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Standard enthalpy of formation values of $C_{6} H_{6}, H_{2} O$ and ${CO}_{2}$ are $+ 52 KJ / mole, - 286 KJ / mole$ and $- 395 KJ / mole$ respectively. Enthalpy of combustion of benzene is

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$- 2095 KJ / mole$

$- 3280 KJ / mole$

$- 5138 KJ / mole$

$- 1676 KJ / mole$

answer is D.

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Detailed Solution

Required equation:

$C_{6} H_{6} (l) + \frac{15}{2} O_{2} (g) ⟶ 6 {CO}_{2} (g) + 3 H_{2} O (1)$

Given equations :

$1) 6 C (graphite) + 3 H_{2} (g) ⟶ C_{6} H_{6} (1); ΔH = + 52 KJ$

$2) C (graphite) + O_{2} (g) ⟶ {CO}_{2} (g); ΔH = - 395 KJ$

$3) H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (1), ΔH = - 286 KJ$

$6 \times (E q 2) + 3 \times (E q 3) - (E q 1) =$ Required equation

$ΔH = 6 (- 395) + 3 (- 286) - (+ 52)$

$= (- 2370) + (- 858) - 52$

$= - 3280 KJ / mole$

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