The energy released when 6 moles of octane is burnt in air will be [Given, Δ H_f^ · for CO_2(g) , H_2 O(g) and C_8 H_ 18 (l) , respectively are -490 , -240 and +160 ~J / mol]

Correct answer: D. -35.5 ~kJ aligned C+O_2 CO_2 ; Δ H=-490 ~J....(i) H_2+(1)/(2) O_2 H_2 O ; Δ H=-240 ~J....(ii) 8 C+9 H_2 C_8 H_ 18 ; Δ H=+160 ~J....(iii) aligned On applying, 8 x Eq. (i) +9 x Eq. (ii)-Eq. (iii), we get aligned & C_8 H_ 18 +(25)/(2) O_2 8 CO_2+9 H_2 O & Δ H_R^0=[8 ×(-490)]+[9 ×(-240)]+160 aligned =-5920 ~J ~mol^ -1 Hence, energy exchange when 6 moles of octane is burnt in air =-5920 × 6=-35520 ~J=-35.5 kJ

JIPMER 2017

Exam: JIPMER
Year: 2017
Subject: Chemistry
Chapter: Thermodynamics (C)

JIPMER 2017 · Chemistry · Thermodynamics (C)

The energy released when 6 moles of octane is burnt in air will be...

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}]

Options

A $-37.4 \mathrm{~kJ}$
B $-20 \mathrm{~kJ}$
C $-6.2 \mathrm{~kJ}$
D $-35.5 \mathrm{~kJ}$

Correct answer

D. $-35.5 \mathrm{~kJ}$

Step-by-step 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

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