Answer:
[tex]Kp^{1000K}=0.141\\Kp^{298.15K}=2.01x10^{-18}[/tex]
[tex]\Delta _rG=1.01x10^5J/mol[/tex]
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]C_2H_6(g)\rightleftharpoons H_2(g)+C_2H_4(g)[/tex]
Thus, Kp for this reaction is computed based on the given molar fractions and the total pressure at equilibrium, as shown below:
[tex]p_{C_2H_6}^{EQ}=2bar*0.592=1.184bar\\p_{C_2H_4}^{EQ}=2bar*0.204=0.408bar\\p_{H_2}^{EQ}=2bar*0.204=0.408bar[/tex]
[tex]Kp=\frac{p_{C_2H_4}^{EQ}p_{H_2}^{EQ}}{p_{C_2H_6}^{EQ}}=\frac{(0.408)(0.408)}{1.184}=0.141[/tex]
Now, by using the Van't Hoff equation one computes the equilibrium constant at 298.15K assuming the enthalpy of reaction remains constant:
[tex]Ln(Kp^{298.15K})=Ln(Kp^{1000K})-\frac{\Delta _rH}{R}*(\frac{1}{298.15K}-\frac{1}{1000K} )\\\\Ln(Kp^{298.15K})=Ln(0.141)-\frac{137000J/mol}{8.314J/mol*K} *(\frac{1}{298.15K}-\frac{1}{1000K} )\\\\Ln(Kp^{298.15K})=-40.749\\\\Kp^{298.15K}=exp(-40.749)=2.01x10^{-18}[/tex]
Finally, the Gibbs free energy for the reaction at 298.15K is:
[tex]\Delta _rG=-RTln(Kp^{298.15K})=8.314J/mol*K*298.15K*ln(2.01x10^{-18})\\\Delta _rG=1.01x10^5J/mol[/tex]
Best regards.
