The difference between heats of reaction at constant pressure and 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

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+ 7.43

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Solution

The correct option is A $- 7.43$
Heat of reaction at constant pressure is ΔH and at constant volume is ΔU.
$Δ H - Δ U = Δ n_{g} R T Δ n_{g} = 12 - 15 = - 3 Δ n_{g} R T = (- 3) . (8.314 \times 10^{- 3}) . (298) Δ n_{g} R T = - 7.43 k J$ .
So, correct answer is option A.

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The difference between heats of reaction at constant pressure and constant volume for the reaction;2C6H6(l)+15O2(g)→12CO2(g)+6H2O(l) at 25oC in kJ is:

The correct option is A −7.43Heat of reaction at constant pressure is ΔH and at constant volume is ΔU.ΔH−ΔU=ΔngRTΔng=12−15=−3ΔngRT=(−3).(8.314×10−3).(298)ΔngRT=−7.43kJ.So, correct answer is option A.

The difference between heats of reaction at constant pressure and 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:

The correct option is A $- 7.43$
Heat of reaction at constant pressure is ΔH and at constant volume is ΔU.
$Δ H - Δ U = Δ n_{g} R T Δ n_{g} = 12 - 15 = - 3 Δ n_{g} R T = (- 3) . (8.314 \times 10^{- 3}) . (298) Δ n_{g} R T = - 7.43 k J$ .
So, correct answer is option A.