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Bond Order and Enthalpy

Calculate hea...

Question

Calculate heat of combustion of ethene (figure) from bond energy data:
$C = C, C - H, O = O, C - O, O - H$
$B E (k J {m o l}^{- 1}) : 619, 414, 499, 724, 460$

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Solution

Heat of combustion of ethene $= 4 (B E o f C - H) + 3 \times (B E o f 0 = 0) + (B E o f C = C) - 4 \times (B E o f C - 0) - 2 \times (B E o f O - H)$
$= (4 \times 414) + (3 \times 499) + (619) - (4 \times 724) - (2 \times 460)$
$∴$ Heat of combustion of ethene $= - 44 K J / m o l$

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Similar questions

C H_{2} = C H_{2} (g) + 3 O_{2} (g) \to 2 C O_{2} (g) + 2 H_{2} O (I)

C = C = 619 k J m o l^{- 1}

C - H = 414 k J m o l^{- 1}

O = O = 499 k J m o l^{- 1}

C = O = 724 k J m o l^{- 1}

O - H = 460 k J m o l^{- 1}

Bond	$C - H$	$C - O$	$O - H$	$O = O$	$C = O$
Bond Enthalpy ( kJ $m o l^{- 1}$	414	351.5	464.5	494	711

C O_{2} = - 143 k J m o l^{- 1}

= 35.5 k J m o l^{- 1}

= 40.6 k J m o l^{- 1}

25^{o} C

C - C

C - H

Δ H_{c o m b u s t i o n o f e t h a n e}^{o} = - 372.0

Δ H_{c o m b u s t i o n o f p r o p a n e}^{o} = - 530.0

Δ H

C_{g r a p h i t e} \to C (g) = + 172.0

H - H

= + 104.0

Δ H_{f}

H_{2} O (l) = - 68.0

Δ H_{f}

{C O}_{2} (g) = - 94.0

{kcal mol}^{- 1}

25^{o} C

C - C

Δ H_{combustion}^{o}

= - 372.0

Δ H_{combustion}^{o}

= - 530.0

Δ H_{combustion}^{o}

\to C (g) = - 172.0

H - H = 104.0

Δ H_{f}^{o}

H_{2} O (l) = 68.0

Δ H_{f}^{o}

C O_{2} (g) = 94.0

C - H = 413.4 k J {m o l}^{- 1}

C - C = 347.0 k J {m o l}^{- 1}

C = O = 728.0 k J {m o l}^{- 1}

O = O = 495.0 k J {m o l}^{- 1}

H - H = 435.8 k J {m o l}^{- 1}

Δ H_{s u b} C (s) = 718.4 k J {m o l}^{- 1}

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