7.1 Bond Energy ad Enthalpy of Reaction

It should be obvious that covalent bonds are stable because molecules exist. However, the bonds can be broken if enough energy is supplied to a molecule. To break most covalent bonds between any two given atoms, a certain amount of energy must be supplied. Although the exact amount of energy depends on the molecule, the approximate amount of energy to be supplied is similar if the atoms in the bond are the same. The approximate amount of energy needed to break a covalent bond is called the bond energy of the covalent bond. Table 7.1 “Bond Energies of Covalent Bonds” lists the bond energies of some covalent bonds.

Table 7.1 Bond Energies of Covalent Bonds
Bond Energy (kJ/mol)
C–C 348
C=C 611
C≡C 837
C–O 351
C=O 799
C–Cl 328
C–H 414
F–F 159
H–Cl 431
H–F 569
H–H 436
N–N 163
N=N 418
N≡N 946
N–H 389
O–O 146
O=O 498
O–H 463
S–H 339
S=O 523
Si–H 293
Si–O 368

A few trends are obvious from Table 7.1. For bonds that involve the same two elements, a double bond is stronger than a single bond, and a triple bond is stronger than a double bond. The energies of multiple bonds are not exact multiples of the single-bond energy; for carbon-carbon bonds, the energy increases somewhat less than double or triple the C–C bond energy, while for nitrogen-nitrogen bonds the bond energy increases at a rate greater than the multiple of the N–N single bond energy. The bond energies in Table 7.1 are average values; the exact value of the covalent bond energy will vary slightly among molecules with these bonds but should be close to these values.

To be broken, covalent bonds always require energy; that is, covalent-bond breaking is always an endothermic process. Thus the ΔH for this process is positive:

[latex]\text{molecule }\ce{OH}\rightarrow \text{molecule O}+\ce{H}\hspace{5mm}\Delta H\approx +463\text{ kJ/mol}[/latex]

However, when making a covalent bond, energy is always given off; covalent-bond making is always an exothermic process. Thus ΔH for this process is negative:

[latex]\text{molecule S}+\ce{H}\rightarrow \text{molecule }\ce{SH}\hspace{5mm}\Delta H\approx -339\text{ kJ/mol}[/latex]

Bond energies can be used to estimate the energy change of a chemical reaction. When bonds are broken in the reactants, the energy change for this process is endothermic. When bonds are formed in the products, the energy change for this process is exothermic. We combine the positive energy change with the negative energy change to estimate the overall energy change of the reaction. For example, in

2H2 + O2 → 2H2O

We can draw Lewis electron dot diagrams for each substance to see what bonds are broken and what bonds are formed:

\begin{matrix} \chemfig{H-H}&&&&\chemfig{H-\Lewis{2:6:,O}-H} \\ &+&\chemfig{\Lewis{2:6:,O}=\Lewis{2:6:,O}}&\rightarrow& \\ \chemfig{H-H}&&&&\chemfig{H-\Lewis{2:6:,O}-H} \\ \end{matrix}

(The lone electron pairs on the O atoms are omitted for clarity.) We are breaking two H–H bonds and one O–O double bond and forming four O–H single bonds. The energy required for breaking the bonds is as follows:

\begin{array}{rl} 2 \ \chemfig{H-H}\text{ bonds:}&2\times(+436\text{ kJ/mol}) \\ 1 \ \chemfig{O=O}\text{ bond:}&\phantom{2\times(}+498\text{ kJ/mol} \\ \hline \text{Total:}&\phantom{2\times(}+1,370\text{ kJ/mol} \end{array}

The energy given off during the formation of the four O–H bonds is as follows:

\begin{array}{rl} 4 \ \chemfig{O-H}\text{ bonds:}&4\times (-463\text{ kJ/mol}) \\ \hline \text{Total:}&-1,852\text{ kJ/mol} \end{array}

Combining these two numbers:

\begin{array}{rl} &+1,370\text{ kJ/mol} \\ +&(-1,852\text{ kJ/mol}) \\ \hline \text{Net Change:}&-482\text{ kJ/ml}\approx \Delta H \end{array}

The actual ΔH is −572 kJ/mol; we are off by about 16% — although not ideal, a 16% difference is reasonable because we used estimated, not exact, bond energies.

Example 7.1

Estimate the energy change of this reaction:

\begin{equation*} \chemfig{C(-[:-135]H)(-[:135]H)=C(-[:-45]H)(-[:45]H)}+\chemfig{H-H}\longrightarrow \chemfig{H-C(-[:90]H)(-[:-90]H)-C(-[:90]H)(-[:-90]H)-H} \end{equation*}

Solution

Here, we are breaking a C–C double bond and an H–H single bond and making a C–C single bond and two C–H single bonds. Bond breaking is endothermic, while bond making is exothermic. For the bond breaking:

\begin{array}{rl} 1 \ \chemfig{C=C}:&+611\text{ kJ/mol} \\ 1 \ \chemfig{H-H}:&+436\text{ kJ/mol} \\ \hline \text{Total:}&+1,047\text{ kJ/mol} \end{array}

For the bond making:

\begin{array}{rl} 1 \ \chemfig{C-C}:&\phantom{2\times}-348\text{ kJ/mol} \\ 2 \ \chemfig{C-H}:&2\times(-414\text{ kJ/mol}) \\ \hline \text{Total: }&-1,176\text{ kJ/mol} \end{array}

Overall, the energy change is +1,047 + (−1,176) = −129 kJ/mol.

Test Yourself

Estimate the energy change of this reaction:

\begin{equation*} \chemfig{H-C~C-H}+\chemfig{2H-H}\longrightarrow \chemfig{H-C(-[:90]H)(-[:-90]H)-C(-[:90]H)(-[:-90]H)-H} \end{equation*}

Answer

−295 kJ/mol

Key Takeaways

  • Covalent bonds can be broken if energy is added to a molecule.
  • The formation of covalent bonds is accompanied by energy given off.
  • Covalent bond energies can be used to estimate the enthalpy changes of chemical reactions.

Exercises

Questions

  1. Estimate the enthalpy change for this reaction. Start by drawing the Lewis electron dot diagrams for each substance.

    N2 + 3H2 → 2NH3

  2. Estimate the enthalpy change for this reaction. Start by drawing the Lewis electron dot diagrams for each substance.

    HN=NH + 2H2 → 2NH3

  3. Estimate the enthalpy change for this reaction. Start by drawing the Lewis electron dot diagrams for each substance.

    CH4 + 2O2 → CO2 + 2H2O

  4. Estimate the enthalpy change for this reaction. Start by drawing the Lewis electron dot diagrams for each substance.

    4NH3 + 3O2 → 2N2 + 6H2O

Answers

1.  −80 kJ

3.  −798 kJ

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