Question

For the following gaseous reversible reaction: 
\( 3 \, \text{A}_{(g)} + \text{B}_{(g)} \rightleftharpoons \, \text{A}_3 \text{B}_{(g)} \) 
(\( \Delta H = -q \, \text{kJ} \)), the amount of product \( \text{A}_3 \text{B}_{(g)} \) is affected by 
\(\underline{\hspace{3cm}}\).

• A. Temperature alone
• B. Pressure alone
• C. Both temperature and pressure
• D. Temperature, pressure and catalyst

Hint:

To quickly determine the effects on equilibrium:

• \textbf{Temperature:} Check if \(\Delta\)H is positive (endothermic) or negative (exothermic). Increasing T favors the endothermic direction.
• \textbf{Pressure:} Calculate \(\Delta n_g\). If \(\Delta n_g \neq 0\), increasing P favors the side with fewer gas moles.
• \textbf{Catalyst:} Never changes the equilibrium position, only the rate at which it is reached.

Answer 1

The Correct Option is C

Step 1: Understanding Le Chatelier's Principle.
Le Chatelier's principle states that if a change of condition (like temperature, pressure, or concentration) is applied to a system in equilibrium, the system will shift in a direction that counteracts the change. We need to see how temperature and pressure changes will affect the equilibrium position.
Step 2: Effect of Temperature.
The reaction is given as exothermic (\(\Delta\)H = -q kJ, meaning heat is released). \[ 3\text{A}_{(g)} + \text{B}_{(g)} \rightleftharpoons \text{A}_3\text{B}_{(g)} + \text{Heat} \] According to Le Chatelier's principle:

• Increasing the temperature will favor the endothermic (reverse) reaction to absorb the added heat. The equilibrium will shift to the left, decreasing the amount of product A\(_3\)B.
• Decreasing the temperature will favor the exothermic (forward) reaction to release heat. The equilibrium will shift to the right, increasing the amount of product A\(_3\)B.

Since temperature changes the equilibrium position, the amount of product is affected by temperature.
Step 3: Effect of Pressure.
The effect of pressure depends on the change in the number of moles of gas (\(\Delta n_g\)). \[ \Delta n_g = (\text{moles of gas products}) - (\text{moles of gas reactants}) \] \[ \Delta n_g = (1) - (3 + 1) = 1 - 4 = -3 \] Since \(\Delta n_g \neq 0\), pressure will affect the equilibrium. According to Le Chatelier's principle:

• Increasing the pressure will favor the side with fewer moles of gas to relieve the pressure. The equilibrium will shift to the right, increasing the amount of product A\(_3\)B.
• Decreasing the pressure will favor the side with more moles of gas. The equilibrium will shift to the left, decreasing the amount of product A\(_3\)B.

Since pressure changes the equilibrium position, the amount of product is affected by pressure.
Step 4: Effect of a Catalyst.
A catalyst increases the rate of both the forward and reverse reactions equally. It helps the system reach equilibrium faster but does not change the position of the equilibrium. Therefore, a catalyst does not affect the amount of product at equilibrium.
Step 5: Final Answer.
The amount of product is affected by both temperature and pressure, but not by a catalyst. Therefore, option (C) is the best answer.