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Introduction

The same reactants can be turned into the same products by more than one route. Consider the following reaction:

C + O2 → CO2

This reaction can take place in a single step, and the energy given out can be measured.
C + O2 → CO2 ΔH = −393 kJmol​​-1

The reaction can also be made to take place in two steps, and the energy given out in each step can be measured.

Step 1 C + ½ O2 → CO ΔH = −110 kJmol​​-1

Step 2 CO + ½ O2 → CO2 ΔH = −283 kJmol​​-1

These two processes, taken together, amount to the same as the first equation. We can add the two equations together, cancelling out things that are on both sides of the equation.


Equation

The COs cancel out, because they appear on both sides of the equation. The ΔH values, added together, give −393 kJmol-1, the overall energy change for the first reaction.

This example illustrates Hess's Law. Hess's Law states that the overall energy change for a reaction is independent of the route taken. No matter how we turn a set of reactants into a set of products, the overall energy change is always the same.

Hess's Law in practice
If we add equations representing chemical reactions, we get an equation that represents a new reaction. If we add the ΔH values in the same way, we get ΔH for the new reaction. If we multiply an equation, the value of ΔH has to be multiplied in the same way. If we reverse an equation, the value of ΔH has to be reversed as well - that is, its sign must be reversed. In this activity, we will look at the energy relationships in reactions involving magnesium.

Task 1: Hess's Law


Model 1

  1. Open Yenka file Model 1. Add the magnesium to the evaporating basin on the tripod and turn on the Bunsen burner using the control on its right. When the magnesium ignites, look at the equation and the enthalpy of reaction in the basin's reaction details. On your paper, write the equation for the combustion of 1 mole of magnesium and the corresponding enthalpy of combustion. (Note that the value for the enthalpy of combustion changes by a small amount during the reaction. An approximate answer is acceptable.)
    Answer
    Mg + ½ O2 → MgO
    ΔH = −604 kJmol​​-1
  2. Reload the model using the F5 key. Repeat the procedure of Q1 using the jar of copper. On your paper, write the equation for the combustion of 1 mole of copper, and the corresponding enthalpy of combustion. (Note that the value for the enthalpy of combustion changes by a small amount during the reaction. An approximate answer is acceptable.)
    Answer
    Cu + ½ O2 → CuO
    ΔH = −155 kJmol​​-1
  3. You will note that the enthalpy of combustion of magnesium is greater than that of copper. Suggest a reason for this difference.
    Answer
    Magnesium is a more reactive metal than copper. It is higher in the Electrochemical Series.
  4. Now write the equation for the reaction of 1 mole of magnesium with copper oxide to give copper and magnesium oxide.
    Answer
    Mg + CuO → MgO + Cu
  5. Combine the two combustion equations in the answers to Q1 and Q2 to arrive at the same equation, and the ΔH for the reaction.
    Answer
    To do this, you must reverse the equation in A2. You must also reverse the sign of ΔH, making it positive, before adding them. The oxygen cancels out, since equal numbers of moles of it appear on both sides of the equation.

  6. Reload the model using the F5 key. Add the magnesium and the copper oxide to the evaporating basin and heat as before. When the reaction starts, note the value for ΔH that appears in the information panel.
    Answer
    You should find that the value is similar to (although not identical with) the value you obtained above.
  7. This is an example of a redox reaction. State which reactant is
    (a) oxidised
    (b) reduced.
    Answer
    (a) Magnesium. (In becoming MgO, the Mg loses electrons. In general, when a substance combines with oxygen, the process is oxidation.)
    (b) Copper oxide. (The Cu2+ ions in CuO have gained electrons. In general, when a metal oxide becomes a metal, the process is reduction.)
  8. Explain whether magnesium is acting as an oxidising agent or a reducing agent.
    Answer
    It is acting as a reducing agent. A reducing agent causes another substance to be reduced, while it itself undergoes oxidation.

Summary

Hess's Law allows us to combine equations for reactions and their corresponding enthalpy values to find the enthalpy values for other reactions. If, in combining equations, it is necessary to reverse an equation, then it is necessary to reverse the sign of ΔH. If it is necessary to multiply equations, then ΔH must be multiplied as well. Chemicals that appear on both sides of the equations may be cancelled out, if there are equal numbers of moles of them.

Teacher Summary

  • This activity can be used to reinforce work on Hess's Law. It is assumed that students will already be familiar with this before they undertake the activity. However, the information given in the introduction would allow an able student to understand the general idea of Hess's Law in practice.