Benzene burns in oxygen according to the following reaction :
$C_{6} H_{6} (l) + \frac{15}{2} O_{2} (g) \to {3 H}_{2} O (l) + {6 C O}_{2} (g)$
If the standard enthalpies of formation of $C_{6} H_{6} (l)$ , $H_{2} O (l)$ and ${C O}_{2} (g)$ are 11.7, -68.1 and -94 kcal/mole, respectively, the amount of heat that will liberate burning 780 g benzene is:

7800 kcal

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780 kcal

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78 kcal

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608.4 kcal

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Solution

The correct option is A 7800 kcal
$C_{6} H_{6} (l) + \frac{15}{2} O_{2} (g) ⟶ 3 H_{2} O (l) + 6 C O_{2} (g)$
enthalpies of formation
$C_{6} H_{6} \to 11.7, H_{2} O \to - 68.1, C O_{2} \to - 94$
Enthalpy of reaction is $3 (- 681) + 3 (- 94) - 1 (11.7)$
$= - 780 K c a l$
This enthalpy is for 1 mol (78 g) of benzene.
for $780 g$ of benzene
$7800 k c a l$ is the amount of heat required.

Suggest Corrections

Similar questions

O_{2}

C_{6} H_{6} (l) + (15 / 2) O_{2} (g) ⟶ 3 H_{2} O (g) + 6 C O_{2} (g)

C_{6} H_{6} (l), H_{2} O (g) a n d C O_{2} (g)

C_{6} H_{6} (l), H_{2} O (l), C O_{2} (g)

11.70, - 68.4

- 94.0

k c a l

l

27^{\circ} C

C_{6} H_{6 (l)} + 7 \frac{1}{2} O_{2 (g)} \to {6 C O}_{2 (g)} + {3 H}_{2} O_{(l)}

{m o l}^{- 1}

27 C

C_{6} {H_{6}}_{(l)} + \frac{15}{2} {O_{2}}_{(g)} \to {6 C O}_{2}_{(g)} + {{3 H}_{2} O}_{(l)}

C_{6} H_{6} (l) + 15 / 2 O_{2} (g) \to 6 {C O}_{2} (g) + 3 H_{2} O (l)

Δ U

25^{o} C = - 3268.12 k J

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