What is the value of [tex]$K$[/tex] for this aqueous reaction at 298 K?

[tex]\[
A + B \rightleftharpoons C + D \quad \Delta G^{\circ}=25.79 \text{ kJ/mol}
\][/tex]

[tex]\[
K = \, ?
\][/tex]



Answer :

Certainly! Let's solve the problem step-by-step.

We are given:

- The standard Gibbs free energy change, [tex]\(\Delta G^\circ = 25.79 \text{ kJ/mol}\)[/tex].
- The temperature, [tex]\(T = 298 \text{ K}\)[/tex].
- The universal gas constant, [tex]\(R = 8.314 \text{ J/(mol·K)}\)[/tex].

First, we need to convert [tex]\(\Delta G^\circ\)[/tex] from kJ/mol to J/mol because the gas constant [tex]\(R\)[/tex] is in units of J/(mol·K).

[tex]\[ \Delta G^\circ = 25.79 \text{ kJ/mol} \times 1000 \frac{\text{J}}{\text{kJ}} = 25790 \text{ J/mol} \][/tex]

Next, we use the relationship between the Gibbs free energy change and the equilibrium constant [tex]\(K\)[/tex], which is given by the equation:

[tex]\[ \Delta G = -RT \ln K \][/tex]

We need to solve for [tex]\(K\)[/tex]. Rearranging the equation gives:

[tex]\[ K = e^{-\Delta G / (RT)} \][/tex]

Substituting in the known values:

[tex]\[ K = e^{-25790 \text{ J/mol} / (8.314 \text{ J/(mol·K)} \times 298 \text{ K})} \][/tex]

Through proper calculation, we find:

[tex]\[ K \approx 3.0148220087706008 \times 10^{-5} \][/tex]

Therefore, the value of the equilibrium constant [tex]\(K\)[/tex] for the given reaction at 298 K is approximately:

[tex]\[ K = 3.0148220087706008 \times 10^{-5} \][/tex]