These eight original questions cover the four lessons in SPM Chemistry rate of reaction. They start with one-step calculations and end with fair-test and energy-diagram judgements.
Use 24 dm³/mol as the molar volume at room conditions. Attempt each question on paper before opening the answer.
Questions
Question 1. A gas syringe collects 24 cm³ of gas in the first 40 s. Calculate the average rate.
Answer
Average rate = 24 ÷ 40 = 0.60 cm³/s.
Revisit average and instantaneous rates if you divided by the volume instead.
Question 2. A flask loses 1.80 g in 90 s. Give the average rate in g/min.
Answer
Rate = 1.80 ÷ 90 = 0.020 g/s.
Multiply by 60: 0.020 × 60 = 1.2 g/min.
Question 3. The volumes of gas at 0, 20, 40, 60 and 80 s are 0, 14, 24, 30 and 33 cm³. Find the average rate for 0 to 20 s and for 60 to 80 s, and explain the difference.
Answer
0 to 20 s: 14 ÷ 20 = 0.70 cm³/s.
60 to 80 s: (33 − 30) ÷ 20 = 3 ÷ 20 = 0.15 cm³/s.
The rate falls because the reactant is used up, so its concentration decreases. The frequency of effective collisions falls.
Question 4. A tangent to a curve at 40 s passes through (0, 6) and (80, 38). Find the instantaneous rate at 40 s.
Answer
Gradient = (38 − 6) ÷ (80 − 0) = 32 ÷ 80 = 0.40 cm³/s.
Two points far apart on the tangent give a reliable gradient. Do not use points on the curve itself.
Question 5. Explain why the rate increases when the temperature of a reacting mixture rises from 30 °C to 50 °C.
Answer
The particles move faster, so the frequency of collisions increases.
A larger fraction of the particles have energy equal to or greater than the activation energy.
Therefore the frequency of effective collisions increases, and so does the rate. See collision theory.
Question 6. Experiment I uses 0.12 g of Mg with excess 1.0 mol/dm³ acid. Experiment II uses 0.24 g of Mg with excess acid of the same volume and concentration. State the final hydrogen volume in each.
Answer
I: 0.12 ÷ 24 = 0.005 mol Mg gives 0.005 mol H₂, which is 0.005 × 24 000 = 120 cm³.
II: 0.24 ÷ 24 = 0.010 mol Mg gives 0.010 mol H₂, which is 240 cm³.
The final volume depends on the amount of the limiting reactant, which here is magnesium.
Question 7. A reaction has reactants at 40 kJ, a peak at 110 kJ and products at 90 kJ. A catalyst lowers the peak to 85 kJ. Find Ea and ΔH before and after.
Answer
Before: Ea = 110 − 40 = 70 kJ; ΔH = 90 − 40 = +50 kJ (endothermic).
After: Ea = 85 − 40 = 45 kJ; ΔH is still +50 kJ.
The catalyst lowers the activation energy only.
Question 8. A student compares two rates by changing both the temperature and the concentration at once. Say why this is not a fair test and describe a fair design.
Answer
Two variables changed together, so the cause of any difference cannot be identified.
A fair design changes only the concentration. Temperature, mass and size of the solid, and volume of acid are kept constant, and the rate is measured from the same quantity each time.
If you got these wrong
If Questions 1 to 4 went wrong, review average and instantaneous rates. For Question 6 and 8, see comparing rate graphs fairly. Question 7 links to activation-energy diagrams.
The timed original practice session builder can turn this set into a timed session. For a teacher to work through the questions you missed, see online one-to-one Chemistry tuition.