From the following data at constant volume for combustion of benzene, calculate the heat of this reaction at constant pressure condition.
$C_{6} H_{6} (l) + 15 / 2 O_{2} (g) \to 6 {C O}_{2} (g) + 3 H_{2} O (l)$
$Δ U$ at $25^{o} C = - 3268.12 k J$

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Solution

$C_{6} H_{6} (l) + 7 \frac{1}{2} O_{2} (g) \to 6 C O_{2} (g) + 3 H_{2} O (l)$
Given, $△ U = - 3268.12 K J$
$T = 298 K$
From the reaction, $△ n_{g} = 6 - 7 \frac{1}{2} = \frac{12 - 15}{2} = - \frac{3}{2}$
Using the reaction, $△ H = - 3268.12 K J - \frac{3}{2} \times 8.314 J / K m o l \times 248 K$
$△ H = - 3268120 J - \frac{3}{2} \times 8.314 J / K m o l \times 298$
$\Rightarrow △ H = - - 3271.84 K J / m o l$

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