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Heat of Reaction

The enthalpy ...

Question

The enthalpy of combustion of one mole of benzene, carbon and hydrogen are $- 3267 k J / m o l, - 393.5 k J / m o l$ and $- 285.8 k J / m o l$ respectively. Calculate the standard enthalpy of formation of benzene.

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Solution

The enthalpy of combustion of one mole of benzene, carbon and hydrogen are $- 3267 k J / m o l, 393.5 K L / m o l$ and $- 285.8 K J / m o l$ respectively.

$C_{6} H_{6} + 15 / 2 O_{2} \to 6 C O_{2} + 3 H_{2} O Δ H = - 3267 k J / m o l$ ......(1)

$C + O_{2} \to C O_{2} Δ H = - 393.5 k J / m o l$ ......(2)

$H_{2} + 1 / 2 O_{2} \to H_{2} O Δ H = - 285.8 k J / m o l$ ......(3)

Reverse equation (1)
$6 C O_{2} + 3 H_{2} O \to C_{6} H_{6} + 15 / 2 O_{2} Δ H = + 3267 k J / m o l$ ......(4)

Multiply equation (2) with 6
$6 C + 6 O_{2} \to 6 C O_{2} Δ H = 6 \times - 393.5 = - 2361 k J / m o l$ ......(5)

Add equations (4) and (5)
$6 C + 3 H_{2} O \to C_{6} H_{6} + 3 / 2 O_{2} Δ H = 3267 - 2361 = 906 k J / m o l$ ......(6)

Multiply equation (3) with 3
$3 H_{2} + 3 / 2 O_{2} \to 3 H_{2} O Δ H = 3 \times - 285.8 = - 857.4 k J / m o l$ ......(7)

Add equation (6) and (7)
$6 C + 3 H_{2} \to C_{6} H_{6}$

The standard enthalpy of formation of benzene.
$Δ H = 960 - 857.4 = 48.6 k J / m o l$

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