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Question: A reaction A + B <==> C has a standard free-...

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A reaction
A + B <==> C
has a standard free-energy change of -3.22 kj/mol at 25oC

What are the concentrations of A, B, and C at equilibrium if at the beginning of the reaction their concentrations are 0.30M, 0.40M and 0M respectively?

How would your answers above change if the reaction had a standard free-energy change of +4.88 kj/mol?
____ there would be no change to the answers.
____ All concentrations would be lower.
____ All concentrations would be higher.
____ There would be less A and B but more C.
____ There would be more A and B but less C.

Answer 1

Go = - 3.22 kJ/mol = - 3220 J/mol

Go = - RT lnK

- 3220 = - 8.314 * 298 * ln K

ln K = 1.29

K = e^1.29

K = 3.63

Now,

                              A               +               B            <==>        C
Initial                  0.30                            0.40                           0
Change                -x                                -x                            +x
At equilibrium   0.30 - x                      0.40 - x                         x

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

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

x = 0.144

Hence, at equilibrium:

[A] = 0.30 - 0.144 = 0.156 M

[B] = 0.40 - 0.144 = 0.256 M

[C] = 0.144 M

If Go = +4.88 kJ/mol, the value of K will be lesser and hence the equilibrium concentration of product will be lesser.

Therefore, there would be more A and B but less C.