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Topic 17 Equilibrium

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

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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. 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

  4. Vaporisation is an endothermic process, it requires energy to break bonds: H2O (l) H2O (g) DHvap>0 Enthalpy of vaporisation, DHvap Stronger bonds => Higher DHvap => Higher boiling point => Lower vapour pressure Whichcompound has the highestboilingpoint?

  5. The liquid will boil when its vapour pressure equals the pressure on its surface (ieathmospheric 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 (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

  7. Water-ethanolsystem- boilingpoints

  8. 17.2 The equilibrium law Solve equilibrium problems using the expression for Kc SO2 (g) + NO2 (g) SO3 (g) + NO (g) A 2.0 dm3 flask was filled with 4.0 mol SO2 and 4.0 mol of NO2. At equilibrium it was found to contain 2.6 mol of NO. Calculate Kc. First we calculate concentrations and fill in the values in the table:

  9. SO2 (g) + NO2 (g) SO3 (g) + NO (g) Then we calculate the changes in concentrations and fill in the values in the table. We calculate the values at equilibrium. Finallyweinsert the values in the Kc expression: Kc=[SO2]*[NO2]/([SO3]*[NO])= 0.7*0.7/(1.3*1.3)= 3.45

  10. 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. • Calculate Kc 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

  11. b. How much ethyl ethanoate had been formed if you had started with 2 mol of ethanol and 1 mol of ethanoic acid? • 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)

  12. Examples to solve In a flask with the volume of 0,500 dm32.00 g of PCl5 is inserted. The flask is heated to 250oC and PCl5 breakesdown and you find the following equilibrium: PCl5(g)PCl3(g) + Cl2(g). At equilibrium you find 0,50 g of Cl2. Kc= ?

  13. 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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