A mixture of carbon monoxide and hydrogen is allowed to reach equilibrium in a closed reaction vessel of volume V V\,V at a constant temperature T T\,T according to the equation:
CO(g)+2H2(g)⇌CH3OH(g)ΔH=−91 kJ mol−1 \text{CO}(g) + 2\text{H}_2(g) \rightleftharpoons \text{CH}_3\text{OH}(g) \quad \Delta H = -91\text{ kJ mol}^{-1} CO(g)+2H2(g)⇌CH3OH(g)ΔH=−91 kJ mol−1The volume of the reaction vessel is then halved to 12V\displaystyle \frac{1}{2}V21V while the temperature is maintained at TTT.
Which of the following statements is correct for the system at the new equilibrium compared to the original equilibrium?
The concentration of H2(g)\text{H}_2(g)H2(g) is lower than in the original equilibrium.
The total number of moles of gas in the vessel is greater than in the original equilibrium.
The rate of the reverse reaction is greater than in the original equilibrium.
The equilibrium constant KcK_cKc is larger because the forward reaction is favored.
23 exam-style questions on OCR A A Level Chemistry Chemical equilibrium. Each one has a worked solution and a mark scheme showing where the marks go.