Adding a catalyst
3 marks · EasyThe concentrations of reactants and products in a reversible reaction at equilibrium are measured every minute. At minute 4, a catalyst is added. The concentrations at each minute are shown below.
| Time (min) | [Reactant] | [Product] |
|---|---|---|
| 0 | 1.00 | 0.00 |
| 1 | 0.85 | 0.15 |
| 2 | 0.75 | 0.25 |
| 3 | 0.71 | 0.29 |
| 4 | 0.70 | 0.30 |
| 5 | 0.70 | 0.30 |
| 6 | 0.70 | 0.30 |
Describe the effect of adding the catalyst on the time taken to reach equilibrium.
Use the data to state the concentrations at equilibrium.
Does adding the catalyst change the equilibrium concentrations? Explain your answer.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a) The catalyst decreases the time taken to reach equilibrium.
(b) [Reactant] = 0.70 mol/dm³; [Product] = 0.30 mol/dm³
(c) No, the equilibrium concentrations remain the same because the catalyst affects the rate, not the position, of equilibrium.
Common mistakes: Stating equilibrium concentrations change
Show mark scheme and worked solution
Marking points
B1 (a) · 1 mark · States that catalyst decreases time to equilibrium/helps reach equilibrium faster.
B2 (b) · 1 mark · Correctly states both equilibrium concentrations from the table ([Reactant] = 0.70 mol/dm³, [Product] = 0.30 mol/dm³).
B3 (c) · 1 mark · Correctly explains that adding the catalyst does not change the equilibrium concentrations.
Worked solution
a) The catalyst causes equilibrium to be reached more quickly after minute 4 (concentrations stay constant from minute 5).
b) The equilibrium concentrations are: [Reactant] = 0.70 mol/dm³, [Product] = 0.30 mol/dm³.
c) No, the catalyst does not change these values because it increases the rate of both forward and reverse reactions equally.