Question

For the reaction \(N_2(g) + 3H_2(g) \rightleftharpoons 2\text{NH}_3(g)\) at 298 K, the enthalpy change \( \Delta H = -92.4 \, \text{kJ/mol} \). What happens to the equilibrium when temperature is increased?

• A. Shifts to the right
• B. Shifts to the left
• C. Remains unchanged
• D. Pressure increases

Hint:

For exothermic reactions (\( \Delta H < 0 \)), increasing temperature shifts the equilibrium towards the reactants (left), as per Le Chatelier’s principle.

Answer 1

The Correct Option is B

- The reaction is: \( \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \), with \( \Delta H = -92.4 \, \text{kJ/mol} \).

- Since \( \Delta H < 0 \), the reaction is exothermic (releases heat).

- According to Le Chatelier’s principle, for an exothermic reaction, increasing the temperature favors the endothermic direction (reverse reaction) to absorb the added heat.

Here, the reverse reaction is: \( 2\text{NH}_3(g) \rightarrow \text{N}_2(g) + 3\text{H}_2(g) \), which is endothermic.

Thus, increasing the temperature shifts the equilibrium to the left (towards reactants).

This matches option (B).