DP Chemistry (last assessment 2024)

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Question 21N.2.sl.TZ0.7

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Date November 2021 Marks available [Maximum mark: 5] Reference code 21N.2.sl.TZ0.7
Level sl Paper 2 Time zone TZ0
Command term Determine, Formulate Question number 7 Adapted from N/A
7.
[Maximum mark: 5]
21N.2.sl.TZ0.7

Alkanes undergo combustion and substitution.

(a)

Determine the molar enthalpy of combustion of an alkane if 8.75 × 10−4 moles are burned, raising the temperature of 20.0 g of water by 57.3 °C.

[2]

Markscheme

«q = mcΔT = 20.0 g × 4.18 J g−1 °C−1 × 57.3 °C =» 4790 «J» ✔

«∆Hc4790 J10008.75×10-4 mol=» –5470 «kJ mol–1» ✔

 

Award [2] for correct final answer.

Accept answers in the range –5470 to –5480 «kJ mol−1».

Accept correct answer in any units, e.g. –5.47 «MJ mol−1» or 5.47 x 106 «J mol−1».

(b)

Formulate equations for the two propagation steps and one termination step in the formation of chloroethane from ethane.

[3]

Markscheme

Cl· + C2H6 → ·C2H5 + HCl ✔

·C2H5 + Cl2 → Cl· + C2H5Cl ✔


·C2H5 + Cl· → C2H5Cl
OR
Cl· + Cl· → Cl2
OR
·C2H5 + ·C2H5 → C4H10 ✔


Do not penalize incorrectly placed radical sign, eg C2H5·.

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