Question

Acetone can be easily converted to isopropyl alcohol by addition of hydrogen to the carbon–oxygen double bond. calculate the enthalpy of reaction using the bond energies given. bond: c=o h–h c–h o–h c–c c–o bond energy (kj/mol): 745 432 413 467 347 358 multiple choice –61 kj +61 kj –366 kj –484 kj +366 kj

Answer 1

Answer:

-484kJ

Explanation:

Carry out bond energy calculation

[6(C-H) + (C=O) +2(C-C)] - [(O-H) +7(C-H)+ (C-O) +2(C-C)]

Substitute the bond energy values = -484kJ

Answer 2

The enthalpy of the reaction using the bond energies is -61 KJ

The enthalpy of a reaction is given by

Enthalpy of reaction = Sum of the bond energies of bonds being broken - Sum of the bond energies of the bonds being formed

That is,

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

NOTE: The equation for the reaction is shown in the attachment below

On the reactant side, we have acetone and hydrogen molecule

The bonds on the reactant side are

6 C-H

2 C-C

1 C=O

1 H-H

∴ Sum of bond energies (B.E) on the reactant side = 6(413) + 2(347) + 745 + 432

Sum of bond energies (B.E) on the reactants side = 4349 KJ/mol

On the product side, we have isopropyl alcohol

The bonds on the product side are

7 C-H

1 C-O

1 O-H

2 C-C

∴ Sum of bond energies (B.E) on the product side = 7(413) + 358 + 467 + 2(347)

Sum of bond energies (B.E) on the product side = 4410 KJ/mol

Now,

Enthalpy of the reaction = 4349 - 4410

Enthalpy of the reaction = -61 KJ

Hence, the enthalpy of the reaction using the bond energies is -61 KJ

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