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S5 Chemistry: Chemical Equilibrium and Le Chatelier Principle (Paper 2: Intermediate)
S5 Chemistry: Chemical Equilibrium and Le Chatelier Principle (Paper 2: Intermediate)
Ten items on comparing Q with Kc, Kc units, pressure and temperature effects and conversion calculations.
Direct answers Answers in order: Kc = [SO3]^2/([SO2]^2[O2]), dm3 mol-1, 1.56 mol dm-3, Q = 2.96 below Kc so forward, left, no shift, Kc rises and shifts right, 60 per cent, 0.80 mol dm-3, 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 Kc expression for 2SO2(g) + O2(g) ⇌ 2SO3(g).
- FoundationWhat is the unit of Kc for that reaction?
- Foundation1.00 mol of H2 and 1.00 mol of I2 are sealed in 1.00 dm3 with Kc = 49. What is [HI] at equilibrium?
- IntermediateAt some temperature [N2] = 0.50, [H2] = 0.30 and [NH3] = 0.20 mol dm-3 while Kc = 5.0. Which way does the reaction proceed?
- IntermediateThe equilibrium 2NO2(g) ⇌ N2O4(g) is disturbed by increasing the volume. How does it shift?
- IntermediateHelium, an inert gas, is added at constant volume to N2 + 3H2 ⇌ 2NH3. Does the equilibrium shift?
- ChallengeAn endothermic reaction is at equilibrium. What happens to Kc and to the position when the temperature is raised?
- Challenge2.00 mol of SO2 and 1.00 mol of O2 are sealed and 0.80 mol of SO2 remains at equilibrium. Find the percentage conversion of SO2.
- ChallengeFor A ⇌ B, Kc = 0.25 and [B] = 0.20 mol dm-3 at equilibrium. Find [A].
- Error redoDrill: a student claims a catalyst raises the equilibrium yield. Is that right?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | Kc = [SO3]^2 / ([SO2]^2 [O2]) | The power of each concentration equals its coefficient in the equation. |
| 2 | dm3 mol-1 | The gas moles fall from 3 to 2, so Kc carries (mol dm-3) to the power minus one, that is dm3 mol-1. |
| 3 | [HI] = 1.56 mol dm-3 | With 2x formed, 2x divided by (1 minus x) equals 7, so x is 7/9 and [HI] is 1.56. |
| 4 | Q = 2.96, below Kc, so it proceeds forward | Calculate Q first, then compare: Q below Kc means more NH3 can still be formed. |
| 5 | To the left, towards NO2 | Lowering the pressure shifts the system towards the side with more gas moles, which is the NO2 side. |
| 6 | No, the partial pressures and concentrations of the reactants are unchanged | Adding an inert gas at constant volume raises only the total pressure, not any reacting concentration. |
| 7 | Kc increases and the equilibrium shifts to the right | Heating favours the endothermic direction, so the forward rate rises more and the product fraction grows. |
| 8 | 60 per cent | Reacted SO2 is 2.00 minus 0.80, that is 1.20 mol, divided by 2.00 and multiplied by 100. |
| 9 | [A] = 0.80 mol dm-3 | Kc equals [B] over [A], so [A] is 0.20 divided by 0.25, which is 0.80. |
| 10 | No, a catalyst does not change the position of equilibrium | It speeds both directions equally and leaves Kc unchanged, so the equilibrium composition stays 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.
Label each item as concentration, pressure or temperature before deciding the direction.Tutor Registration →
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