Equilibrium law
17.1 The equilibrium law (4 hours)
Pause for thought
A simple question, which would seem to be exactly what this topic is about, might be:
What amount (in mol) of ethyl ethanoate will be formed in the equilibrium mixture if 2.0 mol of ethanoic acid and 1.0 mol of ethanol are reacted and allowed to reach equilibrium at 373K? The value for Kc at this temperature is 4.
According to the IB students should be able to deduce the expression which will give them the answer. Assuming the amount is x mol and the volume remains constant then the expression is 4 = x2 / ((2.0 – x) x (1.0 – x)) = x2 / (x2– 3x + 2) i.e. 3x2 – 12x + 8 = 0. Solving this gives the answer 0.85 mol. However the syllabus clearly states that they are not required to solve quadratic equations so a question such as this cannot be asked in the IB exam.
Some universities are concerned that the mathematical ability of students taking pre-university exams such as the IB is not sufficient for them to study degree courses in science (and medicine). A report highlighting this problem has been published by SCORE and is discussed in my blog 'Is IB Chemistry too easy'. Over the years the mathematical content of IB chemistry has decreased considerably. I appreciate that we want to be testing student's knowledge and understanding of chemistry and not their mathematical ability in exams. However they have a calculator in Papers 2 and 3 and surely it is not beyond their ability to solve a quadratic equation since all students have to do mathematics in some form or other as one of their six diploma subjects. Chemistry does involve a considerable amount of mathematics. Shielding students from the necessary mathematics means they are ill-prepared to go on to study science at a higher level.
One area of mathematics that students can be asked about which is relevant to this topic is to solve problems using the equation ΔG = − RT lnKc. It is worth giving your students question 5 on the attached sheet to do. This asks them to calculate Kc for the equilibrium reaction between ethanol and ethanoic acid to form water and ethyl ethanoate using thermodynamic data. If the data from the IB data booklet is used the answer comes out to be 13.2 at 298 K. This is significantly higher than the usual value of 4 that is quoted in the literature. It makes quite a nice Nature of Science question to ask students to comment on this. In fact because of the logarithmic factor in the equation a very small difference in the ΔG value makes a huge difference to the answer. Using the ΔH and ΔS values in the data booklet give a ΔG value of − 6.4 kJ mol-1. If values from other data sources are used this value can change and it only needs to change by 3 kJ mol-1 so that it becomes − 3.4 kJ mol-1 to give the correct answer of 4 for Kc. This makes setting realistic questions for the IB exam quite difficult.
Nature of Science
The syllabus states that this topic provides an example of using quantitative reasoning. It claims that experimentally determined rate expressions for forward and backward reactions can be deduced directly from the stoichiometric
equations. However rate expressions cannot be deduced from stoichiometric equations they must be determined experimentally. Once the equilibrium expression is known then it can be used to explain Le Chatelier's principle as applied to that particular reaction.
Learning outcomesAfter studying this topic students should be able to: Understand
Apply their knowledge to:
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Clarification notesStudents are not expected to be able to derive the expression ΔG = − RT lnKc which is provided in Section 1 of the data booklet. The solving of quadratic equations will not be assessed. International-mindednessNothing is listed on the syllabus under this heading |
Teaching tipsThis can be quite a challenging topic to teach. Students need to have a good grasp of the mole concept to be able to work out equilibrium concentrations of reactants and products from initial concentrations. For straightforward one to one reactions such as esterification it is relatively easy but when the stoichiometry is more difficult then students do have problems. I usually start with esterification and they can see that if you start with a mol of acid and b mol of alcohol then if x mol of ester are formed then the equilibrium amount of the acid will be (a – x) mol, the equilibrium amount of the alcohol will be (b – x) mol and the equilibrium amount of water will be x mol. Stress that it is concentrations not amounts that appear in the equilibrium expression but in this example the volumes cancel out so amounts can be used. I then use other less straightforward examples such as the dissociation of phosphorus(V) chloride to phosphorus(III) chloride and chlorine or the reaction between hydrogen and iodine to form hydrogen iodide. Finally, to see if they really understand the underlying concept, I use 1 mol of nitrogen and 3 mol of hydrogen to give 2x mol of ammonia in a total volume of V dm3 and get them to work out the equilibrium expression for the Haber process. If they work this out correctly the answer is Kc = 4x2V2/27(1 – x)4. |
Give students practice at using the expression ΔG⦵ = − RT lnKc. They will need to be familiar with the ex button on their calculator and also be able to use the expression ΔG⦵ = ΔH⦵ − TΔS ⦵ which is covered in Topic 15. Entropy & spontaneity and separately in Understanding and using ΔG. I have given examples in my study guide, the teaching slides and in the attached Equilibrium law questions and The equilibrium law quiz to help them with this. Note that any other questions you give them should use Kc and not Kp as Kp is not on the syllabus.
Study guide
Page 56 QuestionsFor ten 'quiz' multiple choice questions with the answers explained see MC test: The equilibrium law. For short-answer questions which can be set as an assignment for a test, homework or given for self study together with model answers see Equilibrium law questions. IM, TOK, Utilization etc.See separate page which covers all of Topics 7 & 17 Practical work |
Teaching slides
Teachers may wish to share these slides with students for learning or for reviewing key concepts.
The equilibrium law:
The equilibrium law:
The equilibrium law:
The equilibrium law:
The equilibrium law:
The equilibrium law:
The equilibrium law:
The equilibrium law:
The equilibrium law:
The equilibrium law:
The equilibrium law:
Other resources
1. A resource made for Advance Placement (AP) chemistry but it takes you through the whole of equilibrium ending up with solving some problems.
2. An example of a gaseous equilibrium calculation. Could be worth showing as the chemistry is very much on the IB syllabus but of course the mathematics for obtaining the final solution, which involves solving a quadratic equation, is not.