Answer:
[tex]K^{2000K}=0.774\\\\K^{3000K}=12.56[/tex]
Explanation:
Hello,
In this case, considering the reaction, we can compute the Gibbs free energy of reaction at each temperature, taking into account that the Gibbs free energy for the diatomic element is 0 kJ/mol:
[tex]\Delta _rG=\Delta _fG_{X}-\frac{1}{2} \Delta _fG_{X_2}=\Delta _fG_{X}[/tex]
Thus, at 2000 K:
[tex]\Delta _rG=\Delta _fG_{X}^{2000K}=4.25kJ/mol[/tex]
And at 3000 K:
[tex]\Delta _rG=\Delta _fG_{X}^{3000K}=-63.12kJ/mol[/tex]
Next, since the relationship between the equilibrium constant and the Gibbs free energy of reaction is:
[tex]K=exp(-\frac{\Delta _rG}{RT} )[/tex]
Thus, at each temperature we obtain:
[tex]K^{2000K}=exp(-\frac{4250J/mol}{8.314\frac{J}{mol\times K}*2000K} )=0.774\\\\K^{3000K}=exp(-\frac{-63120J/mol}{8.314\frac{J}{mol\times K}*3000K} )=12.56[/tex]
In such a way, we can also conclude that at 2000 K reaction is unfavorable (K<1) and at 3000 K reaction is favorable (K>1).
Best regards.
