Topic 17 Equilibrium Liquid-vapour equilibrium The equilibrium law

17.1 Liquid-vapour equilibrium Liquid Gas

At equilibrium: Rate of vaporisation = Rate of condensation The vapour pressure is the pressure the gas will give at equilibrium The surface area of the liquid affect the time it takes to reach equilibrium, not the vapour pressure

Vaporisation is an endothermic process, it requires energy to break bonds: H2O (l) H2O (g)  H vap >0 Enthalpy of vaporisation,  H vap Stronger bonds => Higher  H vap => Higher boiling point => Lower vapour pressure Which compound has the highest boiling point?

The liquid will boil when its vapour pressure equals the pressure on its surface (ie athmospheric pressure) Decrease the pressure => decrease the boiling point Higher pressure => Higher boiling point

Distillation A volatile (flyktig) liquid can be separated from a non- volatile liquid by distillation (eg ethanol from water) A mixture of two miscible and volatile liquids will boil when the sum of the two vapour pressures equals the external pressure The gas phase will contain more of the more volatile compound than the liquid phase

Water-ethanol system- boiling points

17.2 The equilibrium law Solve equilibrium problems using the expression for K c SO 2 (g) + NO 2 (g) SO 3 (g) + NO (g) A 2.0 dm 3 flask was filled with 4.0 mol SO 2 and 4.0 mol of NO 2. At equilibrium it was found to contain 2.6 mol of NO. Calculate K c. First we calculate concentrations and fill in the values in the table: SO 2 NO 2 O3O3 NO c (start)4.0/2.0= 2.0 M 00 c (change) c (equlibrium)2.6/2.0= 1.3 M

SO 2 (g) + NO 2 (g) SO 3 (g) + NO (g) Then we calculate the changes in concentrations and fill in the values in the table. We calculate the values at equilibrium. Finally we insert the values in the K c expression: K c =[SO 2 ]*[NO 2 ]/([SO 3 ]*[NO])= 0.7*0.7/(1.3*1.3)= 3.45 SO 2 NO 2 O3O3 NO c (start)4.0/2.0= 2.0 M 00 c (change)-1.3 M M c (equlibrium)0.7 M 1.3 M2.6/2.0= 1.3 M

Ethanol + Ethanoic acid Ethyl ethanoate + water You mix 1 mol of ethanol and 1 mol of ethanoic acid. At equilibrium you find 0,67 mol of ethyl ethanoate. a.Calculate K c Ethanol + Ethanoic acid Ethyl ethanoate + water n start n equ K c = [Ethyl ethanoate]*[water] / [Ethanol]*[Ethanoic acid] = (0.67/V) 2 / (0.33/V) 2 = 4.1

b. How much ethyl ethanoate had been formed if you had started with 2 mol of ethanol and 1 mol of ethanoic acid? a.Ethanol + Ethanoic acid Ethyl ethanoate + water n start n equ 2-x 1-x x x K c = 4.1 = x 2 / (2-x)(1-x) x 1 = 0.85 mol and x 2 =3.1 (not possible)

Examples to solve In a flask with the volume of 0,500 dm g of PCl 5 is inserted. The flask is heated to 250 o C and PCl5 breakes down and you find the following equilibrium: PCl 5(g) PCl 3(g) + Cl 2(g). At equilibrium you find 0,50 g of Cl 2. K c = ?

In a vessel with constant volume and temperature you have an equilibrium mixture of: 0.60 mol SO 3, 0.40 mol NO, 0.10 mol NO 2, 0.80 mol SO 2. Calculate K c for the reaction: SO 2(g) + NO 2(g) SO 3(g) + NO (g) To the vessel 0.78 mol of NO is added. Calculate the number of mol of the compounds there will be at equilibrium. => Need quadratic expression to solve the problem, not in syllabus