Question

has a standard free‑energy change of − 3.59 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?

Answer 1

Answer: The concentrations of A , B , and C at equilibrium are 0.1583 M, 0.2583 M, and 0.1417 M.

Step-by-step explanation:

The reaction equation is as follows.

`A + B \rightarrow C`

Initial : 0.3 0.4 0

Change: -x -x x

Equilbm: (0.3 - x) (0.4 - x) x

We know that, relation between standard free energy and equilibrium constant is as follows.

`\Delta G = -RT ln K`

Putting the given values into the above formula as follows.

`\Delta G = -RT ln K`

`-3.59 kJ/mol = -8.314 * 10^(-3) kJ/mol K ln ((x)/((0.3 - x)(0.4 - x)))`

x = 0.1417

Hence, at equilibrium

• [A] = 0.3 - 0.1417

= 0.1583 M

• [B] = 0.4 - 0.1417

= 0.2583 M

• [C] = 0.1417 M