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Rate of Reaction Paper 3: Challenge
Rate of Reaction Paper 3: Challenge
Combining stoichiometry with rate work: molar volume, unit conversion, tangent slopes, percentages of time, and catalysts against equilibrium.
Direct answers Answers and explanations for all ten questions
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: DSE Chemistry Rate of Reaction |
| Level | Paper 3 (challenge) |
02 Questions (10 items)
- Foundation0.05 mol of magnesium reacts fully with excess dilute acid, taking the molar volume as 24 dm³/mol. What volume of hydrogen forms?
- FoundationIf the 1 200 cm³ of hydrogen is collected in 60 s, what is the average rate in cm³/s and in mol/s?
- Foundation10 g of calcium carbonate (molar mass 100 g/mol) reacts with excess acid. What volume of carbon dioxide forms at 24 dm³/mol?
- IntermediateFor a volume against time curve the tangent at the start has a slope of 4.0 cm³/s and at 60 s a slope of 1.5 cm³/s. How many times the later rate is the initial rate?
- IntermediateThe same amount of calcium carbonate takes 120 s as chips but only 45 s as powder. By what percentage does the time fall?
- IntermediateA reaction takes 240 s at 20 °C when the rate doubles for every 10 °C rise. How long would it take at 50 °C?
- ChallengeWhen a catalyst is added, what happens to activation energy, rate, enthalpy change and final yield?
- ChallengeA reaction proceeds at 0.05 mol/s. What is that in mol/min?
- ChallengeExperiment A runs at 2.4 cm³/s for 100 s and experiment B at 1.6 cm³/s for 150 s. Compare the total gas volumes.
- Error redoError clinic: the claim that a catalyst increases the yield. Using rate and equilibrium, explain the error and say what really changes.
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 0.05 times 24 is 1.2000000000000002 dm³, that is 1200 cm³ | The equation Mg + 2HCl gives MgCl₂ + H₂, a one to one ratio, which is then multiplied by the molar volume. |
| 2 | 20.000000000000004 cm³/s; about 0.000833 mol/s | Convert to moles first and then divide by time; both units describe one rate and must stay consistent. |
| 3 | 10 divided by 100 is 0.1 mol, giving 2400 cm³ | Calcium carbonate and carbon dioxide are in a one to one ratio, so the moles match and the volume follows from the molar volume. |
| 4 | 4.0 divided by 1.5 is about 2.67 times | Tangent slope gives the instantaneous rate, and a flattening curve shows the rate falling as concentration drops. |
| 5 | 62.5% | A fall in time is not the same percentage as a rise in rate; the rate is 120 divided by 45, about 2.67 times. |
| 6 | the rate rises 8 times, so about 30 seconds | Thirty degrees is three doublings, cutting 240 s to 30 s; real values deviate, so treat it as a trend estimate. |
| 7 | activation energy falls and the rate rises; the enthalpy change and the final yield are unchanged | A catalyst offers a lower-energy route without altering the energy difference between reactants and products. |
| 8 | 0.05 times 60 is 3.0 mol/min | Convert the time unit by multiplying: 0.05 mol each second is 3 mol each minute. |
| 9 | A gives 240 cm³ and B gives 240 cm³, so they are equal | Rate times time gives the total, so a slower reaction running longer can deliver exactly the same volume. |
| 10 | a catalyst lowers the activation energy and reaches equilibrium sooner without shifting its position, so the yield is unchanged; what changes is the time needed | Raising the yield needs a change in conditions or in the amount of reactant, not a catalyst. |
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.
The order for rate problems: fix the ratio from the equation, convert moles to volume, then divide by time.Tutor Registration →
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