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Hess's Law

The principle of conservation of energy

Skills to develop

Hess's Law

Germain Henri Hess (1802 - 1850) is important primarily for his thermochemical studies. Hess' Law states that the heat evolved or absorbed in a chemical process is the same whether the process takes place in one or in several steps. This is also known as the law of constant heat summation. To illustrate Hess's law, the thermal equations and the energy level diagrams are shown below.
Thermal equations Hess's law energy level diagram
  • A + B = AB, dH1
  • AB + B = AB2, dH2
    then,
    A + 2 B = AB2, dH1 2 = dH1 + dH2
  •     ======= A + 2 B
         |   | dH1
    dH1 2 | ===== AB + B
         |   | dH2
        ======= 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 dH, dH°, and dHf and their preceding equations.

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

    C(graphite) + 0.5 O2 -> CO,       dH° = -110 kJ/mol. We can write it in any of the following forms: 2 C(graphite) + O2 -> 2 CO,       dH° = -220 kJ/mol (multiplied by 2)
    6 C(graphite) + 3 O2 -> 6 CO,       dH° = -660 kJ/mol (multiplied by 6)

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

    CO -> C(graphite) + 0.5 O2,       dH° = 110 kJ/mol
    2 CO -> 2 C(graphite) + O2,       dH° = 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, but not on the intermediate steps. Energy (enthalpy) changes in chemical reactions are the same, regardless 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 CO and CO2. The direct oxidation of carbon (graphite) into CO2 yields an enthalpy of -393 kJ/mol. When carbon is oxidized into CO and then CO is oxidized to 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: C + 1/2 O2 -> CO,         dH° = -110 kJ/mol
    CO + 1/2 O2 -> CO2,         dH° = -283 kJ/mol.
    Adding the two equations together and cancel out the intermediate, CO, on both sides leads to C + O2 -> CO2,         dH° = (-110)+(-283) = -393 kJ/mol. The real merit is actually to evaluate the enthalpy of formation of 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 CO. However, enthalpy for the oxidation of graphite to CO2 can easily be measured. So can the enthalpy of oxidation of CO to CO2. The application of Hess's law enables us to estimate the enthalpy of formation of CO. Since,

    C + O2 -> CO2,         dH° = -393 kJ/mol
    CO + 1/2 O2 -> CO2,         dH° = -283 kJ/mol.
    Subtracting the second equation from the first gives C + 1/2 O2 -> CO,         dH° = -393 -(-283) = -110 kJ/mol
    The equation shows the standard enthalpy of formation of CO to be -110 kJ/mol.

    Application of Hess's law enables us to calculate dH, dH°, and dHf 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.

    Example 1

    The enthalpy of combustion for H2, C(graphite) and CH4 are -285.8, -393.5, and -890.4 kJ/mol respectively. Calculate the standard enthalpy of formation dHf for CH4.

    Solution:
    Lets interpret the information about enthalpy of formation by writing out the equations:
    dH°f
    /(kJ/mol)
    (1) H2(g) + 0.5 O2(g) -> H2O(l) -285.8
    (2) C(graphite) + O2(g) -> CO2(g) -293.5
    (3) CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l) -890.4
    From the above equations, derive
    C + 2H2 -> CH4
    Answer: C + 2H2 -> CH4 -74.7
    Hint: 2*(1) + (2) - (3), Thus,
    dHf = 2 * (-285.8) + (-393.5) - (-890.4) = ?

    Discussion:
    Three enthalpies of reactions involved in this Example are standard enthalpies of formation, and one is the enthalpy of combustion. The formation of methane from graphite and hydrogen cannot be achieved easily, and its enthalpy of formation is not directly measurable, but the calculations like this provide the data to be included in thermodynamic data. The value of -74.4 kJ/mol has been listed in several data sources.

    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)==========
    

    Example 2

    From the following data, CH4 + 2O2 -> CO2 + 2H2O         dHo = -890 kJ/mol
                    H2O(l) -> H2O(g)         dHo = 44 kJ/mol at 298 K
    Calculate the enthalpy of the reaction CH4 + 2 O2(g) -> CO2(g) + 2 H2O(g)         dHo = ?

    Solution:
    Add the two equations to give the third one:

    CH4(g) + 2O2(g) -> CO2(g) + 2 H2O(l)         dHo = -890 kJ/mol
                    2 H2O(l) -> 2 H2O(g)                     dHo = 88 kJ/mol
    add the equations ----------------------- add the enthalpies
    CH4 + 2 O2(l) -> CO2(g) + 2 H2O(g)           dHo = -802 kJ/mol

    Discussion:
    A higher amount of energy (890 vs 802 kJ/mol) is extracted if the exhaust is condensed to liquid water. The exhaust from high efficiency furnace is at lower temperature, and the water vapour is condensed to liquid. However, there is always some lost in a furnace operation.

    Example 3

    The standard enthalpies of formation of SO2 and SO3 are -297 and -396 kJ/mol respectively. Calculate the standard enthalpy of reaction for the reaction: SO2 + 1/2 O2 -> SO3.

    Solution:
    In order to show how the chemical reactions take place, and for a better appreciation of the technique of problem solving, we write the equations according to the data given:

    SO2(g) -> S(s) + O2(g)       dH = 297 kJ
    S(s) + 3/2 O2 -> SO3           dH = -396 kJ
    Add the two equations to give SO2(g) + 1/2 O2 -> SO3       dH = -99 kJ

    Your turn to work:
    Sketch an energy level diagram for the combinations of substances.

    Example 4

    From the following enthalpies of reactions:
    1. 2 O(g) -> O2(g)         dHo = -249 kJ/mol
    2. H2O(l) -> H2O(g)         dHo = 44 kJ/mol at 298 K
    3. 2 H(g) + O(g) -> H2O(g)         dHo = -803 kJ/mol
    4. C(graphite) + 2 O(g) -> CO2(g)         dHo = -643 kJ/mol
    5. C(graphite) + O2(g) -> CO2(g)         dHo = -394 kJ/mol
    6. C(graphite) + 2 H2(g) -> CH4(g)         dHo = -75 kJ/mol
    7. 2 H(g) -> H2(g)         dHo = -436 kJ/mol
    8. H2O(l) -> H2O(g)         dH = 41 kJ/mol at 373 K, non-standard condition
    Calculate the heat of combustion of methane into gaseous H2O.

    Solution:
    -2(1) + 2(3) + (4) - (6) - 2(7) gives

    CH4(g) + 2 O2(g) -> CO2(g) + H2O(g), and therefore,
    dH = -2*(-249) + 2*(-803) + (-643) - (-75) - 2(-436)
          = -804 kJ/mol

    Discussion:
    Work out the details yourself and check the result. The calculation is rather complicated. Reading it will not be able to master the technique. Data from equations 2, 5 and 8 are not required. Often, you have to select suitable data from a Table of Standard Enthalpy of Formation in problem solving.

    Compare to result of Example 2, this result differ slightly, due to a different set of data being used. The difference is 0.2 %.

    One method is to re-write the key equations as follows and then add them to cancel out undesirable compound on both sides. Practice the cancellation of the formula yourself.

    CH4(g) -> C(graphite) + 2 H2(g)         dHo = 75 kJ/mol
    C(graphite) + 2 O(g) -> CO2(g)         dHo = -643 kJ/mol
    2 O2(g) -> 4 O(g)                               dHo = 498 kJ/mol
    4 H(g) + 2 O(g) -> 2 H2O(g)             dHo = -1606 kJ/mol
    2 H2(g) -> 4 H(g)                               dHo = 872 kJ/mol
    add all equations ---------------------------add all dHs
    CH4 + 2 O2(g) -> CO2(g) + 2 H2O(g)       dHo = -804 kJ/mol

    Confidence Building Questions

    More Hess's law problems

    Problem Generator for the application of Hess's law.

    cchieh@sciborg.uwaterloo.ca