Chemistry Y11 Module 4 — Drivers of Reactions

Module 4 Quiz — Drivers of Reactions

Full module assessment covering all three Inquiry Questions: IQ1 (Energy Changes), IQ2 (Enthalpy & Hess’s Law), IQ3 (Entropy & Gibbs Free Energy). Lessons 01–13.

~90 min total IQ1 Quiz: 5 MC + 2 SA IQ2 Quiz: 5 MC + 2 SA IQ3 Quiz: 5 MC + 2 SA Module Test: 10 MC + 4 ER

Quiz 1

IQ1 — Energy Changes in Chemical Reactions (L01–L05)

5 MC · 5 marks 2 Short Answer · 7 marks Total: 12 marks

Multiple Choice

L01–L02 — Energy Profiles

Q1. A reaction has E_a = 150 kJ mol⁻¹ and ΔH = +60 kJ mol⁻¹. What is the activation energy for the reverse reaction?

A 150 kJ mol⁻¹

B 210 kJ mol⁻¹

C 90 kJ mol⁻¹

D 60 kJ mol⁻¹

Ea(reverse) = Ea(forward) − ΔH = 150 − 60 = 90 kJ mol⁻¹. Products are 60 kJ mol⁻¹ higher than reactants; the transition state peak is 90 kJ mol⁻¹ above products.

L03 — Calorimetry

Q2. 0.800 g of butanol (C₄H₉OH, M = 74.12 g mol⁻¹) is burned, heating 250.0 g of water by 9.3°C. What is the experimental molar enthalpy of combustion? (c = 4.18 J g⁻¹ K⁻¹)

A −901 kJ mol⁻¹

B −2530 kJ mol⁻¹

C −270 kJ mol⁻¹

D −9700 kJ mol⁻¹

n = 0.800/74.12 = 0.01079 mol; q = 250.0 × 4.18 × 9.3 = 9718.5 J = 9.719 kJ; ΔHc = −9.719/0.01079 = −901 kJ mol⁻¹.

L04 — Neutralisation

Q3. Why does strong acid + strong base neutralisation always give approximately the same ΔH_n regardless of which acid or base is used?

A All strong acids have the same molecular formula

B The net ionic equation is always H⁺(aq) + OH⁻(aq) → H₂O(l) — the same bond forms regardless of the counterions

C Spectator ions contribute positive energy that exactly compensates for differences

D The enthalpy of neutralisation is temperature-independent

Strong acids/bases fully dissociate → the reaction is always H⁺ + OH⁻ → H₂O, regardless of counterions. Spectator ions don't participate in the energy step.

L04 — Dissolution Enthalpy

Q4. When NH₄Cl dissolves in water, the solution cools. Which explanation is correct?

A NH₄Cl has a very low molar mass, so few bonds need to be broken

B The lattice energy of NH₄Cl exceeds the hydration energy of NH₄⁺ and Cl⁻ ions (ΔHsoln > 0)

C The hydration energy exceeds the lattice energy

D NH₄Cl reacts with water in an exothermic reaction

Cooling = endothermic dissolution (ΔHsoln > 0). This occurs when the lattice energy (energy to break the lattice) exceeds the hydration energy released when ions interact with water.

L05 — Catalysts

Q5. A platinum catalyst is added to a gas-phase oxidation reaction. Which statement is correct?

A The catalyst lowers both Ea and ΔH, making the reaction more exothermic

B The catalyst lowers Ea only — ΔH, reactants, and products are unchanged

C The catalyst provides energy to overcome the activation barrier

D The catalyst is consumed, appearing in the overall chemical equation

Catalysts provide an alternative pathway with lower Ea. They do not alter ΔH (same reactants and products), do not supply energy, and are regenerated (not consumed).

Short Answer

L03 — Calorimetry Calculation

SA1 (4 marks)
A student burns 0.65 g of ethanol (C₂H₅OH, M = 46.07 g mol⁻¹) in a spirit burner, heating 200.0 g of water from 18.2°C to 29.1°C. (a) Calculate q, the heat absorbed by the water. (2 marks) (b) Calculate the experimental molar enthalpy of combustion. (2 marks)

(a) ΔT = 29.1 − 18.2 = 10.9°C = 10.9 K; q = mcΔT = 200.0 × 4.18 × 10.9 = 9112.4 J = 9.11 kJ (1 mark for substitution, 1 for correct answer with units)

(b) n = 0.65 ÷ 46.07 = 0.01411 mol; ΔHc = −q/n = −9.11 ÷ 0.01411 = −646 kJ mol⁻¹ (1 mark for n, 1 for ΔHc with negative sign). Accepted value = −1367 kJ mol⁻¹; large discrepancy due to heat loss to surroundings in spirit burner setup.

L05 — Catalytic Converter

SA2 (3 marks)
A catalytic converter uses platinum (Pt) to convert CO(g) and NO(g) into CO₂(g) and N₂(g). (a) Identify whether Pt acts as a homogeneous or heterogeneous catalyst. Justify. (2 marks) (b) Explain the effect of Pt on activation energy and ΔH. (1 mark)

(a) Heterogeneous catalyst. (1 mark) Platinum is a solid while CO and NO are in the gas phase — the catalyst and reactants are in different physical phases/states. (1 mark for justification)

(b) Pt lowers the activation energy (Ea) by providing an alternative reaction pathway with lower energy requirements. The enthalpy change (ΔH) is unchanged — the reactants and products are identical in both catalysed and uncatalysed pathways. (1 mark)

Quiz 2

IQ2 — Enthalpy, Bond Energy & Hess’s Law (L06–L10)

5 MC · 5 marks 2 Short Answer · 9 marks Total: 14 marks

Multiple Choice

L06 — Bond Energy

Q1. For H₂(g) + F₂(g) → 2HF(g): BE(H–H) = 436, BE(F–F) = 158, BE(H–F) = 568 kJ mol⁻¹. What is ΔH?

A −542 kJ mol⁻¹

B +542 kJ mol⁻¹

C −1136 kJ mol⁻¹

D +26 kJ mol⁻¹

ΣB(reactants) = 436 + 158 = 594; ΣB(products) = 2 × 568 = 1136; ΔH = 594 − 1136 = −542 kJ mol⁻¹.

L07 — Enthalpy of Formation

Q2. ΔHf°[SO₂(g)] = −297, ΔHf°[SO₃(g)] = −395, ΔHf°[O₂(g)] = 0 kJ mol⁻¹. Calculate ΔH° for 2SO₂(g) + O₂(g) → 2SO₃(g).

A −98 kJ mol⁻¹

B +98 kJ mol⁻¹

C −196 kJ mol⁻¹

D +196 kJ mol⁻¹

ΔH° = 2(−395) − [2(−297) + 0] = −790 − (−594) = −196 kJ mol⁻¹.

L06 — Bond Energy Limitations

Q3. Which statement about the bond energy method is correct?

A It gives exact ΔH values because bond energies are measured precisely

B It gives approximate ΔH values because bond energies are averages and all species are assumed gaseous

C It is more accurate than the enthalpy of formation method for liquid-phase reactions

D The formula is ΔH = ΣB(products) − ΣB(reactants)

Bond energies are average values across different molecular environments, and all species are assumed gaseous — both introduce approximation. Correct formula: ΔH = ΣB(reactants) − ΣB(products).

L08 — Hess’s Law Manipulation

Q4. The equation A → B has ΔH = −180 kJ mol⁻¹. This equation is reversed and multiplied by 3. New ΔH:

A −540 kJ mol⁻¹

B +540 kJ mol⁻¹

C −60 kJ mol⁻¹

D +60 kJ mol⁻¹

Reverse: ΔH = +180 kJ mol⁻¹. Multiply by 3: ΔH = +540 kJ mol⁻¹. Both operations must be applied.

L09 — Photosynthesis & Respiration

Q5. Cellular respiration and photosynthesis are related by Hess’s Law because:

A Both reactions occur in living cells and involve glucose

B Respiration is the exact reverse of photosynthesis, so ΔH values are equal in magnitude and opposite in sign

C Both reactions have the same activation energy

D The enthalpy of formation of glucose is zero

Hess's Law: reversing a reaction changes the sign of ΔH. Respiration is the exact reverse of photosynthesis → ΔH(respiration) = −ΔH(photosynthesis).

Short Answer

L08 — Hess’s Law Three-Equation Cycle

SA1 (5 marks)
Calculate ΔH for: $2\text{C(s)} + \text{H}_2\text{(g)} \to \text{C}_2\text{H}_2\text{(g)}$ (acetylene)
(1) C(s) + O₂(g) → CO₂(g), ΔH₁ = −393.5 kJ mol⁻¹
(2) H₂(g) + ½O₂(g) → H₂O(l), ΔH₂ = −285.8 kJ mol⁻¹
(3) C₂H₂(g) + 5/2 O₂(g) → 2CO₂(g) + H₂O(l), ΔH₃ = −1299.7 kJ mol⁻¹

Target: 2C(s) + H₂(g) → C₂H₂(g)

(1) × 2: 2C(s) + 2O₂(g) → 2CO₂(g), ΔH = 2 × (−393.5) = −787.0 kJ mol⁻¹ (1 mark — scale)

(2) as is: H₂(g) + ½O₂(g) → H₂O(l), ΔH = −285.8 kJ mol⁻¹ (1 mark)

Reverse (3): 2CO₂(g) + H₂O(l) → C₂H₂(g) + 5/2 O₂(g), ΔH = +1299.7 kJ mol⁻¹ (1 mark — flip sign)

Cancel 2CO₂, H₂O, 5/2 O₂ (= ½O₂ + 2O₂) (1 mark) → 2C(s) + H₂(g) → C₂H₂(g) ✓

ΔH = −787.0 + (−285.8) + 1299.7 = +226.9 kJ mol⁻¹ (1 mark — positive ΔHf° confirms endothermic formation)

L10 — Fuel Energy Comparison

SA2 (4 marks)
Compare methanol (CH₃OH, M = 32.04 g mol⁻¹, ΔHc = −726 kJ mol⁻¹) and ethanol (C₂H₅OH, M = 46.07 g mol⁻¹, ΔHc = −1367 kJ mol⁻¹) as fuel additives. (a) Calculate energy per gram for each. (2 marks) (b) Evaluate the claim "ethanol is a more efficient fuel than methanol." (2 marks)

(a) Methanol: 726 ÷ 32.04 = 22.7 kJ g⁻¹; Ethanol: 1367 ÷ 46.07 = 29.7 kJ g⁻¹ (1 mark each)

(b) The claim is supported thermodynamically — ethanol releases more energy per gram (29.7 vs 22.7 kJ g⁻¹) and per mole. (1 mark) However, "efficiency" also depends on combustion completeness, cost, renewable source, and engine octane requirements. Methanol's higher oxygen content can improve combustion in some engines. The thermodynamic data supports the claim, but it is incomplete from an engineering perspective. (1 mark)

Quiz 3

IQ3 — Entropy & Spontaneity (L11–L13)

5 MC · 5 marks 2 Short Answer · 9 marks Total: 14 marks

Multiple Choice

L11 — Predicting ΔS

Q1. For CaCO₃(s) → CaO(s) + CO₂(g), which prediction is correct?

A ΔS < 0, because a solid breaks down into simpler products

B ΔS > 0, because a mole of CO₂(g) is produced from solid reactants, greatly increasing microstates

C ΔS = 0, because the number of moles of solid remains constant

D ΔS < 0, because the product CaO is smaller than CaCO₃

Gas has dramatically higher entropy than solids. Δn(gas) = +1 → ΔS > 0. Production of CO₂(g) from solid reactants creates enormous increase in microstates.

L12 — Third Law

Q2. A student sets S°[C(graphite)] = 0 in a ΔS° calculation. This error will:

A Have no effect — elements have zero contribution to ΔS°

B Give incorrect ΔS° — the Third Law sets S = 0 only at 0 K; at 25°C, S°[C(graphite)] = 5.7 J K⁻¹ mol⁻¹

C Give a more accurate result — the Third Law sets S = 0 for all elements

D Only matter if carbon appears in the products, not the reactants

Third Law: S = 0 only at 0 K for a perfect crystal. At 25°C, graphite has S° = 5.7 J K⁻¹ mol⁻¹. Setting it to zero introduces error.

L13 — Crossover Temperature

Q3. A reaction has ΔH° = +35 kJ mol⁻¹ and ΔS° = +150 J K⁻¹ mol⁻¹. At what temperature does ΔG = 0?

A 23 K

B 2330 K

C 233 K

D 0.23 K

T = ΔH/ΔS = 35 ÷ (150/1000) = 35 ÷ 0.150 = 233 K. Convert ΔS to kJ K⁻¹ mol⁻¹ first. Above 233 K → spontaneous.

L13 — Temperature Dependence

Q4. For ΔH° = −200 kJ mol⁻¹ and ΔS° = −300 J K⁻¹ mol⁻¹, which is correct?

A Always spontaneous because ΔH is negative

B Always non-spontaneous because ΔS is negative

C Spontaneous at low temperatures; non-spontaneous at high temperatures

D Spontaneous at high temperatures; non-spontaneous at low temperatures

ΔH < 0, ΔS < 0 → Row 3. At low T: ΔH term dominates → ΔG < 0 (spontaneous). At high T: −TΔS term grows large and positive → ΔG > 0 (non-spontaneous).

L13 — Never Spontaneous

Q5. Which ΔH/ΔS combination gives a reaction that is never spontaneous at any temperature?

A ΔH < 0, ΔS > 0

B ΔH > 0, ΔS < 0

C ΔH < 0, ΔS < 0

D ΔH > 0, ΔS > 0

ΔH > 0, ΔS < 0: ΔG = (+ve) − T(−ve) = (+ve) + (+ve) = always positive → never spontaneous at any T > 0.

Short Answer

L11–L12 — Entropy Definition & Calculation

SA1 (4 marks)
(a) Define entropy in terms of microstates. (1 mark)
(b) For $2\text{H}_2\text{O(l)} \to 2\text{H}_2\text{(g)} + \text{O}_2\text{(g)}$, predict the sign of ΔS and justify using microstates. (2 marks)
(c) Explain why S°[H₂O(l)] = 70 J K⁻¹ mol⁻¹ is not zero. (1 mark)

(a) Entropy (S) is a measure of the number of microstates available to a system — the dispersal of energy across all possible arrangements of particles. Higher entropy = greater number of microstates. (1 mark)

(b) ΔS > 0. (0.5 mark) The reaction produces H₂(g) and O₂(g) from liquid water. The product gases have vastly more translational freedom and far more accessible arrangements (microstates) than liquid water. Δn(gas) = 3 − 0 = +3; producing 3 moles of gas from 2 moles of liquid represents a very large increase in microstates → positive ΔS. (1.5 marks)

(c) Third Law: S = 0 only for a perfect crystal at 0 K. At 25°C, liquid water has absorbed energy from 0 K, distributing it across molecular thermal motions. This produces a positive S° at 25°C — S° is never zero for real substances at room temperature. (1 mark)

L13 — Industrial ΔG Application

SA2 (5 marks)
For $2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \to 2\text{SO}_3\text{(g)}$: ΔH° = −196 kJ mol⁻¹; ΔS° = −188 J K⁻¹ mol⁻¹
(a) Calculate ΔG° at 25°C. (2 marks)
(b) Find T_cross. (1 mark)
(c) Calculate ΔG° at 450°C and explain why this temperature is used industrially despite the thermodynamic implications. (2 marks)

(a) ΔS° = −188 ÷ 1000 = −0.188 kJ K⁻¹ mol⁻¹; T = 298 K
ΔG° = −196 − (298 × −0.188) = −196 + 56.02 = −140.0 kJ mol⁻¹ (1 mark conversion, 1 mark answer)

(b) T_cross = ΔH°/ΔS° = −196 ÷ (−0.188) = 1043 K (770°C) (1 mark)

(c) T = 450 + 273 = 723 K; ΔG° = −196 − (723 × −0.188) = −196 + 135.9 = −60.1 kJ mol⁻¹ → still spontaneous at 450°C (1 mark). Industrial plants use 450°C because the reaction is thermodynamically spontaneous but too slow at room temperature. Higher temperature provides enough energy to overcome Ea at a practical rate, and a V₂O₅ catalyst lowers Ea further. 1043 K would give faster rates but ΔG would be even less negative. The 450°C compromise balances acceptable rate with sufficient thermodynamic driving force. (1 mark — thermodynamics vs kinetics)

Module Topic Test

Module 4 — Drivers of Reactions: Full Module Test

10 MC · 10 marks 4 Extended Response · 36 marks Total: 46 marks

Multiple Choice — 10 marks

L07 — Formation Enthalpy

1. ΔHf°[CO₂(g)] = −393.5 kJ mol⁻¹ represents:

A Heat released when 1 mol CO₂ decomposes into C and O₂

B Enthalpy change when 1 mol CO₂(g) is formed from C(graphite) and O₂(g) at 25°C and 100 kPa

C The bond energy of the two C=O bonds in CO₂

D The enthalpy change when C(graphite) undergoes complete combustion at any temperature

ΔHf° = enthalpy change forming exactly 1 mol of a compound from elements in standard states at standard conditions (25°C, 100 kPa).

L04 — Calorimetry

2. 50.0 mL of 2.00 mol L⁻¹ HCl mixed with 50.0 mL of 2.00 mol L⁻¹ NaOH. Temperature rises 13.5°C. ΔH_n = ?

A −28.2 kJ mol⁻¹

B −56.4 kJ mol⁻¹

C +56.4 kJ mol⁻¹

D −112.8 kJ mol⁻¹

n(H₂O) = 2.00 × 0.0500 = 0.100 mol; m = 100 g; q = 100 × 4.18 × 13.5 = 5643 J; ΔHn = −5.643/0.100 = −56.4 kJ mol⁻¹.

L13 — ΔG Temperature Dependence

3. A reaction has ΔH < 0 and ΔS < 0. As temperature increases, ΔG will:

A Become more negative — more spontaneous at higher temperatures

B Become less negative (and eventually positive) — less spontaneous at higher temperatures

C Remain constant regardless of temperature

D Become zero at all high temperatures

ΔH < 0, ΔS < 0: ΔG = (−ve) − T(−ve) = (−ve) + T|ΔS|. As T grows, positive −TΔS term grows, making ΔG less negative → eventually positive.

L06 — Bond Energy vs ΔHf°

4. Which statement correctly describes the relationship between the bond energy formula and the enthalpy of formation formula?

A Both use products minus reactants

B Both use reactants minus products

C Bond energy: reactants minus products; enthalpy of formation: products minus reactants

D Bond energy: products minus reactants; enthalpy of formation: reactants minus products

Bond energy: ΔH = ΣB(reactants) − ΣB(products). Formation: ΔH = ΣΔHf°(products) − ΣΔHf°(reactants). Opposite order — a frequent source of sign errors.

L11 — Second Law

5. The Second Law predicts a process is spontaneous when:

A The enthalpy of the system decreases

B The entropy of the system increases

C The total entropy of the universe (system + surroundings) increases

D The Gibbs free energy of the system increases

Second Law: ΔS(universe) > 0 for any spontaneous process. System entropy alone may decrease (e.g. ice forming) as long as surroundings gain more.

L08 — Hess’s Law & Indirect Measurement

6. ΔH for C(s) + ½O₂(g) → CO(g) cannot be measured directly because:

A Carbon does not react with oxygen

B The reaction requires extremely high pressure

C Burning carbon always produces a mixture of CO and CO₂, making isolation of the CO step impossible

D The activation energy for CO formation is unmeasurably high

Burning carbon in controlled O₂ always gives a CO/CO₂ mixture — it is not possible to produce only CO cleanly. Hess's Law is used instead as an indirect route.

L12 — Standard Entropy Values

7. Standard entropy values S° for elements at 25°C are:

A Zero by definition, just like ΔHf° of elements

B Always negative because elements are in their lowest energy state

C Greater than zero because all real substances at 25°C have accumulated entropy from heating above 0 K

D Equal to their standard enthalpy of formation values

Third Law: S = 0 only at 0 K. At 25°C, all substances have S° > 0. This is different from ΔHf°(elements) = 0, which is an arbitrary convention.

L05 — Catalysts

8. A heterogeneous catalyst differs from a homogeneous catalyst because:

A A heterogeneous catalyst is always less effective at lowering Ea

B A heterogeneous catalyst is in a different physical phase from the reactants

C A heterogeneous catalyst changes the enthalpy of reaction

D A heterogeneous catalyst is consumed in the reaction

Homogeneous = same phase as reactants; heterogeneous = different phase. Both lower Ea without being consumed and without changing ΔH.

L11 — Decomposition ΔS

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

A Negative, because a liquid decomposes into simpler molecules

B Positive, because O₂(g) is produced — a mole of gas has far more microstates than liquid

C Zero, because the reaction is in aqueous solution

D Negative, because water has lower entropy than hydrogen peroxide

Δn(gas) = +1 (O₂ produced from liquid reactants only). Production of a mole of gas → dramatically more microstates → ΔS > 0.

L13 — Spontaneous vs Fast

10. A reaction has ΔG = −15 kJ mol⁻¹ at 300 K. Which conclusion is valid?

A The reaction will complete within a few minutes at 300 K

B The reaction is thermodynamically spontaneous but may or may not proceed at an observable rate

C The reaction requires a catalyst to proceed

D The reaction will not occur because ΔG is too small

ΔG < 0 = thermodynamically spontaneous (favourable), but says nothing about rate. A kinetically limited reaction with high Ea may proceed immeasurably slowly despite ΔG < 0.

Extended Response — 36 marks

L03 — Calorimetry Extended

ER1 (8 marks)
Glycine (M = 75.03 g mol⁻¹, accepted ΔHc = −979 kJ mol⁻¹) is burned in a spirit burner. (a) Describe the procedure to collect data for a calorimetry calculation. (3 marks) (b) A student burns 0.300 g of glycine, heating 200.0 g of water by 3.8°C. Calculate experimental ΔHc. (3 marks) (c) Identify two sources of error and their directional effect on the calculated ΔHc. (2 marks)

(a) (3 marks — 1 per step) (i) Weigh the spirit burner before experiment to obtain initial mass. (ii) Measure 200.0 g of water into the calorimeter; record initial temperature with thermometer submerged. (iii) Light the spirit burner; stir continuously; record maximum temperature; cap burner immediately; reweigh to find mass of glycine burned.

(b) n = 0.300/75.03 = 0.003998 mol (1 mark); ΔT = 3.8 K; q = 200.0 × 4.18 × 3.8 = 3176.8 J = 3.177 kJ (1 mark); ΔHc = −3.177/0.003998 = −795 kJ mol⁻¹ with negative sign (1 mark).

(c) (i) Heat loss to surroundings — much combustion heat warms surrounding air rather than water → q(measured) is less than actual → |ΔHc| is underestimated (less negative than true value). (ii) Incomplete combustion — insufficient O₂ produces CO instead of CO₂, releasing less energy per mole → q is less than for complete combustion → |ΔHc| is underestimated. (1 mark each)

L06–L07 — Three-Method Comparison

ER2 (10 marks)
CH₄(g) + 2Cl₂(g) → CCl₄(l) + 2HCl(g).
Bond energies (kJ mol⁻¹): C–H = 414; Cl–Cl = 243; C–Cl = 339; H–Cl = 432.
ΔHf° (kJ mol⁻¹): CH₄(g) = −74.8; CCl₄(l) = −135.4; HCl(g) = −92.3; Cl₂(g) = 0.
(a) Calculate ΔH using bond energy method. (3 marks) (b) Calculate ΔH° using enthalpy of formation method. (3 marks) (c) Explain why the two answers differ. (2 marks) (d) Which method is more accurate and why? (2 marks)

(a) ΣB(reactants): 4(C–H) + 2(Cl–Cl) = 4(414) + 2(243) = 1656 + 486 = 2142 kJ mol⁻¹ (1 mark); ΣB(products): 4(C–Cl) + 2(H–Cl) = 4(339) + 2(432) = 1356 + 864 = 2220 kJ mol⁻¹ (1 mark); ΔH = 2142 − 2220 = −78 kJ mol⁻¹ (1 mark)

(b) ΣΔHf°(products) = 1(−135.4) + 2(−92.3) = −135.4 − 184.6 = −320.0 kJ mol⁻¹ (1 mark); ΣΔHf°(reactants) = 1(−74.8) + 0 = −74.8 kJ mol⁻¹ (1 mark); ΔH° = −320.0 − (−74.8) = −245.2 kJ mol⁻¹ (1 mark)

(c) Bond energy values are averages across many molecular environments — the C–Cl value of 339 kJ mol⁻¹ may not precisely represent C–Cl bonds in CCl₄. (1 mark) Additionally, the bond energy method assumes all species are gaseous; CCl₄ is actually a liquid at standard conditions — the condensation energy (liquid formation from gas) is ignored, systematically underestimating the total energy released. (1 mark)

(d) The enthalpy of formation method is more accurate (0.5 mark) because ΔHf° values are measured experimentally for substances in their actual states at standard conditions — CCl₄(l) is treated as a liquid. This accounts for all phase changes and molecular environment effects. Bond energies are average values that may differ significantly from actual bond enthalpies in specific molecules. (1.5 marks)

L13 — Haber Process Full Analysis

ER3 (10 marks)
N₂(g) + 3H₂(g) → 2NH₃(g): ΔH° = −92.4 kJ mol⁻¹; ΔS° = −198.9 J K⁻¹ mol⁻¹.
(a) Calculate ΔG° at 25°C and determine spontaneity. (3 marks)
(b) Calculate ΔG° at 500°C (773 K). (2 marks)
(c) Calculate T_cross in K and °C. (2 marks)
(d) Explain, using both thermodynamics and kinetics, why industrial plants run at 400–500°C with an iron catalyst. (3 marks)

(a) ΔS° = −198.9/1000 = −0.1989 kJ K⁻¹ mol⁻¹ (1 mark); ΔG° = −92.4 − (298 × −0.1989) = −92.4 + 59.27 = −33.1 kJ mol⁻¹ (1 mark); ΔG° < 0 → spontaneous at 25°C (1 mark)

(b) T = 773 K; ΔG° = −92.4 − (773 × −0.1989) = −92.4 + 153.75 = +61.4 kJ mol⁻¹ → non-spontaneous at 500°C (2 marks)

(c) T_cross = −92.4 ÷ (−0.1989) = 464.6 K ≈ 465 K (1 mark); = 465 − 273 = 192°C (1 mark)

(d) At 25°C: reaction is spontaneous (ΔG = −33.1 kJ mol⁻¹) but the N≡N triple bond has an enormous bond energy (941 kJ mol⁻¹), creating a very high activation energy — the reaction rate is essentially zero at room temperature even with a catalyst. (1 mark) Above T_cross (465 K), ΔG becomes positive — the reaction is thermodynamically non-spontaneous at 400–500°C (673–773 K). (1 mark) The industrial compromise: 400–500°C provides sufficient kinetic energy for a practical rate (molecules can overcome Ea with assistance from the iron catalyst, which lowers Ea); ammonia is continuously removed from the equilibrium mixture (Le Chatelier, Module 5) to drive the reaction forward despite unfavourable thermodynamics. The 400°C lower end of the range is closer to T_cross where ΔG is less positive, providing a better thermodynamic balance. (1 mark)

L12–L13 — Entropy & Spontaneity Combined

ER4 (8 marks)
For $2\text{NO(g)} + \text{O}_2\text{(g)} \to 2\text{NO}_2\text{(g)}$: ΔH° = −114 kJ mol⁻¹.
S°: NO(g) = 210.8; O₂(g) = 205.2; NO₂(g) = 240.1 J K⁻¹ mol⁻¹.
(a) Define standard entropy S° and explain why S°[O₂(g)] = 205.2 J K⁻¹ mol⁻¹ ≠ 0. (3 marks)
(b) Calculate ΔS° for the reaction and verify against qualitative prediction. (2 marks)
(c) Calculate ΔG° at 298 K. Is the reaction spontaneous? (3 marks)

(a) S° is the absolute entropy of 1 mol of a substance in its standard state at 25°C and 100 kPa, in J K⁻¹ mol⁻¹. (1 mark) S°[O₂(g)] = 205.2 ≠ 0 because the Third Law states S = 0 only for a perfect crystal at 0 K. At 25°C, O₂(g) has accumulated entropy through heating from 0 K — real, measurable energy dispersal has occurred. (1 mark) ΔHf°[O₂(g)] = 0 is an arbitrary human convention (relative reference for enthalpy); S°[O₂] = 205.2 is an absolute measurement. These are fundamentally different: entropy has a physical zero; enthalpy has a defined zero. (1 mark)

(b) ΔS° = 2(240.1) − [2(210.8) + 205.2] = 480.2 − [421.6 + 205.2] = 480.2 − 626.8 = −146.6 J K⁻¹ mol⁻¹ (1 mark); Qualitative check: Δn(gas) = 2 − 3 = −1 → ΔS < 0. ✓ Consistent. (1 mark)

(c) Convert: ΔS° = −146.6/1000 = −0.1466 kJ K⁻¹ mol⁻¹; T = 298 K (1 mark for conversions); ΔG° = −114 − (298 × −0.1466) = −114 + 43.69 = −70.3 kJ mol⁻¹ (1 mark); ΔG° < 0 → spontaneous at 25°C. Both the negative ΔH and (at 298 K) the negative TΔS contribution are manageable — the large exothermic term dominates. T_cross = 114 ÷ 0.1466 = 777 K (504°C) — the reaction becomes non-spontaneous only above 504°C. (1 mark)

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Quiz Scores

IQ1 Quiz — MC (5) / 5

IQ1 Quiz — SA1 + SA2 (7) / 7

IQ2 Quiz — MC (5) / 5

IQ2 Quiz — SA1 + SA2 (9) / 9

IQ3 Quiz — MC (5) / 5

IQ3 Quiz — SA1 + SA2 (9) / 9

Module Topic Test

Module Test — MC (10) / 10

ER1 Calorimetry (8) / 8

ER2 Method Comparison (10) / 10

ER3 Haber Process (10) / 10

ER4 Entropy & Spontaneity (8) / 8

Module Test Total— / 46

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Module 4 complete! You have covered all of NSW HSC Chemistry Year 11 Module 4: Drivers of Reactions.