The reaction of cyanamide, $$NH_2CN_{(s)}$$ with oxygen was run in a bomb calorimeter and $$\Delta U$$ was found to be $$-742.24$$ kJ mol$$^{-1}$$. The magnitude of $$\Delta H_{298}$$ for the reaction
$$NH_2CN_{(s)} + \frac{3}{2}O_{2(g)} \to N_{2(g)} + O_{2(g)} + H_2O_{(l)}$$ is ______ kJ. (Rounded off to the nearest integer) [Assume ideal gases and R = 8.314 J mol$$^{-1}$$ K$$^{-1}$$]

We are given the reaction: $$NH_2CN_{(s)} + \frac{3}{2}O_{2(g)} \to N_{2(g)} + O_{2(g)} + H_2O_{(l)}$$ with $$\Delta U = -742.24$$ kJ mol$$^{-1}$$, and we need to find the magnitude of $$\Delta H_{298}$$.

The relationship between enthalpy change and internal energy change is: $$\Delta H = \Delta U + \Delta n_g RT$$, where $$\Delta n_g$$ is the change in the number of moles of gaseous species.

Counting the moles of gaseous species on the product side: $$N_2(g)$$ contributes 1 mol and $$O_2(g)$$ contributes 1 mol, giving a total of 2 moles of gas. Note that $$H_2O$$ is in the liquid state and does not count. On the reactant side, we have $$\frac{3}{2}$$ mol of $$O_2(g)$$. The solid $$NH_2CN$$ does not count.

Therefore, $$\Delta n_g = 2 - \frac{3}{2} = \frac{1}{2}$$.

Substituting the values: $$\Delta H = -742.24 + \frac{1}{2} \times 8.314 \times 10^{-3} \times 298$$

$$\Delta H = -742.24 + \frac{1}{2} \times 2.4776 = -742.24 + 1.2388 = -741.00 \text{ kJ mol}^{-1}$$

The magnitude of $$\Delta H_{298}$$ is $$\mathbf{741}$$ kJ.