The difference between heats of reaction at constatnt pressure and constant volume for the reaction ${2 C}_{6} H_{6} (l) + 150_{2} (g) \to {12 C O}_{2} (g) + {6 H}_{2} O (l) a t 25^{o}$ in ${k J m o l}^{- 1}$ :-

-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
heat of reaction at const. pressure $= Δ_{r} H$
heat of reaction at const. pressure $= Δ_{r} U$
$= Δ_{r} H = Δ_{r} U + Δ n_{g} R T$
$Δ n_{g} = - 3$
$Δ_{r} H - Δ_{r} U = (- 3 R T) = - 7432.7 J / m o l = - 7.432 k J / m o l$

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25^{o} C

2 C_{6} H_{6 (l)} + 15 O_{2 (g)} ⟶ 12 {C O}_{2 (g)} + 6 H_{2} O_{(l)}

2 C_{6} H_{6} (l) + 15 O_{2} (g) \to 12 {C O}_{2} (g) + 6 H_{2} O (l)

25^{o} C

2 C_{6} H_{6} (l) + 15 O_{2} (g) \to 12 C O_{2} (g) + 6 H_{2} O (l)

25^{\circ}

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2 C_{6} H_{6} (I) + 15 O_{2} (g) \to 12 C O_{2} (g) + 6 H_{2} O (I)

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