$C_{6} H_{6} (g) + \frac{15}{2} O_{2} \to 6 C O_{2} (g) + 3 H_{2} O (l)$
The heat liberated on complete combustion of $7.8$ g benzene is $327$ KJ. This heat has been measured at constant volume and at $27^{o}$ C. Calculate heat of combustion of benzene at constant pressure at $27^{o} C$ .

Open in App

Solution

$C_{6} H_{6} (g) + \frac{15}{2} O_{2} (g) \to 6 C O_{2} (g) + 3 H_{2} O (l)$
therefore,
$Δ n = 6 - \frac{15}{2} = - \frac{3}{2}$
$Δ U = \frac{- 327 \times 78}{7.8} = - 3270$
$Δ H = Δ U + Δ n R T$
$Δ H = - 3270 + (- \frac{3}{2}) \times 300 \times 10^{- 3}$
$Δ H = - 3273.735 K J$

Suggest Corrections

Similar questions

Q. The heat liberated on complete combustion of

7.8 g

benzene is

327 k J

. This heat was measured at constant volume and at

27^{o} C

. Calculate the heat of combustion of benzene at constant pressure.

Q. The heat liberated on complete combustion of 7.8 g benzene is 327 kJ. This heat has been measured at a constant volume and at

27^{\circ} C

. Calculate the heat of combustion of benzene at constant pressure.

The heat liberated on complete combustion of $7.8 g$ benzene is $327 K J$ . This heat has been measured at constant volume and at $27^{0} C$ . What is the heat of combustion of benzene at constant pressure at $27^{0} C (R = 8.3 J m o l^{- 1} K^{- 1})$

Q. From the following data at constant volume for combustion of benzene, calculate the heat of this reaction at constant pressure condition.

C_{6} H_{6} (l) + 15 / 2 O_{2} (g) \to 6 {C O}_{2} (g) + 3 H_{2} O (l)

Δ U

25^{o} C = - 3268.12 k J

Q. The combustion of benzene

(l)

gives

C O_{2} (g)

and

H_{2} O (l)

. Given that the heat of combustion of benzene at constant volume is —3263.9

k J m o l^{- 1}

25^{\circ} C

. The heat of combustion

(i n k J m o l^{- 1})

of benzene at constant pressure will be:

(R = 8.314 J K^{- 1} m o l^{- 1})

Join BYJU'S Learning Program

C6H6(g)+152O2→6CO2(g)+3H2O(l)The heat liberated on complete combustion of 7.8g benzene is 327 KJ. This heat has been measured at constant volume and at 27oC. Calculate heat of combustion of benzene at constant pressure at 27oC.

C6H6(g)+152O2(g)→6CO2(g)+3H2O(l)therefore,Δn=6−152=−32ΔU=−327×787.8=−3270ΔH=ΔU+ΔnRTΔH=−3270+(−32)×300×10−3ΔH=−3273.735KJ

$C_{6} H_{6} (g) + \frac{15}{2} O_{2} \to 6 C O_{2} (g) + 3 H_{2} O (l)$
The heat liberated on complete combustion of $7.8$ g benzene is $327$ KJ. This heat has been measured at constant volume and at $27^{o}$ C. Calculate heat of combustion of benzene at constant pressure at $27^{o} C$ .

$C_{6} H_{6} (g) + \frac{15}{2} O_{2} (g) \to 6 C O_{2} (g) + 3 H_{2} O (l)$
therefore,
$Δ n = 6 - \frac{15}{2} = - \frac{3}{2}$
$Δ U = \frac{- 327 \times 78}{7.8} = - 3270$
$Δ H = Δ U + Δ n R T$
$Δ H = - 3270 + (- \frac{3}{2}) \times 300 \times 10^{- 3}$
$Δ H = - 3273.735 K J$