At 25 °C, the equilibrium partial pressures for the following reaction were foun

Question

At 25 °C, the equilibrium partial pressures for the following reaction were found to be PA 4.37 atm, PB 4.90 atm, P 5.63 atm, and PD 5.17 atm. at is the standard change in Gibbs free energy of this reaction at 25 °C? Number kJ mol AO Previous Give Up & View Solution Check Answer Next xit Hin The equilibrium constant is related to the change in Gibbs free energy by the equation AG RT In K where Ris the gas constant, 8.3145 J/(mol. K), Tis the temperature; K is the equilibrium constant, and AGoron is the change in Gibbs free energy. To begin, start by determining the equilibrium constant for the reaction.

Explanation / Answer

delta G0rxn = -RT lnK

T = 298 K

Here K = Kp = p2C x p2D / p3A X p4B

So Kp = (5.63)2 X (5.17)2 / (4.37)3 ( 4.9)4 = 847.223 / 48109.254 = 0.0176

delta G0rxn = -RT lnKp = - 8.314 X 298 X ln (0.0176) = 8.314 X 299 X (-4.039) = 10006.91 Joules / moles

delta G0rxn = 10.006 KJ / moles

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