Benzene bums in $O_{2}$ according to the equation:
$C_{6} H_{6} (l) + (15 / 2) O_{2} (g) ⟶ 3 H_{2} O (g) + 6 C O_{2} (g)$
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 0.5 g benzene in kcal is :

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Solution

$∵$ 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:

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

$Δ H^{\circ} = - 780.6 k c a l$

$∵ 78 g C_{6} H_{6} p r o d u c e s h e a t = 780.6 k c a l$

$∴ 0.5 g C_{6} H_{6} p r o d u c e s h e a t = (780.6 \times 0.5) / 78$
$= 5 k c a l$

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