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Enthalpy of Combustion

The heat libe...

Question

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

-3273.735 KJ

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-3274.735 KJ

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-3275.735 KJ

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-3270.735 KJ

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Solution

The correct option is A
-3273.735 KJ

Let us first write the chemical reaction which is

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

$Δ$ ng = number of moles of gaseous products - number of moles of gaseous reactants

$Δ$ ng = $6 - \frac{15}{2}$ = $- \frac{3}{2}$

Now $Δ U$ per mole = $\frac{- 327 \times 78}{7.8} = - 3270 K J$

$Δ H = Δ U + Δ n R T = - 3270 + (- \frac{3}{2}) (8.3) (300) 10^{- 3}$

= $3270 + (- \frac{3}{2}) (8.3) (300) 10^{- 3}$

$Δ H = - 3273.735 K J$

Suggest Corrections

Similar questions

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

7.8

327

27^{0}

27^{0}

7.8 g

327 k J

27^{o} C

(R = 8.31 J {m o l}^{- 1} K^{- 1})

m o l^{- 1}

25^{0} C

m o l^{- 1}

J K^{- 1} m o l^{- 1}

(l)

C O_{2} (g)

H_{2} O (l)

k J m o l^{- 1}

25^{\circ} C

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

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

{C O}_{2} (g)

H_{2} O (l)

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

25^{o} C

k J

{m o l}^{- 1}

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

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