The difference between heats of reaction at constant pressure and volume for the reaction $2 C_{6} H_{6} (I) + 15 O_{2} (g) \to 12 C O_{2} (g) + 6 H_{2} O (I)$ at $25^{0}$ in $k J m o l^{- 1}$ is :-

-7.43

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3.72

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-3.72

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7.43

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Solution

The correct option is B -7.43
$2 C_{6} H_{6} (l) + 15 O_{2} (g) \leftrightarrow 12 C O_{2} + 6 H_{2} O$ at $25^{o} C$
$Δ H - Δ U = Δ n_{g} R T$
$∵ Δ n_{g} = 12 - 15 = - 3$
$∴ Δ n_{g} R T = (- 3) \times (8.314) \times 10^{- 3} \times 298$
$= - 7.43 k J m o l^{1}$

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