The difference between heats of reaction at constant pressure and at constant volume for the reaction $2 C_{6} H_{6} (l) + 15 O_{2} (g) \to 12 C O_{2} (g) + 6 H_{2} O (l) a t 25^{\circ} C$ at 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 A –7.43
$Δ H - Δ E = Δ n R T = 3 \times 8.314 \times 298$
$= - 7432 J = - 7.43 k J$

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Similar questions

Q. The difference between heat 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)}

25^{\circ} C

in KJ is

The difference between heat of reaction at constant pressure and constant volume for the reaction given below at 25 $^{0}$ C in kJ is: $2 C_{6} H_{6 (l)} + 15 O_{2 (g)} \to 12 C O_{2 (g)} + 6 H_{2} O_{(l)}$

Q. The difference between the heat 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)

25^{\circ} C

k J

is:

Q. The difference between the heats of a reaction at constant pressure and 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°C in kJ is:

Q. For

2 C_{6} H_{6} (l) + 15 O_{2} (g) \to 12 C O_{2} (g) + 6 H_{2} O (g)

the difference in change in enthalpy and change in internal energy at

25^{\circ} C

(kJ) is:

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

The correct option is A –7.43ΔH−ΔE=ΔnRT=3×8.314×298 =−7432 J=−7.43kJ

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

The correct option is A –7.43
$Δ H - Δ E = Δ n R T = 3 \times 8.314 \times 298$
$= - 7432 J = - 7.43 k J$