Question

The equilibrium constant of a reaction doubles on increasing the temperature of the reaction from $$25^oC$$ to $$35^oC$$. Calculate enthalpy change of the reaction, assuming it to be constant in this temperature range.

Answer 1

According to integrated van't Hoff equation
$$\log \dfrac {(K_p)_2}{(K_p)_1}=\dfrac {\Delta H^o}{2.303\ R}\left( \dfrac {T_2-T_1}{T_1T_2}\right)$$
Putting $$(K_p)_2 /(K_p)_1=2, T_1=25^oC=298\ K$$,
$$T_2=35^oC=308\ K$$
$$R=8.314\ JK^{-1}\ mol^{-1}$$, we get
$$\log 2=\dfrac {\Delta H^o}{2.303\times 8.314\ JK^{-1}\ mol^{-1}}\times \dfrac {(308-298)\ K}{298\ K\ 308\ K}$$
or $$\Delta H^o=52898\ J\ mol^{-1}=52.898\ kJ\ mol^{-1}$$