Presentation on theme: "Chemistry. Chemical equilibrium-I Session Objectives."— Presentation transcript:
Session objectives 1.Dynamic nature of equilibrium 2.Equilibrium in physical processes. 3.General characteristics of equilibrium involving physical processes. 4.Equilibrium in chemical process 5.Law of mass action 6.Equilibrium constant in gaseous systems. K P, K C 7.Relation between K P and K C 8.Application of equilibrium constant
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.
Equilibrium in Physical Processes Solid-Liquid Equilibrium This state is represented as: At equilibrium: Rate of melting = Rate of freezing
Liquid-Gas Equilibrium Rate of evaporation = Rate of condensation. Gases in Liquids At equilibrium
Henry’s law Where, m is the mass of gas p is the applied pressure k is the Henry’s constant.
Equilibrium in Chemical Process The state in which both the reactants and products co-exist considered chemical equilibrium. Forward reaction Backward reaction Rate of reaction Time
Equilibrium in Chemical Process Rate of forward reaction = Rate of backward reaction. Rate of formation of HI = Rate of decomposition of HI. H 2 (g) + I 2 (g) 2HI(g) At equilibrium:
Characteristics of chemical equilibrium 1.The concentration of each of the reactants and its products becomes constant at equilibrium. 2.At equilibrium, the rate of forward reaction =rate of backward reaction 3.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. 4. At equilibrium the free energy changes of the system, i.e.,
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. For example, aA + bB cC + dD
Law of Mass Action Rate of forward reaction [A] a [B] b Rate of forward reaction = k f [A] a [B] b Similarly Rate of backward reaction = k b [C] c [D] d k b = rate constant of backward reaction then At equilibrium k f [A] a. [B] b = k b [C] c. [D] d. K C used for molar concentration wherek f = rate constant of forward reaction
Illustrative example 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.
Solution Initial At equ. 1–x 1.5 –2x 2x (moles)
Equilibria in gas-phase reactions( K p ) Let A, B, C and D be gases in the following gaseous equilibrium: Where p A, p B, p C, and p D are the partial pressures of the gases A, B, C and D, respectively in the mixture, at equilibrium.
Relation between K p and K c We know, partial pressure aA + bB cC + dD partial pressure of A, Suppose volume is 1 L then
Relation between K p and K c Similarly for B,C and D n = (No. of gaseous moles of products) – (No. of gaseous moles of reactants)
Application: Equilibrium Constant K = Equilibrium constant (ii) 2A + 2B 2C + 2D
Application : Equilibrium Constant A + B C + D K 3 = K 1 × K 2 (v) And for A – B C – D (iv)A C K 1 B D K 2
Class exercise 1 The equilibrium system was found to contain [SO 2 ] = 0.40 M, [O 2 ] = 0.13 M, [SO 3 ] = 0.70 M. Find K c : (a) 12(b) 23.6 (c) 9(d) 48
Solution At eqm conc. Hence, answer is (b).
Class exercise 2 If N 2 O 4 is 25% dissociated in a 4 litre vessel at given temperature. What is the K c for the reaction ? (a) 1/16(b) 1/4 (c) 1/6(d) 1/12
Solution Initial 1 0 Hence, answer is (d).
Class exercise 3 Calculate the equilibrium constant for the reaction If the equilibrium constant at 1395 K for the following are: at 1395 K. k 1 = 2.1 × 10 –13... (i) k 2 = 1.4 × 10 –12... (ii) (a) 3.26(b) 4.28(c) 1.34(d) 2.58
Solution k 1 = 2.1 × 10 –13... (i) k 2 = 1.4 × 10 –12... (ii) Subtracting equation (ii) from equation (i), we get Divide equation (iii) by equation (ii), we get
Class exercise 4 Equimolar quantities of HI, H 2 and I 2 are brought to equilibrium. If the total pressure in the vessel is 1.5 atm and Kp for the reaction is 49. Which of the following is the correct value for the equilibrium partial pressure of I 2 ? (a) 0.4(b) 0.9(c) 0.7(d) 0.5 Initial p p p Final p – 2x p + x p + x Solution:
solution p + x = 7p – 14x
Class exercise 5 In the following gaseous equilibrium P 1, P 2 and P 3 are partial pressures of X 2, Y 2 and XY 2 respectively. P 1 P 2 P 3 The value of Kp: (b) (P 1 P 2 )P 3 P 1 P 2 P 3 Solution:
Class exercise 6 60 ml of H 2 and 42 ml of I 2 are heated in a closed vessel. At equilibrium the vessel contains 28 ml of HI. Calculate percentage of dissociation of HI. Solution: Initial moles (Given) At equilibrium60 – x 42 – x 2x 2x = – – x = 14 = 46 = 28
Solution On solving
Class exercise 7 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 = ) Solution: Moles at (1 – 0.4) equilibrium Total moles = 1.4 Initial moles 1 0 0
Solution at equilibrium
Class exercise 8 K C for the reaction, is 0.5 mol -2 litre 2 at 400 K. Find K p. Given R = litre atm degree –1 mol –1. = × 10 –4 atm –2 Solution:
Class exercise 9 For a reaction, At equilibrium 7.8 g, g and g of H 2. I 2 and HI were found respectively at equilibrium. Calculate K c. Solution: = 12.8 = 3.9 = 0.8 = 0.019
Class exercise moles of H 2 and 0.5 moles of I 2 react in 10 litre flask at 448°C. The equilibrium constant (K C ) is 50 for (a) What is the value of K p ? (b) Calculate moles of I 2 at equilibrium Initial Final 0.5 – x 0.5 – x 2x Solution:
Solution Moles of I 2 at equilibrium = 0.50 – 0.39 = 0.11 mole