BRAINCAKEFinding the Value of Kc
0.1mol of ethanol is mixed with 0.1mol of ethanoic acid and allowed to reach equilibrium. The total volume of the system is made up to 20.0cm3 with water. By titration it was found that 0.033mol of ethanoic acid was present at equilibrium. At the start,
0.033mol of ethanoic is left at equilibrium so (0.1 - 0.033)mol must have reacted.(0.1 - 0.033) = 0.067 so 0.067mol of ethanoic acid has reacted with 0.067 ethanol to form 0.067mol of water and 0.067mol of ethyl ethanoate since the ratios are 1: 1: 1: 1.so at equilibrium
As the the volume is 20cm3 or 0.02dm3 we can work out the concentrations for the equilibrium expression by dividing the moles by 0.02.
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Key terms in this lesson
- equilibrium constant Kc
- The value, at a given temperature, of the equilibrium expression written in terms of the concentrations of reactants and products.
More in Equilibria
- Dynamic Equilibria
- Homogenous and Heterogenous Equilibria and Kc
- The Effect of Temperature on Kc
- Catalysts and Kc
- Partial Pressures and Kp
All 6 lessons in Equilibria · All OCR A-level Chemistry topics