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Consider the following equilibrium 2 A (g) + 2 B (g) 2 C (g) + D (g) K = 0.

Consider the following equilibrium

2 A(g) + 2 B(g)→ 2 C(g) + D(g)         K = 0.80072

At some point during the reaction the concentrations of the gases were measured and found to be the following

Concentration of A = 1.7

Concentration of B = 9.4

Concentration of C = 5.9

Concentration of D = 4.5

Calculate change in Gibbs Free Energy for the reaction at 298 K in kJ mol-1ΔGrxn298 K  (hint: R = 8.314 J mol-1 K-1)

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delta Go for this reaction is  delta Go = -RTln K delta Go = -8.314 x 298 x ln 0.80072... View the full answer

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delta Go = -RTln K
delta Go = -8.314 x 298 x ln 0.80072
delta Go = +552.85 J delta G rxn = delta Go + RT ln Q
Q = [D]2[C]2/[A]2[B]2 = 5.92 x 4.52/1.72 x 9.42 = 2.76
delta G rxn = +552.85 + 8.314 x...

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