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S5 Chemistry: Chemical Equilibrium and Le Chatelier Principle (Paper 3: Challenge)
S5 Chemistry: Chemical Equilibrium and Le Chatelier Principle (Paper 3: Challenge)
Ten challenge items on combining Kc values, ICE tables for mole amounts and Q comparisons under volume and pressure changes.
Direct answers Answers in order: 0.25, 6.0, 2.0, 2.67 mol, Q becomes one quarter so it shifts right, left, 0.25, nothing else and reverse, 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)
- FoundationIf A ⇌ B has Kc = 4.0 at some temperature, what is Kc for B ⇌ A?
- FoundationGiven Kc = 2.0 for A ⇌ B and Kc = 3.0 for B ⇌ C, find Kc for A ⇌ C.
- FoundationWrite the Kc expression for A(g) ⇌ 2B(g).
- IntermediateN2 + 3H2 ⇌ 2NH3 has Kc = 4.0. What is Kc when the equation is written as 1/2 N2 + 3/2 H2 ⇌ NH3?
- Intermediate2.00 mol of H2 and 2.00 mol of I2 are placed in 2.00 dm3 with Kc = 16. How many moles of HI are present at equilibrium?
- ChallengeThe equilibrium N2 + 3H2 ⇌ 2NH3 is compressed to half its volume. What happens to Q, and which way does it shift?
- ChallengeCaCO3(s) ⇌ CaO(s) + CO2(g) is at equilibrium. Which way does it shift when the pressure is raised?
- Challenge1.00 mol of PCl5 in a 2.00 dm3 vessel is 50 per cent decomposed at equilibrium. Find Kc.
- IntermediateA system has Q greater than Kc at some instant. What else is needed to decide the direction?
- Error redoDrill: a student claims changing a concentration changes the value of Kc. Is that right?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | Kc = 0.25 | The reverse reaction expression is the reciprocal of the forward one. |
| 2 | Kc = 6.0 | When two steps are added, their equilibrium constants are multiplied. |
| 3 | Kc = [B]^2 / [A] | The product B is squared in the numerator while A goes in the denominator. |
| 4 | Kc = 2.0 | Halving every coefficient takes the square root of Kc. |
| 5 | 2.67 mol | Each starts at 1.00 mol dm-3, so x is 2/3, [HI] is 4/3 and multiplying by 2.00 dm3 gives 2.67 mol. |
| 6 | Q becomes 0.25 of the old value, below Kc, so it shifts right | All concentrations double, and Q has four concentration terms in the denominator, so Q falls to one quarter. |
| 7 | To the left, because the left side has zero gas moles | Raising the pressure moves the system towards fewer gas moles, so CO2 is consumed. |
| 8 | Kc = 0.25 | 0.50 mol decomposes, giving 0.25 mol dm-3 of each product and 0.25 mol dm-3 of PCl5 left. |
| 9 | Nothing else: the reaction runs in reverse until Q equals Kc | Q above Kc means too much product, so the reverse reaction must consume it. |
| 10 | No, Kc depends only on temperature | Changing a concentration only moves Q away from Kc, and the system returns Q to the same Kc. |
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.
For each item write the three ICE rows: initial, change, equilibrium.Tutor Registration →
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