In the gas phase at 400 °C, isopropyl alcohol (rubbing alcohol) decomposes to acetone, an important industrial solvent: \[ \text{Isopropyl alcohol: }(\text{CH}_3)_2\text{CHOH(g)} \rightleftharpoons (\text{CH}_3)_2\text{CO(g) + H}_2\text{(g)} \\ \Delta H^\circ = + 57.3 \text{ kJ} \] Does the amount of acetone increase, decrease, or remain the same when an equilibrium mixture of reactants and products is subjected to the following changes? (c) Argon is added. (d) H₂ is added. (e) A catalyst is added.

Verified step by step guidance

1. Step 1: Understand the equilibrium reaction: Isopropyl alcohol decomposes into acetone and hydrogen gas. The reaction is endothermic, as indicated by the positive \( \Delta H^\circ \).
2. Step 2: Consider the effect of adding argon (c): Argon is an inert gas and does not participate in the reaction. Adding argon at constant volume does not change the partial pressures of the reactants or products, so the equilibrium position remains unchanged.
3. Step 3: Analyze the effect of adding H₂ (d): Adding more H₂ increases its partial pressure, shifting the equilibrium to the left according to Le Chatelier's principle, which will decrease the amount of acetone.
4. Step 4: Evaluate the effect of adding a catalyst (e): A catalyst speeds up the rate at which equilibrium is reached but does not affect the position of the equilibrium. Therefore, the amount of acetone remains the same.
5. Step 5: Summarize the effects: (c) Adding argon does not change the amount of acetone, (d) adding H₂ decreases the amount of acetone, and (e) adding a catalyst does not change the amount of acetone.