Bond Enthalpies

Section 5.4

Introduction

Average bond enthalpy: enthalpy change per mole when 1 mole of the same type of covalent bond is broken in the gas phase from many similar molecules

Bond breaking is always endothermic Bond formation is always exothermic

More Good Stuff For H

2 the thermochemical equation

describing the bond enthalpy is: H

2(g) → 2H

(g) ∆Hθ = +436 kJ mol-1

The larger the bond enthalpy, the stronger the covalent bond

Bond enthalpy is inversely proportional to bond length

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

Use of Bond Enthalpies

Are used to determine the enthalpy change for a reaction

The molecules need to be in the gaseous state

Consider the combustion of methane CH

4(g) + 2O

2(g) → CO

2(g) + 2H

2O

(g)

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

Calculations for Bond Breaking

Bond breaking: 4 C-H bonds in CH

4 = 4 x 412 = 1648 kJ

2 O=O bonds in 2O2 = 2 x 496 = 992 kJ

Total amount of energy to break all these bonds = 1648 + 992 = 2640 kJ

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

2O = 4 x 463 =

1852 kJ Total amount of energy released to the

surroundings when these bonds are formed = 1486 + 1852 = 3338 kJ

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

Example Problem Using the average bond enthalpies in the

data booklet, calculate the enthalpy change for this reaction:

H2(g)

+ ½O2(g)

→ H2O

(g)

Breaking bonds: 436 + 496/2 = 684 Forming bonds: 2 x 463 = 926 Breaking – forming = 684 – 926 Enthalpy change = -242 kJ mol-1

Example Problem # 2

Consider the formation of ammonia from nitrogen and hydrogen

N2(g)

+ 3H2(g)

→ 2NH3(g)

Calculate the total enthalpy change using the bond enthalpies in the data booklet

-76 kJ mol-1