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 º = 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). ΔHfº = 1/2 D(H-H).
1/2 H2(g) $\rightarrow$ H(g,atom) ΔHfº = 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 º = 4 D(C-H)
|ΔH º||= ΔHfº(C,g,atom) + 4 ΔHfº(H,g,atom) - ΔHfº(CH4,g)|
|= 716.7 kJ/mol + 4(218.0 kJ/mol) - (-74.5 kJ/mol)|
|= 1663 kJ/mol|
|D(C-H)||= ΔH º/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 º||= 2 ΔHfº(C,g,atom) + 6 ΔHfº(H,g,atom) - ΔHfº(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 º||= D(C-C) + 6 D(C-H)|
|D(C-C)||= ΔH º - 6 D(C-H) = 2826.1 kJ/mol - 6 × 415.8 kJ/mol = 331.3 kJ/mol|
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.
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 º||= 2 ΔHfº(C,g,atom) + 4 ΔHfº(H,g,atom) - ΔHfº(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 º||= D(C=C) + 4 D(C-H)|
|D(C=C)||= ΔH º - 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 º 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)|
The balanced chemical reaction is:
C2H6 + 7/2 O2 $\Rightarrow$ 2 CO2 + 3 H2O
ΔH º = $\sum $D(broken)- $\sum $D(formed)
ΔH º = D(C-C) + 6 × D(C-H) + 3.5 × D(O=O) - 4 × D(C=O) - 6 × D(O-H)
ΔH º = 347 + 6 × 414 + 3.5 × 498.4 - 4 × 804 - 6 × 464
ΔH º = -1425 kJ/mol
Compare this value with that calculated using tabulated values of enthalpy of formation.
ΔH º = 2 × ΔHfº (CO2) + 3 × ΔHfº(H2O, g) - ΔHfº(C2H6) - 3.5 × ΔHfº(O2)
ΔH º = 2(-393.509) + 3(-241.818) - (-84.68) - 0
ΔH º = -1427.79 kJ/mol