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-1, ΔGrxn, 298 K (hint: R = 8.314 J mol-1 K-1)
delta Go for this reaction is delta Go = -RTln K delta Go = -8.314 x 298 x ln 0.80072... View the full answer
This question was asked on Jul 20, 2016 and answered on Jul 20, 2016 for the course CHEM 123 123 at UBC.
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