Benzene bums in $O_{2}$ according to the equation:

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

If standard heat enthalpies of $C_{6} H_{6} (l), H_{2} O (g) a n d C O_{2} (g)$ are 11.7, -68.3 and -94.0 kcal respectively. The amount of heat liberated by burning 1 kg benzene in kcal is:

10007.4

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1000

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500

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None of these

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Solution

The correct option is D 10007.4
$∵$ Standard heat enthalpy of a compound $(H^{\circ}) =$ Standard heat of formation of that compound $(Δ_{f} H)$

$∴ 6 C + 3 H_{2} ⟶ C_{6} H_{6} (l); Δ H = 11.7 k c a l$ ...( i)

$H_{2} + \frac{1}{2} O_{2} ⟶ H_{2} O (g); Δ H = - 68.3 k c a l$ ...(ii)

$C + O_{2} ⟶ C O_{2} (g); Δ H = - 94.0 k c a l$ ...(iii)

By inspection method $3 \times (i i) + 6 \times (i i i) - (i)$ gives:

$C_{6} H_{6} = (15 / 2) O_{2} ⟶ 6 C O_{2} + 3 H_{2} O$ ;

$Δ H^{\circ} = - 780.6 k c a l$

$∵ 78 g C_{6} H_{6}$ produces heat $= 780.6 k c a l$

$∴ 1000 g C_{6} H_{6}$ produces heat $= \frac{(780.6 \times 1000)}{78}$

$= 10007.4 k c a l$

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Similar questions

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

C_{6} H_{6} (l)

H_{2} O (l)

{C O}_{2} (g)

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

Δ H = - 3264.4 k J / m o l

C_{6} H_{6} (l) + 7 \frac{1}{2} O_{2} (g) \to 3 H_{2} O (l) + 6 C O_{2} (g), Δ H = - 781.0 K c a l m o l^{- 1}

156 g o f C_{6} H_{6}

△ H

△ U

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

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