# Covalent Bond Strengths

## Covalent Bond Strengths - Bond Enthalpies

In a chemical reaction we'll have bonds broken, bonds formed, and energy either absorbed or emitted by the reaction. Let's look closer at the enthalpy change associated with each individual bond broken or formed.

ProcessΔH (kJ/mole)
CH4(g) → CH3(g) + H(g) 435
CH3(g) → CH2(g) + H(g) 453
CH2(g) → CH(g) + H(g) 425
CH(g) → C(g) + H(g) 339
total = 1652

If we wanted to know what is the enthalpy change associated with breaking a C-H bond we find that it is slightly dependent on what molecule it is in. Thus, we take an average change in enthalpy when a C-H bond breaks as DC-H = 1652/4 kJ/mole = 413 kJ/mole.

Other average bond enthalpy changes in kJ/mole found this way are...

BondΔH (kJ/mole) Bond ΔH (kJ/mole)
H-H432 C-H413
H-F565 C-C347
H-Cl427 C-N305
C-O358

Notice how multiple bonds are shorter and require more energy to break than single bonds.

Bond ΔH (kJ/mole) Bond Length
C=C6141.37Å
C≡C839 1.20Å
C-N305 1.43Å
C=N6151.38Å
C≡N8911.16Å

We can use these average bond enthalpy changes to calculate the approximate enthalpy change for reactions. For example, to calculate the change in enthalpy for the following reaction:

H2(g) + F2(g) 2 HF(g)

we identify and count all the bonds that are broken (shown in red) and formed (shown in blue).

HH + FF 2 HF

Substituting the average bond enthalpies: DH-H = 432 kJ/mole, DF-F = 154 kJ/mole, DH-F = 565 kJ/mole in the expression

ΔH = Σ (ΔH of bonds broken) - Σ (ΔH of bonds formed)

we obtain

ΔH = [(1 mole) (432 kJ/mole) + (1 mole) (154 kJ/mole)] - [(2 moles) (565 kJ/mole)] = -544 kJ

• Bond Enthalpies:

#### Homework from Chemisty, The Central Science, 10th Ed.

8.65, 8.67, 8.69, 8.71