Question

g what is ΔG for the reaction at 298K, when the partial pressure of C2H4 is 0.275 atm, the partial pressure of H2O is 0.350 atm, and the partial pressure of C2H5OH is 0.100 atm?

Answer 1

`\Delta G=8015(J)/(mol)`

Step-by-step explanation:

Hello.

In this case, for the following chemical reaction in gaseous phase:

`C_2H_4+H_2O\rightleftharpoons C_2H_5OH`

We can write the equilibrium expression as:

`Kp=(p_(C_2H_5OH))/(p_(H_2O)*p_(C_2H_4))`

it means that given the pressures, we can compute Kp as shown below:

`Kp=(0.100atm)/(0.275atm*0.350atm)=1.04`

Next, we compute Kc at 298 K:

`Kc=(Kp)/((RT)^(1-1-1))=(Kp)/((RT)^(-1))`

Whereas 1-1-1 comes from the stoichiometric coefficients of products minus reactants. Therefore, Kc is:

`Kc=(1.04)/((0.08206*298)^(-1))=25.4`

Thus, the Gibbs free energy for this reaction is:

`\Delta G=RTln(kC)=8.314(J)/(mol*K)*298K*ln(25.4)\\ \\\Delta G=8015(J)/(mol)`

Best regards!