1. Calculating K c values CH 3 COOH (aq) + CH 3 CH 2 OH (aq) CH 3 COOCH 2 CH 3 (ag) + H 2 O (l) CH 3 COOHCH 3 CH 2 OHCH 3 COOCH 2 CH 3 H2OH2O Initial conc ab00 From the balanced equation 1 mol of acid reacts with 1 mol of alcohol to form 1 mol of ester and 1 mol of water. This means that….. The amount of the two products formed is the same (x) The amount by which the two reactants are reduced equals the amount of each product formed (x) Equilib conc (a-x)(b-x)xx

2. Calculating K c values CH 3 COOH (aq) + CH 3 CH 2 OH (aq) CH 3 COOCH 2 CH 3 (ag) + H 2 O (l) CH 3 COOHCH 3 CH 2 OHCH 3 COOCH 2 CH 3 H2OH2O Initial conc ab00 K c = [CH 3 COOCH 2 CH 3 ] [H 2 O] [ CH 3 COOH ] [ CH 3 CH 2 OH ] K c = (x/V) x (x/V) (a-x)/V x (b-x)/V Kc = ___x 2 ___ (a-x) x (b-x) Equilib conc (a-x)(b-x)xx Where V = volume in dm 3 If the number of moles is the same on both sides of the equilibrium then the volume cancels, so we can use moles rather than concentrations

3. Calculating K c values CH 3 COOH (aq) + CH 3 CH 2 OH (aq) CH 3 COOCH 2 CH 3 (ag) + H 2 O (l) CH 3 COOHCH 3 CH 2 OHCH 3 COOCH 2 CH 3 H2OH2O Initial moles 1.00.1800 Moles CH 3 CH 2 OH at equilibrium = 0.18 – 0.171 = 0.009 Moles CH 3 COOH at equilibrium = 1.0 – 0.171 Equilib moles 0.171 = 0.829

4. Calculating K c values CH 3 COOH (aq) + CH 3 CH 2 OH (aq) CH 3 COOCH 2 CH 3 (ag) + H 2 O (l) CH 3 COOHCH 3 CH 2 OHCH 3 COOCH 2 CH 3 H2OH2O Initial moles 1.00.3300 K c = 0.171 x 0.171 0.829 x 0.009 Equilib moles 0.8290.0090.171 K c = [CH 3 COOCH 2 CH 3 ] [H 2 O] [ CH 3 COOH ] [ CH 3 CH 2 OH ] = 0.0292 0.829 x 0.009 = 3.92

5. CH 3 COOH (aq) + CH 3 CH 2 OH (aq) CH 3 COOCH 2 CH 3 (ag) + H 2 O (l) CH 3 COOHCH 3 CH 2 OHCH 3 COOCH 2 CH 3 H2OH2O Initial moles 1.0 00 Moles CH 3 COOCH 2 CH 3 at equilibrium = 1.0 – 0.333 = 0.667 Moles CH 3 CH 2 OH at equilibrium = 0.333 Equilib moles 0.333 When 1 mol each of ethanoic acid and ethanol are mixed together at a fixed temperature 0.333mol of acid remain at equilibrium. Calculate K c Moles H 2 O at equilibrium = 1.0 – 0.333 = 0.667

6. Calculating K c values CH 3 COOH (aq) + CH 3 CH 2 OH (aq) CH 3 COOCH 2 CH 3 (ag) + H 2 O (l) CH 3 COOHCH 3 CH 2 OHCH 3 COOCH 2 CH 3 H2OH2O Initial moles 1.0 00 K c = 0.667 x 0.667 0.333 x 0.333 Equilib moles 0.333 0.667 K c = [CH 3 COOCH 2 CH 3 ] [H 2 O] [ CH 3 COOH ] [ CH 3 CH 2 OH ] = 4.01

7. H 2 (g) + I 2g) 2HI (g) H2H2 I2I2 2HI Initial moles 0.2060.1440 Moles I 2 at equilibrium = 0.144 – 0.258/2 = 0.077 Moles H 2 at equilibrium = 0.206 – 0.258/2 Equilib moles 0.258 0.206 mol of hydrogen and 0.144 mol if iodine were heated at 683K. At equilibrium 0.258 mol of HI were present. Calculate K c From the balanced equation 1mol hydrogen reacts with 1 mol iodine to form 2 moles of hydrogen iodide = 0.015

8. H 2 (g) + I 2g) 2HI (g) H2H2 I2I2 2HI Initial moles 0.2060.1440 Equilib moles 0.0770.0150.258 0.206 mol of hydrogen and 0.144 mol if iodine were heated at 683K. At equilibrium 0.258 mol of HI were present. Calculate K c K c = [HI] 2 [ H 2 ] [ I 2 ] K c = (0.258) 2 0.077 x 0.015 K c = 57.63

9. For reactions that involve a change in the number of moles, the volume must be known in order to calculate the concentrations before K c can be calculated as the volume terms will not cancel.

10. N 2 O 4(g) 2NO 2(g) N2O4N2O4 2NO 2 Initial moles 1.00 Equilib moles At 300K 1.0 mol of N 2 O 4 is 20% dissociated in 2.0dm 3 flask. Calculate K c If 20% is dissociated, then 80% remains at equilibrium Mols of N 2 O 4 at equilibriuum = 80/100 x 1.0 = 0.8 Moles of NO 2 at equilibrium = 2 x (1.0 – 0.8) = 0.4

11. N 2 O 4(g) 2NO 2(g) N2O4N2O4 2NO 2 Initial moles 1.00 Equilib moles 0.80.4 At 300K 1.0 mol of N 2 O 4 is 20% dissociated in 2.0dm 3 flask. Calculate K c K c = [NO 2 ] 2 [ N 2 O 4 ] K c = (0.4/V) 2 0.8/V K c = 0.1 moldm -3 K c = (moldm -3 ) 2 moldm -3 = moldm -3 K c = (0.4/2) 2 0.8/2

12. Finding K c Experimentally Known amounts of reagents are used (Initial concentrations known) The system is closed and left until equilibrium is reached The concentration of the products is then analysed, often by titration K c can then be calculated

13. Different initial concentrations of alcohol were used in different experiments Calculate K c for the following experiments All the experiments were performed at 373 k Initial MolesEquilibrium moles KcKc CH 3 COOHCH 3 CH 2 OHCH 3 COOCH 2 CH 3 H2OH2O 11.00.330.171 3.92 21.00.50.425 31.0 0.667 41.02.00.850 51.08.00.967

14. K c is a constant at constant temperature Initial MolesEquilibrium moles KcKc CH 3 COOHCH 3 CH 2 OHCH 3 COOCH 2 CH 3 H2OH2O 11.00.330.171 3.92 21.00.50.425 4.12 31.0 0.667 4.01 41.02.00.850 4.12 51.08.00.967 4.02 Altering the concentration of the reactants will shift the equilibrium position but the value of K c remains constant

15. Factors that Affect Equilibrium Changing the concentration of a reactant or product will shift the position of an equilibrium CH 3 COOH (aq) + CH 3 CH 2 OH (aq) CH 3 COOCH 2 CH 3 (ag) + H 2 O (l) K c = [CH 3 COOCH 2 CH 3 ] [H 2 O] [ CH 3 COOH ] [ CH 3 CH 2 OH ] K c is the ratio of concentrations so… Changing the concentration of a reactant or product will not affect the value of Kc

16. Changing concentration has no effect on K c Changing pressure has no effect on K c Catalysts have no effect on K c Changing temperature will change K c Factors that Affect Equilibrium