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Topic 17 Equilibrium

Topic 17 Equilibrium. Liquid- vapour equilibrium The equilibrium law. 17.1 Liquid-vapour equilibrium Liquid Gas. Equilibrium: Rate of vaporisation = Rate of condensation Vapour pressure: the pressure the gas will give at equilibrium Surface area of the liquid

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Topic 17 Equilibrium

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  1. Topic 17Equilibrium • Liquid-vapour equilibrium • The equilibrium law

  2. 17.1 Liquid-vapour equilibriumLiquid Gas

  3. Equilibrium: Rate of vaporisation = Rate of condensation Vapour pressure: the pressure the gas will give at equilibrium Surface area of the liquid affect the time it takes to reach equilibrium, not the vapour pressure.

  4. Vaporisation is an endothermic process, it requires energy to break bonds. Enthalpy of vaporisation, DHvap. Stronger bonds => Higher DHvap => Higher boiling point => Lower vapour pressure.

  5. The liquid will boil when its vapour pressure = pressure on its surface (egathmospheric pressure) Decrease the pressure => decrease the boiling point Higher pressure => Higher boiling point

  6. Distillation • A volatile (flyktig) liquid can be separated from a non-volatile liquid by distillation. • 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.

  7. Water-ethanol system

  8. 17.2 The equilibrium law Solve equilibrium problems using the expression for Kc. Ethanol + Ethanoicacid 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 Kc. • b. How much ethyl ethanoate had been formed if you had started with 2 mol of ethanol and 1 mol of ethanoic acid?

  9. Ethanol + Ethanoic acid Ethyl ethanoate + water nstart1 1 0 0 nequ1-0.67 1-0.67 0.67 0.67 • Kc= [Ethyl ethanoate] [water] / [Ethanol] [Ethanoic acid] = (0.67/V)2 / (0.33/V)2 = 4.1 • Ethanol + Ethanoic acid  Ethyl ethanoate + water nstart2 1 0 0 nequ2-x 1-x x x • Kc = 4.1 = x2 / (2-x)(1-x) • x1= 0.85 moland x2 =3.1 (not possible)

  10. Examples to solve In a flask with the volume of 0,500 dm3 2.00g of PCl5 is inserted. The flask is heated to 250oC and PCl5 brakes down and you find following equilibrium: PCl5(g)PCl3(g) + Cl2(g). At equilibrium you find 0,50 g of Cl2. Kc= ?

  11. In a vessel with constant volume and temperature you have an equilibrium mixture of: 0.60 mol SO3, 0.40 mol NO, 0.10 mol NO2, 0.80 mol SO2. Calculate Kc for the reaction: SO2(g) + NO2(g)SO3(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

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