Asked by Alexis
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.
2A(g)+2B(g)-->C(g)+ 3D(g)
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.
2A(g)+2B(g)-->C(g)+ 3D(g)
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.
Answers
Answered by
DrBob222
You have the equilibrium pressures, calculate Kp, then
dGo = -R*T*lnK
dGo = -R*T*lnK
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