V. Thermochemistry

Key focus of this chapter: enthalpy change

This chapter focuses on enthalpy change and gives concise summaries of the important things about energy, work by pressure, heat energy, Hess’s law, and entropy change in more detail.

Thermochemistry: The study of heat absorbed or released during chemical reactions.

A. Energy (E)

: The capacity to do work (W) or transfer heat (Q).

ΔE (change in energy) = E_final – E_initial

ΔE = W + Q

W: work

Q: heat

• System: the part of the universe being studied (reactants and products).

• Surroundings: everything outside the system.

Surroundings System

(+) energy

(-) energy

• Work is done on the system

• Work is done by the system

Q (at constant pressure, Q_p = ΔH)

• Heat is absorbed by the system

from the surroundings

• Heat is released by the system

to the surroundings

• Qs/ What is ΔE when the system does 100 J of work and releases 80 J of heat?

ΔE = W + Q

= (-100 J) + (-80 J)

= -180 J

B. Pressure-volume work

1. Pressure

P =

P: pressure (Pa), F: force (N), A: area (m²)

• 1 atm = 760 mmHg (torr)

2. Work (W)

: Work associated with a pressure-volume change.

W = -P_ext ΔV

P_ext: external pressure

ΔV = V_final – V_initial

System

• Work is done on the system

• ΔV < 0 (compression)

• Compression gives W > 0

- W

System

• Work is done by the system

• ΔV > 0 (expansion)

• Expansion gives W < 0

• Qs/ What is the work when the volume expands from 8.0 L to 10.0 L against an external

pressure of 3.0 atm?

W = -P_ext ΔV (work is done by the system)

= -(3.0 atm)(10.0 - 8.0 L)

= -6.0 L·atm

C. Heat energy, Q

Q = C·m·ΔT

C: specific heat capacity

m: mass of substance (g)

ΔT: temperature change

• Calorie: the heat required to raise the temperature of 1 g of water by 1 °C.

** 1 cal = 4.184 J

• Specific heat, C: the heat required to raise the temperature of 1 g of a substance by 1 °C.

C =

** Specific heat of water is 4.18 J/(g·°C).

• Qs/ Calculate the heat required to raise the temperature of 100 g of water from 50 °C to 80 °C.

Q = C·m·ΔT

= 4.18 J/(g·°C) × 100 g × (80 - 50) °C

= 12,540 J

• Qs/ Calculate the specific heat of gold when 650 J of heat raises the temperature of

50 g of gold from 25 °C to 125 °C.

C =

= 0.13 J/(g·°C)

D. Laws of thermodynamics

First law of thermodynamics

• Law of conservation of energy: Energy is neither created nor destroyed;

it is transferred or converted from one form to another.

Second law of thermodynamics

• Second law: For a spontaneous process, the total entropy change of the universe

(system + surroundings) is positive: ΔS_univ > 0.

E. Enthalpy change, ΔH

: ΔH = H_final – H_initial

Exothermic processes, ΔH < 0

Endothermic processes, ΔH > 0

Diagrams

Graph and equation

A + B → C + D (ΔH < 0)

• Reactants have greater enthalpy

than products.

A + B → C + D (ΔH > 0)

• Reactants have lower enthalpy than

products.

System

• Releases heat to the surroundings

• System enthalpy decreases

• Bond formation usually releases energy

• State change: gas → liquid → solid

• Absorbs heat from the surroundings

• System enthalpy increases

• Bond breaking requires energy

• State change: solid → liquid → gas

Surroundings

• Temperature tends to increase

• Temperature tends to decrease

Examples

• Fuel combustion; acid-base

neutralization; glucose catabolism;

mixing water with strong acids

• Melting salts; photosynthesis; cold packs made by dissolving ammonium nitrate in water

** The total energy of the system plus surroundings is conserved in both exothermic and endothermic processes.

F. Hess’s Law

• The ΔH of an overall reaction equals the sum of the ΔH values for the individual steps.

• Enthalpy is a state function, so the enthalpy change depends only on the initial and final states,

not on the reaction pathway.

A + B C + D Δ H 1 = -2 Δ H 3 = -5 E + F Δ H 2 = -3

A + B → C + D, ΔH₁ = -2 ——– (1)

C + D → E + F, ΔH₂ = -3 ——– (2)

—————————————————–

A + B → E + F, ΔH₃ = -5 ——– (3)

By Hess’s law: (1) + (2) = (3); therefore ΔH₁ = ΔH₃ – ΔH₂.

• Qs/ Calculate the enthalpy change, ΔH, for the reaction S(s) + O₂(g) → SO₂(g) from the

following information.

Table A

2S(s) + 3O₂(g) → 2SO₃(g) ΔH₁ = -50 kJ/mol

SO₂(g) + 1/2 O₂(g) → SO₃(g) ΔH₂ = -30 kJ/mol

Sol)

Table A

2S(s) + 3O₂(g) → 2SO₃(g) ΔH₁ = -50 kJ/mol ---- 1st reaction

SO₂(g) + 1/2 O₂(g) → SO₃(g) ΔH₂ = -30 kJ/mol ---- 2nd reaction

1/2 × 1st reaction

S(s) + 3/2 O₂(g) → SO₃(g) ΔH₁ = 1/2(-50) = -25 kJ/mol

SO₂(g) + 1/2 O₂(g) → SO₃(g) ΔH₂ = -30 kJ/mol

Reverse direction of 2^nd reaction

S(s) + 3/2 O₂(g) → SO₃(g) ΔH₁ = -25 kJ/mol

SO₃(g) → SO₂(g) + 1/2 O₂(g) ΔH₂ = +30 kJ/mol

1^st reaction +

2^nd reaction

S(s) + 3/2 O₂(g) → SO₃(g) ΔH₁ = -25 kJ/mol

SO₃(g) → SO₂(g) + 1/2 O₂(g) ΔH₂ = +30 kJ/mol

S(s) + O₂(g) → SO₂(g) ΔH = (-25) + (+30) = +5 kJ/mol

Therefore, the enthalpy change for S(s) + O₂(g) → SO₂(g) is +5 kJ/mol.

G. Determining enthalpy changes, ΔH^°

: Ways to determine ΔH^° for chemical reactions.

Standard enthalpy of formation, ΔH^°_f

ΔH^°_rxn = ΣνΔH^°_f(products) - ΣνΔH^°_f(reactants)

• Use tabulated standard enthalpies of formation to calculate reaction enthalpy

• ΔH^°_f = 0 for an element in its standard state (e.g., O₂(g))

Bond dissociation energy, D

ΔH ≈ ΣD(bonds broken) - ΣD(bonds formed)

• ΔH ≈ ΣD(bonds broken) - ΣD(bonds formed)

• Useful for estimating ΔH when formation-enthalpy data are unavailable

• Breaking a bond requires energy: D > 0

• Forming a bond releases energy

• Greater bond order generally means higher bond dissociation energy

C-C < C=C < C≡C

Qs/ Calculate ΔH^°_rxn for the reaction.

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

Molecules	ΔH^°_f (kJ/mol) at standard state
CH₄(g)	- 75
CO₂(g)	-390
H₂O(l)	-290

Sol) ΔH^°_rxn = ΣνΔH^°_f(products) - ΣνΔH^°_f(reactants)

= [CO₂ + 2H₂O] - [CH₄] ** ΔH^°_f of O₂(g) = 0 kJ/mol

= [(-390) + 2(-290)] – (-75)

= -895 kJ

Qs/ Estimate ΔH for the reaction using the average bond dissociation energies in Table B.

2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(g)

Table B
Bonds	Average bond dissociation energy (kJ/mol)
C-H	410
C-C	350
O=O	500
C=O	730
O-H	460

Sol) Σ(bond energies of bonds broken) - Σ(bond energies of bonds formed)

= [(C-H)×12 + (C-C)×2 + (O=O)×7] - [(C=O)×8 + (O-H)×12]

= [410×12 + 350×2 + 500×7] - [730×8 + 460×12]

≈ -2,240 kJ

H. Entropy change, ΔS

Δ S > 0: entropy of the system increases Δ S < 0: entropy of the system decreases

Entropy change (ΔS)

ΔS > 0

ΔS < 0

• H₂O(s) → H₂O(l) → H₂O(g)

(melting, vaporization, or sublimation)

• C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O

• Diffusion of a gas

• Dissolving a solute often increases dispersal

• H₂O(g) → H₂O(l) → H₂O(s)

(condensation, freezing, or deposition)

• 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂

• ΔS = Σ(entropy of products) - Σ(entropy of reactants)

Qs/ Calculate ΔS.

W + X → Y + Z

	ΔS
W	3
X	2
Y	5
Z	4

ΔS = (Y + Z) - (W + X)

= (5 + 4) - (3 + 2)

= 4

• Entropy change of the surroundings, ΔS_surr

Increasing entropy of surroundings

Decreasing entropy of surroundings

• During an exothermic process, the

entropy of the surroundings increases.

Surroundings System Δ S surr Δ S surr Heat Heat

• During an endothermic process, the

entropy of the surroundings decreases.

System Heat Heat Δ S surr Δ S surr Surroundings

I. Gibbs free energy change, ΔG

: Determines whether a chemical or physical process is spontaneous or nonspontaneous at constant

temperature and pressure.

Δ G < 0 spontaneous reaction Δ G = 0 equilibrium reaction Δ G > 0 non-spontaneous

ΔG = ΔH – TΔS

ΔH: enthalpy change

T: absolute temperature (K)

ΔS: entropy change

• At T > 0 K, ΔH < 0 and ΔS > 0 give ΔG < 0 at all temperatures.

Qs/ Which of the following represents a spontaneous reaction if T = 1?

A. ΔH = -1, ΔS = -2

B. ΔH = +1, ΔS = -2

C. ΔH = +1, ΔS = +2

D. ΔH = -1, ΔS = +2

E. C and D

ANS: E

Sol/ For ΔG = ΔH – TΔS

A. -1 – (-2) = +1

B. +1 – (-2) = +3

C. +1 – (+2) = -1

D. -1 – (+2) = -3

Qs/ ΔH = 6 and ΔS = 3 at equilibrium. Calculate T.

Sol/ At equilibrium, ΔG = 0. Therefore, ΔH – TΔS = 0.

T = ΔH / ΔS

= 6 / 3

= 2

Energy (E)

1. When 75 J of work is done on the system, ΔE is 100 J. Which of the following is the correct

statement about heat transfer?

A. 25 J of heat is absorbed by the system from its surroundings.

B. 25 J of heat is released by the system to its surroundings.

C. 175 J of heat is absorbed by the system from its surroundings.

D. 175 J of heat is released by the system to its surroundings.

E. There is no heat transfer between the system and its surroundings.