The different between heat of reaction at constant pressure and constant volume of 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 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 $- 7.43$
Heat of reaction at constant pressure $= Δ H$

Heat of reaction at constant volume $= Δ E$

As we know that ;

$\Rightarrow Δ H = Δ E + Δ n_{g} R T$

$\Rightarrow Δ H - Δ E = Δ n_{f} R T$

Here $Δ n_{g}$ is difference in number of gaseous products and number of gaseous reactants

$\Rightarrow Δ n_{g} = 12 - 15 = - 3$

$\Rightarrow Δ H - Δ E = - 3 \times 8.314 \times 10^{- 3} \times 298$

$\Rightarrow Δ H - Δ E = - 7.432$ KJ

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