Question

Consider the following reaction at 321 K: 2 A (aq) + 1 B (aq) → 2 C (aq) + 1 D (aq) An experiment was performed with the following intitial concentrations: [A]i = 1.33 M, [B]i = 2.05 M, [C]i = 0.25 M, [D]i = 0.29 M. The reaction was allowed to proceed until equilibrium was reached at which time it was determined that [A] = 0.51 M. What was the maximum amount of work that could have been performed as the reaction began? Maximum amount of work performed as the reaction began (in kJ)=

Answer 1

Maximum work is only achievable if the process is reversible.

In this case, the reaction can yield maximum work if it undergoes an isothermal process only (which the reversible pathway).

At equilibrium, the Gibbs free energy change is zero.

ΔG=ΔGo+RTlnQ

=> 0 =ΔGo+RTlnK (because ΔG= 0 at equlibrium)

=> ΔGo = - RTlnK

To find the equilibrium constant(K), we know [A] = 0.51 M

Therefore A used up = [A]i - [A] = 1.33 - 0.51 = 0.82 M

Therefore B consumed should be half of A consumed (from the balanced chemical equation) = 0.41 M

So, [B] = [B]i - 0.41 = 2.05 - 0.41 = 1.64 M

Also, consumed moles resulted in the products. [D] = 0.41 M and [C] = 0.82 M

Therefore, equilibrium constant K = ([C]²[D])/([A]²[B]) = (0.82² *0.41)/(0.51² * 1.64) = 0.6463.

Since process is assumed isothermal, T = 321 K. R = 8.314 Jmol^-1K^-1 .

Therefore the maximum work done = - (8.314 Jmol^-1K^-1)*(321K)*ln(0.6463) = - (8.314 Jmol^-1K^-1)*(321K)*(- 0.4365) =1164.93 Jmol^-1 = 1.16493 kJmol^-1 (ANSWER)