How to use this review
Use this page after the Unit 6 diagnostic and any topics it flagged. Answer all seven questions before opening any answer.
These are original Marlbridge practice questions, not past exam questions. Substances named with letters, and all experimental data, are fictional. Each rubric is a suggested Marlbridge rubric, not official scoring, and your total gives no predicted score.
Big ideas of the unit
- Name the system first. Exothermic: the system loses energy to the surroundings; endothermic: it gains energy from them (Topic 6.1).
- An energy diagram shows the system only. ΔH = final level − starting level; the hump is the activation energy, not ΔH (Topic 6.2).
- Temperature is average kinetic energy. Collisions pass energy from warmer to cooler until the temperatures are equal (Topic 6.3).
- The thermometer reads the surroundings. So q(process) = −q(surroundings) (Topic 6.4).
- Phase changes move energy at constant temperature. The energy goes into or comes out of attractions: q = n × ΔH (Topic 6.5).
- ΔH is per mole of reaction as written. Use the limiting reactant and its coefficient (Topic 6.6).
- Bonds explain the sign. Breaking needs energy, forming releases it; average values give an estimate for gases (Topic 6.7).
- Formation enthalpies give accurate values. Products minus reactants, with every state symbol and coefficient (Topic 6.8).
- Any route gives the same ΔH. Energy is conserved, so the steps’ ΔH values add up (Topic 6.9).
Key relationships and methods
| Idea | Relationship or method | Topic |
|---|---|---|
| Direction of energy | surroundings warm → exothermic; cool → endothermic | 6.1 |
| Energy diagram | ΔH = E(products) − E(reactants); arrow from start to end | 6.2 |
| Thermal equilibrium | equal average kinetic energy, equal temperature | 6.3 |
| Heating or cooling | q = mcΔT; object: q = CΔT; molar: q = n C_m ΔT | 6.4 |
| Calorimetry | q(lost) + q(gained) = 0; q(process) = −q(solution) | 6.4 |
| Phase change | q = n × ΔH_fus or n × ΔH_vap; reverse → opposite sign | 6.5 |
| Reaction | q = (n ÷ coefficient) × ΔH, using the limiting reactant | 6.6 |
| Bond enthalpies | ΔH ≈ Σ(broken) − Σ(formed) | 6.7 |
| Formation enthalpies | ΔH° = ΣnΔH°f(products) − ΣnΔH°f(reactants); elements = 0 | 6.8 |
| Hess’s law | reverse → change sign; × c → ΔH × c; add equations → add ΔH | 6.9 |
Question 1 (multiple choice · mixed)
Two fictional solutes react: 2D(aq) + E(aq) → F(aq). A student mixes 40.0 mL of 0.500 M D with 60.0 mL of 0.250 M E in an insulated cup. The temperature of the 100.0 g of solution rises by 2.40 °C (c = 4.18 J g⁻¹ °C⁻¹). What is ΔH for the reaction as written?
- (A) +100 kJ mol⁻¹
- (B) −50.2 kJ mol⁻¹
- (C) −100 kJ mol⁻¹
- (D) −66.9 kJ mol⁻¹
Answer and explanation
Answer: (C). q(solution) = 100.0 × 4.18 × 2.40 = +1003 J, so q(reaction) = −1003 J (Topic 6.4). Moles: D 0.0200 mol, E 0.0150 mol. All the D needs only 0.0100 mol E, so D is limiting, and moles of reaction = 0.0200 ÷ 2 = 0.0100 mol (Topic 6.6). ΔH = −1.003 kJ ÷ 0.0100 mol = −100 kJ mol⁻¹.
- (A) gives the solution’s sign, not the reaction’s.
- (B) uses moles of D without dividing by the coefficient 2.
- (D) treats E as limiting because it has fewer moles.
Question 2 (multiple choice · mixed)
A 100.0 g block of aluminium at 90.0 °C is placed in 100.0 g of water at 20.0 °C in an insulated container. Which describes the final state?
- (A) Both reach 55.0 °C, because the masses are equal.
- (B) Both reach 32.4 °C, because the water needs more energy per degree than the aluminium.
- (C) Both reach 77.6 °C, because the water needs less energy per degree than the aluminium.
- (D) Both reach 32.4 °C, and the aluminium atoms then still have a greater average kinetic energy than the water molecules.
Answer and explanation
Answer: (B). Energy lost by aluminium = energy gained by water (Topic 6.4). Heat capacities: aluminium 89.7 J °C⁻¹, water 418.0 J °C⁻¹. T = (89.7 × 90.0 + 418.0 × 20.0) ÷ 507.7 = 32.4 °C. The aluminium falls 57.6 °C but the water rises only 12.4 °C for the same 5.17 kJ.
- (A) assumes equal heat capacities; equal masses are not enough.
- (C) swaps the two specific heat capacities.
- (D) At thermal equilibrium the average kinetic energies are equal (Topic 6.3).
Question 3 (multiple choice · mixed)
For the fictional gas-phase reaction 2XY(g) → X₂(g) + Y₂(g), the bond enthalpies are X–Y 300, X–X 150 and Y–Y 410 kJ mol⁻¹. Which statement is correct?
- (A) ΔH ≈ +40 kJ mol⁻¹; the reaction is endothermic, so the temperature of the surroundings falls.
- (B) ΔH ≈ −40 kJ mol⁻¹; the reaction is exothermic, so the temperature of the surroundings rises.
- (C) ΔH ≈ +40 kJ mol⁻¹; the reaction is endothermic, so the surroundings warm up as they pass energy to the system.
- (D) ΔH ≈ −260 kJ mol⁻¹; the reaction is exothermic, so the temperature of the surroundings rises.
Answer and explanation
Answer: (A). Broken: 2 X–Y = 600 kJ. Formed: X–X + Y–Y = 560 kJ. ΔH ≈ +40 kJ mol⁻¹ (Topic 6.7). The system gains energy from the surroundings, which cool (Topic 6.1).
- (B) uses formed − broken.
- (C) has the right ΔH, but a body that gives away energy cools down.
- (D) counts only one X–Y bond, ignoring the coefficient 2.
Question 4 (constructed response · mixed)
Steam at 100.0 °C is bubbled into 200.0 g of water at 18.0 °C in an insulated flask until the water reaches 46.0 °C. Water: ΔH_vap = 40.7 kJ mol⁻¹ at 100 °C, c = 4.18 J g⁻¹ °C⁻¹, M = 18.016 g mol⁻¹.
(a) Calculate the energy gained by the 200.0 g of water. (b) Calculate the mass of steam that condensed. (c) Explain, in terms of particles, why energy moves from the condensed steam to the cooler water, and why the transfer stops. (d) Calculate the mass of liquid water at 100.0 °C that would give the same final temperature. Explain why it is much larger than (b).
Worked solution and suggested Marlbridge rubric
(a) q = 200.0 × 4.18 × (46.0 − 18.0) = 23 408 J = 23.4 kJ.
(b) Each gram of steam releases energy in two steps (Topics 6.5 and 6.4):
- Condensing at 100.0 °C: 40 700 J mol⁻¹ ÷ 18.016 g mol⁻¹ = 2259.1 J g⁻¹.
- Cooling the condensed water from 100.0 °C to 46.0 °C: 4.18 × 54.0 = 225.7 J g⁻¹.
Total 2484.8 J g⁻¹. Energy lost = energy gained: m = 23 408 ÷ 2484.8 = 9.42 g. Leaving out step 2 gives 10.4 g.
(c) The condensed steam’s molecules have a greater average kinetic energy than the cooler water’s, so in collisions energy passes on average from faster to slower molecules. The net transfer stops at thermal equilibrium (46.0 °C), when the average kinetic energies are equal; collisions continue (Topic 6.3).
(d) m = 23 408 ÷ 225.7 = 104 g, about 11 times the mass of steam. Condensing releases 2259 J g⁻¹ at constant temperature as attractions form: ten times the 226 J g⁻¹ from cooling.
| Point | What earns it |
|---|---|
| 1 | (a) 23.4 kJ (23 408 J) |
| 1 | (b) Condensing energy from ΔH_vap, through moles |
| 1 | (b) Includes cooling the condensed water to 46.0 °C |
| 1 | (b) 9.42 g |
| 1 | (c) Average kinetic energy; collisions transfer energy until the averages are equal |
| 1 | (d) About 104 g, explained by condensation energy at constant temperature |
Total: 6 points.
Question 5 (constructed response · mixed)
Propene burns: C₃H₆(g) + 9/2 O₂(g) → 3CO₂(g) + 3H₂O. Propene is H₂C=CH–CH₃.
Average bond enthalpies (kJ mol⁻¹): C–H 413, C–C 347, C=C 614, O=O 495, C=O (in CO₂) 799, O–H 467. ΔH°f (kJ mol⁻¹): C₃H₆(g) +20.4, CO₂(g) −393.5, H₂O(g) −241.8, H₂O(l) −285.8.
(a) Use bond enthalpies to estimate ΔH when the water forms as a gas. (b) Use enthalpies of formation to calculate ΔH° when the water forms as a gas. Comment on the agreement with (a). (c) Calculate ΔH° when the water forms as a liquid. Use Hess’s law to explain the difference from (b). (d) Describe an energy diagram, with the reactants at 0, that shows your answers to (b) and (c). (e) Calculate the heat released when 7.00 g of propene burns, with the water formed as a liquid.
Worked solution and suggested Marlbridge rubric
(a) Broken: 1 C=C + 1 C–C + 6 C–H + 9/2 O=O = 614 + 347 + 2478 + 2227.5 = 5666.5 kJ. Formed: 6 C=O + 6 O–H = 4794 + 2802 = 7596 kJ. ΔH ≈ 5666.5 − 7596 = −1929.5 kJ mol⁻¹.
(b) Products: 3(−393.5) + 3(−241.8) = −1905.9 kJ. Reactants: +20.4 + 0 = +20.4 kJ. ΔH° = −1905.9 − 20.4 = −1926.3 kJ mol⁻¹. Close agreement (about 3 kJ); (a) is only an estimate because bond enthalpies are averages (Topic 6.7).
(c) Products: 3(−393.5) + 3(−285.8) = −2037.9 kJ. ΔH° = −2037.9 − 20.4 = −2058.3 kJ mol⁻¹. By Hess’s law, split the route: burn to H₂O(g), then condense 3 mol of water, 3 × [(−285.8) − (−241.8)] = −132.0 kJ. −1926.3 + (−132.0) = −2058.3: the extra 132.0 kJ is released as attractions form between water molecules.
(d) Axis “Energy (kJ mol⁻¹)”. Reactants at 0; 3CO₂(g) + 3H₂O(g) at −1926.3; 3CO₂(g) + 3H₂O(l) at −2058.3. Downward arrows from the reactants to each product level, and a short downward arrow (condensing) between them.
(e) M(C₃H₆) = 42.078 g mol⁻¹; n = 7.00 ÷ 42.078 = 0.1664 mol (coefficient 1). q = 0.1664 × (−2058.3) = −342 kJ: 342 kJ released.
| Point | What earns it |
|---|---|
| 1 | (a) Correct bonds, including 9/2 O=O and 6 C=O |
| 1 | (a) −1929.5 kJ mol⁻¹ as broken − formed |
| 1 | (b) −1926.3 kJ mol⁻¹ as products − reactants, with C₃H₆ at +20.4 |
| 1 | (b) Close agreement; bond enthalpies are averages, so (a) is an estimate |
| 1 | (c) −2058.3 kJ mol⁻¹, with 132.0 kJ as a Hess’s law condensation step |
| 1 | (d) Three levels correct, liquid products lowest |
| 1 | (e) 342 kJ, with moles of propene shown |
Total: 7 points. Carry forward from (c) into (e).
Question 6 (constructed response · mixed)
A fictional salt RX (M = 95.0 g mol⁻¹) is tested for a cold pack: ΔH_soln = +18.0 kJ mol⁻¹. Take c = 4.18 J g⁻¹ °C⁻¹ for every solution and ignore energy exchange with the container or air.
(a) Explain, in terms of the attractions broken and formed, why dissolving RX is endothermic. (b) A student dissolves 9.50 g of RX in 90.5 g of water at 22.0 °C. Predict the final temperature. (c) The pack must cool its contents, 150.0 g of solution in total, from 25.0 °C to 5.0 °C. Calculate the mass of RX needed. (d) Describe an energy diagram for dissolving RX. Then explain, with the direction of energy transfer, why a hand holding the pack feels cold.
Worked solution and suggested Marlbridge rubric
(a) Separating the ions, and some water molecules, needs energy; forming ion–water attractions releases it. ΔH is positive, so the attractions formed are weaker overall than those broken (Topic 6.1).
(b) n = 9.50 ÷ 95.0 = 0.100 mol, so q(dissolving) = 0.100 × 18.0 = +1.80 kJ (Topic 6.6). The solution supplies it: q(solution) = −1800 J. Mass of solution = 100.0 g. ΔT = −1800 ÷ (100.0 × 4.18) = −4.31 °C. Final temperature = 17.7 °C.
(c) Energy the solution must lose: 150.0 × 4.18 × 20.0 = 12 540 J = 12.54 kJ. Moles of RX = 12.54 ÷ 18.0 = 0.697 mol. Mass = 0.697 × 95.0 = 66.2 g.
(d) Diagram: RX(s) + water at 0; RX(aq) at +18.0 kJ mol⁻¹; upward arrow labelled ΔH = +18.0 kJ mol⁻¹ (Topic 6.2). The skin’s particles have a greater average kinetic energy than the pack’s, so energy passes by collisions from the hand to the pack (Topic 6.3). The hand loses energy and cools. No “cold” moves.
| Point | What earns it |
|---|---|
| 1 | (a) Breaking needs energy, forming releases it, and those formed are weaker overall |
| 1 | (b) q = +1.80 kJ for 0.100 mol, with q(solution) = −q(dissolving) |
| 1 | (b) 17.7 °C |
| 1 | (c) 12.54 kJ converted to 0.697 mol |
| 1 | (c) 66.2 g |
| 1 | (d) Product level above the start and hand → pack, by average kinetic energy |
Total: 6 points. Carry forward from (b) into (c).
Question 7 (constructed response · mixed)
A student burns ethanol, C₂H₅OH(l), in a spirit burner under a copper can that holds 150.0 g of water. The can has a mass of 120.0 g (c = 0.385 J g⁻¹ °C⁻¹). The water and can warm from 19.5 °C to 47.5 °C, and the burner loses 0.920 g of ethanol.
ΔH°f (kJ mol⁻¹): C₂H₅OH(l) −277.6, CO₂(g) −393.5, H₂O(l) −285.8, H₂O(g) −241.8.
(a) Calculate the standard enthalpy of combustion of ethanol, for C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l). (b) Calculate the energy absorbed by the water and the can. (c) Calculate the experimental enthalpy of combustion in kJ mol⁻¹. (d) What percentage of the expected energy reached the water and the can? (e) The student says: “The difference shows the formation data are wrong.” Evaluate the claim, using ΔH for the reaction with H₂O(g) as the product.
Worked solution and suggested Marlbridge rubric
(a) Products: 2(−393.5) + 3(−285.8) = −1644.4 kJ. ΔH° = −1644.4 − (−277.6) = −1366.8 kJ mol⁻¹ (Topic 6.8).
(b) ΔT = 28.0 °C. Water: 150.0 × 4.18 × 28.0 = 17 556 J. Can: 120.0 × 0.385 × 28.0 = 1293.6 J. Total = 18 849.6 J = 18.8 kJ (Topic 6.4).
(c) M(C₂H₅OH) = 46.068 g mol⁻¹; n = 0.920 ÷ 46.068 = 0.01997 mol. ΔH = −18.85 kJ ÷ 0.01997 mol = −944 kJ mol⁻¹, negative because the fuel lost energy (Topic 6.6). Ignoring the can gives −879.
(d) 943.9 ÷ 1366.8 × 100 = 69.1%.
(e) The claim is not justified. Even with all the water leaving as vapour, ΔH = 2(−393.5) + 3(−241.8) − (−277.6) = −1234.8 kJ mol⁻¹, still far more negative than −944. So the state of the water cannot explain the gap. Most of the missing 8.4 kJ was lost to the air; incomplete combustion (soot) and ethanol evaporating from the wick also lower the result.
| Point | What earns it |
|---|---|
| 1 | (a) −1366.8 kJ mol⁻¹, with coefficients and O₂ = 0 |
| 1 | (b) 18.8 kJ, including the can |
| 1 | (c) Moles of ethanol (0.01997 mol) |
| 1 | (c) −944 kJ mol⁻¹, negative sign included |
| 1 | (d) 69.1% (carry forward) |
| 1 | (e) −1234.8 kJ mol⁻¹ for H₂O(g), still far from the experiment |
| 1 | (e) Rejects the claim, citing heat loss (or incomplete combustion) |
Total: 7 points.
How did you do?
Total: 29 points (3 multiple choice, 26 constructed response). Where you lost points matters more.
- Direction and sign errors (Questions 1, 3, 6): revisit Topics 6.1, 6.2 and 6.4.
- Calorimetry set-ups (Questions 2, 4, 6, 7): revisit Topics 6.4 and 6.5.
- Moles of reaction (Questions 1, 5(e), 7(c)): revisit Topic 6.6.
- Finding ΔH (Questions 3, 5, 7(a)): revisit Topics 6.7 to 6.9.
- Particle explanations (4(c), 6(a), 6(d)): name the attractions, say “average kinetic energy” and give the direction.
If several topics were weak, return to the Unit 6 diagnostic, then tick off each topic checklist: