The effect of concentration
4 marks · Normal| Experiment | [X] (mol/dm^3) added | Rate of forward reaction | Rate of backward reaction | [Z] at equilibrium (mol) |
|---|---|---|---|---|
| 1 | — | fast | fast | 0.70 |
| 2 | 0.20 | faster | fast | 1.00 |
Use the data in the table to answer the questions.
What happens to the equilibrium position when X is added? (1)
Describe the changes in the rates of the forward and backward reactions when X is added. (1)
Suggest an explanation for the change in the amount of Z at equilibrium. (2)
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.
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(a) The equilibrium shifts to the right (towards Z).
(b) The rate of the forward reaction increases; the backward reaction rate stays the same at first.
(c) Adding more X increases its concentration, causing the system to react more X with Y to make more Z, shifting the position of equilibrium to the right and resulting in a greater amount of Z at equilibrium.
Common mistakes: Confusing rates with position; Saying equilibrium does not change
Show mark scheme and worked solution
Marking points
B1 (a) · 1 mark · States that equilibrium shifts to the right/towards Z.
M1 (b) · 1 mark · States that the forward reaction rate increases, while the backward rate stays the same at first.
M2 (c) · 1 mark · Explains that adding X increases its concentration, causing the system to use up more X according to Le Chatelier's principle.
A1 (c) · 1 mark · Relates increase in Z to the shift in equilibrium toward Z.
Worked solution
(a) The equilibrium shifts to the right, so more Z is formed.
(b) The rate of the forward reaction increases because of the higher concentration of X. The backward reaction rate stays the same initially.
(c) Increasing X means more reacts with Y to form Z. The equilibrium shifts to oppose the increase, making more Z, so the amount of Z at equilibrium is higher.