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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:
(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):
Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.
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Answer:
See Explanation
Explanation:
In electrophilic aromatic substitution, the benzene ring undergoes substitution when it is reacted with suitable electrophiles.
The products of electrophilic aromatic substitution depends on the substituents already present on the benzene ring. Some substituents activate the ring towards electrophilic substitution and direct the incoming electrophile to the ortho and para positions on the ring while some substituents deactivate the benzene ring towards electrophilic substitution and direct the incoming electrophlle to the meta position on the ring.
The amide substituent is moderately activating and is an ortho, para director hence the products shown in the mage attached to this answer.
