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Ch 28 - Chemical equilibrium

Ch 28 - Chemical equilibrium. 2Mg + O 2 2MgO. Magnesium + oxygen. Magnesium oxide. This reaction usually goes to completion, and can only go one direction. The Haber process. N 2 + 3H 2 2NH 3 This reaction is reversible – it goes both directions!

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Ch 28 - Chemical equilibrium

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  1. Ch 28 - Chemical equilibrium

  2. 2Mg + O2 2MgO Magnesium + oxygen Magnesium oxide • This reaction usually goes to completion, and can only go one direction.

  3. The Haber process • N2 + 3H2 2NH3 • This reaction is reversible – it goes both directions! The reactants react to form the product…… and at the same time the product is decomposing to form the reactants.

  4. Dynamic equilibrium Dynamic – reaction is still occurring Equilibrium –forward reaction and the backward reaction are going at the same speed. The symbol is used to mean dynamic equilibrium. In a reaction at equilibrium the concentrations of each reactant and product are constant (overall they are not changing!) The man stays in the same place!

  5. Equilibrium constant Kc The equilibrium constant tells us the position of the equilibrium of a reaction … It tells us whether the reaction mixture is made up of mostly reactants or mostly products!

  6. A large equilibrium constant Kc tells us that reaction lies on the right hand side – at equilibrium there is more products than reactants. 2SO2 + O2 2SO3 Say at equilibrium this reaction has an equilibrium constant of 45. This tells us that there is a higher concentration of products compared to reactants at equilibrium!

  7. A small equilibrium constant Kc tells us that reaction lies on the left hand side – at equilibrium there is more reactants than products N2O4 2NO2 Say at equilibrium this reaction has an equilibrium constant of 0.05. This tells us that there is a higher concentration of reactants compared to products at equilibrium!

  8. The formula for Kc • In the reaction aA + bB cC + dD Kc = [C]c [D]d [A]a [B]b

  9. Calculations involving Kc – EXAMPLE 1 • In a reaction between sulfur dioxide and oxygen • 2SO2 + 02 2SO3 It is found that at equilibrium the concentrations of SO2,02 and SO3 were 0.07mol-1L, 0.035mol-1L and 0.03mol-1L. Calculate the value of the equilibrium constant Kc

  10. Calculations involving Kc – EXAMPLE 2 • Mix together 180g of ethanoic acid and 138g of ethanol. At equilibrium there is 46g of ethanol in the reaction. Calculate Kc for the reaction C2H5OH CH3COOC2H5 + CH3COOH + H20

  11. Kc = [CH3COOC2H5 ][H20] [CH3COOH ][C2H5OH] Kc = [2][2] [1][1] Kc = 4

  12. 2007 Q10a • (i) Kc = [NH3]2 [N2][H2]3

  13. Kc = [NH3]2 [N2][H2]3 Kc = [2]2 [2][6]3 Kc = 1 108 Kc = 0.009

  14. 2004 Q9e)Calculations involving Kc – • In an experiment 6 moles of nitrogen and 18 moles of hydrogen were mixed and allowed to come to equilibrium in a 5L sealed vessel at a certain temperature. At equilibrium 6 moles of ammonia were present. Find the value for Kc N2 + 3H2 2NH3

  15. Kc = [NH3]2 [N2][H2]3 Kc = [1.2]2 [0.6][1.8]3 Kc = 0.4 L2/mol2

  16. Calculations involving Kc – 2010 Q7c • In an experiment 208.5g of Phosphorous chloride were placed in a 100L flask. The reaction was allowed to reach equilibrium. The mass of Chlorine gas present was 53.25g. Calculate the equilibrium constant for the reaction at 500K. PCl5 PCl3 +Cl2

  17. Kc = [PCl3] [Cl2] [PCl5] Kc = [0.0075] [0.0075] [0.0025] Kc =0.025 mol/L

  18. 2008 Q7) If 11 moles of hydrogen gas and 11 moles of iodine gas mixed at 764K. At equilibrium 17 moles of HI are present. H2 + I2 2HI

  19. I) Write the expression for Kc • Kc = [HI]2 • [H2][I2]

  20. H2 + I2 2HI

  21. Kc = [HI]2 [H2][I2] Kc = [17]2 [2.5][2.5] Kc = 46.25

  22. Finding equilibrium concentrations when Kc is given 2005 Q9c) The value of Kc is 4.0 at 373K. What mass of ethyl ethanoate would be present in the equilibrium mixture if 15g of ethanoic acid and 11.5g of ethanol were mixed and equilribrium established at this temperature?

  23. Kc = [CH3COOC2H5 ][H20] [CH3COOH ][C2H5OH] 4 = [CH3COOC2H5 ][H20] [CH3COOH ][C2H5OH]

  24. 4 = [CH3COOC2H5 ][H20] [CH3COOH ][C2H5OH] 4 = [x][x] [0.25-x ][0.25-x] If you cannot take the square root of all sides, then you should find the quadratic equation and use 4 = [x] 2 [0.25-x ]2 2 = [x] [0.25-x ]

  25. Sub x back into your table to see equilibrium conc. • 2 [0.25-x ] = [x] • 0.5 – 2x = x • 0.5 = 3x • X = 0.167

  26. Mass of ethyl ethanoate at equilibrium... • 0.167 moles of ethyl ethanoate present • 0167 x RMM = mass present at equilibrium • 0.167x 88g = 14.67g, the mass of ethyl ethanoate at equilibrium

  27. Objectives • Calculating the value of Kc for a reaction • Le Chatelier’s principle. • Effect (if any) on equilibrium position of concentration

  28. Homework • Equilibrium • HL 2012 q11b • Revision – Water and sodium thiosulfate titrations • HL 2007 Q1 • Not on OL syllabus – instead 2011 q6, 2010 q1

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