The difference between the heat of reaction at constant pressure and the heat of reaction at 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 $k J$ is:

- 7.43

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+ 3.72

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- 3.72

No worries! We‘ve got your back. Try BYJU‘S free classes today!

+ 7.43

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Solution

The correct option is A $- 7.43$
$2 C_{6} H_{6} (l) + 15 O_{2} (g) \to 12 C O_{2} (g) + 6 H_{2} O (l)$
$Δ H = Δ E + Δ n_{g} R T$ and

$Δ n_{g} = 12 - 15 = - 3$

Now,
$Δ H - Δ E = Δ n_{g} R T$
$= - 3 \times 8.314 \times 298 J$
$= - 7432 J$
$≃ - 7.43 k J$

Option A is correct.

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