The difference between heat of reaction at constant pressure and constant volume for the reaction given below at 25 $^{0}$ C in kJ is: $2 C_{6} H_{6 (l)} + 15 O_{2 (g)} \to 12 C O_{2 (g)} + 6 H_{2} O_{(l)}$

-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
$Δ H =$ heat of reaction at constant pressure.
$Δ U =$ heat of reaction at constant volume.
$Δ H = Δ U + Δ n_{g} R T$
$Δ H - Δ U = Δ n_{g} R T$
$2 C_{6} H_{6} (l) + 15 O_{2} (g) \to 12 C O_{2} (g) + 6 H_{2} O (l)$
$Δ n_{g} = - 3$
$Δ H - Δ U = - 3 R T$ J
$= - 3 \times 8.314 \times 298$ J
$≃ - 7432$ J
$≃ - 7.43$ kJ

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