Question

Industrial ethanol (CH3CH2OH) is produced by a catalytic reaction of ethylene (CH2═CH2) with water at high pressures and temperatures. Calculate ΔH o rxn for this gas-phase hydration of ethylene to ethanol, using bond energies and then using enthalpies of formation.

Answer 1

using bond energies = - 37 kJ

using enthalpies of formation = - 45.7 kJ

Step-by-step explanation:

From the reaction;

CH2═CH2(g) + H2O(g) -------> CH3CH2OH(g)

From the above reaction; there are 4 (C-H) bonds , 1 ( C=C) and 2 (H-O) of ethene which forms 5(C-H) bonds, 1 (C-C) and 1 (C-O) and 1(H-O) bonds.

Using Bond Energies; the heat of the reaction can be written as:

`\delta H^0_ {rxn}=` ∑ energy of old bond breaking + ∑ energies of the new bond formation.

`\delta H^0_ {rxn}=` ∑
`[(4 * BE_(C-H)) + BE_(C-C)) + + BE_(O-H))]`+ ∑
`[(5 * BE_(C-H)) + BE_(C-C)) + BE_(O-H)+ BE_(C-O))]`

`\delta H^0_ {rxn}=` ∑
`[4*413 kJ)+(614kJ)+(2*647kJ]`+ ∑
`(5*-413kJ)+(-347kJ)+(-467kJ)+(-358kJ)]`

`\delta H^0_ {rxn}=` -37 kJ

To calculate the heats of reaction by using enthalpies of formation; we have:

`\delta H^0_ {rxn}=` ∑
`\delta H^0_ {products}` - ∑
`\delta H^0_ {reactants}`

`\delta H^0_ {rxn}=` ∑
`\delta H^0_ {CH_3CH_2OH}` - ∑
`\delta H^0_ {CH_2=CH_2+H_2O}`

`\delta H^0_ {rxn}=` (-235.1 kJ) - [(+52.47 kJ) + (-241.826 kJ)]

`\delta H^0_ {rxn}=` -45.7 kJ