor of two on both sides of the equation can be canceled to give cell reaction,
H+(aq, 1.000 mol L‒1) → H+(aq, sample solution) ,
for which n = 1.
Applying the Nernst equation to both cells and adding the result gives
Etotal


 [H  , sample solution] 
0.0592 V
[Cl , 1.000 mol L1 ]

log10  
  0.0592 V  log10 


1
1
 [Cl , saturated KCl solution] 
 [H , 1.000 mol L ] 

 Eref  0.0592 V  log10 [H  , sample solution]

 Eref + 0.0592 V  pH .
Here, Eref is a fixed reference potential that is accurately known. It corresponds to the measured
cell potential when pH = 0. In general, the measured potential of the two cells varies linearly with
pH. Specifically, each unit of pH shifts the potential by 0.0592 V.
- 226 -
One might object to the above analysis, by pointing out that there are chloride concentration cells
at the sintered glass membrane (the salt bridge connecting the outer half cell from the sample
solution), and at the gel glass membrane (the salt bridge connecting the sample solution to the
the inner half cell). The sample solution will generally have a different chloride concentration than
either half cell. However, accounting for this observation amounts to splitting the chloride
concentration cell described above into two chloride concentration cells in series – one for each
of the glass membranes. The result is the following two cell reactions in series:
Cl‒(aq, saturated KCl solution) → Cl‒(aq, sample solution) .
and
Cl‒(aq, sample solution) → Cl‒(aq, 1.000 mol L‒1) .
Adding the resulting cell potentials is equivalent to adding the cell reactions. However, chloride in
the sample solution cancels when the two cell reactions are added. Similarly, the dependence of
the two cell potentials on the sample solution chloride concentration cancels when the potentials
are added. Therefore, the potential measured by a pH meter depends only on the hydrogen ion
concentration of the sample solution, and not on the chloride concentration.
The same principles used in the construction of the pH meter are used to make electrochemical
probes sensitive to the concentration of other ions. Such devices utilize ion selective electrodes
that make a measured potential depend on the concentration of a specific ion in a sample
solution. The output of such a device is the logarithm of the concentration of the ion of interest.
The pH meter utilizes the glass membrane electrode, a hydrogen ion specific electrode.
Concentration Cells in Living Systems
Concentration cells occur in living organisms. For example, every cell in our bodies (and those of
most familiar lifeforms) utilize mitochondria to perform the citric acid cycle (also known as Krebs
cycle) of cellular respiration. The energy rich products of this cycle, NADH and FADH2, are
oxidized within mitochondria, with the energy released used to pump hydrogen ions across a
membrane. This process is called the electron transport chain. The result is a hydrogen ion
concentration cell. The energy released by the discharge of this concentration cell is used to
produce ATP for distribution throughout the cell.
Concentration cells are also central to the function of nerve and muscle cells. These cells have
sodium, potassium, chloride and calcium concentration difference across their outer membranes.
This produces a voltage across the membrane. Opening channels in one place on the
membrane allow ions to flow across the membrane at that location. This affects the voltage there
which, in turn causes channels to open in neighboring places on the cell membrane. In this way,
signals are transmitted along nerve and muscle cells. In the brain, our thought processes
correspond to coordinated sequences of signals between brain cells. In muscles, these signals –
which begin in the brain – cause the muscles to contract.
10.6 Corrosion
The oxidation of iron by atmospheric oxygen to form iron oxide (rust) is a spontaneous process.
However, it happens very slowly under dry conditions. This is important because iron (steel) is
used everywhere – e.g., in buildings, highways, cars, appliances and other machines. Since
conditions are not always dry, rusting of steel has a significant economic and environmental
impact. Wet conditions – especially acidic and/or salty wet conditions – greatly accelerate the
rusting of steel. For this reason, steel must be protected using paints (e.g., in the case of cars),
or cathodic protection (e.g., on ships).
Wet conditions facilitate oxidation of iron because water droplets on the surface of iron support
electrochemical cells driven by the oxidation process. Figure 10.6 depicts a water droplet on the
surface of iron, and the electrochemical cell that results due to oxygen in the atmosphere, and
electrolytes in the water droplet that carry current and complete the circuit.
- 227 -
air
Fe2O3.n H2O(s)
a rust deposit
Fe2+(aq) + 3/2 O2(g) + n H2O(l)
Fe2O3.n H2O(s)
water droplet
O2(g) + 4 H+(aq) + 4 eFe(s)
2+
Fe (aq) + 2 e
-
cathode
anode
Fe(s)
2 H2O(l)
eelectrons liberated at the anode
flow through the iron to the
cathode where they are consumed
in the reduction half reaction
Figure 10.6. The corrosion of iron under a water droplet on the iron surface. Oxygen in
the air is reduced at the cathode – anywhere at the edge of the droplet – shown here on
the right with the associated reduction half reaction. Iron is oxidized under the droplet,
with iron II irons entering solution and the electrons liberated flowing through the iron to
the cathode where they are consumed by oxygen to form water. The oxidation half
reaction is shown above the anode, where there is pitting of the iron surface. Iron II ions
are further oxidized by oxygen at the surface of the droplet – shown in the top reaction.
The iron III oxide produced is insoluble. It precipitates as a hydrate with n waters of
hydration (n here is not related to the number of electrons transferred) on the surface of
the drop and accumulates at its edge – shown here on the left.
The half reactions for the corrosion of iron are shown in the figure. Because H+ is a reactant in
the reduction half reaction, we see that corrosion is greatly enhanced if the droplet is acidic – e.g.,
in the case of acid rain. Also, since completion of the circuit requires dissolved electrolytes in the
droplet, corrosion is enhanced by high salt concentration. This is why salting the roads in the
winter leads to enhanced corrosion of cars and the steel bars within the reinforced concrete of
bridges and overpasses.
Iron can be protected from corrosion by paints which prevent water and air from contacting the
iron surface. Corrosion is also prevented by cathodic protection. This is achieved by coating the
iron with a more active metal – i.e., a metal with a lower reduction potential. For example, iron is
typically coated with zinc for this purpose. This coating process is called galvanization. The
result is galvanized steel.
Cathodic protection works because the more active metal is always corroded first. Thus, the iron
of galvanized steel remains intact while the zinc coating gradually disappears due to corrosion.
Until the zinc coating is completely stripped off, the iron does not corrode. The zinc does not
need to completely cover the iron. As long as there is zinc in good electrical contact with the iron
and the water, the iron will not corrode. For example, ships are made of steel and they sit in
water – in the ocean it is salt water with high conductivity. Ships are protected with a large block
of zinc attached to the hull and submerged in the water to make a complete circuit.
If iron were coated with a less active metal (i.e., it has a higher reduction potential), such as
copper, then it is not protected. In fact, the presence of copper on iron makes the iron more
susceptible to corrosion. This is because the copper, or any other less active metal than iron,
provides a better cathode material for the reduction of oxygen. Figure 10.7 depicts cathodic
protection of iron with zinc and the enhanced corrosion of iron when it is in contact with copper.
- 228 -
Cathodic protection of iron with zinc
air
Zn2+(aq) + 2 e-
Zn(s)
O2(g) + 4 H+(aq) + 4 e-
water
anode
Zn(s)
2 H2O(l)
cathode
e-
Fe(s)
Enhanced corrosion of iron with copper
air
O2(g) + 4 H+(aq) + 4 e2 H2O(l)
water
Fe(s)
anode
cathode
2+
Fe (aq) + 2 e
Cu(s)
-
e-
Fe(s)
Figure 10.7. Cathodic protection of iron with more active metal, zinc, and enhanced
corrosion of iron with a less active metal, copper. Zinc attached to iron will corrode
before iron corrodes. Copper attached to iron enhances the rate of iron corrosion.
Example 10.7. Consider the following list of metals: Al, Mg, Sr, Cr, Zn, Cd, Ni, Fe, Cu, Pb and
Ag. Which of these metals are suitable to use as cathodic protection for (a) iron, (b) copper, and
(c) zinc?
Approach: Find the standard reduction potentials to produce these metals from the
associated metal cation. Create an activity series – i.e., order the reduction potentials in
descending order (as in Table 10.1). Only metals below the metal to be protected can be
used as a cathodic protection agent.
The reduction potentials have the following order:
Ag > Cu > Pb > Ni > Cd > Fe > Cr > Zn > Al > Mg > Sr
i.e.,
Ered°(Ag+| Ag) > Ered°(Cu2+| Cu) > Ered°(Pb2+| Pb) > Ered°(Ni2+| Ni) > Ered°(Cd2+| Cd) >
Ered°(Fe2+| Fe) > Ered°(Cr3+| Cr) > Ered°(Zn2+| Zn) > Ered°(Al3+| Al) > Ered°(Mg2+| Mg) >
Ered°(Sr2+| Sr)
(a) Cr, Zn, Al, Mg and Sr have lower reduction potentials than Fe, and can therefore be
used as cathodic protection agents. Since strontium reacts with water, it would never be
used as a cathodic protection agent in any normal application.
- 229 -
(b) Pb, Ni, Cd, Fe, Cr, Zn, Al, Mg and Sr have lower reduction potentials than Cu, and
can therefore be used as cathodic protection agents (with the above caveat regarding
strontium).
(c) Al, Mg and Sr have lower reduction potentials than Zn, and can therefore be used as
cathodic protection agents (with the above caveat regarding strontium).
10.7 Batteries
Electrochemical cells – known as batteries in everyday life – provide convenient energy sources
for portable electric devices. Batteries are divided into two categories, primary and secondary
cells.
Secondary cells are rechargeable, whereas primary cells are not. Recharging of a secondary cell
occurs when the cell is connected to an electric power source that drives current in the reverse of
its natural direction. This process is called electrolysis. It requires applying a potential exceeding
the cell potential, in the reverse direction, so that the reverse cell reaction coupled to the external
power supply is a net spontaneous process. Driving the cell reaction in reverse replenishes
reactants at the expense of products, so that the cell can again be used to deliver portable power.
Primary cells are not rechargeable because attempts to recharge them result in side reactions,
and the reactants are incompletely replenished. In addition, irreversible changes occur within the
cell, including the production of gases such as ammonia (in an NH4Cl dry cell) or hydrogen.
Hydrogen gas results from the reduction of hydrogen ions under the applied voltage. The
production of gases makes these batteries dangerous if one attempts to recharge them – gas
pressure buildup within the battery can lead to battery explosion. In any case, primary batteries
do not perform well after recharging. Their primary advantages are low cost and resistance to
discharge when not in use. Secondary batteries are more expensive, and prone to such
discharge. However, both of these problems are well mitigated by the ability to run the batteries
through many cycles – one cycle corresponds to discharge through use, followed by recharge.
The dry cell (also known as the Leclanché cell) and the alkaline cell are two common primary
cells based on zinc (the reducing agent) and manganese dioxide (the oxidizing agent). The dry
cell is depicted in Fig. 10.8.
C(graphite) cathode
powdered MnO2(s) mixed with
C(graphite) in a water paste
NH4Cl(s) powder in a water paste
Zn(s) anode
Figure 10.8. Cross-section of a dry cell. Sometimes ZnCl2(s) is used instead of
NH4Cl(s).
The dry cell half reactions are
Zn(s) + 2 NH4Cl(aq) → Zn(NH3)2Cl(aq) + 2 H+(aq) + 2 e‒
- 230 -
oxidation
and
MnO2(s) + H+(aq) + e‒ → Mn(OH)O(s) .
reduction
Since neither MnO2(s) nor Mn(OH)O(s) conducts electricity, graphite is used as an inactive
cathode. There is cathode rod at the center of the dry cell which is surrounded with powdered
manganese dioxide mixed with graphite in a water paste. When the cell operates, the MnO2(s)
gradually converts to Mn(OH)O(s).
The anode is a zinc canister which contains the cell. It is separated from the cathode assembly
by a paste made with water and powdered ammonium chloride. The water in the cell is therefore
a saturated solution of ammonium chloride which is a good conductor of electricity. There is just
enough water in the cell to ensure good conductivity throughout the cell. Since the water is
present only in pastes made from manganese dioxide or ammonium chloride, there is no danger
of leakage. This makes the cell more reliable and is the origin of the name, dry cell. Of course,
the cell must still be sealed to prevent evaporation of the water, and it requires internal spacers to
allow for volume changes due to changes in temperature.
A side reaction that can occur at the cathode (or the anode if one is attempting to recharge a dry
cell) is the reduction of hydrogen ions to form hydrogen gas which reduces the power output of
the cell (by increasing internal resistance) and can also be dangerous. One way to address this
problem is to reduce the hydrogen ion concentration – since H+(aq) is the reactant in the side
reaction. This leads to the alkaline cell which replaces the acidic ammonium chloride electrolyte
paste with a basic potassium hydroxide paste. The half reactions for the alkaline cell are
Zn(s) + 2 OH‒(aq) → ZnO(s) + H2O(l) + 2 e‒
oxidation
2 MnO2(s) + H2O(l) + 2 e‒ → Mn2O3(s) + 2 OH‒(aq) .
reduction
and
The net cell reaction for the alkaline cell is
2 MnO2(s) + Zn(s)
→ Mn2O3(s) + ZnO(s) .
Since there are no aqueous or gaseous reactants or products, the cell potential stays at a
constant value – the standard potential – until the limiting reactant is depleted. This is convenient
property for batteries.
The important secondary cells are nickel-cadmium, lithium and lead-acid batteries. Like the
alkaline cell, the nickel-cadmium and lithium cells have net cell reactions with solid reactants and
products. The lead acid cell consumes H+(aq) and SO42‒(aq), and as such its potential is
sensitive to the extent of reaction. However, the lead-acid battery in a car is continually
recharged as the car runs. This keeps the cell potential uniform when battery power is needed.
Fuel cells are cells with reactants continuously fed into the cell, while products are removed. The
hydrogen fuel cell uses hydrogen gas as the reducing agent at the anode and oxygen gas (from
the atmosphere) at the cathode. Because the product is water, using a hydrogen fuel cell – to
power a car, for example – is environmentally friendly (unless environmentally unfriendly power
sources are used to provide the electricity needed to make the hydrogen from water).
Problems:
10.1 Write the cell diagrams for electrochemical cells based on the following net ionic equations.
For parts (a) and (b), draw a picture of the cell similar to that in Fig. 10.2. Indicate the
- 231 -
direction of the flow of ions within the half cells (i.e., towards electrode or salt bridge), and
the flow of electrons through the wires connecting the anode and cathode.
(a) Pb2+(aq) + Cd(s) → Pb(s) + Cd2+(aq)
(b) Ni(s) + 2 H+(aq) → Ni2+(aq) + H2(g)
(c) PbO2(s) + 4 H+(aq) + SO42‒(aq) + Sn2+(aq) → PbSO4(s) + 2 H2O(l) + Sn4+(aq)
10.2 Use Table 10.1 to determine the standard cell potential at 25°C for cells with the following
cell diagrams:
(a)
(b)
(c)
(d)
Cd(s) | Cd2+(aq) || Ni2+(aq) | Ni(s)
Sn(s) | Sn2+(aq) || Cl2(g) | Cl‒(aq) | Pt(s)
Zn(s) | ZnO(s) | OH‒(aq) || OH‒(aq) | MnO2(s) | Mn2O3(s) | C(graphite)
Pb(s) | PbSO4(s) | SO42‒(aq) || H+(aq), SO42‒(aq) | PbO2(s) | PbSO4(s)
Part (d) is the lead acid cell.
10.3 Consider the following table of anode half cells (on the left) and cathode half cells (on the
right). How many cells can be constructed from these half cells, as written (i.e., do not
consider the reverse half reactions)? For each of the possible cells, write the balanced cell
reaction, determine the standard cell potential at 25°C, and draw the cell diagram.
Anode half cells
Cathode half cells
Cr(s) | Cr3+(aq)
C(graphite) | Mn2O3(s) | MnO2(s) | OH‒(aq)
Pt(s) | Fe2+(aq), Fe3+(aq)
OH‒(aq), OCl‒(aq), Cl‒(aq) | Pt(s)
H+(aq), H2(g) | Pt(s)
Pb2+(aq) | Pb(s)
10.4 Determine the cell potential, at 25°C, for
Pb(s) | PbSO4(s) | SO42‒(aq) || H+(aq), SO42‒(aq) | PbO2(s) | PbSO4(s)
under the following conditions:
(a) [H+] = 1.000 mol L‒1, [SO42‒] = 1.000 mol L‒1.
(b) [H+] = 0.100 mol L‒1, [SO42‒] = 0.550 mol L‒1.
(c) [H+] = 0.010 mol L‒1, [SO42‒] = 0.505 mol L‒1.
Part (a) corresponds to a lead acid cell using 1.000 mol L‒1 H2SO4(aq). Note that SO42‒
actually exists as HSO4‒ under these conditions. Part (b) corresponds to the cell after net
cell reaction to 90% completion. Part (c) corresponds to 99% completion of the cell
reaction.
10.5 Determine the molar solubility of lead II sulfate from the reduction potentials of Pb2+(aq) and
PbSO4(s),
Ered°( SO42‒(aq) | PbSO4(s) | Pb(s) ) = ‒0.36 V
and
Ered°( Pb2+(aq) | Pb(s) ) = ‒0.13 V.
10.6 Consider a silver-silver chloride pH meter with chloride concentration in the anode half cell
equal to 4.00 mol L‒1 (saturated potassium chloride).
- 232 -
(a) Find the cell potentials that correspond to samples with pH = 3.00, 5.00, 7.00, 9.00 and
11.00.
(b) If a sample produces a potential of 0.259 V, what is its pH?
Solutions:
10.1 (a) Cd(s) | Cd2+(aq) || Pb2+(aq) | Pb(s)
0.27 V
Voltmeter
e-
esalt bridge
NaNO3(aq) in a porous gel
anode
NO3-
cathode
Na+
Cd2+
NO3-
Cd
Pb
NO3-
Pb2+
1 mol L-1 Cd(NO3)2(aq)
Cd(s)
1 mol L-1 Pb(NO3)2(aq)
Cd2+ + 2 e-
Pb2+ + 2 e-
Pb(s)
Cadmium ions in the anode half cell flow toward the salt bridge, while lead ions flow toward
the cathode. Electrons flow from anode (cadmium metal) to cathode (lead metal).
- 233 -
(b) Ni(s) | Ni2+(aq) || H+(aq) | H2(g) | Pt(s)
0.26 V
Voltmeter
e-
esalt bridge
NaNO3(aq) in a porous gel
anode
cathode
NO3-
1 atm partial
pressure H2(g)
Na+
Ni2+
NO3-
NO3-
H+
Ni
1 mol L-1 Ni(NO3)2(aq)
Ni(s)
Pt
H2
1 mol L-1 HCl(aq)
Ni2+ + 2 e-
2 H+ + 2 e-
H2(g)
Nickel ions in the anode half cell flow toward the salt bridge, while hydrogen ions in the
cathode half cell flow toward the cathode. Electrons flow from anode (nickel metal) to
cathode (platinum metal) through wires.
(c) Pt(s) | Sn2+(aq), Sn4+(aq) || H+(aq), SO42‒(aq) | PbSO4(s) | PbO2(s)
Note that PbO2(s) provides the cathode for this cell. This compound, as written, is not a
metal. However, in practice, there is a deficiency in oxygen – i.e. the chemical formula is
more like PbO1.95(s). The oxygen deficiency gives the material metal properties, including
metal-like electrical conductivity.
10.2 (a) The half reactions for this cell are
Cd(s) → Cd2+(aq) + 2 e‒
oxidation
Ni2+(aq) + 2 e‒ → Ni(s) .
reduction
and
The standard cell potential is therefore
Ecell° = Ered°(Ni2+| Ni) ‒ Ered°(Cd2+| Cd)
= ‒0.26 – (‒0.40) V = 0.14 V .
- 234 -
(b) The half reactions for this cell are
Sn(s) → Sn2+(aq) + 2 e‒
oxidation
Cl2(g) + 2 e‒ → 2 Cl‒(aq) .
reduction
and
The standard cell potential is therefore
Ecell° = Ered°(Cl2| Cl‒) ‒ Ered°(Sn2+| Sn)
= 1.36 – (‒0.14) V = 1.50 V .
(c) The half reactions for this cell are
Zn(s) + 2 OH‒(aq) → ZnO(s) + H2O(l) + 2 e‒
oxidation
2 MnO2(s) + H2O(l) + 2 e‒ → Mn2O3(s) + 2 OH‒(aq) .
reduction
and
The standard cell potential is therefore
Ecell° = Ered°(MnO2| Mn2O3) ‒ Ered°(ZnO| Zn)
= 0.15 – (‒1.26) V = 1.41 V .
(d) The half reactions for this cell are
Pb(s) + SO42‒(aq) → PbSO4(s) + 2 e‒
oxidation
and
PbO2(s) + 4 H+(aq) + SO42‒(aq) + 2 e‒ → PbSO4(s) + 2 H2O(l) .
reduction
The standard cell potential is therefore
Ecell° = Ered°(PbO2| PbSO4) ‒ Ered°(PbSO4| Pb)
= 1.69 – (‒0.36) V = 2.05 V .
10.3 The oxidation and reduction potentials, at 25°C, for anodes and cathodes, respectively, are
given in the following table:
Anode half cell
Eox/V
Cr(s) | Cr3+(aq)
C(graphite) | Mn2O3(s) | MnO2(s) | OH‒(aq)
Pt(s) | Fe2+(aq), Fe3+(aq)
+0.74
‒0.15
‒0.77
Cathode half cell
Ered/V
OH‒(aq), OCl‒(aq), Cl‒(aq) | Pt(s)
H+(aq), H2(g) | Pt(s)
Pb2+(aq) | Pb(s)
+0.81
0.00
‒0.13
- 235 -
The cells that can be constructed from these anodes and cathodes have positive cell
potential. The cell potential is the sum the reduction and oxidation potentials. Therefore,
there are five cells that can be constructed. The corresponding cell diagrams, cell
potentials and balanced cell reactions are listed as follows:
Cr(s) | Cr3+(aq) || OH‒(aq), OCl‒(aq), Cl‒(aq) | Pt(s)
Ecell = 0.81 + 0.74 V = 1.55 V
Half reactions:
Cr(s) → Cr3+(aq) + 3 e‒
OCl‒(aq) + H2O(l) + 2 e‒ → Cl‒(aq) + 2 OH‒(aq)
oxidation
reduction
Balanced cell reaction:
2 Cr(s) + 3 OCl‒(aq) + 3 H2O(l) → 2 Cr3+(aq) + 3 Cl‒(aq) + 6 OH‒(aq)
Cr(s) | Cr3+(aq) || H+(aq), H2(g) | Pt(s)
Ecell = 0 + 0.74 V = 0.74 V
Half reactions:
Cr(s) → Cr3+(aq) + 3 e‒
2 H+(aq) + 2 e‒ → H2(g)
oxidation
reduction
Balanced cell reaction:
2 Cr(s) + 6 H+(aq) → 2 Cr3+(aq) + 3 H2(g)
Cr(s) | Cr3+(aq) || Pb2+(aq) | Pb(s)
Ecell = ‒0.13 + 0.74 V = 0.61 V
Half reactions:
Cr(s) → Cr3+(aq) + 3 e‒
Pb2+(aq) + 2 e‒ → Pb(s)
oxidation
reduction
Balanced cell reaction:
2 Cr(s) + 3 Pb2+(aq) → 2 Co2+(aq) + 3 Pb(s)
C(graphite) | Mn2O3(s) | MnO2(s) | OH‒(aq) || OH‒(aq), OCl‒(aq), Cl‒(aq) | Pt(s)
Ecell = 0.81 ‒ 0.15 V = 0.66 V
Half reactions:
Mn2O3(s) + 2 OH‒(aq) → 2 MnO2(s) + H2O(l) + 2 e‒
OCl‒(aq) + H2O(l) + 2 e‒ → Cl‒(aq) + 2 OH‒(aq)
oxidation
reduction
Balanced cell reaction:
Mn2O3(s) + OCl‒(aq) → 2 MnO2(s) + Cl‒(aq)
Pt(s) | Fe2+(aq), Fe3+(aq) || OH‒(aq), OCl‒(aq), Cl‒(aq) | Pt(s)
Ecell = 0.81 ‒ 0.77 V = 0.04 V
Half reactions:
Fe2+(aq) → Fe3+(aq) + e‒
OCl‒(aq) + H2O(l) + 2 e‒ → Cl‒(aq) + 2 OH‒(aq)
oxidation
reduction
Balanced cell reaction:
2 Fe2+(aq) + OCl‒(aq) + H2O(l) → 2 Fe3+(aq) + Cl‒(aq) + 2 OH‒(aq)
10.4 The half reactions for the cell,
Pb(s) | PbSO4(s) | SO42‒(aq) || H+(aq), SO42‒(aq) | PbO2(s) | PbSO4(s)
- 236 -
are
Pb(s) + SO42‒(aq) → PbSO4(s) + 2 e‒
oxidation
and
PbO2(s) + 4 H+(aq) + SO42‒(aq) + 2 e‒ → PbSO4(s)
+ 2 H2O(l) .
reduction
The balanced cell reaction is (with n = 2)
PbO2(s) + 4 H+(aq) + 2 SO42‒(aq) + Pb(s) → 2 PbSO4(s) + 2 H2O(l) .
The reaction quotient for this reaction is
Q
1
.
[H ] [SO24 ]2
 4
(a) [H+] = 1.000 mol L‒1 and [SO42‒] = 1.000 mol L‒1 corresponds to standard conditions.
In this case, the cell potential is
Ecell° = Ered°(PbO2| PbSO4) ‒ Ered°(PbSO4| Pb)
= 1.69 – (‒0.36) V = 2.05 V .
(b) For [H+] = 0.100 mol L‒1, [SO42‒] = 0.550 mol L‒1 (90% completion of the cell
reaction),
Q

1
[H ] [SO 24 ]2
 4
1
 3.31104 .
2
(1.00 10 ) (0.550)
1 4
In this case, the cell potential is
0.0257 V
ln Q
n
0.0257
 2.05 
ln 3.31104 V
2
=1.92 V .
o
Ecell  Ecell



(c) For [H+] = 0.010 mol L‒1, [SO42‒] = 0.505 mol L‒1 (99% completion of the cell
reaction),
Q

1
[H ] [SO24 ]2
 4
1
 3.9 108 .
2
(1.0 10 ) (0.505)
2 4
In this case, the cell potential is
- 237 -
0.0257 V
ln Q
n
0.0257
 2.05 
ln 3.9 108 V
2
=1.80 V .
o
Ecell  Ecell



10.5 The given reduction potentials correspond to cell potentials with the hydrogen electrode
providing the anode. Specifically,
Ered°( SO42‒(aq) | PbSO4(s) | Pb(s) ) = ‒0.36 V
= Ecell°( H2(g) | H+(aq) || SO42‒(aq) | PbSO4(s) | Pb(s) ) .
The associated balanced cell reaction (labeled reaction 1),
H2(g) + PbSO4(s) → 2 H+(aq) + SO42‒(aq) + Pb(s) ,
(1)
has n = 2 electrons transferred. Therefore, the standard Gibbs free energy of this reaction
is
∆G1° = ‒n F Ecell° = ‒2 × 96485 C mol‒1 × (‒0.36 V) = 69.5 kJ mol‒1 .
Also,
Ered°( Pb2+(aq) | Pb(s) ) = ‒0.13 V
= Ecell°( H2(g) | H+(aq) || Pb2+(aq) | Pb(s) ) .
The associated balanced cell reaction (labeled reaction 2),
H2(g) + Pb2+(aq) → 2 H+(aq) + Pb(s) ,
(2)
has n = 2 electrons transferred. Therefore, the standard Gibbs free energy of this reaction
is
∆G2° = ‒n F Ecell° = ‒2 × 96485 C mol‒1 × (‒0.13 V) = 25.1 kJ mol‒1 .
Reaction 1 minus reaction 2 gives
PbSO4(s) → Pb2+(aq) + SO42‒(aq) .
(3)
The Gibbs free energy of this reaction is
∆G3° = ∆G1° ‒ ∆G2° = 69.5 ‒ 25.1 kJ mol‒1
= 44.4 kJ mol‒1 .
Reaction 3 is the dissolution process of PbSO4. From the Gibbs free energy of this
process, we get the solubility product - the equilibrium constant for reaction 3:
- 238 -
K  Ksp  PbSO4 
 G3o 
 exp  

 RT 


44.4 103 J mol1
 exp  

1
1
 8.314 J K mol  298.15 K 
 1.66 108 .
Constructing an I.C.E. table for the dissolution of PbSO4(s), shows that the extent of
dissolution at equilibrium (the molar solubility of PbSO4) equals (Ksp)1/2. Specifically,
Molar solubility of PbSO4 = (1.66 × 10‒8)1/2 mol L‒1 = 1.3 × 10‒4 mol L‒1 .
10.6 (a) The cell potential of the silver-silver chloride pH meter with chloride concentration in the
anode half cell equal to 4.00 mol L‒1 (saturated potassium chloride) is given by
Etotal  Eref + 0.0592 V  pH ,
where


[Cl , 1.000 mol L1 ]
Eref  0.0592 V  log10  

 [Cl , saturated KCl solution] 
 1.000 
 0.0592 V  log10 
  0.0356 V
 4.00 
We therefore have the following table of potential values for the specified sample pH
values:
pH
3.00
5.00
7.00
9.00
11.00
Ecell (V)
0.2132
0.3316
0.4500
0.5684
0.6868
(b) If a sample produces a potential of 0.259 V, then
0.259 V = 0.0356 V + 0.0592 V × pH
or
pH = ( 0.259 ‒ 0.0356 ) / 0.0592 = 3.77 .
- 239 -
11. Math Skills
Chemistry is a quantitative subject. This is its strength. Only through careful measurement and
calculation are chemists able to reveal the world of atoms and molecules.
11.1 Scientific Notation
You may be asked to do chemical calculations. These calculations frequently involve very large
or very small numbers. It is inconvenient (sometimes ridiculous) to write these numbers in the
usual way – e.g., 0.00071, 0.012, 15, 3700, 450000. Scientific notation was introduced to let us
leave out all the leading or trailing zeros otherwise needed to write numbers. The set of numbers
listed above can be written as 7.1×10−4, 1.2×10−2, 1.5×101, 3.7×103 and 4.5×105. The advantage
of scientific notation is not necessarily evident here – e.g., we would rarely write 15 as 1.5×101.
However, try to write the values 6.022×1023 or 6.626×10−34 without scientific notation!
Writing a number in scientific notation simply amounts to factoring the number into a power of 10
and a number in the range 1.0 to 9.999.... The exponent of 10 is the number of places the
decimal moves to the left or minus the number of places the decimal moves to the right in order to
get the number into the convenient range.
Example 11.1: Write the following numbers in scientific notation: (a) 53400 (b) 0.00618
Approach: How many places do we need to move the decimal to get a number less
than 10 or bigger than (or equal to) 1.0? The number of places moved to the left is the
exponent of 10. Moving the decimal to the right means the exponent is negative.
(a) If the decimal is moved 4 places to the left, we get 5.34. Therefore,
53400 = 5.34 × 104.
(b) If the decimal is moved 3 places to the right, we get 6.18. Therefore,
0.00618 = 6.18 × 10−3.
Scientific notation uses powers of 10 because our number system is the decimal system.
Another way to get to work with more manageable numbers is to change the units. Again,
powers of 10 are convenient for unit conversion. For example, we prefer to measure the distance
between places in km rather than m. 1 km = 103 m. Also, precision machine components must
be accurate to within a micrometer (micron), 1 μm = 10−6 m. Powers of 103 are common unit
conversions in the metric system. Tiny masses might be measured in picograms (1 pg = 10−12 g),
while very large masses measured in teragrams (1 Tg = 1012 g). The range of units spanned
here is shown in the table.
- 240 -
Table 11.1. Units of Distance
Distance measure
unit
relation to m
picometer
nanometer
pm
nm
10−12 m
10−9 m
micrometer
μm
10−6 m
millimeter
mm
10−3 m
centimeter
cm
10−2 m
decimeter
dm
10−1 m
Meter
m
1m
kilometer
km
103 m
megameter
Mm
106 m
gigameter
Gm
109 m
terameter
Tm
1012 m
11.2 Dimensional Analysis
It is important to pay close attention to units in order to do correct calculations. For example, if
you are given density in g mL−1 and a volume of the material in L, then you must convert one of
the given units so that only one volume unit, mL or L, is used. The mass is determined by
multiplying density and volume. The volume units must be the same so that they cancel. for
example,
mass  density

g
units:

volume
1
g mL

mL
g

mL
mL

Attention to units can prevent use of incorrect formulas. The units must be the same on both
sides of a formula. If they are not, the formula is wrong. Here are some examples with correct
units:
amount 
mol

units:

mass
molar mass
g
g mol1
g mol
g
n RT
V
mol  L bar K 1mol1  K
P
units:
bar

L
Prove to yourself that the mol, L and K units all cancel appropriately in the equation above.
- 241 -
The value of the gas constant used in the ideal gas law, above, has the units, L bar K−1 mol−1.
The other form of the gas constant with units of J K−1 mol−1 is useful in work and other
thermodynamic calculations.
Another common unit used in calculations is Avogadro’s number, NA, which has the value and
unit of 6.0223  1023 mol1. We use this unit to scale quantities up to the molar scale, or down
from the molar scale to the scale of individual particles. Carefully paying attention to this unit
helps us use it correctly. We can have Avogadro’s number of anything – atoms, molecules, ions,
etc., so we can make the unit atoms mol−1, molecules mol−1, etc., as needed.
Example 11.2: How many Cu atoms are there in a penny? Let the mass of the penny be 3.35 g
and assume it is pure copper.
Approach: Determine number of moles of copper. Convert to number of atoms.
Amount of Cu (mol) = mass / molar mass = 3.35 g / 63.546 g mol1 = 5.272  102 mol
[Note: the final digit is shown as a subscript to indicate that it is not a significant digit.
Rather, it is an extra digit which we will carry through the next stage of the calculation, in
order to avoid rounding errors.]
To convert to number of atoms,
Number of atoms = 5.272  102 mol  6.0223  1023 atoms mol1 = 3.17  1022 atoms
In this case we needed to multiply by Avogadro’s number, in order to cancel the mol units.
11.3 Solving for a variable
Often in a calculation we are solving for at least one unknown quantity. Basic algebra skills will
help us manipulate equations to isolate the variable we seek.
Example 11.3: A solution with volume 0.500 L and concentration 0.101 mol L1 is diluted to
1.50 L. What is the new concentration?
Approach: The number of moles of solute is unchanged.
Reality check: The new concentration should be less than the initial.
Since mol (initial) = mol (final), and n = cV, then ciVi = cfVf.
0.101 mol L1  0.500 L  cf 1.50 L 
So
cf 
0.101 mol L1  0.500 L
 0.0337 mol L1
1.50 L
This matches our expectation (lower concentration). Notice how the units provide an excellent
way to check that the calculation is set up correctly.
On occasion we may have more than one variable that is unknown. In such cases we will need
more than one equation (or statement of variables) in order to solve for the unknowns. For
example, if we have 2 unknowns, then we will need a minimum of 2 equations involving the
variables, in order to solve for each one.
- 242 -
Example 11.4: We have information that tells us the following relationships:
x + y = 0.295 mol and 0.20 = log [ y / (0.075 + x) ]
What are x and y?
We can use the first equation to write one variable in terms of the second one, then make
a substitution in the second equation, in order to solve.
x = 0.295 – y
Substitute this into the second equation to get
0.20 = log [ y / (0.075 + 0.295 – y) ]
Raise both sides to the power of 10 to remove the log statement on the right side
100.20 = y / (0.370 – y)
1.58 = y / (0.370 – y)
1.58 (0.370 – y) = y
0.5846 – 1.58 y = y
0.5846 = y + 1.58 y
0.5846 = 2.58 y
0.5846/2.58 = y
0.227 = y
Now we can return and solve for x:
x + y = 0.295 mol
x + 0.227 = 0.295
x = 0.295 – 0.227 = 0.068 mol
Note that this last problem required taking the inverse of a log – i.e., we raised both sides of the
equation in the third line of the solution to the power of 10. This is the inverse of a log (antilog).
11.4 Logarithms
Chemistry deals with quantities that can vary over vast ranges, as we have seen. This arises
because many quantities vary exponentially. It is often convenient – and sometimes essential –
to introduce the logarithm function, For example, acidity (or basicity) is generally expressed via
the pH (the “potential of hydrogen”).
pH = −log10( [H+] )
A base 10 logarithm is used here. When we use the natural logarithm, we use the notation, ln.
pH can vary from below 0 to above 14. This corresponds to [H+] varying over more than 14
orders of magnitude.
The key properties of logarithms are as follows:
1. y = 10x is the inverse of the logarithm:
if y = log x, then 10y = x
2. The sign of a log determines the size of the argument relative to 1:
If log x > 0, then x > 1
- 243 -
If log x < 0, then 0 < x < 1.
Note that you cannot take the log of a negative number. Also, the argument of the logarithm
(here, x) must not be negative.
3. The log of a product is the sum of the logarithms:
log(x y) = log x + log y
4. The argument of a log must not have units. The [H+] that appears above in the definition of pH
is the concentration of H+ with the units of mol L−1 divided out. Note that we can take the log of
an equilibrium constant because it has no units – any units of concentration (mol L−1) and
pressure (bar) are divided out.
- 244 -
The Periodic Table of the Elements: Symbols, atomic numbers and atomic masses.
The elements are arranged into the periodic table – elements in the same column (group) have
similar chemical properties. The elements are color-coded with metals (blue-grey), semimetals
(blue) and nonmetals (green – except noble gases are brown) indicated. The atomic number
(above the symbols) is the number of protons in the nucleus of the atom. The atomic mass
(below the symbol) is the average of the isotope masses in amu, with isotopes weighted
according to IUPAC-accepted (2009) natural abundances. For radioactive elements, the mass
number of the most stable isotope is given in parentheses.
It is currently recognized that there is a range of isotope abundances depending on the source of
the material bearing the element. Different processes (e.g., biological versus geological or
atmospheric) generally produce slightly different isotope abundances. Thus, atomic masses of
some elements are now tabulated as ranges – exhibiting the variation found in samples. The
table below provides average atomic masses – rounded in some cases – for ease of use by
students. Data are from CRC Handbook of Chemistry and Physics.
Variation in isotopic abundance between samples is used to identify the source of materials. For
example, the testosterone in an athlete’s blood exhibits isotope rations that are sensitive to
whether the source of testerone is the athlete’s body, or a botanical pathway used to make
synthetic testosterone.
- 245 -
- 246 -
88
Ra
(226)
(223)
91
Pa
231.036
140.116
90
Th
232.038
138.905
89
Ac
(227)
238.029
92
U
144.242
60
Nd
59
Pr
58
Ce
57
La
140.908
(266)
106
Sg
(262)
105
Db
183.84
(261)
104
Rf
180.948
(237)
93
Np
(145)
61
Pm
(244)
94
Pu
150.36
62
Sm
(277)
108
Hs
107
Bh
(272)
190.23
186.207
76
Os
75
Re
(243)
95
Am
151.964
63
Eu
(276)
109
Mt
192.217
77
Ir
(247)
96
Cm
157.25
64
Gd
(281)
110
Ds
195.084
78
Pt
106.42
102.906
(2247)
97
Bk
158.925
65
Tb
(280)
111
Rg
196.967
79
Au
107.868
47
Ag
(251)
98
Cf
162.500
66
Dy
112
Cn
(285)
200.59
80
Hg
112.411
48
Cd
65.409
87
Fr
178.49
74
W
101.07
(98)
46
Pd
45
Rh
63.546
137.327
73
Ta
95.94
44
Ru
58.693
30
Zn
132.905
72
Hf
92.906
43
Tc
58.933
29
Cu
56
Ba
91.224
42
Mo
55.845
28
Ni
55
Cs
88.906
41
Nb
54.938
27
Co
87.62
40
Zr
51.996
26
Fe
85.468
39
Y
50.942
25
Mn
38
Sr
47.867
24
Cr
37
Rb
44.956
23
V
40.078
22
Ti
39.098
21
Sc
20
Ca
(252)
99
Es
164.930
67
Ho
204.383
81
Tl
114.818
49
In
69.723
31
Ga
(257)
100
Fm
167.259
68
Er
207.2
82
Pb
118.710
50
Sn
72.64
32
Ge
(258)
101
Md
168.934
69
Tm
208.980
83
Bi
121.760
51
Sb
(259)
102
No
173.04
70
Yb
(209)
84
Po
127.60
52
Te
78.96
34
Se
33
As
74.922
32.065
30.974
(262)
103
Lr
174.967
71
Lu
(210)
85
At
126.905
53
I
79.904
35
Br
35.453
17
Cl
(277)
86
Rn
131.293
54
Xe
83.798
36
Kr
39.984
18
Ar
20.180
19
K
28.086
26.982
16
S
18.998
24.305
15
P
15.999
22.990
14
Si
13
Al
14.007
12
Mg
12.011
11
Na
10.811
9.012
9
F
6.941
8
O
10
Ne
7
N
4
Be
3
Li
6
C
4.0026
1.008
5
B
2
He
1
H