A reaction A (aq) + B(aq) (aq) has a standard free-energy change of-3.94 kJ/mol

Question

A reaction A (aq) + B(aq) (aq) has a standard free-energy change of-3.94 kJ/mol at 25 degree C. What are the concentrations of A, B, and C at equilibrium if, at the beginning of the reaction, their concentrations are 0.30 M, 0.40 M, and 0 M, respectively? How would your answers above change if the reaction had a standard free-energy change of +3.94 kJ/mol? All concentrations would be higher. There would be less A and B but more C. All concentrations would be lower. There would be no change to the answers. There would be more A and B but less C.

Explanation / Answer

Oops in my previous post, I forgot to use the ln function. That means that the value of x needs to be corrected This should be the corrected equilibrium values: x = 0.117 [A] = 0.3 - 0.1175 = 0.1825 M [B] = 0.4 - 0.117 = 0.2825 M [C] = 0.117 M

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