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]
