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 temperature — the +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