Search any question & find its solution
Question:
Answered & Verified by Expert
Calculate $\Delta \mathrm{H}$ for following reaction, at $25^{\circ} \mathrm{C}$.
Calculate $\Delta \mathrm{H}$ for following reaction, at $25^{\circ} \mathrm{C}$.
$$
\begin{aligned}
& \mathrm{NH}_2 \mathrm{CN}_{(\mathrm{g})}+\frac{3}{2} \mathrm{O}_{2(\mathrm{~g})} \longrightarrow \mathrm{N}_{2(\mathrm{~g})}+\mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_2 \mathrm{O}_{(l)} \\
& \left(\Delta \mathrm{U}=-740.5 \mathrm{~kJ}, \mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)
\end{aligned}
$$
Options:
Calculate $\Delta \mathrm{H}$ for following reaction, at $25^{\circ} \mathrm{C}$.
$$
\begin{aligned}
& \mathrm{NH}_2 \mathrm{CN}_{(\mathrm{g})}+\frac{3}{2} \mathrm{O}_{2(\mathrm{~g})} \longrightarrow \mathrm{N}_{2(\mathrm{~g})}+\mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_2 \mathrm{O}_{(l)} \\
& \left(\Delta \mathrm{U}=-740.5 \mathrm{~kJ}, \mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)
\end{aligned}
$$
Solution:
1747 Upvotes
Verified Answer
The correct answer is:
$-741.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$$
\begin{aligned}
& \Delta \mathrm{n}_{\mathrm{g}}=2-\frac{3}{2}=\frac{1}{2} \mathrm{~mol} \\
& \begin{aligned}
\mathrm{R} & =8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^1 \\
& =8.314 \times 10^{-3} \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}
\end{aligned}
\end{aligned}
$$
Now, using formula,
$$
\begin{aligned}
\Delta \mathrm{H}= & \Delta \mathrm{U}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT} \\
=-740.5 \mathrm{~kJ}+ & \left(\frac{1}{2} \mathrm{~mol}\right) \times 8.314 \\
& \times 10^{-3} \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \times 298 \mathrm{~K}
\end{aligned}
$$
$\begin{aligned} & =-740.5 \mathrm{~kJ}+1.2388 \mathrm{~kJ} \\ & =-739.26 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \approx-741.7 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}$
\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}
$$
$\begin{aligned} & =-740.5 \mathrm{~kJ}+1.2388 \mathrm{~kJ} \\ & =-739.26 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \approx-741.7 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.