Question

Which one of the following conditions will favour maximum formation of the product in the reaction, \( A_{2}(g) + B_{2}(g) \rightleftharpoons X_{2}(g) \), \( \Delta_r H = -x \,\text{kJ} \):

• A. Low temperature and high pressure
• B. Low temperature and low pressure
• C. High temperature and low pressure
• D. High temperature and high pressure
• E.

Hint:

Think of pressure as a "space-saver": High pressure pushes the reaction toward whichever side takes up less "space" (fewer gas moles). For temperature, think of heat as a product in exothermic reactions; adding more "product" (heat) pushes the reaction backward.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The problem asks for conditions that maximize product formation, which means we want to shift the chemical equilibrium to the right (forward direction).
This behavior is governed by Le Chatelier's Principle, which states that if a system at equilibrium is subjected to a change in conditions (temperature, pressure, or concentration), the system will shift its equilibrium position to counteract that change.
Step 2: Key Formula or Approach:
We must analyze two distinct effects according to Le Chatelier's Principle:
1. Effect of Temperature: Look at the sign of the enthalpy change ($\Delta H$).
- Exothermic ($\Delta H<0$): Lowering temperature favors the forward reaction.
- Endothermic ($\Delta H>0$): Raising temperature favors the forward reaction.
2. Effect of Pressure: Look at the change in the number of gaseous moles ($\Delta n_g$).
- $\Delta n_g = \text{moles of gaseous products} - \text{moles of gaseous reactants}$.
- Increasing pressure favors the direction that produces fewer gaseous moles.
Step 3: Detailed Explanation:
Let's analyze the given equilibrium reaction:
\[ A_2(g) + B_2(g) \rightleftharpoons X_2(g) \] Analyzing Temperature:
The enthalpy of reaction is given as $\Delta_rH = -x \text{ kJ}$.
The negative sign definitively indicates that the forward reaction is exothermic (it releases heat).
According to Le Chatelier's Principle, to favor an exothermic reaction (and thus maximize the product $X_2$), we must remove the heat it produces. This is achieved by maintaining a low temperature.
Analyzing Pressure:
First, count the number of moles of gas on each side of the balanced equation:
- Reactants: $1 \text{ mole of } A_2 + 1 \text{ mole of } B_2 = 2 \text{ moles of gas}$ total.
- Products: $1 \text{ mole of } X_2 = 1 \text{ mole of gas}$ total.
The reaction proceeds with a decrease in the number of gaseous moles (from 2 down to 1).
According to Le Chatelier's Principle, an increase in system pressure will cause the equilibrium to shift toward the side with fewer moles of gas in an attempt to reduce that pressure.
Therefore, to shift the reaction forward to the product side (which has fewer moles), a high pressure must be applied.
Combining these two favorable conditions, we need low temperature and high pressure.
Step 4: Final Answer:
The maximum formation of the product is favored by low temperature and high pressure.