Molecular Structure (Cont.)

Bond Dissociation Energies (D)

Readings for this section.

Petrucci: Section 10-9

When bonds are formed energy is released. Electrons are in a more stable arrangement when in a bond (molecular orbital) than when they are unpaired (non-bonded) in atomic orbitals.

H2(g) $\rightarrow$ 2H (g,atom)                         ΔH  = 436.0 kJ/mol

This is a bond-dissociation enthalpy or, bond enthalpy, or just bond energy.

We would write it as D(H-H) = 436.0 kJ/mol

We can use the bond-energies to calculate (approximate) enthalpies of formation for any compound.

take, for example H(g,atom). ΔHf = 1/2 D(H-H).

1/2 H2(g) $\rightarrow$ H(g,atom)                     ΔHf = 218.0 kJ/mol


Consider the bonds in methane CH4. There are four C-H bonds.

CH4(g) $\rightarrow$ C(g,atom) + 4 H(g,atom)             ΔH  = 4 D(C-H)
ΔH  = ΔHf(C,g,atom) + 4 ΔHf(H,g,atom) - ΔHf(CH4,g)
  = 716.7 kJ/mol + 4(218.0 kJ/mol) - (-74.5 kJ/mol)
  = 1663 kJ/mol
D(C-H) = ΔH /4 = 1663 kJ/mol/4 = 415.8 kJ/mol

Now, let's consider the bonds in C2H6. There is one C-C bond and there are 6 C-H bonds.

C2H6(g) $\rightarrow$ 2 C(g,atom) + 6 H(g,atom)
ΔH  = 2 ΔHf(C,g,atom) + 6 ΔHf(H,g,atom) - ΔHf(C2H6,g)
  = 2(716.7 kJ/mol) + 6(218.0 kJ/mol) - (-84.7 kJ/mol)
  = 2826.1 kJ/mol

We assume D(C-H) = 415.8 kJ/mol (same as for CH4).

ΔH  = D(C-C) + 6 D(C-H)
D(C-C) = ΔH  - 6 D(C-H) = 2826.1 kJ/mol - 6 415.8 kJ/mol = 331.3 kJ/mol

Ideally, we could continue like this and build up a complete list of all possible bond energies. From there, we could calculate the exact energies (enthalpies) of every chemical reaction without ever doing a single experiment. (in some ways, this is like the goal of many theoretical chemists; to be able to calculate the energies of the molecules and reactions in chemistry without need for actually doing the chemistry).

Unfortunately, the C-H bond in one compound is never quite the same as it is in any other compound. Therefore, this technique can at best give us rough approximations of certain reaction enthalpies. We wouldn't want to push this idea very far as an analytical tool but it is very useful as a concept to understand reaction energies.


Bond energies from double and triple bonds

Consider the molecule C2H4 . There are four C-H bonds and one C=C bond.

We calculate the D(C=C) as follows:

C2H4(g) $\rightarrow$ 2C(g,atom) + 4 H(g,atom)
ΔH  = 2 ΔHf(C,g,atom) + 4 ΔHf(H,g,atom) - ΔHf(C2H4,g)
  = 2(716.7 kJ/mol) + 6(218.0 kJ/mol) - 52.3 kJ/mol)
  = 2253.1 kJ/mol
We assume D(C=H) = 415.8 kJ/mol (same as for CH4).
ΔH   = D(C=C) + 4 D(C-H)
D(C=C)  = ΔH  - 4 D(C-H) = 2253.1 kJ/mol - 4 415.8 kJ/mol = 589.9 kJ/mol

Note that this is greater than D(C-C) but not twice as great, i.e., a double bond is not twice as strong as a single bond.

Once we have a table of bond energies, we can calculate ΔH  for a reaction by counting bonds broken and bonds formed and taking the algebraic sum of their energies.  Remember bonds formed result in negative energy change for the system (exothermic) while bonds broken cause a positive energy change (endothermic).  This equation would be written as

\[\Delta H^o = \sum_i D(i) - \sum_j D(j)\]

where i is the index of all bonds broken and j is the index of all bonds formed.

A simpler way of writing this is:

\[\Delta H^o = \sum D(broken) - \sum D(formed)\]

Alternatively, we assume all reactants are completly dissoiates in to their atoms (all bonds are broken) and all products are formed from those elements.

\[\Delta H^o = \sum D(reactants) - \sum D(products)\]

This latter form might be confusing for some since normally, we think of products minus reactants and it's reversed here.

For example, estimate the enthalpy of combustion of ethane, given the bond energies below:

bond type Bond Energy / (kJ mol-1)
O-H 464
O=O 498.4
C=O 804
C-H 414
C-C 347

The balanced chemical reaction is:

C2H6 + 7/2 O2 $\Rightarrow$ 2 CO2 + 3 H2O

ΔH  =  $\sum $D(broken)- $\sum $D(formed) 

ΔH  = D(C-C) + 6 D(C-H) + 3.5 D(O=O) - 4 D(C=O) - 6 D(O-H)

ΔH  = 347 + 6 414 + 3.5 498.4 - 4 804 - 6 464

ΔH  = -1425 kJ/mol

Compare this value with that calculated using tabulated values of enthalpy of formation.

ΔH  = 2  ΔHf (CO2) + 3   ΔHf(H2O, g) -  ΔHf(C2H6) - 3.5   ΔHf(O2)

ΔH  = 2(-393.509) + 3(-241.818) - (-84.68) - 0

ΔH  = -1427.79 kJ/mol

 

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Prof. Michael J. Mombourquette.
Copyright © 1997
Revised: August 29, 2013.