Heats of reaction calculated from bond energies and from heats of formation are

Question

Heats of reaction calculated from bond energies and from heats of formation are often, but not always, close to each other. (a) Industrial ethanol (CH_3CH_2OH) is produced by a catalytic reaction of ethylene (CH_2=CH_2) with water at high pressures and temperatures. Calculate DeltaH^0 for this gas-phase hydration of ethylene to ethanol, using bond energies and then using heats of formation. Using bond energies: Using heats of formation: 11267 1225 (b) Ethylene glycol is produced by the catalytic oxidation of ethylene to ethylene oxide, which then reacts with water to form ethylene glycol: The DeltaH^0 for this hydrolysis step, based on heats of formation, is -97 kJ/mol. Calculate DeltaH^0 for rxn the hydrolysis using bond energies.

Explanation / Answer

1) the reaction will be

CH2=CH2 + H2O --> C2H5OH

DeltaH of reaction = Sum of B.E of reactants - Sum of B.E of products

DeltaH of reaction = Sum of B.E. (C=C + 4XC-H + 2XO-H) - B.E. (C-C + 6XC-H + C-O + O-H)

Delta H of reacion = (724 + 4X414 + 2X465) - (368 + 6X414 + 352 + 465)

Delta H of reaction = 3310 - 3669 = -359 KJ / mole

Now from Delta H of formation we may calcualte as

Delta H of reaction = Delta H of formation of products - Delta H of formation of reactants

Delta H of reaction = Delta H formation (C2H5OH) - Delta H of formation (C2H4 + H2O)

Delta H of reaction = 277.0 - ( 52.3 + (-241.8)) = -466.5 KJ / mole

b)

Enthalpy of reaction = B.E of reactants - B.E. of products

Enthalpy of reaction = B.E (4XC-H + C-C + 2X C-O + 2O-H) - ( C-C + 4C-H + 2C C-O + 2XO-H)

Enthalpy of reaction = zero (so theoretically it will be zero
)

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