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

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

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 Moles CH 3 CH 2 OH at equilibrium = 0.18 – = Moles CH 3 COOH at equilibrium = 1.0 – Equilib moles = 0.829

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 K c = x x Equilib moles K c = [CH 3 COOCH 2 CH 3 ] [H 2 O] [ CH 3 COOH ] [ CH 3 CH 2 OH ] = x = 3.92

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 Moles CH 3 COOCH 2 CH 3 at equilibrium = 1.0 – = Moles CH 3 CH 2 OH at equilibrium = Equilib moles 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 – = 0.667

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 K c = x x Equilib moles K c = [CH 3 COOCH 2 CH 3 ] [H 2 O] [ CH 3 COOH ] [ CH 3 CH 2 OH ] = 4.01

H 2 (g) + I 2g) 2HI (g) H2H2 I2I2 2HI Initial moles Moles I 2 at equilibrium = – 0.258/2 = Moles H 2 at equilibrium = – 0.258/2 Equilib moles mol of hydrogen and mol if iodine were heated at 683K. At equilibrium 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

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

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.

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

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 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

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

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

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 Altering the concentration of the reactants will shift the equilibrium position but the value of K c remains constant

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

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