How much energy is released when 6 moles of octane is burnt in air? Given Δ H_f^ for CO_ 2(g) , H_2 O_ (g) and C_8 H_ 18(l) respectively are -490,-240 and +160 ~kJ / mol .

Correct answer: B. -37.4 ~kJ array r C+O_2 arrow CO_2 ; Δ H_f^ =-490 kJ / mol × 8 H_2+(1)/(2) O_2 arrow H_2 O ; Δ H_f^ =-240 kJ / mol × 9 8 C+18 H arrow C_8 H_ 18 ; Δ H_f^ =+160 kJ / mol 8 C+8 O_2+9 H_2+(9)/(2) O_2-8 C-18 H arrow 8 CO_2+9 H_2 O-C_8 H_ 18 Δ H_f^ =-3920-2160-160 C_8 H_ 18 +(25)/(2) O_2 arrow 8 CO_2+9 H_2 O ; Δ H^ =-6240 kJ / mol Δ H^ for 6 moles of octane =-6240 × 6=-37440 kJ / mol . array

AIIMS 2004

Exam: AIIMS
Year: 2004
Subject: Chemistry
Chapter: Thermodynamics (C)

AIIMS 2004 · Chemistry · Thermodynamics (C)

How much energy is released when 6 moles of octane is burnt in air?...

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}

Options

A $-6.2 \mathrm{~kJ}$
B $-37.4 \mathrm{~kJ}$
C $-35.5 \mathrm{~kJ}$
D $-20.0 \mathrm{~kJ}$

Correct answer

B. $-37.4 \mathrm{~kJ}$

Step-by-step solution

\begin{array}{r} \text{ } \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}

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