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S5 Chemistry: Rates of Reaction (Paper 1: basics)
S5 Chemistry: Rates of Reaction (Paper 1: basics)
Ten basic items on finding average rates from gas volume, mass loss and concentration change, the effects of concentration and temperature, surface area and catalysts, and the unit of rate.
Direct answers Answers in order: 1.2 cm3/s, 0.080 g/s, 0.025 mol dm-3 s-1, about twice, about twice, larger surface area, lower activation energy and not used up, volume unchanged, mol dm-3 s-1, and rate only.
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: Rates of Reaction |
| Level | S5 |
02 Questions (10 items)
- FoundationA reaction gives off 48 cm3 of gas in 40 s. What is the average rate in cm3/s?
- FoundationMagnesium reacts with dilute hydrochloric acid and loses 2.0 g of mass in 25 s. What is the average rate in g/s?
- FoundationA solution rises from 0 to 0.50 mol dm-3 in 20 s. What is the average rate in mol dm-3 s-1?
- IntermediateIf the concentration of hydrochloric acid is doubled with other conditions unchanged, what usually happens to the rate?
- IntermediateBy what factor does the rate usually change when the temperature rises by 10 C?
- IntermediateFor the same mass of magnesium, powder reacts faster than a ribbon. What is the main reason?
- ChallengeHow does a catalyst speed up a reaction, and is it used up at the end?
- ChallengeWhen a lump of calcium carbonate is broken into small pieces before reacting with acid, does the final total gas volume change?
- ChallengeWhat is the unit of rate when concentration changes with time?
- Error redoA student claims that adding a catalyst increases the total amount of product. What is the mistake?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 1.2 cm3/s | Rate is the change divided by the time, so 48 divided by 40 gives 1.2 cm3/s. |
| 2 | 0.080 g/s | 2.0 divided by 25 gives 0.080 g/s. |
| 3 | 0.025 mol dm-3 s-1 | 0.50 divided by 20 gives 0.025 mol dm-3 s-1. |
| 4 | The rate increases to about twice the original value | Doubling the concentration doubles the particles per unit volume, so effective collisions roughly double. |
| 5 | About two times | A higher temperature gives particles more kinetic energy, and the rate usually doubles for each 10 C. |
| 6 | Powder has a larger surface area | A larger surface area exposes more particles to collisions, so the rate is higher. |
| 7 | It provides an alternative path with a lower activation energy and is not used up | A catalyst lowers the activation energy so more particles can pass the barrier, and its amount is unchanged afterwards. |
| 8 | The total volume stays the same; only the rate increases | The total volume depends only on the moles of reactant, and particle size affects only the rate. |
| 9 | mol dm-3 s-1 | Concentration is in mol dm-3 and is divided by time in seconds, so the unit is mol dm-3 s-1. |
| 10 | A catalyst only speeds up the reaction and does not change the total amount of product | The total product depends on the moles of reactant, and a catalyst only shortens the time needed to finish. |
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 every item write the formula rate equals change divided by time first, then substitute the numbers and give the unit.Tutor Registration →
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