Answer:
[tex]\Delta G=8015\frac{J}{mol}[/tex]
Explanation:
Hello.
In this case, for the following chemical reaction in gaseous phase:
[tex]C_2H_4+H_2O\rightleftharpoons C_2H_5OH[/tex]
We can write the equilibrium expression as:
[tex]Kp=\frac{p_{C_2H_5OH}}{p_{H_2O}*p_{C_2H_4}}[/tex]
it means that given the pressures, we can compute Kp as shown below:
[tex]Kp=\frac{0.100atm}{0.275atm*0.350atm}=1.04[/tex]
Next, we compute Kc at 298 K:
[tex]Kc=\frac{Kp}{(RT)^{1-1-1}}=\frac{Kp}{(RT)^{-1}}[/tex]
Whereas 1-1-1 comes from the stoichiometric coefficients of products minus reactants. Therefore, Kc is:
[tex]Kc=\frac{1.04}{(0.08206*298)^{-1}}=25.4[/tex]
Thus, the Gibbs free energy for this reaction is:
[tex]\Delta G=RTln(kC)=8.314\frac{J}{mol*K}*298K*ln(25.4)\\ \\\Delta G=8015\frac{J}{mol}[/tex]
Best regards!
