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 $27^{o} C$ . Calculate the heat combustion of benzene at constant pressure.
$(R = 8.31 J {m o l}^{- 1} K^{- 1})$

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

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- 3273.7 k J {m o l}^{- 1}

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

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3273.7 k J {m o l}^{- 1}

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Solution

The correct option is A $- 3273.7 k J {m o l}^{- 1}$
We have,

$C_{6} H_{6} (l) + \frac{15}{2} O_{2} (g) - > 6 C O_{2} + 3 H_{2} O$

Therefore, $Δ n$ = 6 – 15/2 = – 3/2

Also, $Δ U$ per mol $= - (327 \times 78) / 7.8$

= $- 3270 k J$

Therefore, $Δ H$ = $Δ U + Δ n R T$

$= - 3270 + (- 3 / 2) \times 300 \times 10^{- 3}$

Or, $Δ H$ = $- 3273.735$ (B)

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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 $27^{0} C (R = 8.3 J m o l^{- 1} K^{- 1})$