Question

Use bond energies to calculate the enthalpy of reaction for the combustion of ethane. Average bond energies in kJ/mol C-C 347, C=C 614, C=C 839, C-H 413, H-H 432, H-O 467, O-O 204, O=O 498, C=O 745, C=O in CO2 799

2 CH3CH3 + 7 O2 → 4 CO2 + 6 H2O

Answer 1

The enthalpy of reaction for the combustion of ethane 2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O calculated from the average bond energies of the compounds is -2860 kJ/mol.

The reaction is:

2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O  (1)  

The enthalpy of reaction (1) is given by:

$\Delta H = \Delta H_{r} - \Delta H_{p}$   (2)

Where:

r: is for reactants

p: is for products

The bonds of the compounds of reaction (1) are:

• 2CH₃CH₃: 2 moles of 6 C-H bonds + 2 moles of 1 C-C bond

• 7O₂: 7 moles of 1 O=O bond  

• 4CO₂: 4 moles of 2 C=O bonds  

• 6H₂O: 6 moles of 2 H-O bonds

Hence, the enthalpy of reaction (1) is (eq 2):

$\Delta H = \Delta H_{r} - \Delta H_{p}$

$\Delta H = 2*\Delta H_{CH_{3}CH_{3}} + 7\Delta H_{O_{2}} - (4*\Delta H_{CO_{2}} + 6*\Delta H_{H_{2}O})$      

$\Delta H = 2*(6*\Delta H_{C-H} + \Delta H_{C-C}) + 7\Delta H_{O=O} - (4*2*\Delta H_{C=O} + 6*2*\Delta H_{H-O})$  

$\Delta H = [2*(6*413 + 347) + 7*498 - (4*2*799 + 6*2*467)] kJ/mol$  

$\Delta H = -2860 kJ/mol$          

Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.

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