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. Each IQ section is a short practice quiz; the final Module Test is the summative assessment. Work through everything in one ~90-minute sitting, or complete one section at a time.

~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 Ea = 150 kJ mol⁻¹ and ΔH = +60 kJ mol⁻¹. What is the activation energy for the reverse reaction?

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.
L02, Calorimetry

Q2. 0.800 g of butan-1-ol (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)

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.
L03, Neutralisation

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

For dilute strong acids and bases the net ionic reaction is H⁺(aq) + OH⁻(aq) → H₂O(l), so ΔHn per mole of water is approximately the same. Spectator ions do not appear in the net ionic equation (an ideal dilute-solution model).
L04, Dissolution Enthalpy

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

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?

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

L02, 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?

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).

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?

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:

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:

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?

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:

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?

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?

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?

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 Tcross. (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) Tcross = ΔH°/ΔS° = −196 ÷ (−0.188) = 1043 K (770°C), an estimate assuming ΔH° and ΔS° are roughly constant with temperature (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:

ΔHf° = enthalpy change forming exactly 1 mol of a compound from elements in standard states at standard conditions (25°C, 100 kPa).
L03, Neutralisation 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. ΔHn = ?

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:

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?

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:

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:

Burning carbon in controlled O₂ almost always gives a CO/CO₂ mixture, so it is very difficult to obtain a clean, quantitative C → CO reaction to measure directly. 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:

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:

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:

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?

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

L02, Calorimetry Extended

ER1 (8 marks)
Propan-1-ol (M = 60.10 g mol⁻¹, accepted ΔHc = −2021 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 propan-1-ol, heating 200.0 g of water by 9.0°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 propan-1-ol burned.

(b) n = 0.300/60.10 = 0.004992 mol (1 mark); ΔT = 9.0 K; q = 200.0 × 4.18 × 9.0 = 7524 J = 7.524 kJ (1 mark); ΔHc = −7.524/0.004992 = −1507 kJ mol⁻¹ with negative sign (1 mark). Accepted = −2021 kJ mol⁻¹; the difference reflects heat loss in the open spirit-burner setup.

(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 Tcross 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) Tcross = −92.4 ÷ (−0.1989) = 464.6 K ≈ 465 K (estimate, assuming constant ΔH°/ΔS°) (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 Tcross (465 K) the standard ΔG° becomes positive, so the standard equilibrium constant K is small at 400–500°C (673–773 K), this describes K, not a ban on the forward reaction. (1 mark) The industrial compromise: 400–500°C provides a practical rate (an iron catalyst lowers Ea); high pressure (~200 atm) and continuous NH₃ removal keep the reaction quotient Q small, so the actual ΔG = ΔG° + RT ln Q stays negative and ammonia keeps forming (Le Chatelier, Module 5). (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. Tcross = 114 ÷ 0.1466 = 777 K (504°C), the reaction becomes non-spontaneous only above 504°C. (1 mark)

Score Tracker

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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