The Equilibrium Constant, Kc (OCR A Level Chemistry A): Revision Note
Equilibrium Constant Expressions
Equilibrium expression & constant
The equilibrium constant expression is an expression that links the equilibrium constant, K, to the concentrations of reactants and products at equilibrium taking the stoichiometry of the equation into account
So, for a given reaction:
aA + bB ⇌ cC + dD
The corresponding equilibrium constant expression is written as:
Where:
[A] and [B] = equilibrium reactant concentrations (mol dm-3)
[C] and [D] = equilibrium product concentrations (mol dm-3)
a, b, c and d = number of moles of corresponding reactants and products
Solids are ignored in equilibrium constant expressions
The equilibrium constant, K, of a reaction is specific to a given equation
The Kc of a reaction is specific and only changes if the temperature of the reaction changes
Worked Example
Deducing equilibrium expressions
Deduce the equilibrium constant expression for the following reactions
Ag+ (aq) + Fe2+ (aq) ⇌ Ag (s) + Fe3+ (aq)
N2 (g) + 3H2 (g) ⇌ 2NH3 (g)
2SO2 (g) + O2 (g) ⇌ 2SO3 (g)
Answer 1:
[Ag (s)] is not included in the equilibrium constant expression as it is a solid
Answer 2:
Answer 3:
Equilibrium Constant Calculations
Calculations involving Kc
In the equilibrium expression each figure within a square bracket represents the concentration in mol dm-3
The units of Kc therefore depend on the form of the equilibrium expression
Some questions give the number of moles of each of the reactants and products at equilibrium together with the volume of the reaction mixture
The concentrations of the reactants and products can then be calculated from the number of moles and total volume
![](https://cdn.savemyexams.com/cdn-cgi/image/f=auto,width=3840/https://cdn.savemyexams.com/uploads/2021/05/1.7-Equilibria-Equation-for-concentration.png)
Equation to calculate concentration from number of moles and volume
Worked Example
Calculating Kc of ethanoic acid
In the reaction:
CH3COOH (I) + C2H5OH (I) ⇌ CH3COOC2H5 (I) + H2O (I)
ethanoic acid ethanol ethyl ethanoate water
500 cm3 of the reaction mixture at equilibrium contained 0.235 mol of ethanoic acid and 0.035 mol of ethanol together with 0.182 mol of ethyl ethanoate and 0.182 mol of water. Use this data to calculate a value of Kc for this reaction.
Answer
Step 1: Calculate the concentrations of the reactants and products
[CH3COOH (l)] =
[C2H5OH (l)] =
[CH3COOC2H5 (l)] =
[H2O (l)] =
Step 2: Write the equilibrium constant for this reaction in terms of concentration
Kc =
Step 3: Substitute the equilibrium concentrations into the expression
Kc =
Step 4: Deduce the correct units for Kc
Kc =
All units cancel out
Therefore, Kc = 4.03
Examiner Tips and Tricks
Note that the smallest number of significant figures used in the question is 3, so the final answer should also be given to 3 significant figures
Some questions give the initial and equilibrium concentrations of the reactants but not the products
An initial, change and equilibrium table should be used to determine the equilibrium concentration of the products using the molar ratio of reactants and products in the stoichiometric equation
Worked Example
Calculating Kc of ethyl ethanoate
Ethyl ethanoate is hydrolysed by water:
CH3COOC2H5 (I) + H2O (I) ⇌ CH3COOH (I) + C2H5OH (I)
ethyl ethanoate water ethanoic acid ethanol
0.1000 mol of ethyl ethanoate are added to 0.1000 mol of water. A little acid catalyst is added and the mixture made up to 1.00 dm3. At equilibrium, 0.0654 mol of water are present.
Use this data to calculate a value of Kc for this reaction.
Answer
Step 1: Write out the balanced chemical equation with the concentrations of beneath each substance using an initial, change and equilibrium table
![Equilibria Calculating Kc of ethyl ethanoate table](https://cdn.savemyexams.com/cdn-cgi/image/f=auto,width=3840/https://cdn.savemyexams.com/uploads/2020/12/1.7-Equilibria-Calculating-Kc-of-ethyl-ethanoate-table.png)
Step 2: Calculate the concentrations of the reactants and products
[CH3COOC2H5 (l)] =
[H2O (l)] =
[CH3COOH (l)] =
[C2H5OH (l)] =
Step 3: Write the equilibrium constant for this reaction in terms of concentration
Kc =
Step 4: Substitute the equilibrium concentrations into the expression
Kc =
Therefore, Kc = 0.28
Estimating the position of equilibrium
The magnitude of Kc indicates the relative concentrations of reactants and products in the mixture
If Kc is very large (Kc >>1) the equilibrium lies to the RHS so the reaction mixture contains mostly products
If Kc is very small (Kc << 1) the equilibrium lies to the LHS so the reaction mixture contains mostly reactants
If Kc is close to 1 the mixture contains a similar concentration of both reactant and products
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