MCQ
How much ....$kJ$ energy is released when $6$ mole of octane is burnt in air ? Given $\Delta H_f^o$ for $C{O_2}(g),\,{H_2}O(g)$ and ${C_8}{H_{18}}(l)$ respectively are $-490,-240\, and + 160 \,kJ/mol$
- A$-6.2 $
- ✓$-37.4$
- C$-35.5$
- D$-20$
✓
Answer
Correct option: B.
$-37.4$
(b) $C + {O_2} \to C{O_2}\,;\,\,\,\Delta {H_F} = - 490\,KJ/mol$ $-(I)$
${H_2} + 1/2{O_2} \to {H_2}O;\,\,\Delta {H_F} = - 240\,KJ/mol$ $-(II)$
$8C + 18H \to {C_8}{H_{18}};\,\,\Delta {H_F} = + 160\,KJ/mol$ $-(III)$
applying
$(I)$ $ \times \,8 + $ $(II)$ $ \times 9 + $ $(III)$
${C_8}{H_{18}} + \frac{{25}}{2}{O_2} \to 8C{O_2} + 9{H_2}O$
$\Delta {H^o} = - 3920 - 2160 - 160 = 6240\,KJ/mol$
$\Delta {H^o} = $ for $6$ moles of octane $ = 6240 \times 6$
$ = 37440\,KJ/mol = - 37.4\,KJ$
${H_2} + 1/2{O_2} \to {H_2}O;\,\,\Delta {H_F} = - 240\,KJ/mol$ $-(II)$
$8C + 18H \to {C_8}{H_{18}};\,\,\Delta {H_F} = + 160\,KJ/mol$ $-(III)$
applying
$(I)$ $ \times \,8 + $ $(II)$ $ \times 9 + $ $(III)$
${C_8}{H_{18}} + \frac{{25}}{2}{O_2} \to 8C{O_2} + 9{H_2}O$
$\Delta {H^o} = - 3920 - 2160 - 160 = 6240\,KJ/mol$
$\Delta {H^o} = $ for $6$ moles of octane $ = 6240 \times 6$
$ = 37440\,KJ/mol = - 37.4\,KJ$
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