Methanol (CH3OH) is manufactured by the reaction of carbon monoxide with hydrogen in the presence of a Cu/ZnO/Al2O3 catalyst: CO(g) + 2H2(g) ⇌ CH3OH(g) ΔH° = -91 kJ. Does the amount of methanol increase, decrease, or remain the same when an equilibrium mixture of reactants and products is subjected to the following changes? (a) The temperature is increased. (b) CO is added. (c) Helium is added. (d) The catalyst is removed.
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Step 1: Consider the effect of temperature on the equilibrium. Since the reaction is exothermic (ΔH° = -91 kJ), increasing the temperature will shift the equilibrium to favor the endothermic direction, which is the reverse reaction. This will decrease the amount of methanol.
Step 2: Analyze the effect of adding CO. According to Le Chatelier's principle, adding more CO will shift the equilibrium to the right to counteract the change, increasing the amount of methanol produced.
Step 3: Evaluate the impact of adding helium. Adding an inert gas like helium at constant volume does not change the partial pressures of the reactants or products, so the equilibrium position remains unchanged, and the amount of methanol stays the same.
Step 4: Consider the role of the catalyst. Removing the catalyst will slow down the rate at which equilibrium is reached but does not affect the position of the equilibrium. Therefore, the amount of methanol at equilibrium remains the same.
Step 5: Summarize the effects: (a) Methanol decreases with increased temperature, (b) Methanol increases with added CO, (c) Methanol remains the same with added helium, (d) Methanol remains the same if the catalyst is removed.