$\ce{N2O4(g) 2 NO2(g)}$ $\Delta H = +ve$. In the reaction above, an increase in temperature will

Question

$\ce{N2O4(g) <=> 2 NO2(g)}$ $\Delta H = +ve$. In the reaction above, an increase in temperature will

Answer

Explanation: The forward reaction $\ce{N2O4(g) <=> 2 NO2(g)}$ is endothermic ($\Delta H > 0$). Raising the temperature adds “heat,” which—by Le Châtelier’s principle—favours the endothermic (forward) direction, producing more $\ce{NO2}$. For endothermic reactions, increasing temperature increases the equilibrium constant $K$ (van ’t Hoff relation), because the products are favored.

Shift to the left / increase reactants → incorrect (it shifts to the right).
Decrease $K$ → incorrect (endothermic: $K$ increases with $T$).

Correct option: increase the value of the equilibrium constant.

Error

Question from: Chemistry

\(\ce{N2O4(g) <=> 2 NO2(g)}\) \(\Delta H = +ve\). In the reaction above, an increase in temperature will

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