Question:medium

For the chemical equilibrium, \[ 2\text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g); +14.6 \text{ kcal} \] increase in temperature

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For an exothermic reaction (\(\Delta H < 0\)), an increase in temperature shifts the equilibrium to the left (towards reactants).
Updated On: May 24, 2026
  • favours the formation of N\(_2\)O\(_4\)
  • favours the decomposition of N\(_2\)O\(_4\)
  • does not affect equilibrium
  • stops the reaction
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The Correct Option is B

Solution and Explanation

The given chemical equilibrium is:

\(2\text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g); +14.6 \text{ kcal}\)

This equation implies that the formation of \( \text{N}_2\text{O}_4 \) from \( \text{NO}_2 \) is an exothermic process, releasing 14.6 kcal of energy. To understand how changes in temperature affect this equilibrium, we can apply Le Chatelier's Principle, which states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change.

Effect of Temperature: Increasing the temperature adds heat to the system. According to Le Chatelier's Principle, the system will respond by favoring the reaction that absorbs heat (endothermic process).

In this equilibrium, the forward reaction (formation of \(\text{N}_2\text{O}_4\)) is exothermic, which means it releases heat. Conversely, the reverse reaction (decomposition of \(\text{N}_2\text{O}_4\) into \(\text{NO}_2\)) is endothermic, meaning it absorbs heat.

Thus, upon increasing the temperature, the equilibrium position will shift to favor the endothermic reaction, i.e., the decomposition of \(\text{N}_2\text{O}_4\).

Therefore, the correct answer is: favours the decomposition of N\(_2\)O\(_4\).

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