These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Data for every question: H 1.008, C 12.01, Cl 35.45 g mol⁻¹; c = 4.18 J g⁻¹ °C⁻¹ for aqueous solutions. Substances with single-letter names (D, E, F, X, Y, Z, M) are fictional. All reactions are at constant pressure.
Question 1 (multiple choice · foundation)
A fictional solid D decomposes: 2D(s) → 2E(s) + F(g) ΔH = +130 kJ mol⁻¹. Which statement is correct?
- (A) 130 kJ is released for every 2 mol of D that decompose.
- (B) 130 kJ is absorbed for every 1 mol of D that decomposes.
- (C) 130 kJ is absorbed for every 2 mol of D that decompose.
- (D) 130 kJ is absorbed for every 1 g of F produced.
Answer and explanation
Answer: (C). ΔH is positive, so energy is absorbed. It refers to one mole of reaction as written: 2 mol D forming 2 mol E and 1 mol F.
- (A) has the wrong direction: a positive ΔH means absorbed, not released.
- (B) ignores the coefficient 2. For 1 mol of D the energy is 65 kJ.
- (D) treats ΔH as per gram. ΔH is per mole of reaction; 130 kJ goes with 1 mol of F.
Question 2 (multiple choice · core)
Propane burns: C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l) ΔH = −2219 kJ mol⁻¹. How much heat is released when 11.0 g of propane burns completely?
- (A) 50.3 kJ
- (B) 111 kJ
- (C) 554 kJ
- (D) 2.44 × 10⁴ kJ
Answer and explanation
Answer: (C). M(C₃H₈) = 3(12.01) + 8(1.008) = 44.094 g mol⁻¹. n = 11.0 ÷ 44.094 = 0.2495 mol. The coefficient of propane is 1, so q = 0.2495 × (−2219) = −553.6 kJ: 554 kJ released.
- (A) is the energy per gram (2219 ÷ 44.094), not for 11.0 g.
- (B) divides by 5, the coefficient of O₂. The question is about propane, whose coefficient is 1.
- (D) multiplies the mass in grams by ΔH without converting to moles.
Question 3 (multiple choice · core)
For the fictional reaction X₂(g) + 3Y₂(g) → 2XY₃(g), ΔH = −120 kJ mol⁻¹. What is ΔH for 4XY₃(g) → 2X₂(g) + 6Y₂(g)?
- (A) −240 kJ mol⁻¹
- (B) −60 kJ mol⁻¹
- (C) +120 kJ mol⁻¹
- (D) +240 kJ mol⁻¹
Answer and explanation
Answer: (D). The new equation is the original reversed (sign changes to +120) and doubled (× 2): +240 kJ mol⁻¹.
- (A) doubles but forgets to reverse the sign.
- (B) halves instead of doubling and keeps the original sign.
- (C) reverses but forgets to double.
Question 4 (multiple choice · core)
An exothermic reaction is carried out in water in an insulated cup, and the temperature of the water rises. Which statement best explains the temperature rise?
- (A) Breaking the bonds in the reactants releases energy, which heats the water.
- (B) The products have higher potential energy than the reactants, and this energy is passed to the water.
- (C) The products have lower potential energy than the reactants; the difference becomes kinetic energy of the particles, which is transferred to the water by collisions.
- (D) The reaction creates thermal energy, which is then stored in the water.
Answer and explanation
Answer: (C). The drop in chemical potential energy appears as faster-moving particles. These collide with water molecules and transfer energy until thermal equilibrium is reached, so the water warms.
- (A) Breaking bonds always needs energy. The energy is released when new bonds form.
- (B) Higher potential energy in the products describes an endothermic reaction, which would cool the water.
- (D) Energy is conserved; it is converted from potential energy, not created.
Question 5 (calculation · core)
Hydrogen reacts with chlorine: H₂(g) + Cl₂(g) → 2HCl(g) ΔH = −185 kJ mol⁻¹. A mixture of 2.00 g of H₂ and 35.5 g of Cl₂ reacts at constant pressure until the limiting reactant is used up.
(a) Identify the limiting reactant. Show your working. (b) Calculate the heat released. (c) A student claims: “If I double the mass of hydrogen to 4.00 g, twice as much heat will be released.” Evaluate this claim.
Worked solution and suggested Marlbridge rubric
(a) n(H₂) = 2.00 ÷ 2.016 = 0.9921 mol. n(Cl₂) = 35.5 ÷ 70.90 = 0.5007 mol. The ratio is 1 : 1, so Cl₂ (fewer moles) is limiting.
(b) Moles of reaction = 0.5007 mol (coefficient of Cl₂ is 1). q = 0.5007 × (−185) = −92.6 kJ: 92.6 kJ is released.
(c) The claim is incorrect. Hydrogen is already in excess, so adding more does not increase the amount of chlorine that reacts. The moles of reaction stay at 0.5007 mol and the heat released stays at 92.6 kJ.
| Point | What earns it |
|---|---|
| 1 | Moles of both reactants and Cl₂ identified as limiting |
| 1 | −92.6 kJ (or 92.6 kJ released) from the moles of Cl₂ |
| 1 | Rejects the claim because H₂ is in excess and Cl₂ limits the reaction |
Using H₂ in (b) gives −183.5 kJ and does not earn the second point.
Question 6 (constructed response · core)
A student dissolves reactants so that 0.0250 mol of fictional compound Z reacts completely according to Z(aq) + 2W(aq) → products, with W in excess. The 80.0 g of solution in an insulated cup changes from 22.4 °C to 19.1 °C.
(a) Calculate q for the solution. (b) Calculate ΔH for the reaction, in kJ per mole of reaction, with the correct sign. (c) Describe the direction of energy transfer between the reacting particles and the water after the reaction, and explain why the thermometer reading falls.
Worked solution and suggested Marlbridge rubric
(a) ΔT = 19.1 − 22.4 = −3.3 °C. q_soln = 80.0 g × 4.18 J g⁻¹ °C⁻¹ × (−3.3 °C) = −1104 J.
(b) q_rxn = −q_soln = +1104 J. The coefficient of Z is 1, so moles of reaction = 0.0250 mol. ΔH = +1.104 kJ ÷ 0.0250 mol = +44 kJ mol⁻¹ (2 significant figures, from ΔT = 3.3 °C).
(c) The products have higher potential energy than the reactants, so the particles have less kinetic energy and are colder than the water just after reacting. Energy then flows from the water to the reacting particles until thermal equilibrium is reached. The water loses energy, so its temperature falls.
| Point | What earns it |
|---|---|
| 1 | q_soln = −1104 J (or −1.10 kJ) with the sign |
| 1 | Reverses the sign: q_rxn positive |
| 1 | ΔH ≈ +44 kJ mol⁻¹, dividing by 0.0250 mol of reaction |
| 1 | Energy flows from the water to the products, linked to the products being colder (higher potential energy) |
Accept +44.1 kJ mol⁻¹. Do not award the third point for a negative ΔH.
Question 7 (constructed response · stretch)
A self-heating soup can uses a fictional solid M (molar mass 72.0 g mol⁻¹) that reacts with water: M(s) + H₂O(l) → products ΔH = −65.0 kJ mol⁻¹. The can holds 300. g of soup (c = 4.0 J g⁻¹ °C⁻¹). Only 70.0% of the energy released reaches the soup; the rest warms the can and the air.
(a) Calculate the energy the soup must gain to warm by 50.0 °C. (b) Calculate the minimum mass of M needed. Assume water is in excess. (c) The designer says: “The energy that heats the soup was stored in solid M.” Explain, in terms of potential energy and bonds, what is meant by this and why the energy is released.
Model answer and suggested Marlbridge rubric
(a) q = 300. g × 4.0 J g⁻¹ °C⁻¹ × 50.0 °C = 60 000 J = 60.0 kJ.
(b) Energy that must be released by the reaction = 60.0 kJ ÷ 0.700 = 85.71 kJ. Moles of M = 85.71 kJ ÷ 65.0 kJ mol⁻¹ = 1.319 mol. Mass = 1.319 mol × 72.0 g mol⁻¹ = 94.9 g (about 95 g).
(c) M and water together have higher chemical potential energy than the products. When they react, bonds are broken (needing energy) and new, stronger bonds are formed (releasing more energy). The difference becomes kinetic energy of the product particles, so they are hotter than the soup. Energy then flows from them to the soup until thermal equilibrium. The energy was not “stored” in M alone: it is the difference in potential energy between reactants and products.
| Point | What earns it |
|---|---|
| 1 | 60.0 kJ for the soup |
| 1 | Divides by 0.700 to get 85.7 kJ released |
| 1 | 94.9 g of M from the moles of M |
| 1 | Reactants have higher potential energy than products; forming bonds releases more energy than breaking bonds needs |
| 1 | The difference becomes kinetic energy and is transferred to the soup (thermal equilibrium) |
Ignoring the 70.0% gives 66.5 g, which earns the first point only. Allow the third point if a wrong released energy is carried forward correctly.
How did you do?
- Q1 or Q3 wrong: re-read “Scaling and reversing” and the “per mole of reaction” meaning in the study guide.
- Q2 or Q5 wrong: practise Worked examples 1 to 3: convert to moles, divide by the coefficient, use the limiting reactant.
- Q4, Q6(c) or Q7(c) incomplete: use the two-stage story in Figure 1: potential energy → kinetic energy → transfer to the surroundings.
Then tick off the topic checklist.