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Enthalpies of reaction calculated from bond energies and from enthalpies of formation are often, but not always, close to each other.
(a) Industrial ethanol (CH₃CH₂OH) is produced by a catalytic reaction of ethylene (CH₂ = CH₂) with water at high pressures and temperatures. Calculate ΔH° for this gas-phase hydration of ethylene to ethanol, using bond energies and then using enthalpies of formation.

Respuesta :

∆H° of the following reaction H₂O(g) + CH₂ = CH₂→ CH₃CH₂OH is -464kJ/mol.

What is Bond Enthalpy?

The minimum amount of energy which is required to braak down or form the bonds in chemical reaction is known as bond enthalpy.

It can be calculated as:

∆Hrxn = sum of ∆H bond broken - sum. of ∆H of bond formed.

In order to Calculate ∆Hrxn for the given equation we have:

Bond energies in kJ/mol

H—O = 463

C=C = 145

C—O = 1072kJ/mol.

Now, the given reaction is

H₂(g) + I₂(g) → 2HI(g)

Here, 1 mol of H₂ and 1 mole of I₂ breaks to form 2 moles of HI.

Therefore,

We know that,

∆Hrxn = B. E(O—H) + B. E(C=C) - B. E(C—O)

= 463 + 145 - 1072

= 436+ 151 - 590

∆Hrxn = -464kJ/mol.

Thus, from the above conclusion we can say that ∆Hrxn of the reaction H₂O(g) + CH₂ = CH₂→ CH₃CH₂OH is -464kJ/mol.

learn more about Bond energy:

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