The reaction of cyanamide,


The reaction of cyanamide, $\mathrm{NH}_{2} \mathrm{CN}_{(s)}$ with oxygen was run in a bomb calorimeter and $\Delta \mathrm{U}$ was found to be $-742.24 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The magnitude of $\Delta \mathrm{H}_{298}$ for the reaction

$\mathrm{NH}_{2} \mathrm{CN}_{(\mathrm{s})}+\frac{3}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}_{(l)}$

is_______________$\mathrm{kJ}$. (Rounded off to the nearest integer)

[Assume ideal gases and $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ ]


$\Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}$

$=-742.24+\frac{1}{2} \times \frac{8.314}{1000} \times 298$

$=-741 \mathrm{~kJ} / \mathrm{mol}$

Hence answer is (741)

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