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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   = PCPD/PA 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 2 years ago
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