This question is describing the following chemical reaction at equilibrium:
[tex]A\rightleftharpoons B[/tex]
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:
[tex]K_1=\frac{90\%}{10\%}=9\\\\K_2=\frac{20\%}{80\%} =0.25[/tex]
Thus, by recalling the Van't Hoff's equation, we can write:
[tex]ln(K_2/K_1)=-\frac{\Delta H}{R}(\frac{1}{T_2} -\frac{1}{T_1} )[/tex]
Hence, we solve for the enthalpy change as follows:
[tex]\Delta H=\frac{-R*ln(K_2/K_1)}{(\frac{1}{T_2} -\frac{1}{T_1} ) }[/tex]
Finally, we plug in the numbers to obtain:
[tex]\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}[/tex]
Learn more:
- https://brainly.com/question/10038290
- https://brainly.com/question/19671384
