Question

Consider the reaction:

A (aq) ⇌ B (aq)

at 277 K where the initial concentration of A = 1.00 M and the initial concentration of B = 0.000 M. At equilibrium it is found that the concentration of B = 0.363 M. What is the maximum amount of work that can be done by this system when the concentration of A = 0.891 M?

Answer 1

                          A(aq) <--->    B (aq)

Initially                1                0

equilibrium           1-X               X

given at equilibrium [B] = 0.363 = X , , hence [A] = 1-0.363 = 0.637 ,

Keq = [B] /[A] = 0.363 /0.637 = 0.57

dGo = -RT ln Keq = - 8.314 x 277 x ln (0.57) = 1294.6 J/mol

Now system has [A] = 0.891 M , hence [B] = 1-0.891 = 0.109

Q = [B] /[A} = 0.109 /0.891 = 0.12233

dG = dGo + RT ln Q

   = 1294.6 J/mol + ( 8.314 x 277 x ln (0.12233) )

= - 3544 J/mol = -3.544 KJ/mol   ( -ve sign indicates sponatneous )

Thus maximum work that can be done = free energy value ( dG) = 3.544 KJ/mol