The heat of combustion of benzene in a bomb calorimeter i-e- constant volume was found to be 3263-9 kJ mol -1 at 25- C - Calculate the heat of combustion of benzene at constant pressure-A- -4767-6 kJmol -1B- 3667-6 kJ mol -1C- 2767-6 kJ mol -1D- -3267-6 kJ mol -1

The correct option is D 3267.6 kJmol^ 1 nbsp;The reaction is: C_6H_6(l)+15/2O_2(g)⇌ 6CO_2(g)+3H_2O(l) In this reaction, O_2 is the only gaseous reactant an ...

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Standard XII

Chemistry

Enthalpy of Combustion

The heat of c...

Question

The heat of combustion of benzene in a bomb calorimeter (i.e., constant volume) was found to be $3263.9 k J m o l^{- 1}$ at $25^{o} C$ . Calculate the heat of combustion of benzene at constant pressure.

- 4767.6 k J m o l^{- 1}

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3667.6 k J m o l^{- 1}

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- 3267.6 k J m o l^{- 1}

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2767.6 k J m o l^{- 1}

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Solution

The correct option is C $- 3267.6 k J m o l^{- 1}$
The reaction is:
$C_{6} H_{6} (l) + \frac{15}{2} O_{2} (g) ⇌ 6 C O_{2} (g) + 3 H_{2} O (l)$

In this reaction, $O_{2}$ is the only gaseous reactant and $C O_{2}$ is the only gaseous product.
$∴ Δ n_{g} = n_{p} - n_{r} = 6 - \frac{15}{2} = - \frac{3}{2}$
Also, given that $Δ U = - 3263.9 k J m o l^{- 1}, T = 25^{o} C = 25 + 273 = 298 K,$
$R = 8.314 J K^{- 1} m o l^{- 1}$
we know $Δ H = Δ U + Δ n_{g} R T$
$Δ H = - 3263.9 - \frac{3}{2} \times 8.314 \times 10^{- 3} \times 298$
$Δ H = - 3263.9 - 3.7$
$Δ H = - 3267.6 k J m o l^{- 1}$

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Enthalpy of Combustion

Standard XII Chemistry

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The heat of combustion of benzene in a bomb calorimeter (i.e., constant volume) was found to be 3263.9 kJmol−1 at 25oC. Calculate the heat of combustion of benzene at constant pressure.

The heat of combustion of benzene in a bomb calorimeter (i.e., constant volume) was found to be $3263.9 k J m o l^{- 1}$ at $25^{o} C$ . Calculate the heat of combustion of benzene at constant pressure.