Q58 / 180 · JIPMER 2017 · Chemistry
The energy released when 6 moles of octane is burnt in air will be [Given, $\Delta H_f^{\cdot}$ for $\mathrm{CO}_2(g)$, $\mathrm{H}_2 \mathrm{O}(\mathrm{g})$ and $\mathrm{C}_8 \mathrm{H}_{18}(\mathrm{l})$, respectively are -490 , -240 and $+160 \mathrm{~J} / \mathrm{mol}]$
- A$-37.4 \mathrm{~kJ}$
- B$-20 \mathrm{~kJ}$
- C$-6.2 \mathrm{~kJ}$
- Correct answerD$-35.5 \mathrm{~kJ}$
Solution
$\begin{aligned} \mathrm{C}+\mathrm{O}_2 \longrightarrow \mathrm{CO}_2 ; \Delta H=-490 \mathrm{~J}....(i) \\ \mathrm{H}_2+\frac{1}{2} \mathrm{O}_2 \longrightarrow \mathrm{H}_2 \mathrm{O} ; \Delta H=-240 \mathrm{~J}....(ii) \\ 8 \mathrm{C}+9 \mathrm{H}_2 \longrightarrow \mathrm{C}_8 \mathrm{H}_{18} ; \Delta H=+160 \mathrm{~J}....(iii)\end{aligned}$
On applying, $8 x$ Eq. (i) $+9 x$ Eq. (ii)-Eq. (iii), we get
$\begin{aligned}& \mathrm{C}_8 \mathrm{H}_{18}+\frac{25}{2} \mathrm{O}_2 \longrightarrow 8 \mathrm{CO}_2+9 \mathrm{H}_2 \mathrm{O} \\
& \Delta H_R^0=[8 \times(-490)]+[9 \times(-240)]+160
\end{aligned}$
$=-5920 \mathrm{~J} \mathrm{~mol}^{-1}$
Hence, energy exchange when 6 moles of octane is burnt in air $=-5920 \times 6=-35520 \mathrm{~J}=-35.5$ kJ
Hence, the correct answer is option: 'D'.