1. 2 Equilibria
2.1 Chemical Equilibrium (and Equilibrium Constant) 2.2 Calculations using Kc expressions 2.3 Effect of changing conditions on equilibria

2. Chemical Equilibrium Describe the term chemical equilibrium.
Refers to reversible reactions.
Rate of the forward reaction = rate of the reverse reaction
The concentration of reactants and products remains constant (though not always equal).

3. Equilibrium constant, Kc
We can relate the equilibrium concentrations of reactants and products using the equilibrium constant expression.
aA + bB + cC  xX + yY + zZ
𝑋 𝑥 𝑌 𝑦 𝑍 𝑧 𝐴 𝑎 𝐵 𝑏 𝐶 𝑐 = Kc

4. Equilibrium Constant, Kc
Kc is different for each reaction
Kc will vary with temperature
You also have to work out the units of Kc

5. Exam Question Practice
0.10 mol of ethanol is mixed with 0.10 mol ethanoic acid and allowed to reach equilibrium. The total volume of the system is made up to 20 cm3 with water. We find by titration that mol ethanoic acid is present once equilibrium is reached. Calculate the equilibrium constant Kc. C2H5OH + CH3COOH CH3CO2C2H5 + H2O

6. Kc = 4.1

7. 2.2 Calculations using Kc expressions
Learning Objectives:
If we know the Kc, we can…
Calculate the composition of a reaction mixture at equilibrium (when reactant amounts known).
Calculate the amount of reactant needed (to obtain required amount of product).

8. Calculating the composition of reaction mixture
C2H5OH + CH3COOH CH3CO2C2H5 + H2O
How much ethyl ethanoate will be produced by mixing 1 mol ethanol with 1 mol ethanoic acid when Kc = 4.0?
What is the composition of the mixture at equilibrium?

9. 2/3 mol of ethyl ethanoate produced
Composition
Ethanol = 1/3 mol
Ethanoic acid = 1/3 mol
Ethyl ethanoate = 2/3 mol
Water = 2/3 mol

10. Calculating the amount of reactant needed
CH3COCH3 + HCN CH3C(CN)(OH)CH3 Kc = 30.0 mol-1dm3 How much hydrogen cyanide is required to produce 1.00 mol of product if we start with 4.00 mol propanone (carried out in 2.00 dm3 of ethanol)?

11. 1.02 mol hydrogen cyanide

12. 2.3 The effect of changing conditions on equilibria
Learning Objectives:
Use Le Chatelier’s principle to predict how changes in conditions affect the position of equilibrium.
Describe how the equilibrium constant is affected by changing the conditions of the reaction.

13. Le Chatelier’s Principle review
When a system at equilibrium is disturbed, the equilibrium position moves in the direction that will reduce the disturbance.
Temperature ( temp, favors endo : temp, favors exo)
Pressure ( pressure, favors side with less moles gas)
Concentration ( conc of reactant, favors products)

14. Effect of temperature on Kc
Kc is always constant for one particular temperature
Kc = [𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠] [𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠]
What will happen to Kc if forward reaction is favoured?
What will happen to Kc if reverse reaction is favoured?

15. Effect of temperature on Kc
If change favours forward reaction:
Favours products
Kc will increase in value
If change favours reverse reaction:
Favours reactants
Kc will decrease

16. Did you get that? 2SO2 + O2 2SO3 H = -197 kJ/mol
If we increase the temperature:
What effect will this have on the equilibrium position?
What will happen to Kc?
2SO2 + O SO3 H = -197 kJ/mol
N2O NO2 H = +58 kJ/mol
H2 + CO H2O + CO H = +40 kJ/mol

17. Did you get that? 2SO2 + O2 2SO3 H = -197 kJ/mol
If we increase the temperature:
What effect will this have on the equilibrium position?
What will happen to Kc?
2SO2 + O SO3 H = -197 kJ/mol
N2O NO2 H = +58 kJ/mol
H2 + CO H2O + CO H = +40 kJ/mol

18. Effect of concentration on equilibrium
Kc DOES NOT CHANGE if the temperature is constant.
Changes to concentration will balance out so that the equilibrium constant will remain the same
BUT!
The equilibrium composition WILL have changed.

19. Let’s see why
A + B X + Y
Kc = 𝑋 [𝑌] 𝐴 [𝐵]

20. What does Kc tell us about the equilibrium position?
Kc much greater than 1
Products favoured
Equilibrium to the right
Kc much less than 1
Reactants favoured
Equilibrium to the left
Kc = [𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠] [𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠]

21. Much MUCH greater or less than 1
If Kc is greater than 1010, we say that the reaction goes to completion (not a reversible reaction, single headed arrow).
If Kc is less than 10-10, we say that the reaction does not take place (no chemical reaction).

22. What effect do Catalysts have on Kc? Why?
Catalysts have no effect on the equilibrium constant.
This is because the catalysts affects both the forward and reverse reactions equally.