Thermochemistry
Chapters 16 and 17
Thermodynamics – the study of energy and energy transfer.
Thermochemistry – the study of energy involved in chemical
reactions.
Law of Conservation Of Energy
– the total energy in the universe is constant.
- energy cannot be created or destroyed.
∆ Energy Universe = 0
Energy CAN:
- be transferred from one substance to another
- or be converted into various forms.
In order to study energy changes – we need to define what
part of the universe we are dealing with.
System - the part of the universe that is being studied or
observed.
Surroundings – everything else in the universe.
Ex. In this chemical rxn – the system
is the reactants and products in
the flask
- the surroundings is the flask,
the student, the air, and the
hand holding the flask.
Therefore:
Universe = System + Surroundings
∆ E universe = ∆ E system + ∆ E surroundings = 0
This relationship is known as the 1st Law of Thermodynamics:
“Any change in the energy of a system is accompanied
by an equal and opposite change in the energy of
the surroundings”
∆ E system = - ∆ E surroundings
the energy the system gains = the energy the surroundings loses
OR
- ∆ E system = ∆ E surroundings
the energy the system loses = the energy the surroundings gains
Depending on how the system is separated from its
surroundings, systems are defined in three different ways:
Open System – system is open to the surroundings –
both matter and energy can be exchanged between
the two.
Ex. A reaction in an open beaker.
Closed System – a system where energy can move
between the system and the surroundings – however,
matter cannot move.
Ex. A reaction in a stoppered Erlenmeyer flask.
Isolated System – a system is completely isolated from
its surroundings – neither matter nor energy is
exchanged between the two.
Ex. A reaction in a calorimeter.
Types Of Energy
1) Kinetic Energy - energy in motion
(always accompanied by a temperature change)
2) Potential Energy – energy that is stored
(changes in state and chemical rxns)
•
•
•
•
The SI unit for both types of energy is the joule (J).
One joule is equal to 1 kg∙m2/s2
Kinetic Energy is measured in J.
Potential Energy is measured in kJ.
Temperature Change and Heat
• Temperature, T, is a measure of the average kinetic
energy of the particles that make up a system.
• Temperature is measured in either Celsius degrees (oC)
or kelvins (K).
• Temperature change in a substance is an indication of a
change in kinetic energy.
- Symbolized by ∆T
∆T = Tf – Ti
change in temp = temp. final – temp. initial
Transfer of Kinetic Energy
• Heat, q, refers to the transfer of kinetic energy between
objects with different temperatures.
• It is measured in joules.
• When a substance absorbs heat, the average kinetic energy of
the particles increases – therefore the temperature increases.
Reconsidering the 1st Law of Thermodynamics:
q system = -q surroundings
The heat the system gains is equal to the heat the surroundings
loses.
OR
- q system = q surroundings
The heat the system loses is equal to the heat the surroundings
absorbs.
When substances with different temperatures come in
contact, kinetic energy is transferred from the particles of
the warmer substance to the particles of the cooler
substance.
Heat Transfers:
- from hot chocolate to your mouth
- from hot chocolate to the mug
- from the mug to your fingers
- from the hot chocolate to the air
You blow on the hot chocolate
- moves cool air over the hot chocolate
and the hot chocolate transfers heat to the
air.
Factors In Heat Transfer
Factors In Heat Transfer
Temperature and Energy Transfer
• Heat moves from greater to lower temperatures
• The greater the heat difference the greater the energy
transfer
Mass and Energy Transfer
• Mass is directly related to heat transfer.
• Is used in calculations – designated symbol m.
Types of Substances and Energy Transfer
• In calculating heat transfer, we do not use the type of
substance as a variable, but rather we use a variable that
reflects the individual nature of different substances –
the specific heat capacity.
Specific Heat Capacity (c) - the amount of energy
required to change one gram of a substance by
one degree Celsius.
• This reflects how well a substance stores energy.
• Units are J/g∙oC
• A substance with a large specific heat capacity can
absorb and release more energy that a substance with a
smaller specific heat capacity.
• Water has a very large specific heat capacity
= 4.184 J/g∙oC
See Table on Page 632
Calculating Heat Transfer
Three variables determine heat transfer:
1) temperature
2) mass
3) specific heat capacity
Mathematical Relationship:
Complete Practice Problems #1-4 Page 634
Basic Calorimetry
Calorimetry
• the technological process of measuring the changes in
kinetic energy.
• uses a calorimeter
Calorimeter – a basic calorimeter
contains water, a thermometer,
and an isolated system.
We will use a coffee-cup calorimeter or
a Poor Man’s Calorimeter
Basic Principles of Calorimetry
• The system is isolated – no heat is exchanged with the
surroundings outside the calorimeter.
• The amount of heat exchanged with the calorimeter
itself is small enough to be ignored.
• If something dissolves in or reacts with the calorimeter
water, the solution still retains the properties of water.
 specific heat capacity and density
Therefore:
q system = -q surroundings
The heat the system absorbs is equal to the
heat that the surroundings released.
OR
- q system = q surroundings
The heat the system releases is equal to the
heat the surroundings absorbs.
Sample Problem on Page 663
Page 664 - #1-4
16.2 Enthalpy Changes
Now we will consider changes in the potential energy of a
system....
Enthalpy (∆H)
• refers to the potential energy change of a system during
a process such as a chemical or physical change.
• measured at constant pressure
• units are kJ/mol
Enthalpy Changes in Chemical Rxns
• The enthalpy change of a chemical reaction represents
the difference between the potential energy of the
products and the potential energy of the reactants.
• In chemical rxns, potential energy changes result from
the breaking or the forming of bonds.
• A chemical bond is caused by the attraction between
electrons and the nuclei of the atoms. Energy is stored
in bonds.
▫ Breaking a bond will require energy.
▫ Creating a bond will release energy.
Net Absorption of Energy = Endothermic Rxn
Net Release of Energy = Exothermic Rxn
Chemists define the total energy of a substance at
a constant pressure as its enthalpy, H.
They use relative enthalpy of the reactants and
the products to determine the change in enthalpy that
accompanies the reaction.
Consider:
N2
triple bonds
+
O2
→
double bonds
2NO
single bonds
This rxn absorbs energy - less energy is released
when NO bonds are formed than the energy required to
break the double and the triple bonds
Representing Enthalpy Changes
∆H Rxn - enthalpy of reaction
- enthalpy change of a rxn
- dependent on conditions of temperature and
pressure
- it is the energy change associated with the reaction of
1 mole of a compound with another compound.
∆Ho Rxn will represent the enthalpy of rxn at SATP
(SATP = 1 bar, 25oC)
** people will say “heat of reaction”
** O means nought – means standard conditions or
standard state
3 Ways To Represent Enthalpy Changes
1. Thermochemical Equation
- a balanced chemical equation that indicates the amount
of heat absorbed or released in a chemical rxn.
ENDOTHERMIC
- because heat is absorbed in an endothermic rxn, the heat
term is included on the reactant side of the equation.
117.3 kJ + MgCO3(s) → MgO(s) + CO2(g)
EXOTHERMIC
- because heat is released in an exothermic rxn, the heat
term is included on the product side of the equation.
H2 (g) + ½ O2 (g) → H2O (l) + 285.8 kJ
2. Separate Equation Method
- for this method, the heat term is included on the right
side of the equation.
- the sign for the heat term must be included to show
when the equation is endothermic or exothermic.
ENDOTHERMIC
MgCO3(s) → MgO(s) + CO2(g)
∆Ho = +117.3 kJ
EXOTHERMIC
H2 (g) + ½ O2 (g) → H2O (l)
∆Ho = - 285.8 kJ
3. Enthalpy Diagrams
Enthalpy Diagram for an Exothermic Rxn
You should always
include:
• Title
• Correct graph
• reactants
• products
• y-axis
• x-axis
• ∆H
Types of Reactions
• In thermochemistry:
- formation
- combustion
- melting
- freezing
- condensation
- vaporization
- solution
Standard Molar Enthalpy Of Formation
• In a formation reaction, a substance is formed from its
elements in their standard states.
∆Hoform = standard molar enthalpy of formation
- the quantity of energy that is absorbed or
released when 1 mol of a substance is formed directly
from its elements in their *standard states.
- it is a value – a number.
* The standard state of an element is its most stable form
at SATP.
Formation Equation for water (liquid)
Thermochemical Equation:
1 H2 (g) + ½ O2 (g) → 1 H2O (l) + 285.8 kJ
Separate Equation Method:
1 H2 (g) + ½ O2 (g) → 1 H2O (l)
∆Hoform = -285.8 kJ
Standard Molar Enthalpy of Combustion
• In a combustion equation, a substance burns in oxygen
to form carbon dioxide and water.
∆Hocomb = the molar enthalpy of combustion
- the quantity of energy that is absorbed or
released when 1 mol of a compound burns in oxygen to
produce carbon dioxide and water.
- reactants and products should be in their
standard states.
Combustion of Propane
Thermochemical Equation:
1 C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (l) + 2323.7 kJ
Separate Equation Format:
1C3H8 (g) + 5 O2 (g) → 3 CO2 (g)+ 4 H2O (l)
∆Hocomb= -2323.7 kJ
Combustion Of Butane
Thermochemical Equation:
1 C4H10 (g) + 13 O2 (g) → 4 CO2 (g) + 5 H2O (l) + 3003.0 kJ
2
Separate Equation Format:
1C4H10 (g) +
13
O
2 2 (g)
Page 643 #15-18
→ 4 CO2 (g)+ 5 H2O (l)
∆Hocomb= -3003.0 kJ
Calculating Enthalpy Changes
• Use Stoichiometry!!!
How much heat is released when 50.00 g of methane
forms from its elements?
50.00 g CH4
x
1molCH 4
16.05 gCH 4
x
change from
grams to moles
232.4 kJ was released.
74.6kJ
 232.4kJ
1molCH 4
use the molar enthalpy
(obtained from the chart)
How much heat is released when 50.00 g of methane
combusts completely?
50.00 g CH4 x 1molCH 4 x 965.1kJ
16.05 gCH 4
1molCH 4
 3007kJ
Page 645 #19-23
Remember that:
at SATP: 1 mol of a gas = 22.4 L of the gas
1L = 1000 mL
Enthalpy Changes and Changes in State
• Changes in state involve changes in the potential energy
of a system.
• Changes that you know: melting, freezing, etc.
• The temperature of the system undergoing the change
remains constant but because the energy is being
absorbed or released as heat, the temperature of the
surroundings often changes.
• In general, the energy change associated with changes in
state are much smaller than those of chemical changes.
▫ Why? Intermolecular forces (L.D., D.D., H.B.) are easier to
break than intramolecular forces (ionic, covalent, metallic
bonds).
▫ State changes for water animation
▫ Solid changing state
Representing Enthalpy Changes For
Changes In State
• ∆Homelt = the molar enthalpy of melting (fusion)
- the quantity of energy that is absorbed when 1
mol of a compound changes state from a solid to a
liquid.
Ex. ∆Homelt = 27.2 kJ
27.2 kJ + NaCl (s) → NaCl (l )
• ∆Hofre = the molar enthalpy of freezing
- the quantity of energy that is released when 1
mol of a compound changes state from a liquid to a
solid.
Ex. ∆Hofre = - 27.2 kJ
NaCl (l ) → NaCl (s) + 27.2 kJ
• ∆Hovap = the molar enthalpy of vaporization
- the quantity of energy that is absorbed when 1
mol of a compound changes state from a liquid to a gas.
207 kJ + NaCl (l ) → NaCl (g)
• ∆Hocond = the molar enthalpy of condensation
- the quantity of energy that is released when 1
mol of a compound changes state from a gas to a liquid.
NaCl (g) → NaCl (l ) + 207 kJ
Note:
∆Homelt = - ∆Hofre
∆Hovap = - ∆Hocond
See Table 16-4 on Page 647
• ∆Hosoln = the molar enthalpy of solution
- the quantity of energy that is absorbed or
released when 1 mol of a compound dissolves in a
solvent.
• Dissolving can be exothermic or endothermic.
• Manufacturers use this information to make hot packs
and/or cold packs.
• Hot Packs – process of dissolving is exothermic
• Cold Packs – process of dissolving is endothermic
Ex.
25.7 kJ + NH4NO3 (s) → NH4NO3 (aq)
Ex.
CaCl2 (s) → CaCl2 (aq) + 82.8 kJ
Calorimetry for Potential Energy Changes
• We can re-look at the theory for calorimetry when
considering state changes and chemical rxns.
Consider Melting Ice:
When a measured amount of hot water is placed in a
calorimeter and ice is added to it...
• heat travels from the water into the ice.
• the energy the ice absorbs is used to melt the ice.
- q water = + q ice
The heat the water releases is equal to
the heat the ice absorbs
Better:
The heat the water releases is equal to
the enthalpy the ice absorbs
• Water undergoes a kinetic energy change (there is a
temperature change)
we can use q = mc∆T
Then,
- q water = + q ice
• Ice undergoes a potential energy change (no temperature
change)
we can use stoichiometry.
• Note: We do not mass the ice in an analytical balance.
Take the mass of water when the lab is finished and
subtract the initial mass of water and you will obtain the
mass of ice that melted.
Sample Problem:
A student takes 100.0 mL of water at 51.8oC and places it in a
calorimeter. Ice is added to the water until it reaches 0.0oC.
At that point, all the ice is removed and the water is remeasured in a graduated cylinder. The final volume of water
is 165.7 mL. Calculate the molar enthalpy of melting.
qw = mwcw∆Tw
qw = (100.0 g)(4.184 J/g∙oC)(-51.8oC)
= -21673.12 J = -21670 J
- q water = + q ice
-21670 J = +21.67 kJ
∆Hmelt =
kJ  21.67 kJ
moles
3.646mol
 5.94kJ / mol
You need the moles of ice:
165.7 mL – 100.0 mL = 65.7 mL
65.7 mL x 1molH 2O
18.02 gH 2O

3.646molH 2O
Calculating the Molar Enthalpy of Melting
Problems
1. A student adds 7 ice cubes to 101.6 mL of water 38.4oC.
When the ice melted, the temperature of the water is
0.0oC. The final volume of water is 150.3 mL.
Calculate the molar enthalpy of melting of water.
= 6.04 kJ/mol
2. A student takes 225.4 mL of water at 52.6oC and places
it in a calorimeter. Ice is added to the water until it
reaches 0.0oC. At that point, all the ice is removed and
the water is re-measured in a graduated cylinder. The
final volume of water is 371.2 mL. Calculate the molar
enthalpy of melting.
= 6.13kJ/mol
Dissolving A Substance
• When a measured amount of water is placed in a
calorimeter and a solid substance is dissolved in it...
• Energy will travel from the water to the solid or from
the solid to the water.
Exothermic Dissolving
• the heat the water gains is equal to the enthalpy the
process of dissolving releases.
Endothermic Dissolving
• the heat the water loses is equal to the enthalpy the
process of dissolving absorbed.
Exothermic Dissolving
+ qwater = - qsoln
Endothermic Dissolving
- qwater = + qsoln
Use stoichiometry – using the mass of the
substance dissolved – to determine the molar enthalpy
of solution for the substance.
When a 4.25 g sample of ammonium nitrate dissolves in
60.0 g of water in a calorimeter, the temperature drops
from 22.0oC to 16.9oC. Calculate the ∆Hsoln for
ammonium nitrate.
Water
q=mc∆T
= (60.0g)(4.184J/goC)(-5.1oC)
= -1280 J
- qw=qsoln
-1280 J = 1.280 kJ
Ammonia Nitrate
4.25 g x 1 mol NH4NO3 = 0.05309 mol
80.06 g NH4NO3
∆Hsoln = kJ = 1.280 kJ
= 24 kJ/mol
mol
0.05309 mol
Calculating the Molar Enthalpy of Solution
1. When a 1.80 g sample of magnesium hydroxide dissolves in 108.9
g of water in a calorimeter, the temperature rises from 22.0oC to
83.9oC. Calculate the ∆Hsoln for magnesium hydroxide.
= -914 kJ/mol
2. A 9.96 g sample of lithium chloride was dissolved in 101.2 mL of
water at 22.2oC. When the salt is dissolved, the temp. of the soln
was 41.3oC. Calculate the molar enthalpy of solution for lithium
chloride.
= -34.4 kJ/mol
3.
A student dissolves 1.96 g of NaOH in 99.7 mL of water at 23.4oC.
After the NaOH dissolves, the temp. of the water rises to 28.7oC.
What is the molar enthalpy of solution?
= -45 kJ/mol
Calculating Molar Enthalpy of Rxn
• Coffee-cup calorimetry can be used to study changes in
dilute aqueous solutions.
• The water in the calorimeter absorbs (or provides) the
energy that is released (or absorbed) by the chemical
rxn.
+ qsoln = - qrxn or
- qsoln = + qrxn
Same calculations as melting and solution... We just
need to determine the limiting reagent when calculating
the molar enthalpy.
Sample Problem Page 665
• Copper sulfate, CuSO4, reacts with sodium hydroxide, NaOH, in a
double displacement reaction. A precipitate, Cu(OH)2, is formed.
 CuSO4 + 2 NaOH
→ Cu(OH)2 + Na2SO4
50.0 mL of 0.300 mol/L CuSO4 reacts with 51.8 mL of 0.600 mol/L
NaOH. The initial temperature of both solutions is 21.4oC. After
mixing the solutions in a coffee-cup calorimeter, the highest
temperature reached is 24.6oC. Determine the enthalpy change for
the reaction and then write a thermochemical equation.
Page 667
1. Consider the following rxn:
HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)
The chemist uses a coffee-cup calorimeter to neutralize completely 61.1 mL of 0.543
mol/L HCl with 62.5 mL of 0.543 mol/L of NaOH. The initial temperature of both
solutions is 17.8oC. After neutralization, the highest recorded temperature is
21.6oC.
2. Consider the following rxn:
Mg (s) + 2 HCl (aq) → MgCl2 (aq) + H2 (g)
The chemist uses a coffee-cup calorimeter to react completely 0.500 g of magnesium
ribbon with 100.0 mL of 1.00 mol/L of HCl. The initial temperature of the HCl is
20.4oC. After the reaction is complete, the highest recorded temperature is 40.7oC.
3. Consider the following rxn:
HNO3 (aq) + KOH (aq) → KNO3 (aq) + H2O (l)
∆Hrxn = -53.4 kJ/mol
The chemist uses a coffee-cup calorimeter to neutralize completely 55.0 mL of 1.30
mol/L of HNO3 with 58.9 mL of 1.50 mol/L of KOH. The initial temperature of
both solutions is 21.4oC. What is the final temperature of the mixture?
Heating and Cooling Curves
• Chemists use heating and cooling curves to represent
•
•
•
•
•
•
temperature and phase changes in substances.
Temperature goes on the y-axis
Time goes on the x-axis
Key points – boiling points, melting points,
freezing points, etc.
Flat lines - shows a state change
- potential energy changes
- no temperature change
Slanted lines – shows temperature change
- changes in kinetic energy
Curves do not need to show all state changes – you may
select a temp range of your choice.
Heating Curve for Water
Areas 1, 3, 5
- temp. changes
- Kinetic energy
changes
Areas 2 and 4
- no temp. change
- potential energy
changes
Cooling Curve for Water
Heating Curve for Water
Temp. Range -15oC to 120oC
Calculating Energy Changes
• When calculating energy changes that occur through
both potential and kinetic energy – you need to calculate
the changes for each area, convert the different units to
one unit and then add them all together.
• For areas with kinetic energy changes –
Use q = mc∆T
• For areas with potential energy changes –
Use stoichiometry
Page 654
Strategy:
1. Draw a quick curve to determine how many types of
energy are involved.
2. Use q=mc∆T for changes in Kinetic Energy.
3. Use stoichiometry for changes in Potential Energy.
4. Convert the numbers to the same units and add
together.
Sample Problem #2
A sample of solid mercury having a mass of 3.45 g is
heated from -55.7oC to 68.4oC. The melting point of
mercury is -38.8oC. The specific heat capacity of liquid
mercury is 0.140 J/g∙oC and the specific heat capacity of
solid mercury is 0.145 J/g∙oC. What is the total heat
gained by the metal?
= 463 J
Problems #30-34 Page 655
Calculating Enthalpy Changes for Reactions
• For some chemical reactions we cannot use coffee-cup
calorimetry to determine the enthalpy change.
• These reactions may be:
Too explosive
Too slow
Not aqueous in water
• In this case – there are three possible methods to use:
▫ Hess’ Law
▫ Using Enthalpies of Formation
▫ Using Bond Energies
1. Hess’ Law of Summation
• Hess’ Law – the enthalpy change of a physical or chemical
process depends only on the beginning conditions (reactants)
and the end conditions (products). The enthalpy change is
independent of the pathway of the process and the number of
intermediate steps in the process. It is the sum of all the
enthalpy changes of all the individual steps that make
up the process.
• Hess’ Law is valid because enthalpy is a state function.
A state function is a property that is determined only by
the current conditions of the system – it is not dependent
on the path taken by the system to reach those
conditions.
Ex. Carbon dioxide can be formed by two different
pathways:
Pathway #1
C(s) + ½ O2 (g) → CO (g)
then, CO(g) + ½ O2 (g) → CO2 (g)
Pathway #2
C(s) + O2 (g) → CO2 (g)
∆Ho = -110.5 kJ
∆Ho = -283.0 kJ
∆Ho = -393.5 kJ
Adding the two equations from Pathway #1 will result
in the same equation as Pathway #2
Two Ways To Manipulate Equations
• Reverse the equations.... reactants become
products and products become reactants.
You MUST reverse the enthalpy sign.
• Multiply each coefficient by the same integer or
fraction.
You MUST multiply the enthalpy value by the
same integer or fraction.
Page 680
Practice Problems #11-14
2. Using Standard Molar Enthalpies of
Formation to Calculate Enthalpy
Changes for Reactions.
• Use the following formula:
∆Horxn = ∑(∆Hoform
PRODUCTS)
- ∑ (∆Hoform
REACTANTS)
Consider:
CH4(g) + 2 O2 (g) → CO2 (g) + 2 H2O (g)
Practice Problems #20-22 Page 688
3. Using Bond Energies to Calculate
Enthalpy Changes for Reactions.
• Use the following formula:
∆Horxn = ∑(Bond Energies
REACTANTS)
- ∑ (Bond Energies PRODUCTS)
Consider:
CH4(g) + 2 O2 (g) → CO2 (g) + 2 H2O (g)
Practice Problems #23-26 Page 690