Consider the following reaction approaching equilibrium at $$27°$$C and 1 atm pressure
$$\text{A} + \text{B} \underset{K_r=10^2}{\overset{K_f=10^3}{\rightleftharpoons}} \text{C} + \text{D}$$
The standard Gibb's energy change $$(\Delta_r G°)$$ at $$27°$$C is $$(-)$$ ______ kJ mol$$^{-1}$$. (Nearest integer).
(Given: R $$= 8.3$$ J K$$^{-1}$$ mol$$^{-1}$$ and $$\ln 10 = 2.3$$)

We are given the reaction at 27°C (300 K) and 1 atm:

$$A + B \underset{K_r=10^2}{\overset{K_f=10^3}{\rightleftharpoons}} C + D$$

$$K_{eq} = \frac{K_f}{K_r} = \frac{10^3}{10^2} = 10$$

$$\Delta_r G° = -RT \ln K_{eq}$$

$$= -8.3 \times 300 \times \ln(10)$$

$$= -8.3 \times 300 \times 2.3$$

$$= -5727 \text{ J mol}^{-1}$$

$$= -5.727 \text{ kJ mol}^{-1}$$

Rounding to the nearest integer:

$$\Delta_r G° = (-) \boxed{6} \text{ kJ mol}^{-1}$$