MR. SURRETTE
VAN NUYS HIGH SCHOOL
CHAPTER 20: CHEMICAL THERMODYNAMICS
CLASSNOTES
THERMODYNAMICS
Thermodynamics is the study of the energy that accompanies chemical or physical changes. Some
chemical reactions release heat energy. These are called exothermic reactions.
THERMODYNAMICS
Other reactions absorb heat energy and are called endothermic reactions.
HEAT OF REACTION
The discovery that almost all chemical reactions either absorb or release heat led to the idea that all
substances contain heat.
HEAT OF REACTION
Consequently, the heat of reaction is the difference in heat contents between the products and reactants.
ENTHALPY
Chemists use the term enthalpy for the heat content of a substance or the heat of a reaction.
HEAT OF REACTION EQUATION
This equation states that the change in enthalpy during a reaction equals the enthalpy of the products
minus the enthalpy of the reactants.
HEAT OF REACTION EQUATION
H = HP - HR
(H = heat of reaction
HP = enthalpy of products
HR = enthalpy of reactants)
ENTHALPY
Enthalpies can be either positive or negative. In general, substances that release heat when they are
formed have a negative enthalpy.
ENTHALPY
Substances that require heat for their formation have a positive enthalpy.
Example 1. Calculate the heat of reaction for the decomposition of ammonium nitrate:
NH4NO3(s)  N2O(g) + 2 H2O(g).
1|Page
CHEMISTRY
MR. SURRETTE
VAN NUYS HIGH SCHOOL
1A.
(1) Products = N2O(g) + 2 H2O(g)
(2) From the Standard Enthalpies of Formation data table: HP = 19.5 kcal + 2 (- 57.8 kcal)
(3) HP = - 96.1 kcal
(4) Reactants = NH4NO3
(5) From the Standard Enthalpies of Formation data table:
(6) H = HP – HR
(7) H = - 96.1 kcal – (- 87.4 kcal)
(8) H = - 8.7 kcal
(9) This is an exothermic reaction. The overall enthalpy is – 8.7 kilocalories. This means the
decomposition of one mole of ammonium nitrate releases 8.7 kilocalories of heat.
Sign of H
ENTHALPY AND HEAT FLOW
Type of Reaction
Negative
Positive
Exothermic
Endothermic
Heat
Flow
Released
Absorbed
Example 2. Reverse the previous reaction: N2O(g) + 2 H2O(g)  NH4NO3(s) Determine the new
heat of reaction.
2A.
If you reverse the previous reaction, the sign of enthalpy is also reversed: + 8.7 kcal. This reaction is
endothermic.
Example 3. Calculate the quantity of heat released when 100 grams of calcium oxide react with liquid
water to form Ca(OH)2(aq).
3A.
(1) Since calcium is an alkaline metal, it has a +2 charge. Oxygen is assigned a – 2 oxidation number
(O-2).
(2) According to conservation of charge, Ca+2 and O-2 must form CaO (molecular formula for calcium
oxide).
(3) Identify the Reactants: CaO(s) and H2O(l)
(4) Identify the Product: Ca(OH)2(aq).
(5) Balance the equation:
CaO(s) + H2O(l)  Ca(OH)2(aq)
(6) Product = Ca(OH)2(aq)
(7) From Enthalpies data table: HP = - 235.8 kcal
(8) Reactants = CaO(s) + H2O(l)
(9) From Enthalpies data table:
HR = -151.9 kcal + (- 68.3 kcal)
(10) HR = - 220.2 kcal
2|Page
CHEMISTRY
MR. SURRETTE
3A.
(11)
(12)
(13)
VAN NUYS HIGH SCHOOL
(continued...)
H = HP – HR
H = - 235.8 kcal – (- 220.2 kcal)
H = - 15.6 kcal per mole
(14) 100 g CaO
( 1 mol )
---------------- = 1.78 mol CaO
(56.08 g)
(15) 1.78 mol CaO
(- 15.6 kcal)
---------------(1 mol)
(16) H = - 27.8 kcal. This reaction is exothermic.
ENERGY AND ENTROPY
Many powerful equations in thermodynamics are based on a few fundamental principles. These
principles are called the laws of thermodynamics.
ENERGY AND ENTROPY
In this chapter, we will review the first two laws of thermodynamics.
THERMODYNAMICS FIRST LAW
The first law of thermodynamics asserts that energy is conserved during any process.
THERMODYNAMICS SECOND LAW
The second law of thermodynamics involves entropy, which is a statistical measure of the degree of
disorder in a chemical system.
ENTROPY
Chemists have assigned numerical quantities for the entropy of each substance. These are called
standard entropies. Standard entropies are symbolized by So (see Standard Entropies data table).
ENTROPY OF REACTION EQUATION
S = SP - SR
(S = entropy of reaction
SP = entropy of products
SR = entropy of reactants)
3|Page
CHEMISTRY
MR. SURRETTE
VAN NUYS HIGH SCHOOL
Example 4. Calculate the entropy change for the following reaction: H2O(l)  H2O(g).
4A.
(1) S = SP - SR
(2) S = 45.1 cal / deg - 16.7 cal / deg
(3) S = 28.4 cal / deg. The positive S encourages this reaction to happen.
GIBBS FREE ENERGY
The American physicist Josiah Gibbs introduced a thermodynamic quantity combining enthalpy and
entropy into a single value called free energy.
GIBBS FREE ENERGY EQUATION
G = GP - GR
G = change in free energy
GP = free energy of products
GR = free energy of reactants)
(G)
Chemical systems change toward minimum free energy. The sign of G predicts the behavior of a
proposed chemical reaction.
Sign of G
Negative
Zero
Positive
SIGN OF GIBBS FREE ENERGY
Reaction Behavior
Proceeds spontaneously to the right
Is at equilibrium
Will not proceed
Example 5. Calculate the free energy change for the reaction:
CO2(g) + SO2(g)  CO(g) + SO3(g).
5A.
(1) GP = CO(g) + SO3(g)
(2) GP = - 32.8 kcal + (– 88.5 kcal)
(3) GP = - 121.3 kcal
(4) GR = CO2(g) + SO2(g)
(5) GR = - 94.3 kcal + (- 71.8 kcal)
(6) GR = - 166.1 kcal
(7) G = GP – GR
(8) G = - 121.3 kcal – (– 166.1 kcal)
(9) G = + 44.8 kcal. The positive G means the reaction will not proceed to the right.
4|Page
CHEMISTRY
MR. SURRETTE
VAN NUYS HIGH SCHOOL
GIBBS FREE ENERGY
Sometimes free energy values for products and reactants are not available. Gibbs free energy can also
be calculated using three thermodynamic quantities.

GIBBS FREE ENERGY EQUATION
G = H – TS
G = change in free energy
H = change in enthalpy
TS = (temperature)(change in entropy)]
GIBBS FREE ENERGY EQUATION
Notes about Gibbs equation:
(1) Temperature is measured in degrees Kelvin.
(2) Both H and S are expressed in kilocalories.
Example 6. Calculate the free energy change at 400oC for the reaction:
2 CO(g) + O2(g)  2 CO2(g).
6A.
Use the following values from the Enthalpy and Entropy data tables:
(1) CO: Hof = - 26.4 So = 47.3
(2) O2: Hof = 0.0 So = 49.0
(3) CO2: Hof = - 94.0 So = 51.1
(4) HP = 2 CO2(g)
(5) HP = 2 (- 94.0 kcal)
(6) HP = - 188 kcal
(7) HR = 2 CO(g) + O2(g)
(8) HR = 2 (- 26.4 kcal) + (0.0 kcal)
(9) HR = - 52.8 kcal
(10) H = HP – HR
(11) H = - 188 kcal – (- 52.8 kcal)
(12) H = - 135.2 kcal
(13) SP = 2 CO2(g)
(14) SP = 2 (51.1 cal / deg)
(15) SP = 102.2 cal / deg
(16) SR = 2 CO(g) + O2(g)
(17) SR = 2 (47.3 cal / deg) + 49.0 cal / deg
(18) SR = 143.6 cal / deg
(19) S = SP – SR
(20) S = 102.2 cal / deg – 143.6 cal / deg
(21) S = - 41.4 cal/deg = - 0.0414 kcal/deg
(22) T = TC + 273
(23) T = 400 + 273
5|Page
CHEMISTRY
MR. SURRETTE
6A.
(24)
(25)
(26)
(27)
VAN NUYS HIGH SCHOOL
(continued...)
T = 673 K
G = H – TS
G = (- 135.2) – (673 K)(- 0.0414)
G = - 107.3 kcal
Quantity
Enthalpy
Entropy
Free energy
THREE THERMODYNAMIC QUANTITIES
Symbol
Measures
H
S
G
Heat
Disorder
Reactivity
Units
Energy
Energy / deg
Energy
6|Page
CHEMISTRY