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Chemistry

Chemistry. Chemical equilibrium-II. Session Objectives. Session Objectives. Homogeneous equilibria Heterogeneous equilibria Prediction of the direction of a reaction — reaction quotient. Characteristics of equilibrium constant (K). Calculation of equilibrium concentrations.

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Chemistry

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  1. Chemistry

  2. Chemical equilibrium-II

  3. Session Objectives

  4. Session Objectives • Homogeneous equilibria • Heterogeneous equilibria • Prediction of the direction of a reaction — reaction quotient. • Characteristics of equilibrium constant (K). • Calculation of equilibrium concentrations. • Degree of dissociation and vapour density. • Le Chatellier’s principle.

  5. Types: Equilibrium Homogeneous equilibria All the reactants and products of an equilibrium reaction are present in the same phase For example,

  6. Types: Equilibrium Heterogeneous equilibria: All the reactants and products are present in more than one phases For example The concentration of pure solids and pure liquidsis taken as 1.

  7. and Where is partial pressure of carbon dioxide. Types: Equilibrium Heterogeneous equilibria:

  8. Prediction of direction of reaction – reaction quotient When, Qc > Kc — (Backward reaction will occur, the reaction will proceed in direction of reactants) When, Qc < Kc — (Forward reaction will occur, the reaction will proceed in direction of products) When, Qc = Kc — The reaction will be in equilibrium

  9. Characteristics of Equilibrium constant (K) • Its value is independent of original concentration of reactants, pressure or presence of a catalyst. • It is independent of the direction from which equilibrium is attained. • Its value is constant at certain temperature and would change with temperature. • The larger the value of ‘K’, greater will be the value of products as compared to reactants. • Its value, however, depends on the coefficients of balanced reactants and products in the chemical equation.

  10. Calculation of Equilibrium concentration Considering the equilibrium Let the reaction starts with ‘a’ moles of H2 and ‘b’ moles of I2 taken in a container of volume V litres. ‘x’ moles of H2 have reacted when the equilibrium is attained, then ‘x’ moles of I2 will also react and 2x moles of HI are produced at T K temperature.

  11. Calculation of Equilibrium concentration Initial moles 1 1 0 Moles at eqm a – x b – x 2x

  12. Calculation of Equilibrium concentration Suppose the total pressure at equilibrium is ‘P’ atm. and total moles at equilibrium = a – x + b – x + 2x = a + b

  13. Calculation of Equilibrium concentration 1. KP = KC. 2. No volume or pressure term is involved in the expression for KP or KC. 3. Pressure has no effect on the state of equilibrium.

  14. Reactions in which Suppose initially one mole of PCl5 is present in a vessel of V litres and x moles of PCl5 dissociates at equilibrium at T K temperature. As per stoichiometry of the reaction, ‘x’ moles of PCl5 dissociates to give ‘x’ moles of PCl3 and ‘x’ moles of Cl2 at equilibrium, Initial moles 1 0 0

  15. Reactions in which If the total equilibrium pressure is ‘p’ and total moles at equilibrium are 1 – x + x + x = 1+ x, The partial pressure of each gas is

  16. If the degree of dissociation is very small then Hence KP = x2·P Reactions in which 1. Equilibrium is affected by the pressure. An increase in the value of ‘P’ will prefer backward reaction and the value of ‘x’ decreases. 2. Increase in the conc. of PCl5 favours the forward while the increase in the concentration of PCl3 or Cl2 favours the backward reaction.

  17. Characteristics of Equilibrium constant (K) For example in the reaction If the same reaction is written as i.e, for this reaction which can be balanced in two ways

  18. PCl5 PCl3 + Cl2 Degree of dissociation and vapour Density Initial conc. 1 0 0Conc. at eqm. (1 – a) aa Where a = degree of dissociation Total number of moles at equilibrium = 1 + aFrom ideal gas equation, PV = nRT =

  19. \ a Degree of dissociation and vapour Density \ Vapour density = [M = 2 × V.D.] Again, Vapour density = \ Vapour density Let initial and equilibrium vapour density are D and d respectively, then This is valid for those equilibrium where Kp exists

  20. Question

  21. dissociated. Illustrative example 1 The vapour density of PCl5 at 250° C at equilibrium is found to be 57.9. Calculate percentage dissociation at this temperature. Solution: Degree of dissociation is related with vapour density as

  22. Free energy change and equilibrium constant DG° = – RT ln K Where K is equilibrium constant, R is gas constant and T is absolutely temperature.

  23. Question

  24. Free energy change and equilibrium constant DG° = – RT ln K Where K is equilibrium constant, R is gas constant and T is absolute temperature.

  25. Le Chatelier’s principle According to this principle, ‘When the equilibrium is disturbed in a chemical reaction by changing any external factor such as, concentration, pressure, temperature etc. then the equilibrium will be shifted in a direction to minimize the effect of the change. Considering the following example. Let a, b and c moles of PCl5, PCl3, Cl2 respectively are present at equilibrium and P is eqm. Pressure.

  26. Le Chatelier’s principle

  27. Le Chatelier’s principle (Ideal gas equation.) Effect of concentration Increase in concentration of reactants and decrease in product the reaction moves to forward direction and vice-versa.

  28. Effect of pressure/volume If the pressure decreases (increasing the volume) The reaction moves to more no. of moles of gas With increased pressure Effect will be opposite

  29. Effect of inert gas Inert gas at constant volume there will be no effect on the equilibrium Inert gas at constant pressure For reactions Dn > 0, the equilibrium shifts to greater number of moles and vice versa. If Dn = 0, there will be no effect.

  30. Effect of temperature For an endothermic reaction, (DH is +ve) K2 > K1, i.e. the reaction will move towards forward direction. Favoured by high temperature While for an exothermic reaction (DH is –ve), thenK2 < K1 and the reaction moves toward backward direction.Favoured by low temperature

  31. Class exercise

  32. In the following gaseous equilibrium p1, p2 and p3 are partial pressures p1 p2 p3 The value of Kp: Class exercise 1 Solution: Hence, the answer is (c).

  33. The equilibrium constant for the reaction is 9. If one mole of each of W and X are mixed and there is no change in volume, number of moles of Y formed is (a) 0.10 (b) 0.50 (c) 0.75 (d) 0.90 Class exercise 2

  34. Solution 1 1 0 0 1 – x 1 – x x x x = 3 – 3x x = 0.75 Hence, the answer is (c).

  35. The equilibrium constant at 323°C is 1000. What would be the value in the presence of a catalyst for the following reaction (a) 1000 (Concentration of catalyst)(b) 1000 (c) 1000/Catalyst (d) Cannot be determined Class exercise 3

  36. Solution A catalyst only speeds up or slows down the rate of reaction in a particular direction. Its only the temperature that can change the value of KC. Hence, the answer is (b).

  37. The Keq for the dissociation of iodine If the equilibrium concentration of atomic iodine is What is the concentration of molecular iodine? (a) 0.8M (b) 0.4 M (c) 0.3 M (d) 0.2 M Class exercise 4

  38. Solution = 0.4 M Hence, the answer is (b).

  39. NO2 associates (or dimerises) asthe apparent molecular weight of a sample of NO2 calculated from the vapour density under certain conditions was 60, the mole fraction of dimer is (a) 14/46 (b) 16/46 (c) 28/46 (d) 46/60 Class exercise 5

  40. Solution Initial 1 0 At eqm. 1–2x x (moles)

  41. Solution Hence, answer is (a).

  42. Class exercise 6 At 700 K, hydrogen and bromine react to form hydrogen bromide. The value of equilibrium constant for this reaction is 5 108. Calculate the amount of H2, Br2, and HBr at equilibrium if a mixture of 0.6 mol of H2 and 0.2 mol of Br2 is heated to 700 K.

  43. Solution Such a large value indicates near completion reaction H2 = 0.6 – 0.2 = 0.4 moles Br2 = 0 moles HBr = 0.4 moles

  44. The Kp for the reaction is 640 mm of Hg at 775 K. Calculate the percentage dissociation of N2O4 at equilibrium pressure of 160 mm of Hg . At what pressure, the dissociation will be 50% Class exercise 7 Solution:

  45. solution

  46. Class exercise 8 The degree of dissociation of N2O4 into NO2 at one atmosphere and 40°C is 0.310. Calculate its Kp at 40°C. Also report the degree of dissociation at 10 atm pressure and same temperature. Solution: Total number of moles = 1.31 moles

  47. Solution

  48. In an equilibrium A and B are mixed in a vessel at temperature T. The initial concentration of A was twice the initial concentration of B. After the equilibrium has reached, concentration of C was thrice the equilibrium concentration of B. Calculate KC. Class exercise 9

  49. Solution Initial 2x x 0 0 Final 2x – y x – y y y

  50. What will be the effect on the reaction equilibrium of (i) increased temperature (ii) decreased pressure (iii) presence of a catalyst (iv) lower concentration of N2 Class exercise 10

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