Reversible reactions in a sealed container
4 marks · NormalThe table shows the number of blue and yellow squares in a sealed container at intervals during a reversible reaction.
| Time (s) | Number of blue squares | Number of yellow squares |
|---|---|---|
| 0 | 64 | 0 |
| 1 | 48 | 16 |
| 2 | 36 | 28 |
| 3 | 27 | 37 |
| 4 | 20 | 44 |
Describe the trend in the number of blue squares over time.
Describe how the number of yellow squares changes over time.
Predict what will happen to the numbers of blue and yellow squares if the reaction continues.
Explain how these trends are related to reversible reactions reaching equilibrium.
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) Number of blue squares decreases over time
(b) Number of yellow squares increases over time
(c) Numbers of blue and yellow squares will eventually stay constant / stop changing
(d) At equilibrium, the rates of forward and reverse changes are equal, so the numbers of both squares remain constant.
Common mistakes: Missing equilibrium point; Assuming numbers go to zero or maximum
Show mark scheme and worked solution
Marking points
B1 (a) · 1 mark · Describes decrease in blue squares over time
B2 (b) · 1 mark · Describes increase in yellow squares over time
B3 (c) · 1 mark · Predicts that numbers will stop changing and stay constant at equilibrium
M1 (d) · 1 mark · Explains that equilibrium is reached when forward and reverse rates are equal and numbers do not change further
Worked solution
a) The number of blue squares falls steadily each second.
b) The number of yellow squares increases each second.
c) Eventually, both numbers will stop changing and reach a steady value.
d) This happens because, at equilibrium in a reversible reaction, the rates of the forward (blue to yellow) and reverse (yellow to blue) changes become equal, so no further net change occurs in the numbers.