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Intros
Lessons
  1. Another method of calculating enthalpy: Bond enthalpy.
  2. Bond enthalpy: introduction.
  3. Using mean bond enthalpy to find enthalpy change.
  4. Example: Using bond enthalpy to find enthalpy change.
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Examples
Lessons
  1. Use the mean bond enthalpy values to find the enthalpy change in chemical reactions.
    The combustion of propane to make carbon dioxide and water vapour is described in this equation:

    C3H8 (g) + O2 (g)  \,   \, CO2 (g) + H2O (g)

    The table below shows relevant bond enthalpy data:

    Bond:

    Mean bond enthalpy1 (kJ mol-1)

    C-C

    348

    C-H

    412

    O=O

    497

    C=O

    743

    O-H

    463



    Balance the equation and find the enthalpy change of the reaction.
    1. Calculate the mean bond enthalpy from enthalpy of reaction data.
      The combustion reaction of butene is described below, with the enthalpy of reaction provided:

      C4H8 (g) + 6O2 (g)  \,   \, 4CO2 (g) + 4H2O \qquad △H \triangle H r = -2062 kJ mol-1

      Bond:

      Mean bond enthalpy1 (kJ mol-1)

      C-C

      348

      C-H

      412

      O=O

      497

      C=O

      x

      O-H

      463



      Use this information and the data to find the mean bond enthalpy of the C=O bond.
      Topic Notes
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      In this lesson, we will learn:

      • To understand what is meant by mean bond enthalpy.
      • To use mean bond enthalpy as a method to calculate the enthalpy change of a reaction.
      • To understand the limitation of using it to find reaction enthalpies precisely.

      Notes:

      • When a chemical reaction begins, what actually happens?
        Recall from Introduction to enthalpy or Enthalpy that in every chemical change there are two steps:
        • The bonds of the reactants are broken, which requires energy.
        • The bonds of the products are formed, which releases energy in the form of heat.

        In short, bond enthalpy finds the enthalpy change of a reaction by:

        • Finding the energy needed to break all the reactant bonds that need breaking. This is the endothermic part of the reaction.
        • Finding the energy released from all the product bonds that get made. This is the exothermic part of the reaction.

        ... and finding the difference between them! There are two possible outcomes of doing this:

        • If the energy needed to break bonds is greater than the energy released when forming them, it is an overall endothermic reaction.
        • If the energy released is greater than the energy needed, it is an overall exothermic reaction.

      • Mean bond enthalpy is the energy required to break one mole of a chemical bond in the gas phase, taken as an average across different compounds.
        • Because the definition is about breaking bonds, which requires energy, mean bond enthalpy is ALWAYS a positive value. Dont use negative values, you will get a wrong answer in the calculations!

        As well as the mean bond enthalpy, you will need to know (or be given) the structure of the reactants and products; you need to know what bonds are being broken and made!

        With mean bond enthalpy values, you can find the enthalpy change of a reaction with the following equation:

        △H \triangle H r = ∑\sum H(bonds broken) - ∑\sum H(bonds formed)

        You can also draw an enthalpy cycle, like in Calculating enthalpy: Hesss Law to find the total enthalpy change of a reaction, plotting a chemical detour to break reactant bonds and form product ones.
        Again, this works because energy isnt being lost or created, only transferred into or out of the substances in exchange with the surroundings!
        Like using Hesss law, this calculation gives you information based on the sign:
        • A negative enthalpy change shows an exothermic reaction, as the chemical substances gave out more energy than they absorbed.
        • A positive enthalpy change shows an endothermic reaction, as the chemical substances absorbed more energy than they gave out.


      • WORKED EXAMPLE:
        The reaction of ethene with hydrogen to produce ethane is described by the equation below:

        C2H4 (g) + H2 (g)  \,   \, C2H6 (g)

        The mean bond enthalpy data1 for the bonds in these compounds are as follows:
        C-C: 348 kj mol-1; C=C: 612 kJ mol-1; C-H: 412 kJ mol-1; H-H: 436 kJ mol-1;

        Use this data to find the enthalpy change of the reaction, △H \triangle Hr.


        We first need to look at the structure of the molecules and see what bonds are in the reactants and products. See the image below:

        Draw out the structure of the molecules involved so you can see what bonds are being broken and made. Sum the bond enthalpies in two categories: bonds broken and bonds made, remembering to multiply by the number of bonds there.

        Bond type:

        Bonds broken

        Bonds formed

        C-C

        0

        348 * 1

        C=C

        612 * 1

        0

        C-H

        412 * 4

        412 * 6

        H-H

        436 * 1

        0

        TOTAL:

        2696 kJ mol-1

        2820 kJ mol-1



        Then apply the equation:

        △H \triangle H r = ∑\sum H(bonds broken) - ∑\sum H(bonds formed)

        Which gives us the calculation:

        △H \triangle H r = 2696 - 2820 = -124 kJ mol-1

      • Using bond enthalpy to find the enthalpy change of reaction is not totally accurate. This is for a few reasons:
        • Mean bond enthalpy is a mean an average value as the bond is found in similar molecules containing the same bond. In any compound with more than one of this bond, an average of all of those are taken as they break individually.
          • For example, ammonia, (NH3), has three identical N-H bonds, but they wont have three identical bond enthalpies. When an N-H bond is broken from NH3 the molecule will be changed, and breaking N-H from NH3 costs a different amount of energy to breaking N-H from the remaining NH2 fragment.
          • In the same way, with a different electronic environment the N-H bonds in NH3 (ammonia) will have slightly different strength than the lone N-H bond in (CH3)2NH (dimethylamine).

        • When using mean bond enthalpies, reactants and products must be in the gas phase. Solids and liquids have significant intermolecular forces, and this complicates things when we try to measure the energy needed to break the bonds because we also have to overcome intermolecular forces. Dont use bond enthalpies alone for liquids or solids; we need another enthalpy term we will learn later. Even gases do have intermolecular forces we just assume they are negligible for bond enthalpy calculations.

        All of this means that using mean bond enthalpy is not completely accurate. It is an approximation of the enthalpy change.


        1 Source for mean bond enthalpy data: ATKINS, P. W., & DE PAULA, J. (2006). Atkins' Physical chemistry. Oxford, Oxford University Press.