At 25 °C, the equilibrium partial pressures for the following reaction were found to be PA...

At 25 °C, the equilibrium partial pressures for the following reaction were found to be PA = 5.65 atm, PB = 5.98 atm, PC = 5.27 atm, and PD = 4.83 atm. What is the standard change in Gibbs free energy of this reaction at 25 °C?

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Expert Solution

assume A(g) + 2B(g) ------------------> C(g) + D(g)

Kp = P_CP_D/P_A P²B

= 5.27*4.83/5.65*(5.98)^2

= 0.126

G⁰ = -RTlnKp

G⁰ = -8.314*298ln0.126

= -8.314*298*-2.0717

= 5132.78J >>>>answer

queen_honey_blossom answered 1 year ago

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