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

Question

A reaction A(aq) + B(aq) doubleheadarrow C(aq) has a standard free-energy change of -3.39 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 +3.39 kJ/mol? There would be less A and B but more C. All concentrations would be higher. 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

Standard free energy chane = - 3.39 kJ/mol

It is known that,

deltaG0 = - R T lnK

lnK = - 3.39 / (-0.008314 * 298)

lnK = 1.37

K = 3.94

A (aq.) + B (aq.) = C (aq.)

0.30      0.40         0.00 M              at initial

0.30-x    0.40-x       x M                  at equilibrium

Equilibrium constant expression can be written as,

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

3.94 = x / (0.30-x)(0.40-x)

3.94(0.12-0.70x+x2) = x

3.94x2 - 3.76x + 0.473 = 0

applying quadratic equation,

x = [-(-3.76)+/-sqrt.((-3.76)2-(4*3.94*0.473))] / (2*3.94)

x = 0.149 M

So, at equilibrium,

[C] = 0.149 M

[A] = 0.30 - 0.149 = 0.151 M

[B] = 0.40 - 0.149 = 0.251 M

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