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:
[tex] \Delta H = \Delta H_{r} - \Delta H_{p} [/tex] (2)
Where:
r: is for reactants
p: is for products
The bonds of the compounds of reaction (1) are:
Hence, the enthalpy of reaction (1) is (eq 2):
[tex] \Delta H = \Delta H_{r} - \Delta H_{p} [/tex]
[tex] \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}) [/tex]
[tex] \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}) [/tex]
[tex] \Delta H = [2*(6*413 + 347) + 7*498 - (4*2*799 + 6*2*467)] kJ/mol [/tex]
[tex] \Delta H = -2860 kJ/mol [/tex]
Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.
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