Answer:
[tex]\Delta G=-48.2kJ/mol[/tex]
Explanation:
Hello,
In this case, considering that the relationship between Kp and K is:
[tex]K=\frac{Kp}{RT^{\Delta \nu}}[/tex]
Whereas the change in the number of moles (stoichiometric coefficients) is:
[tex]\Delta \nu=1-2-1=-2[/tex]
The equilibrium constant is:
[tex]K=\frac{6.25x10^{-3}}{(8.314\frac{J}{mol\times K}*500.0K)^{-2}}\\\\K=1.08x10^{5}[/tex]
In such a way, the Gibbs free energy of reaction is:
[tex]\Delta G=-RTln(K)=-8.314\frac{J}{mol\times K}*500.0K*ln(1.08x10^5)\\\\\Delta G=-48.2kJ/mol[/tex]
Regards.
