Question

In an exothermic reaction, a $10{}^\circ C$ rise in temperature will:
A. Decrease the value of equilibrium constant
B. Double the value of equilibrium constant
C. Not produce any change in equilibrium constant
D. Produce some increase in equilibrium constant.

Answer 1

Hint: Equilibrium constant is affected by temperature according to the Le Chatelier’s principle which states that the equilibrium will shift in the direction to undo the change that you have made. In an exothermic reaction , the heat is liberated.

Complete step by step answer:
If we take a typical exothermic reaction, let’s say:
${{H}_{2}}+{{I}_{2}}\rightleftharpoons 2HI$
$\Delta H=-10.4KJmo{{l}^{-1}}$
This is an exothermic reaction, if we increase the temperature then, the equilibrium will try to move in such a way to undo the changes that we have made. In this reaction , the equilibrium will try to undo the effect by moving the reaction to that side where the heat is absorbed. The forward side is heat releasing therefore, the reaction will be moving to the left side where the heat can be absorbed which means that there will be more formation of more hydrogen gas and iodine gas which are unreacted, therefore if we look at the equilibrium constant of this reaction:
${{K}_{p}}=\dfrac{pH{{I}^{2}}}{p{{H}_{2}}\times p{{I}_{2}}}$
If there is an increase in the denominator then, the numerator is decreasing which means that lower value of the equilibrium constant.

Therefore, the answer of this question is A. There will be a decrease in the value of equilibrium constant when we will increase the temperature by $10{}^\circ C$.

Note: Some students think that forward reaction is always exothermic and backward is endothermic which is not true, always look at the $\Delta H$ of the reaction before you see its effect on the equilibrium constant. Also, the equilibrium constant is affected by other properties too such as effect of catalyst, change in pressure and change in concentration.