Posted by Alexis on .
At 25 °C, the equilibrium partial pressures for the following reaction were found to be PA = 4.10 bar, PB = 5.60 bar, PC = 1.20 bar, and PD = 9.70 bar.
What is the standard change in Gibbs free energy of this reaction at 25 °C.
I don't really get how to even start the problem because I never seen partial pressure in the Gibb free energy. I just know that delta G = -RTlnK.
You have the equilibrium pressures, calculate Kp, then
dGo = -R*T*lnK