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Hess's Law: The Principle of Conservation of Energy

  • Page ID
    32650
  • . To illustrate Hess's law, the thermal equations and the energy level diagrams are shown below.

    Thermal equations Hess's law energy level diagram
    • \(\mathrm{A + B = AB},\: \Delta H_1\)
    • \(\mathrm{AB + B = AB_2},\: \Delta H_2\)
      then,
      \(\mathrm{A + 2 B = AB_2},\: \Delta H_{1\: 2} = \Delta H_1 + \Delta H_2\)
        ======= A + 2 B
         |   | \Delta H1
    \Delta H1 2 | ===== AB + B
         |   | \Delta H2
        ======= AB2
    

    Chemical energy and Hess's law

    The standard enthalpy of reaction and standard enthalpy of formation introduced in Chemical Energy are very useful chemical properties. We have already mentioned some basic rules regarding the quantities \Delta H, \Delta H°, and \Delta Hf and their preceding equations.

    If both sides of the equations are multiplied by a factor to alter the number of moles, \Delta H, \Delta H°, or \Delta Hf for the equation should be multiplied by the same factor, since they are quantities per equation as written. Thus, for the equation

    \(\mathrm{C_{\large{(graphite)}} + 0.5\, O_2 \rightarrow CO, \hspace{20px} \mathit{\Delta H}^\circ = -110\: kJ/mol}\).

    We can write it in any of the following forms:

    \(\mathrm{
    2 C_{\large{(graphite)}} + O_2 \rightarrow 2 CO, \hspace{20px} \mathit{\Delta H}^\circ = -220\: kJ/mol\: (multiplied\: by\: 2) \\
    6 C_{\large{(graphite)}} + 3 O_2 \rightarrow 6 CO, \hspace{20px} \mathit{\Delta H}^\circ = -660\: kJ/mol\: (multiplied\: by\: 6)}\)

    For the reverse reaction, the signs of these quantities are changed (multiply by -1). The equation implies the following:

    \(\mathrm{
    CO \rightarrow C_{\large{(graphite)}} + 0.5 O_2, \hspace{20px} \mathit{\Delta H}^\circ = 110\: kJ/mol\\
    2 CO \rightarrow 2 C_{\large{(graphite)}} + O_2, \hspace{20px} \mathit{\Delta H}^\circ = 220\: kJ/mol}\)

    Hess's law states that energy changes are state functions. The amount of energy depends only on the states of the reactants and the state of the products, not on the intermediate steps. Energy (enthalpy) changes in chemical reactions are the same, regardless of whether the reactions occur in one or several steps. The total energy change in a chemical reaction is the sum of the energy changes in its many steps leading to the overall reaction.

    For example, in the diagram below, we look at the oxidation of carbon into \(\ce{CO}\) and \(\ce{CO2}\). The direct oxidation of carbon (graphite) into \(\ce{CO2}\) yields an enthalpy of -393 kJ/mol. When carbon is oxidized into \(\ce{CO}\) and then \(\ce{CO}\) is oxidized to \(\ce{CO2}\), the enthalpies are -110 and -283 kJ/mol respectively. The sum of enthalpy in the two steps is exactly -393 kJ/mol, same as the one-step reaction.

    
           0 kJ ------------ C(graphite) + O2
                  |       |
          -110 kJ |       |
                  V       |
     CO + 0.5 O2 -----    |
                  |       | -393 kJ
                  |       |
          -283 kJ |       |
                  |       |
                  V       V
                ------------ CO2
    

    The two-step reactions are:

    \(\mathrm{
    C + \dfrac{1}{2} O_2 \rightarrow CO, \hspace{20px} \mathit{\Delta H}^\circ = -110\: kJ/mol\\
    CO + \dfrac{1}{2} O_2 \rightarrow CO_2, \hspace{20px} \mathit{\Delta H}^\circ = -283\: kJ/mol}\)

    Adding the two equations together and canceling out the intermediate, \(\ce{CO}\), on both sides leads to

    \(\mathrm{C + O_2 \rightarrow CO_2, \hspace{20px} \mathit{\Delta H}^\circ = (-110)+(-283) = -393\: kJ/mol}\)

    The real merit is actually to evaluate the enthalpy of formation of \(\ce{CO}\) as we shall see soon.

    Application of Hess's Law

    Hess's law can be applied to calculate enthalpies of reactions that are difficult to measure. In the above example, it is very difficult to control the oxidation of graphite to give pure \(\ce{CO}\). However, enthalpy for the oxidation of graphite to \(\ce{CO2}\) can easily be measured. So can the enthalpy of oxidation of \(\ce{CO}\) to \(\ce{CO2}\). The application of Hess's law enables us to estimate the enthalpy of formation of \(\ce{CO}\). Since,

    \(\mathrm{
    C + O_2 \rightarrow CO_2, \hspace{20px} \mathit{\Delta H}^\circ = -393\: kJ/mol\\
    CO + \dfrac{1}{2} O_2 \rightarrow CO_2, \hspace{20px} \mathit{\Delta H}^\circ = -283\: kJ/mol}\)

    Subtracting the second equation from the first gives

    \(\mathrm{C + \dfrac{1}{2} O_2 \rightarrow CO, \hspace{20px} \mathit{\Delta H}^\circ = -393 -(-283) = -110\: kJ/mol}\)

    The equation shows the standard enthalpy of formation of \(\ce{CO}\) to be -110 kJ/mol.

    Application of Hess's law enables us to calculate \Delta H, \Delta H°, and \Delta Hf for chemical reactions that impossible to measure, providing that we have all the data of related reactions.

    Some more examples are given below to illustrate the applications of Hess Law.

    From these data, we can construct an energy level diagram for these chemical combinations as follows:

       ===C(graphite) + 2 H2(g) + 2 O2(g)===
    - 74.7 kJ |              |
       == CH4 (g) + 2 O2(g)== |
              |              |
              |              |
              |              |
              |              |-965.1 kJ
    -890.4 kJ |              | [(-2*285.8-393.5) kJ]
              |              |
              |              |
              |              |
              |              |
              V              V
       ==========CO2(g) + 2 H2O(l)==========
    

    \mathrm{CH_4 + 2 O_{2\large{(g)}} \rightarrow CO_{2\large{(g)}} + 2 H_2O_{\large{(g)}}} \hspace{35px} \Delta H^\circ = \mathrm{-804\: kJ/mol}\hspace{5px}\)

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