The difference between heats of Reaction at constant pressure and constant volume for the Reaction $2 C_{6} H_{6} + 15 O_{2} \to 12 C O_{2} + 6 H_{2} O$ at $25^{\circ} C$ . in $k J$ 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 A
-7.43

$q_{p} - q_{v} = Δ H - Δ E = Δ n R T$

$(∵ q_{v} = Δ E, W = 0) (Δ n = 12 - 15 = - 3)$

= $- 3 \times 8.314 \times 10^{- 3} \times 298$

= $- 7.43 k J$

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