DP Chemistry (first assessment 2025)

Question 22M.2.HL.TZ1.c(ii)

Date	May 2022	Marks available	[Maximum mark: 2]	Reference code	22M.2.HL.TZ1.c(ii)
Level	HL	Paper	2	Time zone	TZ1
Command term	Calculate	Question number	c(ii)	Adapted from	N/A

c(ii).

[Maximum mark: 2]

22M.2.HL.TZ1.c(ii)

Calculate the value of the equilibrium constant, K, at 298 K. Use sections 1 and 2 of the data booklet.

[2]

Markscheme

$\ln K = ⟨ ⟨ \frac{∆ G}{R . T} = ⟩ ⟩ - \frac{- 33000}{8.31 x 298} / « + » 13.3$ ✔

$K = 6.13 \times 10^{5}$ ✔

Award [2] for correct final answer.

Accept answers in the range 4.4×10⁵ to 6.2×10⁵ (arises from rounding of ln K).

Examiners report

Good performance; for lnK calculation in the equation ΔG = RTlnK, ΔG unit had to be converted from kJ to J. This led to an error of 1000 in the value of lnK for some.

Syllabus sections

Reactivity 2. How much, how fast and how far? » Reactivity 2.3—How far? The extent of chemical change » Reactivity 2.3.7—The equilibrium constant and Gibbs energy change, ΔG, can both be used to measure the position of an equilibrium reaction.

Question 22M.2.HL.TZ1.c(ii)

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