Thermochemistry

Chapters 3, 8, & 12 in
Introductory Chemistry
Thermochemistry

● Thermochemistry: the study of energy changes that occur during phase

changes and chemical reactions
Crash Course Chemistry: Energy & Chemistry
Part 1: Energy
Sections 3.8-3.9 in Introductory Chemistry
Energy
● Energy: the ability to do work
○ Work: the result of a force acting
on a distance

Energy is important because it drives the

behavior of matter.

● Ex:
○ Light
○ Heat
○ Electricity
Law of Conservation of Energy
● Like matter, energy is conserved.

● Law of Conservation of Energy:

states that energy cannot be
created or destroyed

● However, energy can be changed

from one form to another.
Forms of Energy
Kinetic Energy
● Energy of motion or movement
Potential Energy
● Energy due to physical position or
chemical composition

● Chemical potential energy: energy

stored within the chemical bonds of a
substance.

This energy is released or reapportioned in chemical

reactions.
Electrical Energy
● Energy due to flow of electrical charge
Thermal Energy
● The total amount of kinetic energy
in all of the particles in an object
or substance

The kinetic energy comes from particle

movement!

● A substance’s temperature is a
measure of its thermal energy.
Thermal Energy
● The hotter an object is…
○ The greater the random
motion of the atoms
and particles that
compose it
○ The more thermal
energy it contains
○ The higher its
temperature will be
Energy in Chemical Reactions

● In a chemical reaction, either:

1. Potential energy is converted to thermal energy

● Heat is given off
● Feels hot

2. Thermal energy is converted to potential energy

● Heat is absorbed
● Feels cool
Energy in Chemical Reactions
Exothermic reaction: energy Endothermic reaction:
is released energy is absorbed
Units of Energy
● joule (J): the SI unit for energy
● calorie (cal): the amount of
energy required to raise the
temperature of 1 g of pure water
by 1°C
● Calorie (Cal): a nutritional
calorie. 1 Cal is equal to 1000 cal
(so 1 Cal is also a kilocalorie)
Don’t forget your metric prefixes!
● kilowatt-hour (kWh): a unit of
kilo = 1000
energy by which electricity is
sold
Worked Example: Energy Unit Conversion

A candy bar contains 225 Cal of energy. How many joules (J) of energy does it
contain?

GIVEN: 225 Cal CONVERSION FACTORS:

FIND: J
Practice
How many calories are in 287 J?
Practice
How many kJ are equal to 1478 cal?
Part 2: Heat
Sections 3.11-3.12 in Introductory Chemistry
Heat
● Heat is not the same thing as
temperature!

● Temperature: a measure of the

thermal energy of matter (not
the exchange of energy)
Heat (q)
● Heat (q): the transfer or exchange of
thermal energy caused by a
temperature difference.

Heat always moves from a hotter object to a

colder object.
Heat & Temperature Changes
● All substances change temperature when they are heated.

● However, some substances are more resistant to temperature changes than

others.

● There are 2 properties that affect temperature changes:

1. Heat Capacity
2. Specific Heat
1. Heat Capacity
● Heat capacity: the amount of heat
(usually in J) needed to raise the
temperature of a substance by 1°C or
1 K.

● The higher the heat capacity of a

substance, the more energy must be
put in to change its temperature.
1. Heat Capacity
● Extrinsic property - depends on mass of the substance

100 g H2O
50 g H2O
2. Specific Heat (C)
● Specific heat (C): the amount of heat needed
to raise the temperature of exactly 1 g of a
substance by 1°C
○ Unit: J/g•°C

Specific heat is like heat capacity but just specific for 1 g

of the substance!

● The higher the specific heat of a substance,

the more energy must be put in to change its
temperature.
2. Specific Heat (C)
● Intrinsic property - depends only on the substance

100 g H2O
50 g H2O
Heat Calculations
● When a substance changes temperature, the change is affected by three
factors:
1. Amount of heat absorbed or released
2. Mass of the substance
3. Composition of the substance (specific heat)
Heat Calculations
● With heat calculations, q can be a positive or a negative value.
○ +q: heat is absorbed from the surroundings
○ -q: heat is released to the surroundings
Worked Example: Heat Calculation
Gallium is a solid metal at room temperature but melts at 29.9°C. If you hold gallium in your
hand, it melts from your body heat. How much heat must 2.5 g of gallium absorb from your
hand to raise the temperature of the gallium from 25.0°C to 29.9°C? The specific heat of
gallium is 0.372 J/g•°C.

GIVEN
m = 2.5 g
Ti = 25.0°C
Tf = 29.9°C
C = 0.372 J/g•°C

FIND q
Practice q = m•C•ΔT
How much heat (in J) is needed to raise the temperature of 27.0 g of water from
10°C to 90°C? (Cwater = 4.184 J/g•°C)
Practice q = m•C•ΔT
The temperature of a sample of iron with a mass of 10 g changed from 50.4°C to
25°C with the release of 114 J. What is the specific heat of iron?
Calorimetry
● Calorimetry: the accurate and precise measurement of heat change for
chemical and physical processes.

● Calorimeter: a device used to measure the amount of heat absorbed or

released during these processes.
“Coffee Cup” Calorimeter
● Constant-pressure calorimeter

● The thermometer records temperature

change as the substances in the water
react.

● The temperature change is then

mathematically converted into units of
energy - heat!
Crash Course Chemistry: Calorimetry
Part 3: Enthalpy
Sections 8.7, 12.4, & 12.5 in Introductory Chemistry
Crash Course Chemistry: Enthalpy
Enthalpy (H)
● System: the specific part of the
universe that contains the reaction
or process you wish to study

● Surroundings: everything else in the

universe other than the system

● Enthalpy (H): the heat content of a

system at constant pressure
Enthalpy Change (ΔH)
● We can’t measure the actual energy or
enthalpy of a substance.

● However, we can measure the change

in enthalpy, or ΔH. This is the heat
absorbed or released during:
○ Phase changes
○ Chemical reactions
Phase Changes & ΔH
● The sign of ΔH (positive or negative)
depends on the direction in which heat
flows when the phase change takes place.

● Exothermic phase changes: freezing,

condensation
○ Energy flows from the system to the
surroundings (out) when the phase
change takes place
○ -ΔH
Phase Changes & ΔH
● Endothermic phase changes: melting,
vaporization
○ Energy flows from the
surroundings to the system (in)
when the phase change takes
place
○ +ΔH
Phase Change Problems Using ΔH
● For a phase change, ΔH will be given in units of
kJ per mol (kJ/mol).

This means that for 1 mol of a substance to undergo the

phase change, a certain amount of heat (in kJ) will be needed.

● Ex: ΔH for ice to melt is 6.01 kJ/mol. For 1 mol

of ice to melt, it has to absorb 6.01 kJ of heat.
Phase Change Problems Using ΔH
● To calculate the amount of heat involved during a phase change for a
particular sample, we use dimensional analysis to convert moles (or grams) of a
substance to kJ.

● We can do this by using the ΔH provided as a conversion factor that relates

moles to kJ!
Worked Example: Phase Change Problem
How much heat is absorbed when 3.20 mol of ammonia (NH3) is converted from a
liquid to a gas at its boiling point? The ΔH for this phase change is 23.3 kJ/mol.

GIVEN: RELATIONSHIP USED: 1 mol = 23.3 kJ

3.20 mol NH3

FIND:
kJ

3.20 mol x 23.3 kJ = 74.5 kJ

1 1 mol
Practice
How much heat is released when 10.0 g of water is converted from a liquid to a
solid at its freezing point? The ΔH for this phase change is -6.01 kJ/mol.
Chemical Reactions & ΔH
● ΔHrxn: the amount of heat that is emitted or absorbed when a chemical
reaction occurs at a constant pressure

● The sign of ΔHrxn (positive or negative) depends on the direction in which

heat flows when the reaction takes place.
Chemical Reactions & ΔH
● Exothermic reactions:
A + B → C + Energy
○ Reactants contain more
stored energy than the
products
○ Energy flows from the system
to the surroundings (out)
when the reaction takes place
○ -ΔHrxn
Chemical Reactions & ΔH
● Endothermic reactions:
A + Energy → B + C
○ Products contain more
stored energy than the
reactants
○ Energy flows from the
surroundings to the system
(in) when the reaction takes
place
○ +ΔHrxn
Thermochemical Equations
● Balanced equations that include physical state and ΔHrxn

● Indicate whether a reaction is exothermic or endothermic

● Thermochemical equations can be written in 2 ways:

1. kJ written as a reactant
or product

2. ΔHrxn out to the side

Practice
Determine whether the thermochemical equations shown below indicate an
exothermic or endothermic reaction.
Thermochemical Equations & Stoichiometry
● We can create ratios between moles of reactants and products and kJ of
energy using balanced thermochemical equations.

CH4 + 2O2 → CO2 + 2H2O ΔH = -891 kJ

● This allows us to convert between moles and kJ - or even grams and kJ!
Worked Example: Thermochemical Equation
Stoichiometry
CH4 + 2O2 → CO2 + 2H2O ΔH = -891 kJ
Calculate the amount of heat (in kJ) associated with burning 24.0 g of CH4.

GIVEN: 24.0 g CH4 RELATIONSHIPS USED: 1 mol CH4 = 16.042 g CH4

FIND: kJ 1 mol CH4 = -891 kJ

24.0 g CH4 1 mol CH4 -891 kJ

x x = -1.34 x 103 kJ
1 16.042 g CH4 1 mol CH4
Practice
2KClO3 + 84.9 kJ → 2KCl + 3O2
Calculate the amount of heat (in kJ) required to break down 152.0 g of KClO3.
The Organic Chemistry Tutor: Thermochemical Equations Practice Problems
Hess’s Law
● Hess’s Law: says that the enthalpy change for a reaction is the sum of the
enthalpy changes for a series of reactions that add up to the overall reaction.

This law allows chemists to calculate ΔH for reactions that occur in steps.
Part 4: Reaction
Spontaneity, Entropy, &
Free Energy
Crash Course: Entropy
Entropy (S)
● Entropy (S): the degree of disorder
or randomness in a system

● Entropy is influenced by:

○ Number of particles in a
substance
○ Disorder (movement) of
particles in a substance
Entropy (S)
● The more particles a substance has, the greater its entropy (S).

● Ex: NaCl vs C6H12O6

C6H12O6 has greater entropy because there are more particles (more atoms).
Entropy (S)
● The more disordered the particles in a substance are (the more they move),
the greater its entropy (S).

● Ex: H2O(s) vs. H2O(g)

H2O(g) has greater entropy because in the gas state, the particles have more freedom of movement.
Changes in Entropy (ΔS)
● Processes that increase entropy (+ΔS) include:
1. Phase changes that lead to increased particle movement
■ Melting (solid → liquid)
■ Vaporizing (liquid → gas)
■ Sublimation (solid → gas)
2. Dissolving a solid or liquid in solution (or in water)
■ Ex: NaCl(s) → Na+(aq) + Cl-(aq)
3. Increasing the temperature of a substance
4. Increasing number of particles through a chemical reaction
Changes in Entropy (ΔS)
● Processes that decrease entropy (-ΔS) include:
1. Phase changes that lead to decreased particle movement
■ Freezing (liquid → solid)
■ Condensation (gas → liquid)
■ Deposition (gas → solid)
2. Dissolving a gas in solution (or in water)
■ Ex: CO2(g) → CO2(aq)
3. Decreasing the temperature of a substance
4. Decreasing the number of particles through a chemical reaction
Gibbs Free Energy (Gsystem)

● Gibbs free energy (Gsystem) : the amount of

energy in a system that is available to do
work

● More often, we’re actually interested in the

free energy change of a system: ΔGsystem
Calculating ΔGsystem

● ΔGsystem is the difference between the system’s change in enthalpy (ΔHsystem)

and the product of the Kelvin temperature (T) and the change in entropy
(ΔSsystem)

ΔGsystem= ΔHsystem -
TΔSsystem
Worked Example: Calculating ΔGsystem
For a process, the enthalpy change of a system is -91.8 kJ and the entropy change
is -0.197 kJ/K. Calculate the free energy change of the system at a temperature of
290 K.

GIVEN: ΔHsystem = -91.8 kJ ΔGsystem= ΔHsystem -

T = 290 K
TΔSsystem
ΔGsystem = -91.8 kJ - (290 K)(-0.197 kJ/K)
ΔSsystem = -0.197 kJ/K
= -34.67 kJ
FIND: ΔGsystem
Practice ΔGsystem= ΔHsystem -
TΔSsystem
For a process that happens at 382 K, ΔHsystem = 145 kJ and ΔS = 0.322 kJ/K. What
is the ΔG?
Test Question Practice!
CH4 + 2O2 → CO2 + 2H2O + 891 kJ
● Is this exothermic or endothermic?
● What is the ΔH?
● How many moles of CH4 are needed to release 891 kJ?
CH4 + 2O2 → CO2 + 2H2O + 891 kJ
● How many moles of O2 are needed to release 891 kJ?

● How much energy is released by:

○ 3.12 mol CH4?

○ 78.3 g O2?
ClF(g) + F2(g) → ClF3(g)
How does the entropy change for the reaction above?

Why?