Equilibrium Constant Kc for Homogeneous Systems (Oxford AQA International A Level Chemistry)
Revision Note
Written by: Richard Boole
Reviewed by: Stewart Hird
Kc Expressions & Calculations
Equilibrium expression and constant
The equilibrium expression is an expression that links the equilibrium constant, Kc, 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
Kc is defined as:
Where:
[A] and [B] are the equilibrium concentrations of A and B, in mol dm-3
[C] and [D] are the equilibrium concentrations of C and D, in mol dm-3
a, b, c and d are the respective number of moles of each reactant and product
Solids are ignored in equilibrium expressions
The Kc of a reaction is specific and only changes if the temperature of the reaction changes
Worked Example
Deduce the equilibrium 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)
Answers:
Kc =
Kc =
Kc =
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 using:
Worked Example
At equilibrium, 500 cm3 of the following reaction mixture contains 0.235 mol of ethanoic acid, 0.0350 mol of ethanol, 0.182 mol of ethyl ethanoate and 0.182 mol of water.
CH3COOH (l) + C2H5OH (l) CH3COOC2H5 (l) + H2O (l)
Use this information to calculate a value of Kc for this reaction.
Answer:
Step 1: Calculate the concentrations of the reactants and products:
[CH3COOH (l)] = = 0.470 mol dm-3
[C2H5OH (l)] = = 0.070 mol dm-3
[CH3COOC2H5 (l)] = = 0.364 mol dm-3
[H2O (l)] = = 0.364 mol dm-3
Step 2: Write out the balanced chemical equation with the calculated concentrations beneath each substance:
CH3COOH (l) | + | C2H5OH (l) | CH3COOC2H5 (l) | + | H2O (l) | |
0.470 mol dm-3 |
| 0.070 mol dm-3 |
| 0.364 mol dm-3 |
| 0.364 mol dm-3 |
Step 3: Write the equilibrium constant for this reaction in terms of concentration:
Kc =
Step 4: Substitute the equilibrium concentrations into the expression:
Kc =
Kc = 4.03
Step 5: Deduce the correct units for Kc:
Kc =
All units cancel out
Therefore, Kc = 4.03
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 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
Ethyl ethanoate is hydrolysed in water.
CH3COOC2H5 (l) + H2O (l) CH3COOH (l) + C2H5OH (l)
0.1000 mol of ethyl ethanoate are added to 0.1000 mol of water. A little acid catalyst is added and the mixture is made up to 1 dm3.
At equilibrium, 0.0654 mol of water are present.
Use this information to calculate a value of Kc for this reaction.
Answer:
Step 1: Write the balanced chemical equation, with the concentrations beneath each substance, into an initial, change and equilibrium (ICE) table:
| CH3COOC2H5 (l) | + H2O (l) | CH3COOH (l) | + C2H5OH (l) | |
---|---|---|---|---|---|
Initial moles | 0.1000 | 0.1000 |
| 0 | 0 |
Change | –0.0346 | –0.0346 |
| +0.0346 | +0.0346 |
Equilibrium moles | 0.0654 | 0.0654 |
| 0.0346 | 0.0346 |
Step 2: Calculate the concentrations of the reactants and products:
[H2O (l)] = = 0.0654 mol dm-3
[CH3COOC2H5 (l)] = = 0.0654 mol dm-3
[C2H5OH (l)] = = 0.0346 mol dm-3
[CH3COOH (l)] = = 0.0346 mol dm-3
Step 3: Write the equilibrium constant for this reaction in terms of concentration:
Kc =
Step 4: Substitute the equilibrium concentrations into the expression:
Kc =
Kc = 0.28
Step 5: Deduce the correct units for Kc:
Kc =
All units cancel out
Therefore, Kc = 0.28
Changes that Affect the Equilibrium Constant
Changes in concentration
If all other conditions stay the same, the equilibrium constant Kc is not affected by any changes in concentration of the reactants or products
For example, the decomposition of hydrogen iodide:
2HI ⇌ H2 + I2
The equilibrium expression is:
Kc = = 6.25 x 10–3
Adding more HI makes the ratio of [ products ] to [ reactants ] smaller
To restore equilibrium, [H2] and [I2] increase and [HI] decreases
Equilibrium is restored when the ratio is 6.25 x 10-3 again
Changes in pressure
A change in pressure only changes the position of the equilibrium (see Le Chatelier’s principle)
If all other conditions stay the same, the equilibrium constant Kc is not affected by any changes in the pressure of the reactants and products
Changes in temperature
Changes in temperature change the equilibrium constant Kc
For an endothermic reaction such as:
2HI (g) H2 (g) + I2 (g) Kc =
With an increase in temperature:
[H2] and [I2] increases
[HI] decreases
Because [H2] and [I2] increase and [HI] decreases, the equilibrium constant Kc increases
For an exothermic reaction such as:
2SO2 (g) + O2 (g) 2SO3 (g) Kc =
With an increase in temperature:
[SO3] decreases
[SO2] and [O2] increases
Because [SO3] decreases and [SO2] and [O2] increase, the equilibrium constant Kc decreases
Examiner Tips and Tricks
You need to be able to predict the qualitative effects of changes of temperature on the value of Kc.
To do this:
Write the Kc expression
Use the balanced chemical equation and enthalpy change information to talk about the concentrations of each species in the expression
Then talk about the overall effect of a temperature change on the value of Kc.
Presence of a catalyst
If all other conditions stay the same, the equilibrium constant Kc is not affected by the presence of a catalyst
A catalyst speeds up both the forward and reverse reactions at the same rate so the ratio of [ products ] to [ reactants ] remains unchanged
Worked Example
Factors affecting Kc
An equilibrium is established in the following reaction:
AB (aq) + CD (aq) AC (aq) + BD (aq) ΔH = +180 kJ mol-1
Which factors would affect the value of Kc in this equilibrium?
Answer:
Only a change in temperature will affect the value of Kc
Any other changes in conditions would result in the position of the equilibrium moving to oppose this change
Adding a catalyst increases the rate of reaction meaning the state of equilibrium will be reached faster but has no effect on the position of the equilibrium and, therefore, Kc is unchanged
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