Then draw the structural formulas for each molecule showing single and double bonds.
C₂H₅OH has one C-C bond, five C-H bonds, one C-O bond and one O-H bond add up the bond energies for a mole of each of these single bonds from your bond energy chart.
Now in O₂ we have O=O double bond so add up the total bond energy for three moles of O₂. (three times the doubly bonded oxygen bond energy)
CO₂ is formed O=C=O, and you have two moles of CO₂, and four moles of C=O bonds so multiply the C=O bond energy by 4
and for H₂O, bonded H-O-H you will have 6 moles of H-O bonds, so multiply the H-O bond energy by six.
Now to get ΔH for this reaction, simply subtract the sum of the reactant bond energies from the sum of the product bond energies and you have it.
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satria31x5
you can calculate it, by {(C-C) + 5(C-H) + (C-O) + (O-H) + 3(O=O)} - {(4(C=O) + 6(H-O)} = .... Kj/mc
Then draw the structural formulas for each molecule showing single and double bonds.
C₂H₅OH has one C-C bond, five C-H bonds, one C-O bond and one O-H bond
add up the bond energies for a mole of each of these single bonds from your bond energy chart.
Now in O₂ we have O=O double bond so add up the total bond energy for three moles of O₂. (three times the doubly bonded oxygen bond energy)
CO₂ is formed O=C=O, and you have two moles of CO₂, and four moles of C=O bonds so multiply the C=O bond energy by 4
and for H₂O, bonded H-O-H you will have 6 moles of H-O bonds, so multiply the H-O bond energy by six.
Now to get ΔH for this reaction, simply subtract the sum of the reactant bond energies from the sum of the product bond energies and you have it.