Hydrogen reacts with iodine vapour to form hydrogen iodide in a reversible reaction.
H2(g)+I2(g)⇌2HI(g)\mathrm{H_{2}(g)} + \mathrm{I_{2}(g)} \rightleftharpoons \mathrm{2HI(g)}H2(g)+I2(g)⇌2HI(g)
Iodine vapour is purple. Hydrogen and hydrogen iodide are colourless.
A scientist carries out two experiments at the same temperature in sealed containers of the same size.
Experiment 1 starts with 1.00 mol of hydrogen and 1.00 mol of iodine.
Experiment 2 starts with 2.00 mol of hydrogen iodide only.
The table shows the amount of hydrogen iodide in each container over time.
| Time / min | Amount of HI in experiment 1 / mol | Amount of HI in experiment 2 / mol |
|---|---|---|
| 0 | 0.00 | 2.00 |
| 10 | 0.90 | 1.80 |
| 20 | 1.40 | 1.62 |
| 30 | 1.56 | 1.56 |
| 40 | 1.56 | 1.56 |
| 50 | 1.56 | 1.56 |
Give the time at which experiment 1 first reached equilibrium.
Calculate the amount, in moles, of iodine in the container in experiment 1 at equilibrium.
Describe what the scientist sees in the container in experiment 2 during the 50 minutes.
Explain what the results of the two experiments show about this reaction.
Predict the effect of increasing the pressure on the amount of hydrogen iodide at equilibrium. Give a reason for your answer.
45 exam-style questions on Edexcel GCSE Chemistry 5.2 Reversible reactions and equilibria, covering 5.2.1 Reversible reactions and dynamic equilibrium, 5.2.2 The Haber process, and 5.2.3 Changing the position of a dynamic equilibrium. Each one has a worked solution and a mark scheme showing where the marks go.