Calculate \(\Delta \mathrm{H}\) for following reaction, at \(25^{\circ} \mathrm{C}\). Calculate \(\Delta \mathrm{H}\) for following reaction, at \(25^{\circ} \mathrm{C}\).
\(\mathrm{NH}_2 \mathrm{CN}_{(\mathrm{g})}+\frac{3}{2} \mathrm{O}_{2(\mathrm{~g})} \longrightarrow \mathrm{N}_{2(\mathrm{~g})}+\mathrm{CO}_{2(\mathrm{~g})}+\) \(\mathrm{H}_2 \mathrm{O}_{(l)} \ (\Delta \mathrm{U}=-740.5 \mathrm{~kJ}, \mathrm{R}=\) \(8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1})\)

\(
\begin{aligned}
& \Delta \mathrm{n}_{\mathrm{g}}=2-\frac{3}{2}=\frac{1}{2} \mathrm{~mol} \\
& \begin{aligned}
\mathrm{R} & =8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^1 \\
& =8.314 \times 10^{-3} \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}
\end{aligned}
\end{aligned}
\)
Now, using formula,
\(
\begin{aligned}
\Delta \mathrm{H}= & \Delta \mathrm{U}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT} \\
=-740.5 \mathrm{~kJ}+ & \left(\frac{1}{2} \mathrm{~mol}\right) \times 8.314 \\
& \times 10^{-3} \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \times 298 \mathrm{~K}
\end{aligned}
\)
\(\begin{aligned} & =-740.5 \mathrm{~kJ}+1.2388 \mathrm{~kJ} \\ & =-739.26 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \approx-741.7 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}\)