Using the bond energies provided, calculate the enthalpy of the reaction (∆Hrxn, in kJ) for the hydrogenation of propene to propane, shown below?
CH₂CHCH₃(g) + H₂(g) → CH₃CH₂CH₃(g)

Answers

Answer 1

The enthalpy of reaction for the hydrogenation of propene to propane is -926 kJ/mol.

The bond energy of a molecule is obtained as the sum of the bond energies of all the bonds between atoms in the molecule. The heat of reaction can be obtained using the relation;

∆Hrxn = ∑Bonds being broken - ∑Bonds being formed

Hence, we have;

Where all the energies are quoted in KJ/mol

C=C = 598

C - C = 346

H - H = 436

C - H = 416

For the reaction;

CH₂CHCH₃(g) + H₂(g) → CH₃CH₂CH₃(g)

∆Hrxn = ∑[6(C - H) + 1(C=C) + 1(C - C)] -  ∑[3(C - C) + 8(C - H)

∆Hrxn = [6(416) + 1(598) + 1(346)] - [3(346) + 8(416)]

∆Hrxn =3440 - 4366 = -926 kJ/mol

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