Answer
$$\begin{aligned} \Delta G^{\circ \prime} &=5.7 \mathrm{kJ} \cdot \mathrm{mol}^{-1} \end{aligned}$$
$\text{Since $\Delta G^{\circ \prime}$ is positive, the reaction is endergonic under standard conditions.}$
Work Step by Step
We find:
$\begin{aligned} \Delta G^{\circ \prime} &=-R T \ln K_{\mathrm{eq}}=-R T \ln ([\mathrm{C}][\mathrm{D}] /[\mathrm{A}][\mathrm{B}]) \\ &=-\left(8.314 \mathrm{J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1}\right)(298 \mathrm{K}) \ln \left[\left(3 \times 10^{-6}\right)\left(5 \times 10^{-6}\right) /\right.\\ &\left(10 \times 10^{-6}\right)\left(15 \times 10^{-6}\right) ] \\ &=5700 \mathrm{J} \cdot \mathrm{mol}^{-1}=5.7 \mathrm{kJ} \cdot \mathrm{mol}^{-1} \end{aligned}$
$\text{Since $\Delta G^{\circ \prime}$ is positive, the reaction is endergonic under standard conditions.}$
