Solutions
– Solution, A homogeneous mixture in
which all of the material is in the same
state.
– Substances present in lesser amounts,
called solutes, are dispersed uniformly
throughout the substance in the greater
amount, the solvent
– Aqueous solution — a solution in
which the solvent is water
– Nonaqueous solution — any substance
other than water is the solvent
– Many of the chemical reactions that are
essential for life depend on the
interaction of water molecules with
dissolved compounds.
Aqueous Solutions
Polar Substances
 An individual water molecule consists of two hydrogen atoms bonded to an
oxygen atom in a bent (V-shaped) structure.
 The oxygen atom in each O–H covalent bond attracts the electrons more
strongly than the hydrogen atom.
 O and H nuclei do not share the electrons equally.
– Hydrogen atoms are electron-poor
compared with a neutral hydrogen atom and
have a partial positive charge, indicated by
the symbol δ+.
– The oxygen atom is more electron-rich
than a neutral oxygen atom and has a partial
negative charge, indicated by the symbol 2δ-.
 Unequal distribution of charge creates a
polar bond, which makes them good
solvents for ionic compounds.
 Individual cations and anions are called
hydrated ions.
Aqueous Solutions
 Electrolyte — any compound that can form ions when
it dissolves in water
– When strong electrolytes dissolve, constituent
ions dissociate completely, producing aqueous
solutions that conduct electricity very well.
– When weak electrolytes dissolve, they
produce relatively few ions in solution.
Aqueous solutions, of weak electrolytes do not
conduct electricity as well as solutions of strong
electrolytes.
– Nonelectrolytes dissolve in water as neutral
molecules and have no effect on conductivity.
CH3CO2H(aq) → CH3CO2-(aq) +H+(aq)
Aqueous Solutions
Molarity





Most common unit of concentration
Most useful for calculations involving the stoichiometry of
reactions in solution
Molarity of a solution is the number of moles of solute
present in exactly 1 L of solution:
Units of molarity — moles per liter of solution (mol/L),
abbreviated as M
Relationship among volume, molarity, and moles is
expressed as:
number of moles of solute
molarity 
number of liters of solution
moles
M
L
mmol
M
mL
Preparation of Solutions
Calculating Volume from Mass
Calculating Moles from Volume
Preparation of Solutions
Problem: What mass of oxalic acid, H2C2O4, is required to
make 250. mL of a 0.0500 M solution?
Preparation of Solutions
0.250 L of water was used to make 0.250 L of solution.
Notice the water left over.
Preparation of Solutions
(Vs) (Ms) = moles of solute = (Vd) (Md).
Preparation of Solutions
PROBLEM: You have 50.0 mL of 3.0 M NaOH and you
want 0.50 M NaOH. What do you do?
Ion Concentrations in Solution
 Classify each compound as either a strong electrolyte or a
nonelectrolyte.
 If a compound is a nonelectrolyte, the concentration is the same
as the molarity of the solution.
 If a compound is a strong electrolyte, determine the number of
each ion contained in one formula unit and find the
concentration of each species by multiplying the number of each
ion by the molarity of the solution.
Preparation of Solutions
PROBLEM: Dissolve 5.00 g of NiCl2•6 H2O in enough water to
make 250 mL of solution. Calculate molarity of the solution
and the concentration of each of the ions.
SOLUTION STOICHIOMETRY
• Zinc reacts with acids to
produce H2 gas.
• Have 10.0 g of Zn
• What volume of 2.50 M
HCl is needed to
convert the Zn
completely?
SOLUTION STOICHIOMETRY
Zinc reacts with acids to produce H2 gas. If you have 10.0 g of Zn,
what volume of 2.50 M HCl is needed to convert the Zn completely?
Limiting Reactants in Solutions
 The concept of limiting reactants applies to reactions that are
carried out in solution and reactions that involve pure
substances.
 If all the reactants but one are present in excess, then the amount
of the limiting reactant can be calculated.
 When the limiting reactant is not known, one can determine
which reactant is limiting by comparing the molar amounts of
the reactants with their coefficients in the balanced chemical
equation.
 Use volumes and concentrations of solutions of reactants to
calculate the number of moles of reactants.
Ionic Equations
• Chemical equation for a reaction in solution can be written
in three ways:
1. Overall equation — shows all of the substances present in their
undissociated form
2. Complete ionic equation — shows all of the substances present in
the form in which they actually exist in solution
3. Net ionic equation
– Derived from the complete ionic equation by omitting all
spectator ions, ions that occur on both sides of the equation
with the same coefficients
– Demonstrate that many different combinations of reactants can give the
same net chemical reaction
Ionic Equations
There are three ways to write reactions in aqueous solutions.
Molecular equation:
Show all reactants & products in molecular or ionic form
Zn (s) + CuSO 4(aq)  ZnSO 4(aq) + Cu(s)
Total ionic equation:
Show the ions and molecules as they exist in solution
Zn (s) + Cu
2
aq 
+ SO
24 aq 
 Zn
2
aq 
+ SO
24 aq 
+ Cu (s)
Net ionic equation:
Shows ions that participate in reaction and removes spectator ions.
Spectator ions do not participate in the reaction.
Zn (s) + Cu
2
aq 
 Zn
2
aq 
+ Cu (s)
Net Ionic Equations
Mg(s)+ 2HCl(aq)→ H2(g) + MgCl2(aq)
The molecular formula above can be written as the total ionic formula
Mg(s)+ 2H+(aq)+ 2Cl-(aq)→ H2(g)+ Mg2+(aq)+ 2Cl-(aq)
The two Cl- ions are SPECTATOR IONS — they do not
participate. Could have used NO3- for the spectator ion as salts of
nitrates are all soluble.
By leaving out the spectator ions out you get the net ionic reaction
Mg(s) + 2 H+(aq) ---> H2(g) + Mg2+(aq)
Classifying Chemical Reactions
1. Exchange reactions
a. Single Displacement Reactions – one element displaces another from a
compound – AB + C  AC + B
b. Metathesis Reactions - Exchange Reactions – AB + CD  AD + CB
i.
Precipitation: products include an insoluble substance which
precipitates from solution as a solid
ii. Acid-base neutralization: product is a salt and water
iii. Gas formation – primarily the reaction of metal carbonates
2. Condensation reactions (and the reverse, cleavage reactions)
a. Combination Reactions (Condensation) – More than one reactant, one
product. Some condensation reactions are redox reactions. – A + B  AB
b. Decomposition Reactions (Cleavage) – Single reactant, more than one
product – AB  A + B
3. Redox (Oxidation Reduction Reactions) – Electrons are transferred between
reactants. Oxidation numbers of some elements change; at least one element must
increase and one must decrease in oxidation number.
a) Single Displacement Reactions are always Redox reactions as well.
b) oxidant + reductant  reduced oxidant + oxidized reductant
Aqueous Chemical Reactions: Metathesis
• EXCHANGE REACTIONS
The anions exchange places between cations.
AX + BY
AY + BX
Precipitation
Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2KNO3(aq)
Pb2+(aq) + 2 I-(aq) → PbI2(s)
Neutralization:
NaOH(aq) + HCl(aq)→ NaCl(aq)+ H2O(l)
OH-(aq) + H+(aq) → H2O(lq)
Gas Formation
MgCO3(s)+ 2HCl(aq) → 2Mg(Cl)2(aq)+ H2O(l) + CO2(g)
Precipitation Reactions
• A reaction that yields an insoluble
product, a precipitate, when two
solutions are mixed
• Are a subclass of exchange reactions
that occur between ionic compounds
when one of the products is insoluble
• Used to isolate metals that have been
extracted from their ores and to
recover precious metals for recycling
Water Solubility of Ionic Compounds
If one ion from the “Soluble
Compound” list is present in a
compound, the compound is
water soluble.
Precipitation Reactions
In addition to understanding solubility. It is equally important to
know if a reaction will occur.
K2Cr2O7(aq) + AgNO3(aq) → 2KNO3(aq) + Ag2Cr2O7(s)
KBr(aq) + NaCl(aq) → KCl(aq) + NaBr(aq)
Neutralization Reactions
 A reaction in which an acid and a
base react in stoichiometric amounts
to produce water and a salt
 Strengths of the acid and base
determine whether the reaction goes
to completion
1. Reactions that go to completion
a.Reaction of any strong acid with any strong
base
b.Reaction of a strong acid with a weak base
c.Reaction of weak acid with a weak base
2. Reaction that does not go to
completion is a reaction of a weak acid
or a weak base with water
Neutralization Reactions
A brief history of
Acid-Base
Identification Systems
System
Arrhenius
BrönstedLowry
Lewis
Acid (HCl)
Base (NaOH)
ACIDS
A Brönsted-Lowry Acid → H+ in water
Strong Brönsted-Lowry acids are strong electrolytes
HCl
H2SO4
HClO4
HNO3
hydrochloric
sulfuric
perchloric
nitric
HNO3
Weak Brönsted-Lowry acids are weak electrolytes
CH3CO2H
H2CO3
H3PO4
HF
acetic acid (CH3COOH)
carbonic acid
phosphoric acid
hydrofluoric acid
Carbonic Acid
Acetic acid
Polyprotic Acids
• Acids differ in the number of hydrogen ions they can
donate.
– Monoprotic acids are compounds capable of
donating a single proton per molecule.
– Polyprotic acids can donate more than one hydrogen ion
per molecule.
BASES
Brönsted-Lowry Base → OH- in water
NaOH(aq) → Na+(aq) + OH-(aq)
NaOH is a strong base
Ammonia, NH3 an Important weak
Base
ACIDS
Nonmetal oxides can be acids
CO2(aq) + H2O(l) → H2CO3(aq)
SO3(aq) + H2O(l) → H2SO4(aq)
NO2(aq) + H2O(l) → HNO3(aq)
Acid Rain is an example of nonmetal oxides behaving as
acids. This process can result from burning coal and oil.
BASES
Metal oxides can be bases
CaO(s)+H2O(l) → Ca(OH)2(aq)
CaO in water. Phenolphthalein indicator shows
a of calcium oxide solution is basic.
pH, a Concentration Scale
pH: a way to express acidity -- the concentration of H+ in
solution.
Low pH: high [H+]
Acidic solution
Neutral
Basic solution
High pH: low [H+]
pH < 7
pH = 7
pH > 7
The pH Scale
pH = log (1/ [H+]) = - log [H+]
In a neutral solution,
[OH-] = 1.00 x 10-7 M at 25 oC
pH = - log [H+] =
If the [H+] of soda is 1.6 x 10-3 M, the pH is ____.
If the pH of Coke is 3.12, it is _____.
[H+] =
Acid-Base Strength
Identification
You should know the
strong acids & bases
Oxidation-Reduction Reactions in
Solution
• Oxidation-reduction reactions —
electrons are transferred from one
substance or atom to another.
• Oxidation-reduction reactions that
occur in aqueous solution are
complex, and their equations are very
difficult to balance.
• Two methods for balancing
oxidation-reduction reactions in
aqueous solution are:
1. Oxidation states — overall reaction is
separated into an oxidation equation
and a reduction equation
2. Half-reaction
Oxidation-Reduction Reactions
 The term oxidation was first used to
describe reactions in which metals react
with oxygen in air to produce metal oxides.
– Metal acquires a positive charge by
transferring electrons to the neutral oxygen
atoms of an oxygen molecule.
– Oxygen atoms acquire a negative
charge and form oxide ions (O2-).
– Metals lose electrons to oxygen and
have been oxidized—oxidation is the loss of
electrons.
– Oxygen atoms have gained electrons
and have been reduced—reduction is the
gain of electrons.
Oxidation-Reduction Reactions
• Oxidation and reduction reactions are now characterized by a change in the
oxidation states of one or more elements in the reactants.
• Oxidation states of each atom in a compound is the charge that atom would
have if all of its bonding electrons were transferred to the atom with the
greater attraction for electrons. Atoms in their elemental form are assigned
an oxidation state of zero.
• Oxidation-reduction reactions are called redox reactions, in which there is a
net transfer of electrons from one reactant to another. The total number of
electrons lost must equal the total number of electrons gained.
• Oxidants and reductants
– Oxidants – Compounds that are capable of accepting electrons are called
oxidants, or oxidizing reagents, because they can oxidize other compounds.
• An oxidant is reduced in the process of accepting electrons.
– Reductants – Compounds that are capable of donating electrons are
called reductants, or reducing agents, because they can cause the
reduction of another compound.
• A reductant is oxidized in the process of donating electrons.
OXIDATION NUMBERS
NH3
ClOH3PO4
MnO4Cr2O72-
Recognizing a Redox Reaction
2 Ag(s) + Cu2+(aq) → Al+(aq) + Cu(s)
Hydrogen Fuel Cell
2 H2(g) + O2(g) → 2H2O(l)
Thermite reaction
Fe2O3(s) + 2Al(s) → 2 Fe(s) + Al2O3(s)
Balancing Redox Equations Using
Oxidation States
Will be covered
in Chem 102
Quantitative Analysis: Titrations
• Quantitative analysis — used to determine the amounts or
concentrations of substances present in a sample by using a
combination of chemical reactions and stoichiometric calculations
• Titration
– A method in which a measured volume of a solution of known concentration,
called the titrant, is added to a measured volume of a solution containing a
compound whose concentration is to be determined (the unknown)
– Reaction must be fast, complete, and specific (only the compound of interest
should react with the titrant)
– Equivalence point — point at which exactly enough reactant has been added for
the reaction to go to completion (computed mathematically)
Acid-Base Titrations
 Most common acids and bases are not intensely colored
Rely on an indicator
–
 Endpoint — point at which a color change is observed, which is
close to the equivalence point in an acid-base titration
Standard Solutions
 A solution of a primary standard whose
concentration is known precisely
 A primary standard is non-hygroscopic, has a
high mass, is fairly inexpensive compound, is
of known reactive ability that can be accurately
weighed for use as a titrant.
 A standard solution is used to determine the
concentration of the titrant.
 Accuracy of any titration analysis depends on
accurate knowledge of the concentration of
the titrant.
 Most titrants are first standardized—their
concentration is measured by titration with a
standard solution.
Titration
1. Add titrant solution from the
buret.
2. Reagent (base) reacts with
compound (acid) in solution
in the flask.
3. Indicator shows when exact
stoichiometric reaction has
occurred.
4. Net ionic equation
H+ + OH- → H2O
5. At equivalence point
moles H+ = moles OH-
ACID-BASE REACTIONS
Titrations
H2C2O4(aq) + 2 NaOH(aq) → Na2C2O4(aq) + 2 H2O(l)
acid
base
Carry out this reaction using a TITRATION.
Oxalic acid,
H2C2O4
Quantitative Analysis: Titrations
PROBLEM: Standardize a solution of NaOH — i.e., accurately
determine its concentration. 1.065 g of H2C2O4 (oxalic acid) requires
35.62 mL of NaOH for titration to an equivalence point. What is the
concentration of the NaOH?
Quantitative Analysis: Titrations
PROBLEM : Use standardized NaOH to determine the amount of an
acid in an unknown. Apples contain malic acid, C4H6O5. 76.80 g of apple
requires 34.56 mL of 0.663 M NaOH for titration.
What is weight % of malic acid?
HOOCCH2COHCOOH(aq) + 2NaOH(aq) → Na2C4H4O5(aq) + 2H2O(l)
Energy & Chemistry
2H2(g) + O2(g) → 2H2O(g) + heat and light
This can be set up to provide
ELECTRIC ENERGY in a fuel cell.
Oxidation:
2 H2 → 4 H+ + 4 e-
Reduction:
4 e- + O2 + 2 H2O → 4 OHH2/O2 Fuel Cell Energy
Energy & Chemistry
ENERGY is the capacity to do work (w) or transfer heat (q).
HEAT is the thermal energy that can be transferred from an
object at one temperature to an object at another temperature
– Net transfer of thermal energy stops when the two
objects reach the same temperature.
Other forms of energy —
1. Radiant (light) — energy in light, microwaves,
and radio waves
2. Thermal (kinetic and potential) — results from
atomic and molecular motion – Temperature of an
object is a measure of the thermal energy content
3. Chemical — results from the particular arrangement
of atoms in a chemical compound; radiant and thermal
energy produced in this reaction due to energy released
during the breaking and reforming of chemical bonds
4. Nuclear — radiant and thermal energy released when
particles in the nucleus of the atoms are rearranged
5. Electrical — due to the flow of electrically charged
particles
Potential & Kinetic Energy
Kinetic energy
rotate
vibrate
translate
 Potential energy – Energy stored in an object because of
positions or orientations of its components –

PE = Fd = mad = mgh = work
Kinetic energy – Energy due to the motion of an object –
KE = ½ mv2
the relative
Potential & Kinetic Energy
Internal Energy (E)
• PE + KE = Internal energy (E or U)
• Internal Energy of a chemical system depends
on
• number of particles
• type of particles
• temperature
• The higher the T the higher the internal energy
• So, use changes in T (∆T) to monitor changes in E
(∆E).
Internal Energy (E)
heat transfer in
(endothermic), +q
heat transfer out
(exothermic), -q
SYSTEM
∆E = q + w
w transfer in
(+w)
w transfer out
(-w)
Energy & Chemistry
All of thermodynamics depends on the law of
CONSERVATION OF ENERGY.
• The total energy is unchanged in a chemical reaction.
• If PE of products is less than reactants, the difference must
be released as KE.
Energy Change in Chemical Processes
PE
Reactants
Kinetic
Energy
Products
Potential Energy of system
dropped. Kinetic energy increased.
Therefore, you often feel a
Temperature increase.
Thermodynamics -Enthalpy
• Thermodynamics – the science of heat (energy) transfer.
Heat transfers until thermal equilibrium is
established. ∆T measures energy transferred.
• SYSTEM – The object under study
1. Open system — can exchange both matter and
energy with its surroundings
2. Closed system — can exchange energy but not
matter with its surroundings
3. Isolated system — exchanges neither energy nor
matter with the surroundings; total energy of the
system plus the surroundings is constant
• SURROUNDINGS – Everything outside the system
FIRST LAW OF
THERMODYNAMICS
heat energy transferred
∆E = q + w
work done by the system
energy
change
Energy is conserved!
The First Law of
Thermodynamics
• Exothermic reactions
generate specific
amounts of heat.
• This is because the
potential energies of the
products are lower than
the potential energies of
the reactants.
The First Law of Thermodynamics
•
1.
2.
•
There are two basic ideas of importance for
thermodynamic systems.
Chemical systems tend toward a state of minimum
potential energy.
Chemical systems tend toward a state of maximum
disorder.
The first law is also known as the Law of Conservation
of Energy.
– Energy is neither created nor destroyed in chemical
reactions and physical changes.
State of a System
•
•
The state of the system is a complete description of the system at a
given time, including its temperature and pressure, the amount of
matter it contains, its chemical composition, and the physical state of
the matter Some examples of state functions are:
– T (temperature), P (pressure), V (volume), E (change in energy), H (change
in enthalpy – the transfer of heat), and S (entropy)
Examples of non-state functions are:
– n (moles), q (heat), w (work)
∆H along one path = ∆H along another path
•
•
•
•
This equation is valid because ∆H is a STATE FUNCTION
These depend only on the state of the system and not how it got there.
V, T, P, energy — and your bank account!
Unlike V, T, and P, one cannot measure absolute H. Can only measure
∆H.
Standard States and Standard Enthalpy Changes
•Thermochemical standard state conditions
–The thermochemical standard T = 298.15 K.
–The thermochemical standard P = 1.0000 atm.
•Be careful not to confuse these values with STP.
•Thermochemical standard states of matter
–For pure substances in their liquid or solid phase the standard
state is the pure liquid or solid.
–For gases the standard state is the gas at 1.00 atm of pressure.
•For gaseous mixtures the partial pressure must be 1.00 atm.
–For aqueous solutions the standard state is 1.00 M
concentration.
∆Hfo = standard molar enthalpy of formation
• the enthalpy change when 1 mol of compound is formed from
elements under standard conditions.
State Function
• The properties of a system that depend only on the state of the
system are called state functions.
– State functions are always written using capital letters.
• The value of a state function is independent of pathway.
• An analog to a state function is the energy required to climb a
mountain taking two different paths.
–
–
–
–
–
E1 = energy on the first floor of Heldenfels
E1 = mgh1
E2 = energy on the fourth floor of Heldenfels
E2 = mgh2
E = E2-E1 = mgh2 – mgh1 = mg(h)
ENTHALPY
Most chemical reactions occur at constant P, so
Heat transferred at constant P = qp
qp = ∆H
where H = enthalpy
and so ∆E = ∆H + w (and w is usually small)
∆H = heat transferred at constant P ≈ ∆E
∆H = change in heat content of the system
∆H = Hfinal - Hinitial
Directionality of Heat Transfer
• Heat always transfer from hotter object to cooler
one.
• EXOthermic: heat transfers from SYSTEM to
SURROUNDINGS.
T(system) goes down
T(surr) goes up
Directionality of Heat Transfer
• Heat always transfers from hotter object to cooler
one.
• ENDOthermic: heat transfers from
SURROUNDINGS to the SYSTEM.
T(system) goes up
T (surr) goes down
ENTHALPY
∆H = Hfinal - Hinitial
If Hfinal > Hinitial then ∆H is positive
Process is ENDOTHERMIC
If Hfinal < Hinitial then ∆H is negative
Process is EXOTHERMIC
USING ENTHALPY
Consider the formation of water
H2(g) + 1/2 O2(g) → H2O(g) + 241.8 kJ
Exothermic reaction — heat is a “product” and ∆H = – 241.8 kJ
USING ENTHALPY
Example of HESS’S LAW— Making liquid H2O from H2 + O2
involves two exothermic steps.
H2 + O2 gas
H2O vapor
Liquid H2O
Making liquid H2O from H2 involves two steps.
H2(g) + 1/2 O2(g) → H2O(g) + 242 kJ
H2O(g) → H2O(l) + 44 kJ
H2(g) + 1/2 O2(g) → H2O(l) + 286 kJ
If a rxn. is the sum of 2 or more others, the net ∆H is the sum of the
∆H’s of the other rxns.
Enthalpy Values
Depend on how the reaction is written and on
phases of reactants and products
H2(g) + 1/2 O2(g) → H2O(g)
∆H˚ = -242 kJ
2H2(g) + O2(g) → 2H2O(g)
∆H˚ = -484 kJ
H2O(g) → H2(g) + 1/2 O2(g)
H2(g) + 1/2 O2(g) → H2O(l)
∆H˚ = +242 kJ
∆H˚ = -286 kJ
Standard Molar Enthalpies of Formation, Hfo
•
•
The standard molar enthalpy of formation is defined as the
enthalpy for the reaction in which one mole of a substance is
formed from its constituent elements.
– The symbol for standard molar enthalpy of formation is Hfo.
The standard molar enthalpy of formation for MgCl2 is:
Standard Molar Enthalpies of Formation, Hfo
•
•
•
Standard molar enthalpies of formation have been determined for many
substances and are tabulated in Table 15-1 and Appendix K in the text.
Standard molar enthalpies of elements in their most stable forms at
298.15 K and 1.000 atm are zero.
Example 15-4: The standard molar enthalpy of formation for
phosphoric acid is -1281 kJ/mol. Write the equation for the reaction
for whichHorxn = -1281 kJ.
P in standard state is P4
Phosphoric acid in standard state is H3PO4(s)
Standard Molar Enthalpies of Formation, Hfo
H2(g) + ½ O2(g) → H2O(g) ∆Hf˚ (H2O, g)= -241.8 kJ/mol
C(s) + ½ O2(g) → CO(g)
∆Hf˚ of CO = - 111 kJ/mol
By definition, ∆Hfo = 0 for elements in their standard states.
Use ∆H˚’s to calculate enthalpy change for
H2O(g) + C(graphite) → H2(g) + CO(g)
Standard Molar Enthalpies of Formation, Hfo
• Calculate the enthalpy change for the reaction of one mole of
H2(g) with one mole of F2(g) to form two moles of HF(g) at 25oC
and one atmosphere.
Standard Molar Enthalpies of Formation, Hfo
• Calculate the enthalpy change for the reaction in which 15.0 g of
aluminum reacts with oxygen to form Al2O3 at 25oC and one
atmosphere.
Hess’s Law – Enthalpies of
Formation and Reaction

Standard enthalpies of reaction (ΔH°xn)
– The enthalpy that occurs when a reaction is carried out with all
reactants and products in their standard states
– General reaction: aA + bB  cC + dD where A, B, C, and D are
chemical substances and a, b, c, and d are their stoichiometric
coefficients
– Magnitude of ΔH°rxn is the sum of the standard enthalpies of
formation of the products, each multiplied by its appropriate
coefficient , minus the sum of the standard enthalpies of formation
of the reactants, multiplied by their coefficients
ΔH°rxn = [cΔH°f(C) + dΔH°f (D)]  [aΔH°f(A) + bΔH°f(B)]
or
ΔH°rxn= mΔH°f(product)  nΔH°f(reactants)
where  means “sum of” and m and n are the stoichiometric
coefficients of each of the products and reactants, respectively
Hess’s Law & Energy
Level Diagrams
Forming H2O can occur in a single
step or in a two steps. ∆Htotal is the
same no matter which path is
followed.
Enthalpies of Formation

Standard enthalpies of formation
– The magnitude of ΔH for a reaction depends on the physical
states of the reactants and products, the pressure of any gases
present, and the temperature at which the reaction is carried
out.
– A specific set of conditions under which enthalpy changes
are measured are used to ensure uniformity of reaction
conditions and data.
– Standard conditions: a pressure of 1 atmosphere (atm) for
gases and a concentration of 1 M for species in solution (1
mol/L), each pure substance must be in its standard state, its
most stable form at a pressure of 1 atm at a specified
temperature.
– Enthalpies of formation measured under standard conditions
are called standard enthalpies of formation (ΔH°f).
– Standard enthalpy of formation of any element in its
standard state is zero.
Enthalpies of Reaction
 Hess’s law allows the calculation of the enthalpy change for
any conceivable chemical reaction by using a relatively
small set of tabulated data
1. Enthalpy of combustion, ΔHcomb—enthalpy change that
2.
3.
4.
5.
occurs when a substance is burned in excess oxygen
Enthalpy of fusion, ΔH fus—enthalpy change that
accompanies the melting, or fusion, of 1 mol of a substance
Enthalpy of vaporization, ΔHvap—the enthalpy change
that accompanies the vaporization of 1 mol of a substance
Enthalpy of solution, ΔHsoln—enthalpy change when a
specified amount of solute dissolves in a given quantity of
solvent
Enthalpy of formation, ΔHf—enthalpy change for the
formation of 1 mol of a compound from its component
elements
Hess’s Law
• Hess’s Law of Heat Summation, Hrxn = H1 +H2 +H3 + ...,
states that the enthalpy change for a reaction is the same whether it
occurs by one step or by any (hypothetical) series of steps.
– Hess’s Law is true because H is a state function.
• If we know the following Ho’s
Hess’s Law
•
•
For example, we can calculate the Ho for reaction [1] by
properly adding (or subtracting) the Ho’s for reactions [2] and
[3].
Notice that reaction [1] has FeO and O2 as reactants and Fe2O3
as a product.
– Arrange reactions [2] and [3] so that they also have FeO and
O2 as reactants and Fe2O3 as a product.
• Each reaction can be doubled, tripled, or multiplied by
half, etc.
• The Ho values are also doubled, tripled, etc.
• If a reaction is reversed the sign of the Ho is changed.
Hess’s Law
• Given the following equations and Ho values
calculate Ho for the reaction below.
Hess’s Law
• Calculate the H o298 for the following reaction:
Hess’s Law
•
•
Application of Hess’s Law and more algebra allows us to calculate the
Hfo for a substance participating in a reaction for which we know
Hrxno , if we also know Hfo for all other substances in the reaction.
Given the following information, calculate Hfo for H2S(g).
HEAT CAPACITY
The heat required to raise an object’s T by 1 ˚C.
Which has the larger heat capacity?
Thermal energy cannot be measured easily.
Temperature change caused by the flow of thermal energy between objects or
substances can be measured.
Calorimetry is the set of techniques employed to measure enthalpy changes in
chemical processes using calorimeters.
Specific Heat Capacity
How much energy is transferred due to
Temperature difference?
The heat (q) “lost” or “gained” is
related to
a) sample mass
b) change in T and
c) specific heat capacity
Specific heat capacity = heat lost or gained by substance (J)
(mass, g) (T change,K)
Table of specific heat
capacities
 Heat capacity of an object depends on both its
mass and its composition.
 Molar heat capacity (Cp) — amount of energy
needed to increase the temperature of 1 mol of a
substance by 1°C; units of Cp are J/(mol•°C)
 Specific heat (Cs) — amount of energy needed
to increase the temperature of 1 g of a substance
by 1°C, units are J/(g•°C)
 The quantity of a substance, the amount of heat
transferred, its heat capacity, and the temperature
change are related in two ways:
q = nCpΔT where n = number of moles of
substance
q = mCsΔT where m = mass of substance in
grams
Cp J mol-1 K-1
Phase
Air (typ. room
conditions)
gas
1.012
29.19
Aluminium
solid
0.897
24.2
Argon
gas
0.5203
20.7862
Copper
solid
0.385
24.47
Diamond
solid
0.5091
6.115
Ethanol
liquid
2.44
112
Gold
solid
0.1291
25.42
Graphite
solid
0.710
8.53
Helium
gas
5.1932
20.7862
Hydrogen
gas
14.30
28.82
Iron
solid
0.450
25.1
Lithium
solid
3.58
24.8
Mercury
liquid
0.1395
27.98
Nitrogen
gas
1.040
29.12
Neon
gas
1.0301
20.7862
Oxygen
gas
0.918
29.38
Uranium
solid
0.116
27.7
gas (100 °C)
2.080
37.47
liquid (25 °C)
4.1813
75.327
2.114
38.09
Water
solid (0 °C)
Aluminum
cp J g-1 K-1
Substance
All measurements are at 25 °C unless noted. Notable minimums and
maximums are shown in maroon text.
Specific Heat Capacity
If 25.0 g of Al cool from 310 oC to 37 oC, how
many joules of heat energy are lost by the Al?
Heat/Energy Transfer
No Change in State
q transferred = (sp. ht.)(mass)(∆T)
Heat Transfer
• Use heat transfer as a way to find specific heat
capacity, Cp
• 55.0 g Fe at 99.8 ˚C
• Drop into 225 g water at 21.0 ˚C
• Water and metal come to 23.1 ˚C
• What is the specific heat capacity of the metal?
Heating/Cooling Curve for Water
Note that T is
constant as ice
melts or water
boils
Thermochemical Equations
• Thermochemical equations are a balanced chemical reaction plus the H
value for the reaction.
– For example, this is a thermochemical equation.
• The stoichiometric coefficients in thermochemical equations must be
interpreted as numbers of moles.
• 1 mol of C5H12 reacts with 8 mol of O2 to produce 5 mol of CO2, 6 mol of
H2O, and releasing 3523 kJ is referred to as one mole of reactions.
Thermochemical Equations
•
This is an equivalent method of writing thermochemical equations.
•
•
H < 0 designates an exothermic reaction.
H > 0 designates an endothermic reaction
Thermochemical Equations
• Write the thermochemical equation for the reaction for
CuSO4(aq) + 2NaOH(aq)
Cu(OH)2(s) + Na2SO4(aq)
50.0mL of 0.400 M CuSO4 at 23.35 oC
Tfinal 25.23oC
50.0mL of 0.600 M NaOH at 23.35 oC
Density final solution = 1.02 g/mL
CH2O = 4.184 J/goC
Gas Laws
At the macroscopic level, a complete physical description of a sample
of a gas requires four quantities:
1. Temperature (expressed in K)
2. Volume (expressed in liters)
3. Amount (expressed in moles)
4. Pressure (given in atmospheres)
• These variables are not independent — if the values of any three of
these quantities are known, the fourth can be calculated.
Standard Temperature and Pressure
• Standard temperature and pressure is given the symbol STP.
– It is a reference point for some gas calculations.
• Standard P  1.00000 atm or 101.3 kPa
• Standard T  273.15 K or 0.00oC
– Gas laws must use the Kelvin scale to be correct.
• Relationship between Kelvin and centigrade.
Boyle’s Law: The Volume-Pressure Relationship
•
•
•
•
•
•
V  1/P or V= k (1/P) or PV = k
P1V1 = k1 for one sample of a gas.
P2V2 = k2 for a second sample of a gas.
k1 = k2 for the same sample of a gas at the same T.
Mathematically we write Boyle’s Law as P1V1 = P2V2
This relationship between pressure and volume is
known as Boyle’s law which states that at constant
temperature, the volume of a fixed amount of a gas is inversely
proportional to its pressure.
Robert Boyle (16271691). Son of Earl of
Cork, Ireland.
Charles’ Law: The Volume-Temperature Relationship;
The Absolute Temperature Scale
Charles’s and Gay-Lussac’s findings can be stated as:
At constant pressure, the volume of a fixed amount of a gas
is directly proportional to its absolute temperature
(in K).
This relationship is referred to as Charles’s law and is
stated mathematically as
V = (constant) [T (in K)] or V  T (in K, at
constant P).
Jacques Charles (17461823). Isolated boron and
studied gases. Balloonist.
Charles’ Law: The Volume-Temperature Relationship;
The Absolute Temperature Scale
Volume (L)
35
30
25
20
15
10
5
0
Gases liquefy
before reaching 0K
0
50
100 150 200 250 300 350 400
Temperature (K)
absolute zero = -273.15 0C
• Gay-Lussac showed that a plot of V versus T was a straight line that could be extrapolated to –
273.15ºC at zero volume, a theoretical state.
• The slope of the plot of V versus T varies for the same gas at different pressures, but the
intercept remains constant at –273.15ºC.
• Significance of the invariant T intercept in plots of V versus T was recognized by Thomson
(Lord Kelvin), who postulated that –273.15ºC was the lowest possible temperature that could
theoretically be achieved, and he called it absolute zero (0 K).
The Combined Gas Law Equation
• Boyle’s and Charles’ Laws combined into one statement
is called the combined gas law equation.
– Useful when the V, T, and P of a gas are changing.
Avogadro’s Law and
the Standard Molar Volume
• Avogadro’s Law states that at the same temperature and pressure,
equal volumes of two gases contain the same number of molecules (or
moles) of gas.
• If we set the temperature and pressure for any gas to be STP, then one
mole of that gas has a volume called the standard molar volume.
• Stated mathematically: V = (constant) (n) or V  n (at constant T
and P)
• The standard molar volume is 22.4 L at STP.
– This is another way to measure moles.
– For gases, the volume is proportional to the number of moles.
11.2 L of a gas at STP = 0.500 mole
44.8 L of a gas at STP = ? Moles
Boyle’s Law:
The Volume-Pressure Relationship
• At 25oC a sample of He has a volume of 4.00 x 102 mL under a
pressure of 7.60 x 102 torr. What volume would it occupy under
a pressure of 2.00 atm at the same T?
Charles’ Law:
The Volume-Temperature Relationship;
The Absolute Temperature Scale
• A sample of hydrogen, H2, occupies 1.00 x 102 mL at 25.0oC and
1.00 atm. What volume would it occupy at 50.0oC under the same
pressure?
The Combined Gas Law Equation
• A sample of nitrogen gas, N2, occupies 7.50 x 102 mL at 75.00C
under a pressure of 8.10 x 102 torr. What volume would it
occupy at STP?
Avogadro’s Law and the
Standard Molar Volume
• One mole of a gas occupies 36.5 L and its density is 1.36 g/L at a
given temperature and pressure. (a) What is its molar mass? (b)
What is its density at STP?
Summary of Gas Laws:
The Ideal Gas Law
•
What volume would 50.0 g of ethane, C2H6, occupy at 1.40 x
102 oC under a pressure of 1.82 x 103 torr?
Summary of Gas Laws:
The Ideal Gas Law
• Calculate the number of moles in, and the mass of, an 8.96 L
sample of methane, CH4, measured at standard conditions?
Dalton’s Law of Partial Pressures
1. The ideal gas law assumes that all gases behave identically. If V, T, and
n for each gas in a mixture are known, the pressure of each gas, its
partial pressure, can be calculated. Holding V and T constant, P is
directly proportional to n:
P = n(RT/V) = n(constant)
2. Generally, for a mixture of i components, the total pressure is given by
Pt = (n1 + n2 + n3 + - - - +ni) (RT/V).
3. The above equation makes it clear that, at constant T and V, P depends
on only the total number of moles of gas present, whether the gas is a
single chemical species or a mixture of gaseous species. Nothing in the
equation depends on the nature of the gas, only on the quantity.
4. The total pressure exerted by a mixture of gases is the sum of the partial pressures John Dalton 1766-1844
of component gases. This law is known as Dalton’s law of partial pressures and
can be written mathematically as
Pt = P1 + P2 + P3 - - - + Pi
where Pt is the total pressure and the other terms are the partial
pressures of the individual gases.
• Vapor Pressure is the pressure exerted by a
substance’s vapor over the substance’s liquid at
equilibrium.
• Partial pressure is the pressure the gas would exert if
it were the only one present (at the same temperature
and volume).
Dalton’s Law of Partial Pressures
• If 1.00 x 102 mL of hydrogen, measured at 25.0 oC and 3.00 atm
pressure, and 1.00 x 102 mL of oxygen, measured at 25.0 oC and
2.00 atm pressure, were forced into one of the containers at 25.0
oC, what would be the pressure of the mixture of gases?
Dalton’s Law of Partial Pressures
• A sample of hydrogen was collected by displacement of water at
25.0 oC. The atmospheric pressure was 748 torr. What pressure
would the dry hydrogen exert in the same container?
Mass-Volume Relationships
in Reactions Involving Gases
•In this section we are looking at reaction stoichiometry, like in
Chapter 3, just including gases in the calculations.
2 mol KClO3
2(122.6g)
yields
yields
2 mol KCl and 3 mol O2
2 (74.6g)
and 3 (32.0g)
Those 3 moles of O2 can also be thought of as:
3(22.4L)
or
67.2 L at STP
Mass-Volume Relationships in
Reactions Involving Gases
• What volume of oxygen measured at STP, can be produced by the
thermal decomposition of 120.0 g of KClO3?
Real Gases: Deviations from Ideality
The Ideal Gas Law ignores both the volume occupied
by the molecules of a gas and all interactions between
molecules, whether attractive or repulsive.
In reality, all gases have a volume and the molecules
of real gases interact with one another.
For an ideal gas, a plot of PV/nRT versus P gives a
horizontal line with an intercept of 1 on the PV/nRT
axis.
Real gases behave ideally at ordinary temperatures and
pressures. At low temperatures and high pressures real
gases do not behave ideally.
The reasons for the deviations from ideality are:
• The molecules are very close to one another,
thus their volume is important.
• The molecular interactions also become
important.
J. van der Waals, 1837-1923, Professor
of Physics, Amsterdam. Nobel Prize
1910.
Real Gases: Deviations from Ideality
The van der Waals’ equation
(P + an2/V2) (V – nb) = nRT
accounts for the behavior of real gases at low temperatures and high pressures.
The van der Waals constants a and b are empirical constants that
differ for each gas that take into account two things:
Pressure term, P + (an2/V2)
a corrects for intermolecular attractive forces that tend to
reduce the pressure from that predicted by the ideal gas
law
For nonpolar gases the attractive forces are London Forces
For polar gases the attractive forces are dipole-dipole
attractions or hydrogen bonds.
n2/V2 represents the concentration of the gas (n/V) squared
because it takes two particles to engage in the pairwise
intermolecular interactions
Volume term, V – nb, corrects for the volume occupied by the
gaseous molecules
b accounts for volume of gas molecules
At large volumes a and b are relatively small and the van der Waal’s
equation reduces to the ideal gas law at high temperatures and low
pressures.
Real Gases:
Deviations from Ideality
• Calculate the pressure exerted by 84.0 g of ammonia, NH3, in a 5.00 L
container at 200. oC using the ideal gas law.
PV = nRT
P = nRT/V n = 84.0g * 1mol/17 g T = 200 + 273
P = (4.94mol)(0.08206 L atm mol-1 K-1)(473K)
(5 L)
P = 38.3 atm
Real Gases: Deviations from Ideality
• Calculate the pressure exerted by 84.0 g of ammonia, NH3, in a 5.00 L container at
200. oC using the van der Waal’s equation. The van der Waal's constants for
ammonia are: a = 4.17 atm L2 mol-2 b =3.71x10-2 L mol-1
n = 84.0g * 1mol/17 g T = 200 + 273
P = (4.94mol)(0.08206 L atm mol-1 K-1)(473K)
5 L – (4.94 mol*3.71E-2 L mol-1)
P = 39.81 atm – 4.07 atm = 35.74
P = 38.3 atm
7% error
(4.94 mol)2*4.17 atm L2 mol-2
(5 L)2