A reaction A(aq) + B(aq) C(aq) has a standard free-energy change of -3.37 kJ/mol

Question

A reaction A(aq) + B(aq) C(aq) has a standard free-energy change of -3.37 kJ/mol at 25degreeC. 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 +3.37 kJ/mol? concentrations would be lower. All concentrations would be higher. There would be no change to the answers. There would be more A and 8 but less C. There would be less A and B but more C.

Explanation / Answer

At equilibrium K = product / reactants

A+       B--- > C

I                       0.30     0.40         0.0

E                      0.30-X 0.40 –X   X

K = X/(0.30-X)(0.40-X)

Here dG = -3.37 kJ/ mol or -3.73*10^3 J/mol

delta G = -RT*ln(K)

8.3145

JK1mol1

-3.73*10^3 = -298*R*ln(x/(0,3-x)(0,4-x))
x = 0.157

At equilibrium

A :0.30- X= 0.30- 0.157

=0.143

B :0.40- X= 0.40- 0.157

=0.243

C=X=0.157

If the sign of free energy changed then reaction will reverse means there would be more A and B and less C.

8.3145

JK1mol1

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