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:
Hope this helps :D
Explanation:
Metal: Aluminum and Copper
Non-Metal: Hydrogen and Flourine
Acid: Sulfuric Acid and Phosphoric Acid
Alkali Metals: Hydrogen and Lithium
Compounds: Water and Carbon Dioxide
Elements: Carbon and Oxygen
Answer:
Explanation:
b is the most stable( noble gas ) since it has an octet valance shell and can't loses or gains any more of electrons
A. →
B. →
C. →
What is a balanced chemical equation?
An equation that has an equal number of atoms of each element on both sides of the equation is called a balanced chemical equation.
A. →
B. →
C. →
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Answer:
C
Explanation:
it belong to that group as it needs 1 electron to be chemically stable