Chapter 6.8 - Application of Bond Enthalpy

In the previous section, we saw Hess's law and bond enthalpy. In this section, we will see some applications of bond enthalpy.

• The bond enthalpy values given in the data book can be used to find ΔH^⊖_r
• This can be explained in 11 steps using an example
1. Consider the following reaction:
CO(g) + H₂O(g) → CO₂(g) + H₂(g)
• Steam is made to react with carbon monoxide.
2. Let us see the making and breaking of bonds in this reaction:
• When the reaction proceeds:
    ♦ Bonds in the reactant side are broken.
    ♦ As a result, the reactant molecules separate into individual atoms.
    ♦ Those individual atoms rearrange and make new bonds.
    ♦ The new bonds result in product molecules.
3. We want answers to these questions:
• Which all bonds are broken in the reactant side?
    ♦ How many bonds of each type are broken?
• Which all bonds are made in the product side?
    ♦ How many bonds of each type are made?
4. It will be easier to find the answers to the above questions, if we  rewrite the equation in (1) using structural formulae. This is shown below:
C≡O + H-O-H → O=C=O + H-H
5. It is clear that:
• On the reactant side:
    ♦ One C≡O bond is broken.
    ♦ Two O-H bonds are broken.
• On the product side:
    ♦ Two C=O bonds are made.
    ♦ One H-H bond is made.
6. We know the following two facts:
(i) When bonds are broken, energy is absorbed.
    ♦ So, we need to supply energy to break the bonds in the reactant side.
(ii) When bonds are made, energy is released.
    ♦ So we will receive energy when bonds are made in the product side.
7. Let us calculate each of the above two items in our present case.
• From the data book, we have:
    ♦ C≡O has a bond enthalpy of 1072 kJ mol^-1
    ♦ O-H has a bond enthalpy of 463 kJ mol^-1
    ♦ C=O has a bond enthalpy of 799 kJ mol^-1
    ♦ H-H has a bond enthalpy of 436 kJ mol^-1
• So we get:
(i) Total energy required for breaking all the bonds in the reactant side
= (1 × 1072) + (2 × 463) = 1998
(ii) Total energy released when making all the bonds in the product side
= (2 × 799) + (1 × 436) = 2034
8. We see that:
   ♦ Energy required to break the bonds in the reactant side
   ♦ is lesser than
   ♦ Energy released when bonds are made in the product side.
• So we receive a net energy of (2034-1998) = 36 kJ mol^-1
• Since energy is released, it is an exothermic reaction. The enthalpy value should be given a negative sign.
9. Now we can write the thermochemical equation:
CO(g) + H₂O(g) → CO₂(g) + H₂(g); ΔH^⊖_r = -36.0 kJ mol^-1
10. Based on the above discussion, we can write a general equation to find ΔH^⊖_r
Eq.6.10: $\mathbf\small{\rm{\Delta {H^{\ominus }}_r=\sum{Bond\;Enthalpies_{reactants}}-\sum{Bond\;Enthalpies_{products}}  }}$
11. Consider the two terms on the right side of Eq.6.10
• If the first term is larger, it means that, the bonds in the reactant side have greater energy.
   ♦ We will have to supply a net energy to break the bonds.
   ♦ ΔH^⊖_r will have a positive sign.
   ♦ The reaction will be endothermic.
• If the second term is larger, it means that, the bonds in the product side have greater energy.
   ♦ We will receive a net energy.
   ♦ ΔH^⊖_r will have a negative sign.
   ♦ The reaction will be exothermic.
• In our present case, the second term is larger.
• We see that, the Eq.6.10 automatically gives us the appropriate sign.

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Next we will see a simple case. It is called 'simple case' because, only a very few bonds are broken and made. It can be written in steps:
1. Consider the reaction shown in fig.6.11 below:

• It is the reaction between propene and hydrogen to give propane.
2. Let us see the making and breaking of bonds in this reaction:
• When the reaction proceeds:
    ♦ Two bonds in the reactant side are broken.
          ✰ They are marked with red color in fig.6.12 below.
    ♦ Three bonds are newly formed in the product side.
          ✰ They are marked with green color in the fig.6.12

 

Fig.6.12

3. Since only a very few bonds are broken and made, we can easily answer our earlier questions:
• Which all bonds are broken in the reactant side?
    ♦ How many bonds of each type are broken?
Answer:
One C=C bond and one H-H bond are broken.
• Which all bonds are made in the product side?
    ♦ How many bonds of each type are made?
Answer:
One C-C bond and two C-H bonds are made.
4. We know the following two facts:
(i) When bonds are broken, energy is absorbed.
    ♦ So, we need to supply energy to break the bonds in the reactant side.
(ii) When bonds are made, energy is released.
    ♦ So we will receive energy when bonds are made in the product side.
7. Let us calculate each of the above two items in our present case.
• From the data book, we have:
    ♦ C=C has a bond enthalpy of 614 kJ mol^-1
    ♦ H-H has a bond enthalpy of 436 kJ mol^-1
    ♦ C-C has a bond enthalpy of 348 kJ mol^-1
    ♦ C-H has a bond enthalpy of 413 kJ mol^-1
• So we get:
(i) Total energy required for breaking bonds in the reactant side
= (1 × 614) + (1 × 436) = 1050
(ii) Total energy released when making all the bonds in the product side
= (1 × 348) + (2 × 413) = 1174
8. We see that:
   ♦ Energy required to break the bonds in the reactant side
   ♦ is lesser than
   ♦ Energy released when bonds are made in the product side.
• So we receive a net energy of (1174-1050) = 124 kJ mol^-1
• Since energy is released, it is an exothermic reaction. The enthalpy value should be given a negative sign.
9. Now we can write the thermochemical equation:
C3H6(g) + H₂(g) → C3H₈(g); ΔH^⊖_r = -124.0 kJ mol^-1
10. Based on the above discussion, we can write a general equation to find ΔH^⊖_r for simple cases
Eq.6.11:
$\mathbf\small{\rm{\Delta {H^{\ominus }}_r=\sum{Broken \; Bond\;Enthalpies}-\sum{Made \; Bond\;Enthalpies}}}$
11. Consider the two terms on the right side of Eq.6.11.
• If the first term is larger, it means that, the 'total energy required to break bonds' is greater.
   ♦ We will have to supply a net energy to break the bonds.
   ♦ ΔH^⊖_r will have a positive sign.
   ♦ The reaction will be endothermic.
• If the second term is larger, it means that, the 'total energy released when bonds are made' is greater.
   ♦ We will receive a net energy.
   ♦ ΔH^⊖_r will have a negative sign.
   ♦ The reaction will be exothermic.
• In our present case, the second term is larger.
• We see that, the Eq.6.11 automatically gives us the appropriate sign.

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• Now we have a basic idea about the application of bond enthalpy
• In the next section, we will see lattice enthalpy

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