A reaction A(aq) + B(aq) equilibrium C(aq) has a standard free-energy change of

Question

A reaction A(aq) + B(aq) equilibrium C(aq) has a standard free-energy change of -3.79 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 +3.79 kJ/mol? There would be no change to the answers. All concentrations would be lower. There would be more A and B but less C. All concentrations would be higher. There would be less A and B but more C.

Explanation / Answer

dGo = -RTlnK

-3790 = -8.314 x 298 lnK

K = 4.6

let x be the change at equilibrium in the concentration of reactants

K = [C]/[A][B]

4.6 = x/(0.3-x)(0.4-x)

4.6x^2 - 4.22x + 0.55 = 0

x = 0.16 M

So at equilbrium,

[A] = 0.3 - 0.16 = 0.14 M

[B] = 0.4 - 0.16 = 0.24 M

[C] = 0.16 M

Now, If standard free energy change was instead +3.79 kJ/mol,

There would be more A and B but less C

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