Q11 / 160 · TS EAMCET 2022 (19 Jul Shift 1) · Chemistry
$$
\begin{aligned}
& \text { Given } \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l) ; \Delta \mathrm{H}=-285 \mathrm{~kJ} \\
& \mathrm{~N}_2 \mathrm{O}_5(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(l) \rightarrow 2 \mathrm{HNO}_3(l) ; \Delta \mathrm{H}=-76.6 \mathrm{~kJ} \\
& \mathrm{~N}_2(\mathrm{~g})+3 \mathrm{O}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{HNO}_3(l) ; \Delta \mathrm{H}=-348.2 \mathrm{~kJ} \\
& \text { Calculate the } \Delta \mathrm{H} \text { of } 2 \mathrm{~N}_2(\mathrm{~g})+5 \mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{~N}_2 \mathrm{O}_5(\mathrm{~g})
\end{aligned}
$$
Calculate the $\Delta \mathrm{H}$ of $2 \mathrm{~N}_2(\mathrm{~g})+5 \mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{~N}_2 \mathrm{O}_5(\mathrm{~g})$.
- A$572 \mathrm{~kJ}$
- B$419 \mathrm{~kJ}$
- C$14.5 \mathrm{~kJ}$
- Correct answerD$26.8 \mathrm{~kJ}$
Solution
Given that;
$$
\begin{gathered}
\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{l}) ; \\
\Delta \mathrm{H}=-285 \mathrm{KJ} \\
\mathrm{N}_2 \mathrm{O}_5(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \longrightarrow 2 \mathrm{HNO}_3(\mathrm{l}) ; \\
\Delta \mathrm{H}=-76.6 \mathrm{KJ} \\
\mathrm{N}_2(\mathrm{~g})+3 \mathrm{O}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{HNO}_3(\mathrm{l}) ; \\
\Delta \mathrm{H}=-348.2 \mathrm{KJ}
\end{gathered}
$$
Reverse equation (i) \& (ii) and multiply by 2 .
$$
\begin{aligned}
2 \mathrm{H}_2 \mathrm{O}(\ell) \underset{\Delta \mathrm{H}}{\longrightarrow}=+285 \times 2=570 \mathrm{HJ} & 2 \mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) ;
\end{aligned}
$$
$$
\begin{array}{r}
4 \mathrm{HNO}_3(\ell) \longrightarrow 2 \mathrm{~N}_2 \mathrm{O}_5(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \\
\Delta \mathrm{H}=+76.6 \times 2=153.2 \mathrm{KJ}
\end{array}
$$
Multiply equation (iii) by 2 .
$$
\begin{array}{r}
2 \mathrm{~N}_2(\mathrm{~g})+3 \mathrm{O}_2(\mathrm{~g})+2 \mathrm{H}_2(\mathrm{~g}) \longrightarrow 4 \mathrm{HNO}_3(\mathrm{l}) ; \\
\Delta \mathrm{H}=-348.2 \times 2=696.4 \mathrm{KJ}
\end{array}
$$
Add equation (iv), (v) \& (vi) to get,
$$
\begin{aligned}
& 2 \mathrm{~N}_2(\mathrm{~g})+5 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{~N}_2 \mathrm{O}_5 \\
& \Delta \mathrm{H}=570+153.2+696.4=26.8 \mathrm{KJ}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.