Chemistry Y11 Module 4, Drivers of Reactions

Checkpoint 3, IQ3: Entropy & Gibbs Free Energy

Covering Lessons 11–13: entropy definition & prediction, calculating ΔS°, the Third Law, Gibbs free energy, spontaneity, and the Haber process paradox.

~25 min 10 MC · 4 Short Answer Lessons 11–13

What’s Covered

L11

Entropy, Definition & ΔS

Entropy as microstates
Units: J K⁻¹ mol⁻¹
Predicting ΔS sign (Δn(gas))
Second Law of Thermodynamics

L12

Calculating ΔS°

Third Law: S = 0 at 0 K
S°(elements) ≠ 0
ΔS° = ΣS°(products) − ΣS°(reactants)
J → kJ conversion for L13

L13

Gibbs Free Energy

ΔG = ΔH − TΔS
ΔG < 0 → spontaneous
Four ΔH/ΔS combinations
T_cross = ΔH/ΔS

Multiple Choice, 10 marks

L11, Entropy Definition

1. Entropy is best defined as:

A The heat released by a chemical reaction at constant pressure

B A measure of the dispersal of energy across the available microstates of a system

C The degree of visual disorder or messiness in a chemical system

D The maximum work a system can do at constant pressure

L11, Predicting ΔS

2. For $2\text{H}_2\text{O}_2\text{(l)} \to 2\text{H}_2\text{O(l)} + \text{O}_2\text{(g)}$, the sign of ΔS is:

A Negative, because water has lower entropy than hydrogen peroxide

B Zero, because the reaction produces both liquid and gas products

C Positive, because 1 mol of O₂(g) is produced where no gas existed in the reactants

D Negative, because the reaction is exothermic

L11, Second Law

3. An endothermic reaction can be spontaneous if:

A The reaction is carried out at very low temperature

B The enthalpy change is small enough

C The increase in entropy of the system is large enough so that ΔS(universe) > 0

D The activation energy is exceeded

L12, Third Law & S° Values

4. A student writes: "S°[N₂(g)] = 0 because N₂ is an element in its standard state." This is incorrect because:

A N₂ is not in its standard state at 25°C

B Only noble gases have S° = 0 at standard conditions

C ΔHf°(elements) = 0 applies only to enthalpy of formation, not standard entropy

D Both C and the underlying reason: S° of elements ≠ 0 because the Third Law reference is S = 0 at 0 K, not at 25°C, at 298 K, all real substances have accumulated entropy

L12, Calculating ΔS°

5. For $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \to 2\text{NH}_3\text{(g)}$:
S°[N₂(g)] = 191.6, S°[H₂(g)] = 130.7, S°[NH₃(g)] = 192.4 J K⁻¹ mol⁻¹. What is ΔS°?

A −129.5 J K⁻¹ mol⁻¹

B +198.9 J K⁻¹ mol⁻¹

C −198.9 J K⁻¹ mol⁻¹

D +129.5 J K⁻¹ mol⁻¹

L13, Gibbs Free Energy

6. For $\Delta G = \Delta H - T\Delta S$, which unit conversion must be performed before substituting ΔS°?

A Convert ΔS from kJ K⁻¹ mol⁻¹ to J K⁻¹ mol⁻¹ by multiplying by 1000

B Convert temperature from K to °C by subtracting 273

C Convert ΔS from J K⁻¹ mol⁻¹ to kJ K⁻¹ mol⁻¹ by dividing by 1000, and T from °C to K by adding 273

D No conversion is needed, all thermodynamic quantities use the same units

L13, Spontaneity Classification

7. A reaction has ΔH° = −45 kJ mol⁻¹ and ΔS° = +120 J K⁻¹ mol⁻¹. This reaction is:

A Always spontaneous at any temperature (ΔG always negative)

B Never spontaneous at any temperature

C Spontaneous only at low temperatures

D Spontaneous only at high temperatures

L13, Crossover Temperature

8. A reaction has ΔH° = +120 kJ mol⁻¹ and ΔS° = +200 J K⁻¹ mol⁻¹. What is the crossover temperature (T where ΔG = 0)?

A 1.67 K

B 600 K

C 6000 K

D 0.6 K

L13, Haber Paradox

9. The Haber process (N₂ + 3H₂ → 2NH₃, ΔH < 0, ΔS < 0) is run at 400–500°C. At this temperature, ΔG is:

A Negative, the reaction is spontaneous at high temperature because ΔH < 0

B Zero, the system is at equilibrium at industrial temperatures

C Positive, the reaction is thermodynamically non-spontaneous above T_cross ≈ 465 K

D Not meaningful, ΔG only applies to reactions at equilibrium

L13, Spontaneous vs Fast

10. Diamond is thermodynamically spontaneous to convert to graphite at 25°C (ΔG < 0). This means:

A Diamond converts to graphite rapidly at room temperature

B Diamond will never convert to graphite under any conditions

C Diamond and graphite are in thermodynamic equilibrium at 25°C

D Thermodynamically the conversion is favourable, but it is kinetically frozen, the activation energy is so high that the reaction is essentially undetectable at room temperature

Short Answer, 20 marks

L11, Predicting & Explaining ΔS

Q11 (4 marks)
For each reaction, predict the sign of ΔS and give a brief justification:

(a) $\text{CaCO}_3\text{(s)} \to \text{CaO(s)} + \text{CO}_2\text{(g)}$

(b) $2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \to 2\text{H}_2\text{O(l)}$

(d) $\text{N}_2\text{O}_4\text{(g)} \to 2\text{NO}_2\text{(g)}$

(a) ΔS > 0, Δn(gas) = 1 − 0 = +1. CO₂ gas is produced from solid reactant only → large increase in microstates.

(b) ΔS < 0, Δn(gas) = 0 − (2+1) = −3. Three moles of gas → zero moles of gas (liquid water) → major decrease in microstates and translational freedom.

(c) ΔS > 0, ionic solid dissolves → NH₄⁺ and NO₃⁻ ions dispersed freely in aqueous solution. No gas change, but increased freedom of ions in solution → more microstates.

(d) ΔS > 0, Δn(gas) = 2 − 1 = +1. One mole of gas → two moles of gas → more particles, more microstates.

L12, Calculating ΔS° & Third Law

Q12 (5 marks)
(a) Explain why $S°[\text{O}_2\text{(g)}] = 205.2 \text{ J K}^{-1}\text{mol}^{-1}$, while $\Delta H_f°[\text{O}_2\text{(g)}] = 0$. Reference the Third Law in your answer.

(b) Calculate ΔS° for: $\text{C(graphite)} + \text{O}_2\text{(g)} \to \text{CO}_2\text{(g)}$
S°: C(graphite) = 5.7, O₂(g) = 205.2, CO₂(g) = 213.8 J K⁻¹ mol⁻¹

(a) The Third Law of Thermodynamics states that the entropy of a perfect crystal at absolute zero (0 K) is exactly zero. This provides an absolute reference point, entropy can be measured as it accumulates from 0 K to any temperature. At 25°C (298 K), O₂(g) has been "warmed up" from 0 K through every energy input (heating as a solid, melting at 54 K, vaporising at 90 K, further heating as a gas to 298 K), it has accumulated 205.2 J K⁻¹ mol⁻¹ of real entropy. In contrast, ΔHf°[O₂(g)] = 0 is a human convention, we arbitrarily define the enthalpy of elements in their standard state as zero (a relative reference). Entropy has an absolute physical zero; enthalpy does not. These are fundamentally different conventions.

(b) ΔS° = ΣS°(products) − ΣS°(reactants)

ΔS° = 213.8 − (5.7 + 205.2) = 213.8 − 210.9 = +3.0 J K⁻¹ mol⁻¹

Qualitative check: Δn(gas) = 1 − 1 = 0 → ΔS ≈ 0. The small positive value is consistent, no net change in moles of gas, but the very low S° of graphite (ordered solid, S° = 5.7) means the products have slightly more entropy overall. ✓

L13, Full ΔG Calculation

Q13 (6 marks)
For the decomposition of dinitrogen tetroxide: $\text{N}_2\text{O}_4\text{(g)} \to 2\text{NO}_2\text{(g)}$
ΔH° = +57.2 kJ mol⁻¹; ΔS° = +175.6 J K⁻¹ mol⁻¹
(a) Identify the ΔH/ΔS combination category and predict whether this reaction is spontaneous at low or high temperature.
(b) Calculate ΔG° at 25°C. Is the reaction spontaneous at 25°C?
(c) Calculate ΔG° at 100°C. Is the reaction spontaneous at 100°C?
(d) Calculate T_cross and convert to °C.

(a) ΔH > 0, ΔS > 0 → Row 4 (endothermic, more disorder). Prediction: spontaneous only at high temperaturethe +TΔS term must grow large enough to outweigh the positive ΔH. Not spontaneous at low T; becomes spontaneous above T_cross.

(b) Convert: T = 298 K; ΔS = +175.6 ÷ 1000 = +0.1756 kJ K⁻¹ mol⁻¹
ΔG° = 57.2 − (298 × 0.1756) = 57.2 − 52.33 = +4.87 kJ mol⁻¹ > 0 → non-spontaneous at 25°C
Consistent with prediction, just below T_cross, so ΔG is barely positive.

(c) T = 100 + 273 = 373 K
ΔG° = 57.2 − (373 × 0.1756) = 57.2 − 65.50 = −8.3 kJ mol⁻¹ < 0 → spontaneous at 100°C
Consistent with prediction, above T_cross, so ΔG is negative. ✓

(d) T_cross = ΔH°/ΔS° = 57.2 / 0.1756 = 325.7 K = 52.7°C ≈ 53°C
Above 53°C: spontaneous. Below 53°C: non-spontaneous. This explains why the brown NO₂ gas is more prevalent at higher temperatures (e.g. in a warm flask vs cold flask, familiar from Module 5 equilibrium demonstrations).

L11–L13, Extended Response

Q14 (5 marks)
A student states: "A reaction with ΔH < 0 and ΔS < 0 will never be spontaneous because the negative entropy change works against it."
Evaluate this claim. In your answer: (i) identify the ΔH/ΔS category; (ii) explain using $\Delta G = \Delta H - T\Delta S$ when this reaction is spontaneous; (iii) give a real chemical example.

Evaluation: The student's claim is incorrect.

(i) ΔH < 0, ΔS < 0 corresponds to Row 3 in the four-combination table: spontaneous at low temperature; becomes non-spontaneous above T_cross.

(ii) Using ΔG = ΔH − TΔS: when ΔH < 0 and ΔS < 0, the TΔS term is negative (since ΔS < 0), so −TΔS is positive. The equation becomes: ΔG = (−ve) − (−ve) = (−ve) + (+ve). At low temperatures, the TΔS term is small (small T × |ΔS|), so the dominant term is the negative ΔH → ΔG < 0 (spontaneous). At high temperatures, T × |ΔS| becomes large, the positive contribution eventually exceeds |ΔH|, and ΔG becomes positive (non-spontaneous). The crossover temperature T_cross = ΔH/ΔS marks this transition.

(iii) Real example: The Haber process, N₂(g) + 3H₂(g) → 2NH₃(g), has ΔH° = −92.4 kJ mol⁻¹ and ΔS° = −198.9 J K⁻¹ mol⁻¹. At 25°C (298 K), ΔG° = −33.1 kJ mol⁻¹ → spontaneous. Above T_cross = 465 K (192°C), ΔG° becomes positive → non-spontaneous. The reaction is spontaneous, just not at high temperatures, directly disproving the student's claim. The negative ΔS "works against" spontaneity at high T but does not prevent spontaneity altogether at low T.

Score Tracker

Record Your Marks

MC (Q1–Q10), 10 marks / 10

Q11 Predicting ΔS, 4 marks / 4

Q12 Third Law & ΔS°, 5 marks / 5

Q13 Full ΔG Calculation, 6 marks / 6

Q14 Extended Response, 5 marks / 5

Total/ 30

If you scored below 20/30, review Lessons 11–13 before attempting the Module Quiz.

I've completed Checkpoint 3, IQ3: Entropy & Gibbs Free Energy

IQ3 complete, move on to the Module 4 Quiz