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.

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.

      $\Delta G = -RT ln K$

      $-3.59 kJ/mol = -8.314 \times 10^{-3} kJ/mol K ln (\frac{x}{(0.3 - x)(0.4 - x)})$

                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