Question

Consider the reaction: A (aq) ⇌ B (aq) at 253 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 = 0.357 M. What is the maximum amount of work that can be done by this system when the concentration of A = 0.867 M? (in kJ)

Answer 1

Answer:

The maximum amount of work that can be done by this system is -2.71 kJ/mol

Explanation:

Maximum amount of work denoted change in gibbs free energy $(\Delta G)$ during the reaction.

Equilibrium concentration of B = 0.357 M

So equilibrium concentration of A = (1-0.357) M = 0.643 M

So equilibrium constant at 253 K, $K_{eq}= \frac{[B]}{[A]}$

[A] and [B] represent equilibrium concentrations

$K_{eq}=\frac{0.357}{0.643}=0.555$

When concentration of A = 0.867 M then B = (1-0.867) M = 0.133 M

So reaction quotient at this situation, $Q=\frac{0.133}{0.867}=0.153$

We know,  $\Delta G=RTln(\frac{Q}{K_{eq}})$

where R is gas constant and T is temperature in kelvin

Here R is 8.314 J/(mol.K), T is 253 K, Q is 0.153 and $K_{eq}$ is 0.555

So, $\Delta G=8.314\times 253\times ln(\frac{0.153}{0.555})mol/K$

                           = -2710 J/mol

                            = -2.71 kJ/mol