For a given chemical reaction A $$\to$$ B at 300 K the free energy change is -49.4 kJ mol$$^{-1}$$ and the enthalpy of reaction is 51.4 kJ mol$$^{-1}$$. The entropy change of the reaction is ___ JK$$^{-1}$$ mol$$^{-1}$$.

The relationship between Gibbs free energy, enthalpy, and entropy at constant temperature is given by:

$$\Delta G = \Delta H - T\Delta S$$

Given: $$\Delta G = -49.4 \text{ kJ mol}^{-1}$$, $$\Delta H = 51.4 \text{ kJ mol}^{-1}$$, and $$T = 300 \text{ K}$$. Rearranging to solve for $$\Delta S$$:

$$\Delta S = \frac{\Delta H - \Delta G}{T}$$

$$\Delta S = \frac{51.4 - (-49.4)}{300} \text{ kJ K}^{-1}\text{mol}^{-1}$$

$$\Delta S = \frac{51.4 + 49.4}{300} = \frac{100.8}{300} = 0.336 \text{ kJ K}^{-1}\text{mol}^{-1}$$

Converting to JK$$^{-1}$$ mol$$^{-1}$$:

$$\Delta S = 0.336 \times 1000 = 336 \text{ J K}^{-1}\text{mol}^{-1}$$

The entropy change of the reaction is $$\mathbf{336}$$ J K$$^{-1}$$ mol$$^{-1}$$.