The problems are on the enthalpy changes worksheet. Try them on your own first, then compare with the answers below.
Quick answer: An enthalpy change ΔH is the heat transferred at constant pressure; it is negative for exothermic reactions. These worked answers correct the 20 problems of the Chemistry 11-12 worksheet on enthalpy changes, from name the enthalpy change to spot the errors: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the formula, the substitution and the result with its unit and sign. Compare your cycles and bond counts line by line, because most lost marks come from signs and from forgetting how many moles take part.
Level: Chemistry grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB R1.1-R1.2
Problem 1 Name the enthalpy change
1. 2C(graphite) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l).2. H₂(g) + ½O₂(g) → H₂O(l).
3. 100 kPa and 298 K (25 °C).
Problem 2 Exothermic or endothermic?
1. X is exothermic (products lower); Y is endothermic.2. X: 120 − 200 = −80 kJ/mol. Y: 170 − 100 = +70 kJ/mol.
3. Forward Ea of X = 260 − 200 = 60 kJ/mol. Reverse Ea = 260 − 120 = 140 kJ/mol.
Problem 3 Heat in a water bath
1. ΔT = 27.5 − 18.0 = 9.5 K.2. q = 150 × 4.18 × 9.5 = 5956.5 J = 5.96 kJ.
3. ΔT = 5956.5 / (300 × 4.18) = 4.75 K.
Problem 4 Dissolving sodium hydroxide
1. n = 2.00 / 40.0 = 0.0500 mol.2. q = 100 × 4.18 × 5.3 = 2215 J.
3. ΔH = −2215 / 0.0500 = −44 308 J/mol = −44.3 kJ/mol.
Problem 5 Neutralisation
1. q = 50.0 × 4.18 × 6.8 = 1421 J.2. n = 0.0250 × 1.00 = 0.0250 mol.
3. ΔH = −1421 / 0.0250 = −56 848 J/mol = −56.8 kJ/mol.
Problem 6 Energy per gram and in kilocalories
1. 2803 / 180.0 = 15.57 kJ/g.2. 25.0 × 15.57 = 389 kJ (389.3 kJ).
3. 389.3 / 4.184 = 93.0 kcal.
Problem 7 Carbon, carbon monoxide and Hess
1. The total enthalpy change is the same by any route between the same reactants and products.2. −393.5 = ΔH₁ + (−283.0), so ΔH₁ = −393.5 + 283.0 = −110.5 kJ/mol.
3. CO(g) + ½O₂(g) → CO₂(g): the enthalpy of combustion of carbon monoxide.
Problem 8 Combining equations
1. N₂(g) + 2O₂(g) → 2NO₂(g).2. 180.5 + (−114.1) = +66.4 kJ.
3. 66.4 / 2 = +33.2 kJ/mol.
Problem 9 Formation of butane
1. 4C(s) + 5H₂(g) → C₄H₁₀(g).2. Σ ΔHc(reactants) − Σ ΔHc(products) = 4(−393.5) + 5(−285.8) − (−2877) = −1574.0 − 1429.0 + 2877 = −126 kJ/mol.
Problem 10 Reducing iron ore
1. 2(0) + 3(−393.5) = −1180.5 kJ.2. −824.2 + 3(−110.5) = −1155.7 kJ.
3. ΔH = −1180.5 − (−1155.7) = −24.8 kJ.
Problem 11 Hydrogen chloride
1. 436 + 243 = 679 kJ.2. 2 × 432 = 864 kJ.
3. ΔH = 679 − 864 = −185 kJ; per mole of HCl: −92.5 kJ/mol.
Problem 12 Hydrogenation of ethene
1. Broken: 1 C=C, 1 H–H. Formed: 1 C–C and 2 C–H.2. Broken = 612 + 436 = 1048. Formed = 347 + 2(413) = 1173. ΔH = 1048 − 1173 = −125 kJ/mol.
3. Yes, ΔH is negative.
Problem 13 Spirit burner
1. q = 200 × 4.18 × 24.0 = 20 064 J.2. n = 0.80 / 60.0 = 0.0133 mol; ΔHc = −20 064 / 0.013333 = −1.505 × 10⁶ J/mol = −1505 kJ/mol.
3. (2021 − 1505) / 2021 × 100 = 25.5 %. Heat escapes to the air and the can, and combustion is incomplete.
Problem 14 Bond enthalpy versus data
1. Broken: 3(413) + 358 + 464 + 1.5(498) = 2808 kJ. Formed: 2(805) + 4(464) = 3466 kJ. ΔH = −658 kJ/mol.2. ΔH = (−393.5 + 2(−241.8)) − (−201.0) = −877.1 + 201.0 = −676.1 kJ/mol.
3. Difference = 18.1 kJ/mol (2.7 %). Mean bond enthalpies are averages, not values for these exact molecules.
Problem 15 Zinc and copper sulfate
1. The cooling line is T = 45.8 − 1.0(t − 4), so at 4 min T = 45.8 °C. ΔT = 45.8 − 20.0 = 25.8 K.2. q = 50.0 × 4.18 × 25.8 = 5392 J.
3. n = 0.0500 × 0.500 = 0.0250 mol; ΔH = −5392 / 0.0250 = −216 kJ/mol.
Problem 16 Hydration of copper sulfate
1. Route A: anhydrous solid + water → hydrate → solution (ΔH_hyd then +11.7). Route B: anhydrous solid → solution (−66.1).2. ΔH_hyd + 11.7 = −66.1, so ΔH_hyd = −77.8 kJ/mol.
3. It is hard to add exactly five moles of water and stop at the hydrate without partial dissolving, so the enthalpy cannot be measured by simple calorimetry.
Problem 17 Formation of ethyne
1. 2C(s) + H₂(g) → C₂H₂(g).2. ΔHf = 2(−393.5) + (−285.8) − (−1300) = −1072.8 + 1300 = +227.2 kJ/mol.
3. Positive: ethyne is higher in enthalpy than its elements, so it is energy-rich and releases a lot of energy when it burns.
Problem 18 Mean C–H bond enthalpy
1. Route: CH₄ → C(s) + 2H₂ (−ΔHf), then C(s) → C(g), then 2H₂ → 4H(g).2. ΔH = 74.8 + 716.7 + 4(218.0) = +1663.5 kJ.
3. 1663.5 / 4 = 416 kJ/mol (415.9).
Problem 19 Hydration of ethene
1. Broken: C=C + O–H = 612 + 464 = 1076. Formed: C–C + C–H + C–O = 347 + 413 + 358 = 1118. ΔH = −42 kJ/mol.2. ΔH = −235.3 − (52.3 − 241.8) = −235.3 + 189.5 = −45.8 kJ/mol.
3. 3.8 / 45.8 × 100 = 8.3 %. Mean bond enthalpies are averages and ignore differences between particular molecules.
Problem 20 Spot the errors
1. The heated mass is the solution (100 g), not the solid: q = 100 × 4.18 × 5.3.2. A rise in temperature means heat was released, so ΔH is negative: −44.3 kJ/mol.
3. The answer is in J/mol; divide by 1000: −44.3 kJ/mol.
4. It must be Σ(bonds broken) − Σ(bonds formed).
