S5 Chemistry: Chemical Equilibrium And Le Chatelier Principle (Paper 1: Basics)

S5 Chemistry: Chemical Equilibrium and Le Chatelier Principle (Paper 1: Basics)

2026-11-12

S5 Chemistry: Chemical Equilibrium and Le Chatelier Principle (Paper 1: Basics)

Ten items building the basics of Kc expressions, dynamic equilibrium, catalysts and changes in concentration, pressure and temperature.

Direct answers　Answers in order: Kc = [HI]^2/([H2][I2]), 64, equal but non-zero rates, Kc unchanged and no shift, to the right, Kc falls and shifts left, forward, to the right, 0.267, no.

01　Paper details (check answers afterwards)

Item	Detail
Questions	10
Suggested time	20
Difficulty mix	Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1
Chapter	Chapter: S5 Chemistry: Chemical Equilibrium and Le Chatelier Principle
Level	S5

02　Questions (10 items)

FoundationWrite the equilibrium constant Kc expression for H2(g) + I2(g) ⇌ 2HI(g).
FoundationIn a 1.0 dm3 vessel at equilibrium [H2] = 0.20, [I2] = 0.20 and [HI] = 1.60 mol dm-3. Find Kc.
FoundationChemical equilibrium is described as dynamic. What does that mean?
IntermediateA catalyst is added to a system already at equilibrium. What happens to Kc and to the position of equilibrium?
IntermediateFor N2(g) + 3H2(g) ⇌ 2NH3(g) at constant temperature, which way does equilibrium shift when the pressure is raised by reducing the volume?
IntermediateAn exothermic reaction is at equilibrium. How do Kc and the position of equilibrium change when the temperature is raised?
ChallengeFor a reaction Q = 4.0 while Kc = 64 at the same temperature. Which way does the reaction proceed?
ChallengeHI is continuously removed from the equilibrium H2(g) + I2(g) ⇌ 2HI(g). How does the equilibrium shift?
Challenge1.00 mol of PCl5 is placed in a 1.00 dm3 vessel and 0.40 mol of Cl2 is present at equilibrium. Find Kc.
Error redoDrill: a student says that the reaction has stopped once equilibrium is reached. Is that correct?

03　Answers and one-line notes

#	Answer	One-line note
1	Kc = [HI]^2 / ([H2][I2])	Products go in the numerator and coefficients become powers; pure solids and pure liquids are left out.
2	Kc = 64	Substituting gives 1.60 squared divided by 0.20 times 0.20, which is 64.
3	The forward and reverse rates are equal but not zero, so both directions continue.	Particles keep reacting in both directions even though there is no net change in concentration.
4	Kc is unchanged and the position does not shift; equilibrium is simply reached faster.	A catalyst lowers the activation energy of both directions equally, so both rates speed up by the same factor.
5	To the right, because the number of gas moles falls from 4 to 2	Raising the pressure shifts the system towards the side with fewer gas moles.
6	Kc decreases and the equilibrium shifts to the left	Heating favours the endothermic direction, so the exothermic forward reaction is suppressed and Kc falls.
7	In the forward direction, because Q is smaller than Kc	Q below Kc means products are still short, so the system makes more products until Q reaches Kc.
8	It shifts to the right until Q equals Kc again	Removing a product lowers Q, so the forward reaction replaces the product.
9	Kc = 0.267	PCl3 is also 0.40 mol and PCl5 left is 0.60 mol, so Kc is 0.40 times 0.40 divided by 0.60.
10	No, equilibrium is dynamic and both reactions continue	If reactions had stopped, isotope labelling experiments would show no exchange, but they do.

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.

Write the Kc expression first, decide the direction second, and only then check the answer.Tutor Registration →

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