And provides the relative amounts of both A and B at 25 °C and 75 °C, this means the equilibrium expressions and equilibrium constants can be written as:

$K_1=\frac{90\%}{10\%}=9\\\\K_2=\frac{20\%}{80\%} =0.25$

Thus, by recalling the Van't Hoff's equation, we can write:

$ln(K_2/K_1)=-\frac{\Delta H}{R}(\frac{1}{T_2} -\frac{1}{T_1} )$

Hence, we solve for the enthalpy change as follows:

$\Delta H=\frac{-R*ln(K_2/K_1)}{(\frac{1}{T_2} -\frac{1}{T_1} ) }$

Finally, we plug in the numbers to obtain:

$\Delta H=\frac{-8.314\frac{J}{mol*K} *ln(0.25/9)}{[\frac{1}{(75+273.15)K} -\frac{1}{(25+273.15)K} ] } \\\\\\\Delta H=4,785.1\frac{J}{mol}$

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