Calculate $\Delta \mathrm{H}$ for following reaction, at $25^{\circ} \mathrm{C}$. Calculate $\Delta…
Calculate $\Delta \mathrm{H}$ for following reaction, at $25^{\circ} \mathrm{C}$.
Calculate $\Delta \mathrm{H}$ for following reaction, at $25^{\circ} \mathrm{C}$.
$\begin{aligned}
& \mathrm{NH}_2 \mathrm{CN}_{(\mathrm{g})}+\frac{3}{2} \mathrm{O}_{2(\mathrm{~g})} \longrightarrow \mathrm{N}_{2(\mathrm{~g})}+\mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_2 \mathrm{O}_{(l)} \\
& \left(\Delta \mathrm{U}=-740.5 \mathrm{~kJ}, \mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)
\end{aligned}$
- $-708.4 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- $-789.4 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- $-741.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- $-863.9 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Solution
$\begin{aligned}
& \Delta \mathrm{n}_{\mathrm{g}}=2-\frac{3}{2}=\frac{1}{2} \mathrm{~mol} \\
& \begin{aligned}
\mathrm{R} & =8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^1 \\
& =8.314 \times 10^{-3} \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}
\end{aligned}
\end{aligned}$
Now, using formula,
$\begin{aligned}
\Delta \mathrm{H}= & \Delta \mathrm{U}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT} \\
=-740.5 \mathrm{~kJ}+ & \left(\frac{1}{2} \mathrm{~mol}\right) \times 8.314 \\
& \times 10^{-3} \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \times 298 \mathrm{~K}
\end{aligned}$
$\begin{aligned} & =-740.5 \mathrm{~kJ}+1.2388 \mathrm{~kJ} \\ & =-739.26 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \approx-741.7 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}$
Asked in: MHT CET 2023 (14 May Shift 1)
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