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.
|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...
|Bond||ΔH (kJ/mole)||Bond||ΔH (kJ/mole)|
Notice how multiple bonds are shorter and require more energy to break than single bonds.
|Bond||ΔH (kJ/mole)||Bond Length|
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).
H—H + F—F → 2 H—F
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)
ΔH = [(1 mole) (432 kJ/mole) + (1 mole) (154 kJ/mole)] - [(2 moles) (565 kJ/mole)] = -544 kJ
Chemisty, The Central Science, 10th Ed.
8.65, 8.67, 8.69, 8.71
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