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

- 7.43

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3.72

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 3.72

No worries! We‘ve got your back. Try BYJU‘S free classes today!

7.43

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $- 7.43$
$Δ H = Δ U + Δ n_{g} R T$
Using this relation where
$Δ H \to$ Heat of relation at constant pressure.
$Δ U \to$ Heat of reaction at constant volume.
$2 C_{6} H_{6} (l) + 15 O_{2} (g) \to 12 {C O}_{2} (g) + 6 H_{2} O (l) T = 25^{\circ} T = 298.15 K Δ n_{g} = 12 - 15 = - 3 Δ H - Δ U = - 3 \times 8.314 \times 298.15 = - 7.436 K J$

Suggest Corrections

Similar questions

Q. 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

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 heats of Reaction at constant pressure and constant volume for the Reaction $2 C_{6} H_{6} + 15 O_{2} \to 12 C O_{2} + 6 H_{2} O$ at $25^{\circ} C$ . in $k J$ 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:

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

Join BYJU'S Learning Program

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

The correct option is A −7.43ΔH=ΔU+ΔngRTUsing this relation whereΔH→ Heat of relation at constant pressure.ΔU→ Heat of reaction at constant volume.2C6H6(l)+15O2(g)→12CO2(g)+6H2O(l)T=25∘T=298.15KΔng=12−15=−3ΔH−ΔU=−3×8.314×298.15=−7.436KJ

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