How much energy is released when 6 moles of octane is burnt in air? Given $\Delta H_f^{\circ}$ for…

How much energy is released when 6 moles of octane is burnt in air? Given $\Delta H_f^{\circ}$ for $\mathrm{CO}_{2(g)}, \mathrm{H}_2 \mathrm{O}_{(\mathrm{g})}$ and $\mathrm{C}_8 \mathrm{H}_{18(l)}$ respectively are $-490,-240$ and $+160 \mathrm{~kJ} / \mathrm{mol}$.
  1. $-6.2 \mathrm{~kJ}$
  2. $-37.4 \mathrm{~kJ}$
  3. $-35.5 \mathrm{~kJ}$
  4. $-20.0 \mathrm{~kJ}$

Solution

$\begin{array}{l} \mathrm{C}+\mathrm{O}_2 \rightarrow \mathrm{CO}_2 ; \Delta H_f^{\circ}=-490 \mathrm{~kJ} / \mathrm{mol} \times 8 \\ \mathrm{H}_2+\frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{H}_2 \mathrm{O} ; \Delta H_f^{\circ}=-240 \mathrm{~kJ} / \mathrm{mol} \times 9 \\ 8 \mathrm{C}+18 \mathrm{H} \rightarrow \mathrm{C}_8 \mathrm{H}_{18} ; \Delta H_f^{\circ}=+160 \mathrm{~kJ} / \mathrm{mol} \\ 8 \mathrm{C}+8 \mathrm{O}_2+9 \mathrm{H}_2+\frac{9}{2} \mathrm{O}_2-8 \mathrm{C}-18 \mathrm{H} \\ \rightarrow 8 \mathrm{CO}_2+9 \mathrm{H}_2 \mathrm{O}-\mathrm{C}_8 \mathrm{H}_{18} \\ \Delta H_f^{\circ}=-3920-2160-160 \\ \mathrm{C}_8 \mathrm{H}_{18}+\frac{25}{2} \mathrm{O}_2 \rightarrow 8 \mathrm{CO}_2+9 \mathrm{H}_2 \mathrm{O} ; \Delta H^{\circ}=-6240 \mathrm{~kJ} / \mathrm{mol} \\ \Delta H^{\circ} \text { for } 6 \mathrm{moles} \text { of octane }=-6240 \times 6=-37440 \mathrm{~kJ} / \mathrm{mol} . \end{array}$

Asked in: NEET 2010 (Mains)

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