A reaction A(aq) + B(aq) rightarrow has a standard free-energy change of -4.26 k

Question

A reaction A(aq) + B(aq) rightarrow has a standard free-energy change of -4.26 kJ/mol at 25 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? [A] = Number M [B] Number M [C] Number M How would your answers above change if the reaction had a standard free-energy change of +4.26 kJ/mol? There would be no change to the answers. 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 concentrations would be lower.

Explanation / Answer

Equilibrium constant (K) and standard gibbs free energy are related as deltaG0 = -RT lnK

lnK = -deltaG0/RT = 4.26*1000/(8.314*298)= 1.72, K= 5.58

ICE table

                                           [A]                              [B]                                  [C]

Initial                                0.3                               0.4                                   0

Change                            -x                                   -x                                   +x

Equilibrium                    0.3-x                            0.4-x                                x

For the given reaction K= [C] /[A][B] = x/(0.3-x)*(0.4-x)= 5.58, when solved using excel, x =0.169

At equilibrium [A] =0.3-0.169=0.131, [B] =0.4-0.169=0.231, [C] =0.169

When deltaG is +ve, the reaction is reactants favoured. This leads to more A and B, less C as compared to the earlier case since K= -4.26*1000/(8.314*298), K= 0.179

K= 0.179= x/(0.3-x)*(0.4-x) ,x =0.019 [A] =0.3-0.019 =.281, [B] = .381, [C] =0.019 ( D is correct)

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