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 $25^{\circ}$ in $k J m o l^{- 1}$ 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 $- 7.43$
Heat of reaction at constant pressure is $Δ H$ and at constant volume is $Δ U$ .

$Δ H - Δ U = Δ n_{g} R T$

$Δ n_{g} = 12 - 15 = - 3$

$Δ n_{g} R T = (- 3) \times (8.314 \times 10^{- 3} K J / m o l / K) \times (298 K)$

$= - 7.43$

Hence,option A is correct.

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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 25∘ in kJ mol−1 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)×(8.314×10−3KJ/mol/K)×(298K) =−7.43Hence,option A is correct.

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 $25^{\circ}$ in $k J m o l^{- 1}$ is:

The correct option is A $- 7.43$
Heat of reaction at constant pressure is $Δ H$ and at constant volume is $Δ U$ .

$Δ H - Δ U = Δ n_{g} R T$

$Δ n_{g} = 12 - 15 = - 3$

$Δ n_{g} R T = (- 3) \times (8.314 \times 10^{- 3} K J / m o l / K) \times (298 K)$

$= - 7.43$

Hence,option A is correct.