Steam is passed over hot carbon in a closed reactor and an equilibrium forms: C(s)+H2O(g)⇌CO(g)+H2(g)\mathrm{C(s) + H_{2}O(g) \rightleftharpoons CO(g) + H_{2}(g)}C(s)+H2O(g)⇌CO(g)+H2(g) The forward reaction is endothermic. The pressure is increased at constant temperature. What happens to the position of equilibrium?
It does not move, because each side of the equation has two particles
It moves to the right, because higher pressure favours the endothermic direction
It moves to the right, because more collisions with the carbon speed up the forward reaction
It moves to the left, because the left-hand side has fewer molecules of gas
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.