Chapter 19

Chemical Thermodynamics

Chemical Thermodynamics

• The study of the energy transformations that accompany chemical and physical changes.

The Driving Forces

• All reactions (changes) in nature occur because of the interplay of two driving forces:

(1) The drive toward lower energy.

Decrease in enthalpy

(2) The drive toward increased disorder.

Increase in entropy

Enthalpy

• Enthalpy is the internal energy of a system at constant pressure.

• We cannot measure enthalpy directly but we can measure changes in the enthalpy of a system.

• Changes in enthalpy normally are in the form of heat.

Enthalpy (Energy) Change

• Exothermic reactions

–∆H

• Endothermic reactions

+∆H

Energy Entropy

Table 19.1 page 613

P4O10 + 6H2O(l) → 4H3PO4

Use the information below to determine the ∆H.

4P + 5O2 → P4O10 ∆H = -2984 kJ/mol

H2 + ½ O2 → H2O(l) ∆H = -285.83 kJ/mol

3/2 H2 + P + 2O2 → H3PO4 ∆H = -1267kJ/mol

∆H = -369 kJ/mol

2NH3 + 3O2 + 2CH4 → 2HCN + 6H2OUse the information below to determine the ∆H.

½ N2 + 3/2 H2 → NH3 ∆H = - 46 kJ/mol C + 2H2 → CH4 ∆H = -75 kJ/mol

½ H2 + C + ½ N2 → HCN ∆H = +135.1 kJ/mol

H2 + ½ O2 → H2O ∆H = -242 kJ/mol

∆H = -940 kJ/mol

Another method to determine ∆H for a reaction

∆H = ∑∆Hf(products) ─ ∑∆Hf(reactants)

Appendix I in your notebook has ∆Hf values

∆H = ∑∆Hf(products) ─ ∑∆Hf(reactants)

Determine ∆H for the reaction using the above formula.Na(s) + O2(g) + CO2(g) → Na2CO3(s)∆Hf for Na2CO3(s) = -1130.8kJ/mol

∆H = -737.3 kJ/mol

∆H = ∑∆Hf(products) ─ ∑∆Hf(reactants)

Determine ∆H for the reaction using the above formula.

C2H5OH(g) + 3O2(g)→ 2CO2(g) + 3H2O(g)

∆H = -1277.4 kJ/mol

Calculate the amount of energy released when 100.0g of C2H5OH is burned.

C2H5OH(g) + 3O2(g)→ 2CO2(g) + 3H2O(g)

∆H = -1277.4 kJ

-2772 kJ = 2772 kJ released

Change in Entropy (∆S)

(+∆S) = increase in disorder (entropy)

(-∆S) = decrease in disorder (entropy)

An Increase in Entropy (+∆S) Is a Force that Drives Physical and

Chemical Changes.

Second Law of Thermodynamics• Every time a change occurs it

increases the entropy of the universe.

Second Law of Thermodynamics• The second law implies that whenever

a change occurs, some of the energy will be wasted which will lead to an increase in the disorder of the universe.

Entropy • Entropy is a complicated concept to truly

understand.• It may be best to think of entropy as the degree of

dispersion.– As matter or energy disperses (becomes more free to

move or becomes more spread out) entropy will increase.

Entropy 1 Video ≈ 2:25

Entropy increases when matter is dispersed.

• Production of liquid or gas from a solid “or” production of gas from a liquid results in the dispersal of matter.

• The individual particles become more free to move and generally occupy a larger volumewhich causesentropy toincrease.

Increase in Entropy (+∆S)• Expansion of a gas.

Increase in Entropy (+∆S)

• Formation of a mixture.

Increase in Entropy (+∆S)• Dissolution of a crystalline solid in water.

Increase in Entropy (+∆S)

• More particles are created.

Increase in Entropy (+∆S)

• More particles are created.

Decrease in Entropy (-∆S)• Is simply a reverse of the previous processes.

A gas dissolves in a liquid A precipitate forms

Third Law of Thermodynamics

The entropy of any pure substance at 0 K is zero.

We can interpret this to mean that as temperature increases entropy increases.

A more complex molecule has a higher entropy.

Calculating Entropy Change

∆S = ∑S(products)

─ ∑S(reactants)

Determine ∆S for the reaction using the above formula.

C2H5OH(g) + 3O2(g) → 2CO2(g) + 3H2O(g)

∆S = +95.64 J/mol K

“Gibbs” Free Energy

• Allows use to determine whether a reaction is spontaneous or not spontaneous.

• We will say a process is or is not “thermodynamically favored” instead of using the terms spontaneous or not spontaneous.

• This avoids common confusion with associating the term spontaneous with the idea that something must occur immediately or without cause.

“Gibbs” Free Energy

• If a process is “thermodynamically favored” means the products are favored at equilibrium.

• The term “not thermodynamically favored” means the reactants are favored at equilibrium.

Change in Free Energy (ΔG)

• It is the change in free energy (ΔG) that determines whether a reaction is thermodynamically favorable or not.

∆G = ∑∆Gf(products) ─ ∑∆Gf(reactants)

Determine ∆G for the reaction using the above formula.

C2H5OH(g) + 3O2(g) → 2CO2(g) + 3H2O(g)

∆G = -1306 kJ/mol

(-∆G) vs (+∆G) • If -∆G the process is “thermodynamically

favored” and the products are favored at equilibrium.

• If +∆G the process is “not thermodynamically favored” and the reactants are favored at equilibrium.

Understanding ΔG

• Just because a process is spontaneous or “thermodynamically favored” (-∆G) does not mean that it will proceed at any measurable rate.

• A reaction that is thermodynamically favored, (spontaneous), may occur so slowly that in practice it does not occur at all.

Understanding ΔG

• Thermodynamically favored reactions that do not occur at any measurable rate are said to be under “kinetic control”.

• These reactions often have a very high activation energy.

• The fact that a process does not proceed at a noticeable rate does not mean that the reaction is at equilibrium.

• We would conclude that such a reaction is simply under kinetic control.

Example

The reaction of diamond (pure carbon) with oxygen in the air to

form carbon dioxide is thermodynamically

favored (spontaneous) but has an extremely

slow rate.

Understanding ΔG

• A reaction with a with a +ΔG may be forced to occur with the application of energy or by coupling it to thermodynamically favorable reactions.

+ΔG

• Energy can be used to cause a process to occur that is not thermodynamically favored.

• Using electricity to charge a battery.

+ΔG

• Energy can be used to cause a process to occur that is not thermodynamically favored.

• Photoionization of an atom by light.

+ΔG

• A thermodynamically unfavorable reaction may be made favorable by coupling it to a favorable reaction or series of favorable reactions.

• This process involves a series of reactions with common intermediates, such that the reactions add up to produce an overall reaction that is thermodynamically favorable (–ΔG).

C2H5OH(g) + 3O2(g) → 2CO2(g) + 3H2O(g)

∆H = -1277.3 kJ/mol

∆S = +95.64 J/mol K

∆G = -1306 kJ/mol

∆G = 0

Reactants ↔ Products

-∆G = thermodynamically favored = products favored

+∆G = not thermodynamically favored = reactants favored

∆G is 0 = equilibrium

∆G = ∆H – T∆S

∆G = ∆H – T∆S

C2H5OH(g) + 3O2(g) → 2CO2(g) + 3H2O(g)Determine ∆G given:

∆H = -1277.3 kJ/mol ∆S = +95.64 J/mol K

∆G = -1306 kJ/mol or -1,306,000J/mol

∆G = ∑∆Gf(products) ─ ∑∆Gf(reactants)

Determine ∆G for the reaction using the above formula.

C2H5OH(g) + 3O2(g) → 2CO2(g) + 3H2O(g)

∆G = -1306 kJ/mol

∆G = ∆H – T∆S

C2H5OH(g) + 3O2(g) → 2CO2(g) + 3H2O(g)Determine ∆G given:

∆H = -1277.3 kJ/mol ∆S = +95.64 J/mol K

∆G = -1306 kJ/mol

Solve for ∆H: ∆G = ∆H – T∆S

∆H = ∆G + T∆S

Solve for ∆S: ∆G = ∆H – T∆S

T

GHS

T

HGS

or

Solve for T: ∆G = ∆H – T∆S

S

GHT

S

HGT

or

CaO(s) + SO3(g) → CaSO4(s)

Calculate ∆S at 298K using the data given below.

∆H = -401.5 kJ/mol

∆G = -345.0 kJ/mol

∆S = -0.1896 kJ/mol K or -189.6 J/mol K

∆G = ∆H – T∆S• Can be used to explain the driving forces

and thermodynamic favorability.

C2H5OH(g) + 3O2(g) → 2CO2(g) + 3H2O(g)

∆H = -1277.3 kJ/mol (exothermic)

∆S = +95.64 J/mol K (increased disorder)

∆G = ∆H – T∆S

∆G = (-) – [+(+)]

negative – positive = always negative

∆G = -1305.2 kJ/mol (thermodynamically favored)

∆G = ∆H – T∆SWhat are the other possibilities?

∆G = ∆H – T∆S+∆H = endothermic-∆S = decreased entropy

∆G = ∆H – T∆S∆G = (+) – [+(-)]

positive – negative = always positive

+∆G = not thermodynamically favored

∆G = ∆H – T∆S+∆H = endothermic+∆S = increased entropy

∆G = ∆H – T∆S∆G = (+) – [+(+)]

positive – positive = ?

+ or - ∆G = may be thermodynamically favored

∆G = ∆H – T∆S-∆H = exothermic-∆S = decreased entropy

∆G = ∆H – T∆S∆G = (-) – [+(-)]

negative – negative = ?

+ or - ∆G = may be thermodynamically favored

What is the determining factor for thermodynamic favorability when the

factors ∆H and T∆S contradict?

∆G = ∆H – T∆S

∆G = (+) – [+(+)]

Temperature

∆G = ∆H – T∆S

∆G = (-) – [+(-)]

A process may be thermodynamically favored at one temperature and not at another.

• Is it thermodynamically favored for ice to melt or water to freeze?

CaO(s) + SO3(g) → CaSO4(s)The data given below are for 298K. Calculate ∆G using the ∆H and ∆S

values below at 2463K.

∆H = -401.5 kJ/mol∆G = -345.0 kJ/mol∆S = -189.6 J/mol K

∆G = +65.5 kJ/mol at 2463K

Conclusion: ∆S becomes a larger factor as temperature increases

CaO(s) + SO3(g) → CaSO4(s)Estimate the temperature at which the reaction

becomes spontaneous.∆H = -401.5 kJ/mol∆S = -189.6 J/mol K

T = 2118K = 1845°C

CaO(s) + SO3(g) → CaSO4(s)Estimate the temperature at which the reaction

becomes spontaneous.∆H = -401.5 kJ/mol∆S = -189.6 J/mol K

T = 2118K = 1845°C

Why is this temperature only an estimate of the temperature?

CaO(s) + SO3(g) ↔ CaSO4(s)Calculate ∆S at 298K using the data given below.

∆H = -401.5 kJ/mol∆G = -345.0 kJ/mol

∆S = -0.1896 kJ/mol K

• As the temperature increases above ≈ 2118K the reaction becomes nonspontaneous as ∆G becomes positive. Why?

Boiling & Equilibrium• Boiling is a reversible reaction:

H2O(l) ↔ H2O(g)

• What is the value of ∆G for the

boiling water?

Estimate the boiling point of ethanol (C2H5OH) .

BP = 350K = 77°C

Estimate the boiling point of ethanol (C2H5OH) .

Estimated BP = 350K = 77°C

Formula Names Boiling Point

C2H5OH Ethanol

Ethyl alcohol

Hydroxyethane

Grain alcohol

78.4°C

Why is there a discrepancy in the boiling point?

Thermodynamic Favorability (∆G) and

Equilibrium (K)

∆G K Reaction

Negative >1 Thermodynamically favored (products favored)

0 =1 At equilibrium

Positive <1 Not thermodynamically favored (reactants favored)

∆G° = -R T lnK

• R = 8.314 J/K

• Used to determine:(1) K when given ∆G°

(2) ∆ G° when a reaction is at equilibrium.

∆G° = -R T lnK

• Calculate ∆G° for a reaction at equilibrium where K = 1.

∆G° = -R T lnK

• If K ≠ 1 then ∆G° ≠ 0.

• This means the reactions is moving to the right towards products or moving to the left towards reactants and is not at equilibrium.

• LeChatlier’s Principle should help you to understand this.

• What does changing the temperature do to an equilibrium system.

What does changing the temperature do to an equilibrium system?

The reaction shifts right or left favoring the products or reactants

and raising or lowering K.

Every reaction has a temperature at which K = 1 and the reaction is neither

favoring products nor reactants

Reaction Kelvin

TemperatureK

The equilibrium constant for a reaction at 0°C is 1.657 x 10-5. What is ∆G°?

∆G° = 24980J/mol = 24.98 kJ/mol

Calculate the equilibrium constant for a reaction at 298K if ∆Gº = -228.59 kJ.

K = 1.17 x 1040

K vs ∆G

• Compare K and ∆G from the last two problems.

• ln K = -11.0079 and K = 1.657 x 10-5 and ∆G° = 24.98 kJ/mol

• ln K = 92.264 and K = 1.17 x 1040 and ∆G° = -228.59 kJ/mol

5 Formulas

1. ∆H = ∑∆Hf(products) ─ ∑∆Hf(reactants)

2. ∆S = ∑S(products)

─ ∑S(reactants)

3. ∆G = ∑∆G(products)

─ ∑∆G(reactants)

4. ∆G = ∆H – T∆S

5. ∆G° = -RTlnK

Entropy 2 Video ≈ 8:21