The difference between the heats of reaction at constant pressure and constant volume for the reaction.
$2 C_{6} H_{6} (l) + 15 O_{2} (g) \to 12 C O_{2} (g) + 6 H_{2} O (l) a t 25^{\circ} C$ in kJ 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 B $- 7.43$
The difference between heats of reaction at constant pressure and constant volume for the following reaction at 25 degree Celsius in kJ
$2 C_{6} H_{6} (l) + 15 O_{2} (g) \to 12 C O_{2} (g) + 6 H_{2} O$
$Δ n = 12 - 15 = - 3$
$Δ H = Δ E + R T Δ n$
$Δ H - Δ E = R T Δ n$
$= (- 3) \times 8.314 \times 10^{- 3} \times 298$
$= - 7.43 k J / m o l$

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The difference between the heats of reaction at constant pressure and constant volume for the reaction.2C6H6(l)+15O2(g)→ 12CO2(g)+6H2O(l)at25∘C in kJ is:

The correct option is B −7.43The difference between heats of reaction at constant pressure and constant volume for the following reaction at 25 degree Celsius in kJ2C6H6(l)+15O2(g)→12CO2(g)+6H2OΔn=12−15=−3ΔH=ΔE+RTΔnΔH−ΔE=RTΔn =(−3)×8.314×10−3×298 =−7.43 kJ/mol

The difference between the heats of reaction at constant pressure and constant volume for the reaction.
$2 C_{6} H_{6} (l) + 15 O_{2} (g) \to 12 C O_{2} (g) + 6 H_{2} O (l) a t 25^{\circ} C$ in kJ is: