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March 28, 2017

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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.

  • Chemistry - ,

    You have the equilibrium pressures, calculate Kp, then
    dGo = -R*T*lnK

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