1. Bond Enthalpies Section 5.4

2. Introduction

3. More Good Stuff For H 2 the thermochemical equation describing the bond enthalpy is: H 2(g) → 2H (g) ∆H θ = +436 kJ mol -1 The larger the bond enthalpy, the stronger the covalent bond Bond enthalpy is inversely proportional to bond length

4. Average Bond Enthalpies Many bond enthalpies are average bond enthalpies, so some error occurs with calculations (won't be the exact experimental value) See Table 10 in the data booklet for some average bond enthalpies

5. Use of Bond Enthalpies

6. Methane Combustion The reaction can be thought of in 2 steps 1. all the bonds of the reactants have to be broken (endothermic) 2. bond formation for the products (exothermic) Use the data booklet for the average bond enthalpies

8. Calculations for Bond Breaking Bond breaking: 4 C-H bonds in CH 4 = 4 x 412 = 1648 kJ 2 O=O bonds in 2O 2 = 2 x 496 = 992 kJ Total amount of energy to break all these bonds = 1648 + 992 = 2640 kJ

9. Calculations for Bond Formation Bond making: Making 2 C=O bonds in CO 2 = 2 x 743 = 1486 kJ Making 4 O-H bonds in 2H 2 O = 4 x 463 = 1852 kJ Total amount of energy released to the surroundings when these bonds are formed = 1486 + 1852 = 3338 kJ

10. Calculating the Enthalpy Change of the Reaction ∆H = Σ (energy required to break bonds) - Σ (energy released when bonds are formed) For the combustion of methane: ∆H = 2640 – 3338 = -698 kJ mol -1 Note that the overall reaction is exothermic

11. Example Problem Using the average bond enthalpies in the data booklet, calculate the enthalpy change for this reaction: H 2(g) + ½O 2(g) → H 2 O (g) Breaking bonds: 436 + 496/2 = 684 Forming bonds: 2 x 463 = 926 Breaking – forming = 684 – 926 Enthalpy change = -242 kJ mol -1

12. Example Problem # 2 Consider the formation of ammonia from nitrogen and hydrogen N 2(g) + 3H 2(g) → 2NH 3(g) Calculate the total enthalpy change using the bond enthalpies in the data booklet -76 kJ mol -1