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Chemistry

Chemistry. Chemical equilibrium-I. Session Objectives. Session objectives. Dynamic nature of equilibrium Equilibrium in physical processes. General characteristics of equilibrium involving physical processes. Equilibrium in chemical process Law of mass action

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Chemistry

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  1. Chemistry

  2. Chemical equilibrium-I

  3. Session Objectives

  4. Session objectives • Dynamic nature of equilibrium • Equilibrium in physical processes. • General characteristics of equilibrium involving physical processes. • Equilibrium in chemical process • Law of mass action • Equilibrium constant in gaseous systems. KP, KC • Relation between KP and KC • Application of equilibrium constant

  5. Dynamic equilibrium At equilibrium, the reaction does not stop. moles of reactant converted into product = moles of product converted into reactant in the same time.It is considered as dynamic equilibrium.

  6. Equilibrium in Physical Processes Solid-Liquid Equilibrium This state is represented as: At equilibrium: Rate of melting = Rate of freezing

  7. Liquid-Gas Equilibrium At equilibrium Rate of evaporation = Rate of condensation. Gases in Liquids

  8. Henry’s law Where, m is the mass of gas p is the applied pressure k is the Henry’s constant.

  9. Forward reaction Rate of reaction Backward reaction Time Equilibrium in Chemical Process The state in which both the reactants and products co-exist considered chemical equilibrium.

  10. H2(g) + I2 (g) 2HI(g) Equilibrium in Chemical Process At equilibrium: Rate of forward reaction = Rate of backward reaction. At equilibrium: Rate of formation of HI = Rate of decomposition of HI.

  11. 4. At equilibrium the free energy changes of the system, i.e., Characteristics of chemical equilibrium • The concentration of each of the reactants and its products becomes constant at equilibrium. • At equilibrium, the rate of forward reaction =rate of backward reaction • A chemical equilibrium can be attained from either direction, i.e., from the direction of the reactants as well as from the direction of products.

  12. For example, aA + bB cC + dD Law of Mass Action Law of mass action states that rate of a chemical reaction is proportional to the product of the active masses of the reactants raise to the power of stoichiometric coefficients of balanced chemical equation.

  13. Rate of forward reaction [A]a [B]b Rate of forward reaction = kf [A]a [B]b Law of Mass Action where kf = rate constant of forward reaction Similarly Rate of backward reaction = kb [C]c [D]d kb = rate constant of backward reaction then At equilibrium kf [A]a . [B]b = kb [C]c . [D]d. KC used for molar concentration

  14. Determine the value of the equilibrium constant for the reaction if 1.0 mol of A and 1.5 mol of B are placed in a 2.0-litre vessel and allowed to come to equilibrium. The equilibrium concentration of C is 0.35 mol L-1. Illustrative example

  15. Solution Initial 1 1.5 0 At equ. 1–x 1.5 –2x 2x(moles)

  16. Equilibria in gas-phase reactions(Kp) Let A, B, C and D be gases in the following gaseous equilibrium: Where pA, pB, pC, and pD are the partial pressures of the gases A, B, C and D, respectively in the mixture, at equilibrium.

  17. aA + bB cC + dD Relation between Kp and Kc We know, partial pressure partial pressure of A , Suppose volume is 1 L then

  18. Relation between Kp and Kc Similarly for B,C and D Dn = (No. of gaseous moles of products) – (No. of gaseous moles of reactants)

  19. (ii) 2A + 2B 2C + 2D Application: Equilibrium Constant K = Equilibrium constant

  20. (iv) A C K1 B D K2 A + B C + D K3 = K1 × K2 (v) And for A – B C – D Application : Equilibrium Constant

  21. Questions

  22. The equilibrium system was found to contain [SO2] = 0.40 M, [O2] = 0.13 M, [SO3] = 0.70 M. Find Kc: (a) 12 (b) 23.6(c) 9 (d) 48 Class exercise 1

  23. Solution At eqm. 0.4 0.13 0.7conc. Hence, answer is (b).

  24. If N2O4 is 25% dissociated in a 4 litre vessel at given temperature. What is the Kc for the reaction ? (a) 1/16 (b) 1/4(c) 1/6 (d) 1/12 Class exercise 2

  25. Solution Initial 1 0 Hence, answer is (d).

  26. Calculate the equilibrium constant for the reaction at 1395 K. If the equilibrium constant at 1395 K for the following are: k1 = 2.1 × 10–13... (i) k2 = 1.4 × 10–12 ... (ii) (a) 3.26 (b) 4.28 (c) 1.34 (d) 2.58 Class exercise 3

  27. Solution k1 = 2.1 × 10–13... (i) k2 = 1.4 × 10–12... (ii) Subtracting equation (ii) from equation (i), we get Divide equation (iii) by equation (ii), we get

  28. Solution

  29. Equimolar quantities of HI, H2 and I2are brought to equilibrium. If the total pressure in the vesselis 1.5 atm and Kp for the reaction is 49. Which of the following is the correct value for the equilibrium partial pressure of I2? (a) 0.4 (b) 0.9 (c) 0.7 (d) 0.5 Class exercise 4 Solution: Initial p p p Final p – 2x p + x p + x

  30. solution p + x = 7p – 14x

  31. In the following gaseous equilibrium P1, P2 and P3 are partial pressures of X2, Y2 and XY2 respectively. P1 P2 P3 The value of Kp: (b) (P1P2)P3 Class exercise 5 Solution: P1 P2 P3

  32. Class exercise 6 60 ml of H2 and 42 ml of I2 are heated in a closed vessel. At equilibrium the vessel contains 28 ml of HI. Calculate percentage of dissociation of HI. Solution: Initial moles 60 42 0 (Given) At equilibrium 60 – x 42 – x 2x 2x = 28 60 – 14 42 – 14 28 x = 14 = 46 = 28

  33. Solution Again

  34. Solution On solving

  35. The percentage of dissociation is 0.4 at 400 K and 1.0 atm for the gaseous reaction . Assuming ideal behaviour of all the gases, calculate the density of equilibrium mixture at 400 K and 1.0 atmosphere. (Given observed molecular mass at equilibrium = 148.92) Class exercise 7 Solution: Initial moles 1 0 0 Moles at (1 – 0.4) 0.4 0.4equilibrium Total moles = 1.4

  36. Solution at equilibrium

  37. KC for the reaction, is 0.5 mol-2 litre2 at 400 K. Find Kp. Given R = 0.082 litre atm degree–1 mol–1. Class exercise 8 Solution: = 4.636 × 10–4 atm–2

  38. For a reaction,At equilibrium 7.8 g, 203.2 g and 1638.4 g of H2 . I2 and HI were found respectively at equilibrium. Calculate Kc. Class exercise 9 Solution: = 12.8 = 3.9 = 0.8 = 0.019

  39. 0.5 moles of H2 and 0.5 moles of I2 react in 10 litre flask at 448°C. The equilibrium constant (KC) is 50 for (a) What is the value of Kp? (b) Calculate moles of I2 at equilibrium Class exercise 10 Solution: Initial 0.5 0.5 0 Final 0.5 – x 0.5 – x 2x

  40. Solution Moles of I2 at equilibrium = 0.50 – 0.39 = 0.11 mole

  41. Thank you

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