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
Learn more: https://brainly.com/question/13440572
