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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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Similar questions

3263.9 k J m o l^{- 1}

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(I)

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- 3263.9 k J m o l^{- 1}

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