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?
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?
Show Hint
- Shifts to the right
- Shifts to the left
- Remains unchanged
- Pressure increases
The Correct Option is B
Solution and Explanation
- 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).
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