Why does inert gas at constant volume not shift the equilibrium?

The total pressure rises, but the partial pressures and concentrations of the reacting gases do not, so the position stays.

What does a very small Kc mean in practice?

Most of the reactant is unchanged at equilibrium, so the yield is low and conditions must be changed instead.

S6 Chemistry Chemical Equilibrium, Paper 3: size of Kc, Le Chatelier and equilibrium calculations

2026-11-02

Tutor Prep Guide · Primary English

Ten graded items from the meaning of a small Kc through pressure and inert gas shifts to ICE calculations and an esterification, each with an answer.

Focus　A Kc far below one lies left; pressure pushes to fewer gas moles; inert gas at constant volume does nothing; take the square root before solving for x.

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)

FoundationA reaction has Kc = 1 x 10^-4 at a given temperature. What does that mean?
FoundationWhich way does 2SO2 + O2 <=> 2SO3 shift when the pressure is raised?
FoundationWhat happens to Kc when an exothermic reaction is heated?
FoundationHelium is added to a reaction at constant volume. What happens to the position of equilibrium?
IntermediateFor H2 + I2 <=> 2HI with Kc = 64.0 and both starting at 0.10 mol dm^-3, find [HI] at equilibrium.
Intermediate0.500 mol of PCl5 is sealed in 1.00 dm^3 and at equilibrium [Cl2] = 0.100 mol dm^-3. Find Kc for PCl5 <=> PCl3 + Cl2.
IntermediateAt some moment Q = 4.0 while Kc = 1.0. Which way does the reaction proceed?
Challenge0.200 mol of N2 and 0.400 mol of H2 are sealed in 1.00 dm^3 and at equilibrium [NH3] = 0.100 mol dm^-3. Find Kc for N2 + 3H2 <=> 2NH3.
ChallengeFor CH3COOH + C2H5OH <=> CH3COOC2H5 + H2O with Kc = 4.0 and 1.00 mol of each acid and alcohol in 1.00 dm^3, find the equilibrium concentration of the ester.
Error redoError repair: a student claims a catalyst raises the equilibrium yield. What is wrong?

03　Answers and one-line notes

#	Answer	One-line note
1	The equilibrium lies far to the left, mostly reactants	Kc is the product to reactant ratio, so a value far below one puts the position well to the left.
2	To the right, as the gas moles fall from three to two	Raising pressure favours the side with fewer gas moles, and the right has only two.
3	Kc falls and the equilibrium shifts left	Heating favours the endothermic direction, which for an exothermic reaction is the reverse, so Kc falls.
4	Nothing; the partial pressures are unchanged	At constant volume an inert gas raises only the total pressure; the partial pressures of the reactants are unchanged.
5	0.16 mol dm^-3	Let x react; 2x over (0.10 – x) equals 8, giving x = 0.08 and so 0.16 mol dm^-3 of HI.
6	Kc = 0.0250	0.100 mol decomposes leaving 0.400 mol of PCl5, so Kc is 0.100 times 0.100 over 0.400.
7	To the left, since there is too much product	Q above Kc means too much product, so the reverse reaction dominates until Q reaches Kc.
8	Kc = 4.27	Making 0.100 mol of ammonia uses 0.050 mol of nitrogen and 0.150 mol of hydrogen, leaving 0.150 and 0.250; 0.0100 over that product gives 4.27.
9	0.667 mol dm^-3	Let x form; x squared over (1 – x) squared equals 4.0, so x over (1 – x) is 2 and x is 0.667.
10	It only shortens the time to reach equilibrium; the position is unchanged	A catalyst speeds both directions equally, so Kc and the equilibrium yield stay the same.

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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