Work through these eight questions in order. They cover all four lessons in rates of reaction.
The timed practice session builder can turn the set into a timed run.
Questions
Questions 1 to 4 use this table of gas collected from a reaction.
| Time (s) | 0 | 10 | 20 | 30 | 40 | 50 |
|---|---|---|---|---|---|---|
| Gas (cm³) | 0 | 25 | 40 | 48 | 50 | 50 |
Question 1 (2 marks). Calculate the average rate in the first 10 s.
Answer
25 ÷ 10 = 2.5 cm³ per second.
Question 2 (2 marks). Calculate the average rate between 20 s and 30 s.
Answer
Change = 48 − 40 = 8 cm³ in 10 s. Rate = 8 ÷ 10 = 0.8 cm³ per second.
Question 3 (2 marks). Explain why the rate in Question 2 is lower than in Question 1.
Answer
As the reaction goes on, the reactants are used up, so fewer particles are left to collide. Collisions become less frequent, and the rate falls from 2.5 to 0.8 cm³ per second.
Question 4 (2 marks). At about what time did the reaction finish? Give the evidence.
Answer
At about 40 s. The reading is 50 cm³ at 40 s and stays at 50 cm³ at 50 s, so no more gas forms.
Question 5 (3 marks). Four experiments each use 1 g of zinc and 50 cm³ of acid.
| Experiment | Zinc form | Temperature | Acid |
|---|---|---|---|
| P | Granules | 30 °C | Dilute |
| Q | Powder | 30 °C | Dilute |
| R | Granules | 45 °C | Dilute |
| S | Powder | 45 °C | Dilute |
Which two experiments should be compared to show the effect of temperature only? Give a reason.
Answer
Compare P and R (or Q and S). Each pair has the same zinc form and the same acid, and only the temperature differs. Comparing P with S would change two factors at once, so the result could not be blamed on temperature.
Question 6 (3 marks). Explain, using particles, why the rate in Experiment R is higher than in Experiment P.
Answer
At 45 °C the particles move faster and have more energy. They collide more often, and more of the collisions have enough energy to react. The rate of reaction is therefore higher.
Question 7 (3 marks). Experiment T gives 60 cm³ of gas in 120 s. Adding a catalyst in Experiment U gives 60 cm³ in 40 s. Calculate both rates and state how many times faster U is.
Answer
T: 60 ÷ 120 = 0.5 cm³ per second. U: 60 ÷ 40 = 1.5 cm³ per second.
U is 1.5 ÷ 0.5 = 3 times faster. The total volume is the same, so the catalyst changed only the speed.
Question 8 (4 marks). A student writes: “The catalyst made 20 cm³ more gas because the curve was steeper.” Use the data in Question 7 to evaluate the claim.
Answer
The claim is wrong. Both experiments formed 60 cm³ of gas, so the catalyst did not change the total amount.
A steeper curve means a faster rate, not more product. The catalyst gives a route needing less energy, so the same 60 cm³ formed in 40 s instead of 120 s.
If you got these wrong
Questions 1 to 4 test reading rate graphs and calculating a rate from a supplied table. Question 5 tests comparing factors affecting rate. Questions 3 and 6 test explaining collision ideas, and questions 7 and 8 use both.
Log each slip in the mistake log and paper error review. To work through the gaps with a teacher, see online one-to-one Science tuition.