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

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).

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

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

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 0.033 mol ethanoic acid is present once equilibrium is reached. Calculate the equilibrium constant Kc. C2H5OH + CH3COOH CH3CO2C2H5 + H2O

Kc = 4.1

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).

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?

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

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)?

1.02 mol hydrogen cyanide

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.

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)

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?

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

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 + O2 2SO3 H = -197 kJ/mol N2O4 2NO2 H = +58 kJ/mol H2 + CO2 H2O + CO H = +40 kJ/mol

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 + O2 2SO3 H = -197 kJ/mol N2O4 2NO2 H = +58 kJ/mol H2 + CO2 H2O + CO H = +40 kJ/mol

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.

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

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]

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).

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.