What comes before calculating Kc?

Subtract the amount reacted from the initial amount so all three equilibrium concentrations are known.

Which factors change Kc?

Only temperature; concentration, pressure and catalysts affect position or timing only.

S6 Chemistry Chemical Equilibrium, Paper 1: Kc expressions, Le Chatelier and calculations

2026-11-02

Ten graded items from writing Kc through equilibrium calculations to pressure and temperature effects, each with an answer.

Focus　Match each power to its coefficient; subtract the change from the initial amount before calculating Kc; neither catalysts nor dilution alter Kc.

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)

FoundationFor aA + bB <=> cC + dD, what is the expression for Kc?
FoundationFor an exothermic reaction, how do the position of equilibrium and Kc change when temperature rises?
FoundationWhat effect does a catalyst have on the position of equilibrium and on Kc?
FoundationWhen pressure is increased in a gaseous equilibrium, which way does it shift?
IntermediateFor H2 + I2 <=> 2HI, equilibrium concentrations are 0.20, 0.20 and 1.60 mol dm^-3. Find Kc.
IntermediateIn N2 + 3H2 <=> 2NH3, ammonia is continuously removed. Which way does the equilibrium shift?
IntermediateFor A <=> B with Kc = 4.0, find [B] when [A] is 0.50 mol dm^-3 at equilibrium.
Challenge0.100 mol of PCl5 is placed in a 1.00 dm^3 vessel and 0.040 mol decomposes at equilibrium. Find Kc for PCl5 <=> PCl3 + Cl2.
ChallengeFor 2SO2 + O2 <=> 2SO3, equilibrium concentrations are 0.20, 0.10 and 0.40 mol dm^-3. Find Kc.
Error redoError repair: a student says diluting an equilibrium mixture lowers Kc. What is wrong?

03　Answers and one-line notes

#	Answer	One-line note
1	Kc = [C]^c[D]^d / ([A]^a[B]^b)	Product concentrations go on top and reactants below, each raised to its coefficient.
2	It shifts to the left and Kc decreases	Heating favours the endothermic direction, here the reverse, so products fall and Kc decreases.
3	Neither changes; only the time to reach equilibrium shortens	A catalyst speeds both directions equally, so the position and Kc are unchanged.
4	Towards the side with fewer moles of gas	Raising the pressure drives the system towards fewer gas molecules to oppose the change.
5	Kc = 64.0	1.60 squared is 2.56, and dividing by 0.040 from the two reactants gives 64.0.
6	To the right, forming more ammonia	Removing a product lowers its concentration, so the system shifts forward to replace it.
7	2.0 mol dm^-3	Kc is the concentration of B over A, so B is 4.0 times 0.50, which is 2.0.
8	Kc = 0.0267	At equilibrium 0.060 mol of PCl5 remains and each product is 0.040 mol, so 0.0016 over 0.060 gives 0.0267.
9	Kc = 40	The top is 0.40 squared, or 0.16, and the bottom is 0.040 times 0.10, or 0.0040, giving 40.
10	Kc depends only on temperature; dilution only shifts the position	Changing concentrations moves the position but Kc stays constant at a fixed temperature.

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.

Book an S6 Chemistry tutor Tutor Registration →

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.