The difference between heats of reaction at constant pressure and constant volume for the reaction,
$2 C_{6} H_{6} (l) + 15 O_{2} (g) \to 12 {C O}_{2} (g) + 6 H_{2} O (l)$ at $25^{o} C$ in kJ is:

+ 7.43

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+ 3.72

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- 7.43

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- 3.72

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Solution

The correct option is A $- 7.43$
The heat of reaction is basically $Δ H$ at constant pressure and is $Δ U$ at constant volume.
Since, for any reaction the relation between this heats is given by equation,
$Δ H$ = $Δ U$ + ( $Δ$ $n_{g}$ ) $R T$
where, $Δ$ $n_{g}$ $= change in number of gaseous moles in any reaction.$
Since, in the above reaction the only gaseous species are $C O_{2}$ and $O_{2}$ , so,
$Δ$ $n_{g}$ $= 12 - 15 = - 3.$
Hence, using $Δ H$ - $Δ U$ = ( $Δ$ $n_{g}$ ) $R T$
$Δ H$ - $Δ U$ $= (- 3) R T = (- 3) \times 8.314 \times 298.15 J$ ( $T$ in Kelvin)
=> $Δ H$ - $Δ U$ $= 7430 J = 7.43 K J$
Hence, answer is option $C$ .

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