For combustion of one mole of magnesium in an open container at 300 K and 1 bar pressure, $$\Delta_C H^\ominus = -601.70$$ kJ mol$$^{-1}$$, the magnitude of change in internal energy for the reaction is ______ kJ. (Nearest integer)
(Given: R = 8.3 J K$$^{-1}$$ mol$$^{-1}$$)

We need to find the magnitude of change in internal energy for the combustion of one mole of magnesium. The standard enthalpy of combustion is $$\Delta_C H^\ominus = -601.70 \text{ kJ mol}^{-1}$$, the temperature is 300 K, and the gas constant is R = 8.3 J K$$^{-1}$$ mol$$^{-1}$$.

The combustion reaction for two moles of magnesium is $$2\text{Mg}(s) + \text{O}_2(g) \rightarrow 2\text{MgO}(s)$$, so for one mole of Mg it is $$\text{Mg}(s) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{MgO}(s)$$.

The change in moles of gaseous species is $$\Delta n_g = \text{moles of gaseous products} - \text{moles of gaseous reactants} = 0 - \frac{1}{2} = -\frac{1}{2}$$.

Using $$\Delta H = \Delta U + \Delta n_g RT$$, we have $$\Delta U = \Delta H - \Delta n_g RT = -601.70 - \left(-\frac{1}{2}\right) \times 8.3 \times 10^{-3} \times 300 = -601.70 + \frac{1}{2} \times 2.49 = -601.70 + 1.245 = -600.455 \text{ kJ}$$.

The magnitude is $$|\Delta U| = 600.455 \approx 600 \text{ kJ}$$.

Hence, the answer is 600.