Why does Kc depend only on temperature?

The equilibrium position, and hence the concentration ratio, is fixed by temperature for a given reaction; pressure and catalysts do not alter it.

Where do students lose marks on a decomposition question?

They substitute moles straight into Kc without dividing by the vessel volume, so every term is wrong.

S6 Chemistry Chemical Equilibrium, Paper 4: Kc calculations, reaction quotient and shifts

2026-11-02

Tutor Prep Guide · Primary English

Ten graded items on Kc expressions, temperature effects, pressure shifts, equilibrium concentrations and a decomposition percentage, each with an answer.

Focus　Kc depends only on temperature, compare Q with Kc for direction, pressure favours fewer gas moles and divide by volume before substituting.

01　Paper details (check answers afterwards)

Item	Detail
Questions	10
Suggested time	25
Difficulty mix	Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1
Chapter	Chapter: S6 Chemistry: Chemical Equilibrium
Level	Four basic, three advanced, two challenge and one error-repair item

02　Questions (10 items)

FoundationWrite the equilibrium constant expression for N2(g) + 3H2(g) ⇌ 2NH3(g).
FoundationFor an endothermic reaction at constant conditions, what happens to Kc when the temperature is raised?
FoundationWhat effect does a catalyst have on the position of equilibrium and on Kc?
FoundationN2O4(g) ⇌ 2NO2(g) is at equilibrium. At constant temperature the pressure is raised. Which way does it shift?
IntermediateAt equilibrium [H2] = 0.20, [I2] = 0.20 and [HI] = 0.80 mol dm^-3 for H2 + I2 ⇌ 2HI. Find Kc.
IntermediateFor A(g) + B(g) ⇌ 2C(g), Kc = 4.0. At equilibrium [A] = [B] = 0.50 mol dm^-3. Find [C].
IntermediateA mixture gives a reaction quotient Q = 8.0 while Kc = 16 at that temperature. Which way does the reaction proceed?
ChallengeFor 2SO2(g) + O2(g) ⇌ 2SO3(g), Kc = 2.8 x 10^2 at a certain temperature. At equilibrium [SO2] = 0.10 and [O2] = 0.20 mol dm^-3. Find [SO3].
Challenge0.500 mol of PCl5 is sealed in a 2.00 dm^3 vessel and 40% decomposes at equilibrium. Find Kc for PCl5(g) ⇌ PCl3(g) + Cl2(g).
Error redoError repair: a student writes that adding a catalyst raises the equilibrium yield of ammonia. What is wrong?

03　Answers and one-line notes

#	Answer	One-line note
1	Kc = [NH3]^2 / ([N2][H2]^3)	Products go on top raised to their coefficients and reactants below, so ammonia is squared and hydrogen is cubed.
2	Kc increases	Heating favours the endothermic direction, so the equilibrium shifts right and the ratio of products to reactants rises.
3	Neither changes; only the time to reach equilibrium is shortened	A catalyst speeds both directions equally, so the position and Kc, which depend on temperature, are unchanged.
4	To the left, towards N2O4	Raised pressure favours the side with fewer gas moles, which is the single mole of dinitrogen tetroxide on the left.
5	16	The numerator is 0.80 squared, or 0.64, and the denominator is 0.20 times 0.20, or 0.040, so Kc is 16.
6	1.0 mol dm^-3	The square of [C] equals 4.0 times 0.50 times 0.50, which is 1.0, so [C] is 1.0.
7	To the right, forming more product	Q below Kc means products are relatively scarce, so the forward reaction dominates and the system shifts right.
8	0.75 mol dm^-3	The square of [SO3] equals 280 times 0.010 times 0.20, or 0.56, whose square root is about 0.75.
9	0.0667	0.200 mol decomposes, leaving 0.300 mol of PCl5 and 0.200 mol each of PCl3 and Cl2, which divide by 2.00 dm^3 before substitution.
10	A catalyst does not shift the equilibrium; it only shortens the time taken to reach it	A catalyst speeds both directions equally, so the equilibrium concentrations and Kc are unchanged and the yield does not rise.

04　What to do after marking

After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
Repeat the same item two days later; move to the next paper only when both attempts are correct.
Parents only need the answer and note columns; there is no need to explain every item.

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Questions and answers are written originally by the TutorZone team and are not taken from any school, publisher or examination authority paper. Every value is checked by computer.