The difference between the heat of reaction at constant pressure and constant volume for the reaction given below at $25^{o} C$ in KJ is:
$2 C_{6} H_{6 (l)} + 15 O_{2 (g)} ⟶ 12 {C O}_{2 (g)} + 6 H_{2} O_{(l)}$

- 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 D $- 7.43$
The difference between heats of reaction at constant pressure and constant volume for the following reaction at $25^{o} C$ in kJ
$2 C_{6} H_{6_{(} l)} + 15 O_{2_{(} g)} ⟶ 12 C O_{2_{(} g)} + 6 H_{2} O_{(l)}$
$Δ n = 12 - 15 = - 3$
$Δ H = Δ E + R T Δ n$
$Δ H - Δ E = R T Δ n$
$= (- 3) \times 8.314 \times 10^{- 3} \times 298$
$= - 7.443 k J / m o l e$

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The difference between the heat of reaction at constant pressure and constant volume for the reaction given below at 25oC in KJ is: 2C6H6(l)+15O2(g)⟶12CO2(g)+6H2O(l)

The correct option is D −7.43The difference between heats of reaction at constant pressure and constant volume for the following reaction at 25oC in kJ2C6H6(l)+15O2(g)⟶12CO2(g)+6H2O(l)Δn=12−15=−3ΔH=ΔE+RTΔnΔH−ΔE=RTΔn =(−3)×8.314×10−3×298 =−7.443 kJ/mole

The difference between the heat of reaction at constant pressure and constant volume for the reaction given below at $25^{o} C$ in KJ is:
$2 C_{6} H_{6 (l)} + 15 O_{2 (g)} ⟶ 12 {C O}_{2 (g)} + 6 H_{2} O_{(l)}$