Performing Equilibria Calculations
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Students process and present information from secondary sources to calculate K from a mixture of initial and equilibrium conditions.
Equilibrium calculations
The nylon equilibrium is complicated, and we take a simpler example: the hydrolysis of an ester, methyl methanoate, to methanol and methanoic (formic) acid:
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For simplicity, we consider equilibrium with all substances dilute in a solvent in which reactants and products are all soluble.
Exercise:
A 0.5 M solution of both water and methyl methanoate in a particular solvent was taken and allowed to come to equilibrium. The final concentration of methanol was measured as 0.003 M. Find the equilibrium constant K.
Answer
Look at the initial and final concentrations (all molar):
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| initial: | 0.5 | 0.5 | 0 | 0 | |||
| final: | 0.5-0.003 | 0.5-0.003 | 0.003 | 0.003 |
(the final concentrations are worked out by conservation of mass: if we started out with 0.5 mol of ester and finish with 0.003 mol of methanol, then there must be 0.5 - 0.003 mol of ester remaining). Hence the equilibrium constant, K is given by:
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Notes for teachers:
- Conditions for this example have been chosen to avoid the problem of discussing the activity of water being 1: because everything is at low concentration in a solvent, the quantities which appear in the equilibrium constant are actual concentrations.
- The equilibrium constant for this exercise is obtained using the standard free energy values at 25 °C of reactants and products calculated from the tabulation given in the SI Data Book (Aylward and Findlay, 3rd edn., John Wiley).
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