# The reaction of cyanamide,

Question:

The reaction of cyanamide, $\mathrm{NH}_{2} \mathrm{CN}_{(\mathrm{s})}$ 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 $\mathrm{NH}_{2} \mathrm{CN}_{(\mathrm{s})}+\frac{3}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}+\mathrm{H}_{2} \mathrm{O}_{(1)}$ is__________ $\mathrm{kJ}$. (Rounded off to the nearest integer)

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

Solution:

(741)

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

$\Delta \mathrm{ng}=(1+1)-\frac{3}{2}=\frac{1}{2}$

$\Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{ngRT}$

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

$=-742.24+1.24$

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