• # question_answer Equilibrium constant K, for the reaction, $2HI(g){{H}_{2}}(g)+{{I}_{2}}(g)$ at room temperature is 2.85 and that at 698 K is $1.4\times {{10}^{-2}}.$ This implies that the forward reaction is: A)  exothermic      B)        endothermic C) exergonic       D)        unpredictable

Key Idea: Write expression for equilibrium constant and apply Le-Chateliers principle $2HI{{H}_{2}}(g)+{{I}_{2}}(g)$ $k=\frac{[{{H}_{2}}][{{I}_{2}}]}{{{[HI]}^{2}}}$ When temperature increases, ${{K}_{c}}$decreases. $\therefore$ Concentration of product decreases. It means forward reaction is not favoured at higher temperature. This is true for exothermic reactions. So, the forward reaction is exothermic.