Chemistry

1. 1.

   Explain why crushing a solid reactant can increase its reaction rate with a solution without changing the mass used.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Reaction occurs where phases meet. Smaller fragments create more interfaces for the same total amount, increasing contact opportunities rather than particle energy.

   Marking points

   • Crushing increases exposed surface area.
   • More solid particles are accessible to solution particles.
   • Successful collisions at the interface occur more frequently.

   Examiner tip: Surface area does not lower activation energy.

2. 2.

   On a product-volume versus time graph, explain how to estimate the initial rate and why final volume is not a rate.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. The curve's instantaneous slope is steepest early in many reactions. A tangent estimates that slope; a chord to the final point averages over the whole run instead.

   Marking points

   • Draw a tangent at the start.
   • Use its gradient, change in volume divided by change in time.
   • Final volume measures total product, without a time denominator.

   Examiner tip: Use a large triangle on the tangent, not two arbitrary points on the curve.

3. 3.

   Doubling [A] with [B] and all other conditions constant doubles the initial rate; doubling [B] with [A] and all other conditions constant quadruples it. Determine the rate law, overall order and units of k when rate is mol dm^-3 s^-1.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. A concentration factor of two gives 2^order in the rate. Orders are one and two, and dividing rate units by three concentration factors gives the units of k.

   Marking points

   • Rate = k[A][B]^2.
   • Overall order is 3.
   • k units are dm^6 mol^-2 s^-1.

   Examiner tip: Orders are experimental; do not take them from the balanced equation.

4. 4.

   For rate = k[A]^2, rate = 0.0080 mol dm^-3 s^-1 when [A] = 0.20 mol dm^-3. Calculate k and the rate at [A] = 0.30 mol dm^-3 at the same temperature.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Rearrange k = 0.0080/(0.20)^2, then use 0.20(0.30)^2. The concentration ratio is 1.5, so the rate ratio is 2.25, an independent scaling check.

   Marking points

   • k = 0.20 dm^3 mol^-1 s^-1.
   • Substitute the new concentration squared.
   • New rate = 0.018 mol dm^-3 s^-1.

   Examiner tip: Keep k fixed only because the temperature and reaction conditions are unchanged.

5. 5.

   An observed rate law is rate = k[A][B]. A proposed mechanism starts with slow A + B -> X, followed by fast X + B -> P. Assess its consistency and explain why the rate law does not prove this mechanism uniquely.

   [4 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Add the steps and cancel X to check material balance. The slow step matches the measured concentration dependence, which supports but does not uniquely establish the pathway.

   Marking points

   • The slow elementary step predicts dependence on [A][B].
   • X is formed then consumed and is an intermediate.
   • The net reaction is A + 2B -> P.
   • Other mechanisms can yield the same law; intermediate detection or further kinetic evidence is needed.

   Examiner tip: Only elementary-step molecularity directly supplies powers under this simple mechanism assumption.

6. 6.

   A student says higher temperature speeds a reaction only because collisions happen more often. Evaluate this explanation using an energy distribution and activation energy.

   [4 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Keep the activation threshold fixed and compare areas beyond it. Temperature changes the population above the barrier, unlike a catalyst which changes the barrier height.

   Marking points

   • The distribution spreads towards higher energies as temperature rises.
   • The fraction with energy at least Ea increases.
   • A larger fraction of collisions can overcome the barrier.
   • Collision frequency also increases, but the claim omits the key increase in energetic success; Ea itself need not change.

   Examiner tip: The total area stays constant for a fixed number of particles.

Original, unofficial practice questions from SubjectScout — not official past-paper questions.