The value of log K for the reaction A $$\rightleftharpoons$$ B at 298 K is ______. (Nearest integer)
Given: $$\Delta H° = -54.07$$ kJ mol$$^{-1}$$, $$\Delta S° = 10$$ J K$$^{-1}$$ mol$$^{-1}$$
(Taken $$2.303 \times 8.314 \times 298 = 5705$$)

We have $$\Delta H° = -54.07\;\text{kJ mol}^{-1} = -54070\;\text{J mol}^{-1}$$, $$\Delta S° = 10\;\text{J K}^{-1}\text{mol}^{-1}$$, and $$T = 298\;\text{K}$$.

The standard Gibbs free energy change is:

$$\Delta G° = \Delta H° - T\Delta S° = -54070 - 298 \times 10 = -54070 - 2980 = -57050\;\text{J mol}^{-1}$$

Now, using the relationship $$\Delta G° = -2.303RT\log K$$, we solve for $$\log K$$:

$$\log K = \frac{-\Delta G°}{2.303RT} = \frac{57050}{5705} = 10.0$$

Hence, the value of $$\log K$$ is $$10$$.