\( \mathrm{N_2O_4(aq)} \rightleftharpoons 2\mathrm{NO_2(g)} \quad \triangle H = +\mathrm{ve} \) In the reaction above, an increase in temperature will

The given chemical equation represents the equilibrium reaction between N2O4 (aq) and 2NO2 (g) with the release of heat (H=+ve). When the temperature of the reaction mixture is increased, the system will try to counteract the stress by shifting the equilibrium position in a direction that will partially alleviate the stress. According to Le Chatelier's principle, an increase in temperature favors an endothermic reaction, which means that the equilibrium position will shift towards the direction that absorbs heat. In this case, the forward reaction (N2O4 → 2NO2) is endothermic, meaning that it absorbs heat. Therefore, an increase in temperature will shift the equilibrium position to the right, favoring the formation of more NO2 (g) from N2O4 (aq). As a result, the concentration of NO2 (g) will increase, while the concentration of N2O4 (aq) will decrease. So, the correct answer to the question is that an increase in temperature will increase the production of the product NO2 (g) by shifting the equilibrium position to the right. It will not affect the equilibrium constant (Kc), which is a constant at a given temperature and does not change with the changes in concentration or reaction conditions.