This set practises reading rates from data, explaining the effect of concentration and surface area, describing a catalyst, and judging fair comparisons. The questions get harder as you go. All data is invented for practice.
Write each answer on paper first, including units. The final section sends each kind of error back to the right lesson.
Part A: Reading rates from data
Q1. A gas is collected from a reaction. The invented readings are:
| Time (s) | 0 | 20 | 40 | 60 | 80 |
|---|---|---|---|---|---|
| Volume (cm³) | 0 | 30 | 48 | 57 | 60 |
Find the average rate from 0 to 20 s.
Show answer
Change in volume = 30 cm³, change in time = 20 s. Rate = 30 ÷ 20 = 1.5 cm³/s.
Q2. Using the table in Q1, find the average rate from 40 s to 60 s and state whether it is faster or slower than 0 to 20 s.
Show answer
Change in volume = 57 − 48 = 9 cm³. Rate = 9 ÷ 20 = 0.45 cm³/s. It is slower, because the reactants have been partly used up so their concentration is lower.
Q3. A tangent at t = 0 to a different invented curve passes through (0 s, 0 cm³) and (15 s, 33 cm³). Find the initial rate.
Show answer
Gradient = 33 ÷ 15 = 2.2 cm³/s.
Q4. A flask of reacting mixture loses gas. Its mass falls from 120.00 g to 119.40 g in 30 s. Find the average rate and give the unit.
Show answer
Change in mass = 0.60 g. Rate = 0.60 ÷ 30 = 0.020 g/s.
Part B: Concentration and surface area
Q5. 20.0 cm³ of acid of concentration 2.0 mol/dm³ is diluted with water to a total of 80.0 cm³. Calculate the new concentration.
Show answer
Moles = 2.0 × 0.0200 = 0.040 mol. New volume = 0.0800 dm³. Concentration = 0.040 ÷ 0.0800 = 0.50 mol/dm³.
Q6. Explain why a reaction with the diluted acid in Q5 is slower than with the original, everything else being equal.
Show answer
The diluted acid has fewer particles per unit volume, so collisions are less frequent. Fewer successful collisions happen each second, so the rate is lower. The particle energy has not changed.
Q7. 2.0 g of calcium carbonate (CaCO₃, Mr = 100) reacts completely with excess hydrochloric acid: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. Calculate the volume of CO₂ at room temperature and pressure (24 dm³ per mole). Then say how the final volume would compare if the solid were lumps instead of powder.
Show answer
Moles of CaCO₃ = 2.0 ÷ 100 = 0.020 mol. CO₂ = 0.020 mol (1 : 1 ratio). Volume = 0.020 × 24 = 0.48 dm³ = 480 cm³. The final volume for lumps is the same 480 cm³. Only the time taken differs, because powder has a larger surface area.
Q8. A 3 cm cube is cut into twenty-seven 1 cm cubes. Calculate the factor by which the surface area increases.
Show answer
Before: 6 × 3 × 3 = 54 cm². After: 27 × 6 = 162 cm². Factor = 162 ÷ 54 = 3.
Part C: Catalysts and fair comparison
Q9. In an invented supervised laboratory comparison, 0.040 mol of hydrogen peroxide decomposes: 2H₂O₂ → 2H₂O + O₂. A catalyst makes it faster. Calculate the oxygen volume (24 dm³ per mole) and state whether the catalyst changes it.
Show answer
O₂ = 0.040 ÷ 2 = 0.020 mol. Volume = 0.020 × 24 = 0.48 dm³ = 480 cm³. The catalyst does not change it. It provides an alternative pathway with a lower activation energy, so the same amount of product forms sooner, and the catalyst is not used up.
Q10. Invented runs, each timed to collect 40 cm³ of gas:
| Run | Acid (mol/dm³) | Solid (1.0 g) | Temperature (°C) | Time (s) |
|---|---|---|---|---|
| A | 1.0 | powder | 30 | 20 |
| B | 1.0 | lumps | 30 | 45 |
| C | 1.0 | lumps | 40 | 30 |
(a) Which two runs give a fair test of surface area? (b) A student repeated run B and got 45 s, 47 s, 61 s and 46 s. Give the mean, excluding any anomaly.
Show answer
(a) Runs A and B: only the form of the solid differs. Powder was faster (20 s against 45 s). (b) 61 s is far from the others, so it is anomalous and excluded. Mean = (45 + 47 + 46) ÷ 3 = 138 ÷ 3 = 46 s.
If you got these wrong
| Error type | Go to |
|---|---|
| Divided one reading by time, or forgot units (Q1 to Q4) | Extract a rate from a supplied graph |
| Said particles have more energy, or left out “more frequent” (Q5, Q6) | Explain concentration effects using collision ideas |
| Thought powder gives more product (Q7, Q8) | Compare powder and lumps without changing amount |
| Said a catalyst is used up or changes the yield (Q9) | Describe a catalyst’s role |
| Paired runs that differ in two ways, or kept an anomaly (Q10) | Evaluate fair comparison in a fictional dataset |
Log each error in the mistake log and retest queue, and use the rate and energy graph interpreter to redo the graph questions. The mole and equation-ratio tutor checks the volume calculations. Back in the rates of reaction module you can reread the route.
Some students can explain each idea alone but lose marks when two appear in one question. Our teachers work through that in online one-to-one Chemistry tuition.