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S5 Chemistry: Rates of Reaction (Paper 3: challenge)
S5 Chemistry: Rates of Reaction (Paper 3: challenge)
Ten challenge items on the gas volume and consumption rate of hydrogen peroxide decomposition, converting to concentration rates, percentage completion with calcium carbonate, repeated doubling per ten degrees, rate ratios between two runs, and the activation energy myth.
Direct answers Answers in order: 1200 cm3, 1.0 x 10-3 mol/s, 4.0 x 10-3 mol dm-3 s-1, 60.0 percent, 80.0 percent, 4.8 cm3/s, 8 times, 0.0625 mol dm-3 s-1, 5 to 1, and heating does not change the activation energy.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 20 |
| Difficulty mix | Foundation 3 · Challenge 3 · Intermediate 3 · Error redo 1 |
| Chapter | Chapter: S5 Chemistry: Rates of Reaction |
| Level | S5 |
02 Questions (10 items)
- Foundation0.10 mol of hydrogen peroxide decomposes completely by 2H2O2 to 2H2O plus O2. How many cm3 of oxygen form at 24 dm3/mol?
- FoundationThe same reaction uses up 0.050 mol of hydrogen peroxide in 50 s. What is the average consumption rate in mol/s?
- FoundationIf the solution in the previous item has a volume of 250 cm3, what is the average rate in mol dm-3 s-1?
- Challenge2.0 g of calcium carbonate should give 480 cm3 of carbon dioxide with excess acid but only 288 cm3 is collected. What percentage of the reaction has taken place?
- ChallengeA reaction gives off 96 cm3 of gas in 20 s and the final volume is 120 cm3. What percentage is complete at 20 s?
- IntermediateWhat is the average rate over the first 20 s in the previous item, in cm3/s?
- IntermediateA reaction is heated from 300 K to 330 K and the rate doubles for every 10 K. By what factor does the rate change?
- IntermediateIn 120 cm3 of solution a reactant falls by 0.30 mol in 40 s. What is the average rate in mol dm-3 s-1?
- ChallengeThe same reaction finishes in 25 s in one experiment and 125 s in another. What is the ratio of their average rates?
- Error redoA student says a temperature rise speeds up a reaction because the activation energy falls. What is the mistake?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 1200 cm3 | 0.10 mol of hydrogen peroxide gives 0.050 mol of oxygen, and times 24000 gives 1200 cm3. |
| 2 | 1.0 x 10-3 mol/s | 0.050 divided by 50 gives 1.0 x 10-3 mol/s. |
| 3 | 4.0 x 10-3 mol dm-3 s-1 | 0.050 divided by 0.250 dm3 gives 0.20 mol dm-3, and dividing by 50 s gives 4.0 x 10-3. |
| 4 | 60.0 percent | 288 divided by 480 gives 0.60, which is 60.0 percent. |
| 5 | 80.0 percent | 96 divided by 120 gives 0.80, which is 80.0 percent. |
| 6 | 4.8 cm3/s | 96 divided by 20 gives 4.8 cm3/s. |
| 7 | 8 times | 30 K is three intervals of 10 K, and 2 to the power 3 gives 8. |
| 8 | 0.0625 mol dm-3 s-1 | The 0.30 mol in 0.120 dm3 is 2.5 mol dm-3, and dividing by 40 s gives 0.0625. |
| 9 | 5 to 1, with the faster run first | The times are in the ratio 1 to 5 and rate is inversely proportional to time, so the rates are in the ratio 5 to 1. |
| 10 | A temperature rise does not change the activation energy; it gives more particles energy above that barrier | The activation energy is fixed by the reaction, while heating raises kinetic energy and the fraction of effective collisions. |
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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