Physical Chemistry
8 marks · NormalA student investigates the combustion of ethanol in air, measuring temperature rise to calculate enthalpy change and evaluating the reliability of results. She also experiments with the rate of a reaction between magnesium and hydrochloric acid under changed pressure.
In a combustion experiment, 0.90 g of ethanol (C₂H₅OH, Mᵣ = 46.0) is burned to heat 120.0 g water. The temperature of the water rises from 20.0°C to 47.0°C. Assume c = 4.18 J/g°C.
Calculate the heat absorbed by the water in joules.
Calculate the number of moles of ethanol burned.
Calculate the experimental molar enthalpy change (ΔH), in kJ/mol, for ethanol combustion. Use the correct sign and 2 significant figures.
Suggest two sources of error in the calorimetry method that could explain why the experimental ΔH is less negative than expected.
The student repeats a reaction between magnesium and hydrochloric acid, but increases the pressure (with all chemicals in solution). Describe and explain what effect, if any, changing the pressure will have on the rate of this reaction.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a(i)) 13,507 J
(a(ii)) 0.0196 mol
(a(iii)) –690 kJ/mol · must show negative sign, rounded to 2 SF
(b) Heat lost to the surroundings; incomplete combustion of ethanol; (also accept: evaporation of ethanol, not all heat absorbed by the water only) · must show two suitable errors
(c) No effect; changing pressure has little effect on the rate of reactions in solution because liquids are nearly incompressible. · must show no effect, reason (liquids nearly incompressible)
Common mistakes: Forgets negative sign for ΔH; Uses wrong number of moles; States pressure will increase rate when it does not for solutions
Show mark scheme and worked solution
Marking points
A1 (a(i)) · 2 marks · Correct calculation of Q: Q = 120 × 4.18 × 27 = 13,507 J.
A2 (a(ii)) · 1 mark · Moles of ethanol: 0.90 / 46.0 = 0.0196 mol (2 SF, accept 0.01957).
A3 (a(iii)) · 2 marks · ΔH = –13,507 J / 0.0196 mol = –689,130 J/mol = –690 kJ/mol (2 SF, negative sign present). (follow-through)
E1 (b) · 2 marks · Names two valid sources of error (heat loss, incomplete combustion, evaporation, etc.).
E2 (c) · 1 mark · States no effect on rate for liquid reactions, with reason (liquids nearly incompressible).
Worked solution
a(i) Q = mcΔT = 120.0 g × 4.18 J/g°C × (47.0–20.0)°C = 120 × 4.18 × 27 = 13,507 J.
a(ii) Moles ethanol = 0.90 g / 46.0 g/mol = 0.0196 mol.
a(iii) ΔH = –Q / moles = –13,507 / 0.0196 = –689,130 J/mol = –690 kJ/mol (2 SF, negative sign for exothermic).
b) Two sources of error: Heat is lost to the surroundings (not all heats the water); incomplete combustion of ethanol (some remains unburnt).
c) Increasing pressure has almost no effect on the rate of a reaction between substances in solution because liquids are (almost) incompressible. So, there will be no significant effect.