For a chemical reaction, the standard Gibbs energy change, $\Delta G^{\circ}$ is $-7.64 \times 10^4…
For a chemical reaction, the standard Gibbs energy change, $\Delta G^{\circ}$ is $-7.64 \times 10^4 \mathrm{~J} \mathrm{~mol}^1$. What is the value of equilibrium constant (K)?
$K=1$
$k>1$
$K < 1$
$K=0$
Solution
Standard Gibbs free energy,
$$
\Delta G^{\circ}=-2.303 R T \log K
$$
Given, $\Delta G^{\circ}=-7.64 \times 10^4 \mathrm{~J} / \mathrm{mol}$
$$
\begin{aligned}
\log K & =-\frac{\Delta G^{\circ}}{2.303 R T} \\
K & =\operatorname{antilog}-\left(\frac{\Delta G^{\circ}}{2.303 R T}ight) \\
K & =\text { antilog }-\left(\frac{(-) 7.64 \times 10^4}{2.303 \times 8.314 \times 298}ight)
\end{aligned}
$$
Here, $K$ is greater than one.
Thus, option (2) is correct.
.