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EQUILIBRIUM 179 Calculate K cor the reaction, the value 0.194 should be neglected because it will give concentration of the reactant PCl 5 PCl 3 Cl 2 which is more than initial concentration. Solution Hence the equilibrium concentrations are, The equilibrium constant K c for the above [CO ] = [H ] = x = 0.067 M reaction can be written as, 2 2- PCl Cl 1 59 ▯ [CO] = [H O] = 0.1 – 0.067 = 0.033 M K ▯ ▯ 3▯ 2▯▯ ▯ ▯ 1 79 2 c ▯Cl 5▯ ▯ 41 ▯ Problem 6.5 Problem 6.4 For the equilibrium, The value of K = 4.24 at 800K for the 2NOCl(g) 2NO(g) + Cl (g) c 2 reaction, the value of the equilibrium constant, K c is CO (g) + H 2 (g) CO (2) + H (2) 3.75 × 10 –6at 1069 K. Calculate the K for CalculateequilibriumconcentrationsofCO , 2 the reaction at this temperature? p H , CO and H O at 800 K, if only CO and H O are present initially at concentrations Solution 2 of 0.10M each. We know that, ∆n Solution Kp= K cRT) For the reaction, For the above reaction, CO (g) + H O (g) CO (g) + H (g) ∆n = (2+1) – 2 = 1 2 2 2 K = 3.75 ×10 (0.0831 × 1069) Initial concentration: p Kp = 0.033 0.1M 0.1M 0 0 Let x mole per litre of each of the product be formed. 6.5 HETEROGENEOUS EQUILIBRI A At equilibrium: Equilibrium in a system having more than one (0.1-x) M (0.1-x) M x M x M phase is called heterogeneous equilibrium. where x is the amount of CO and H at The equilibrium between water vapour and 2 2 liquid water in a closed container is an equilibrium. Hence, equilibrium constant can be written example of heterogeneous equilibrium. H O(l) H O(g) as, 2 2 K c x /(0.1-x) = 4.24 In this example, there is a gas phase and 2 2 a liquid phase. In the same way, equilibrium x = 4.24(0.01 + x -0.2x) x2 = 0.0424 + 4.24x -0.848x between a solid and its saturated solution, 2 2+ – 3.24x – 0.848x + 0.0424 = 0 Ca(OH) 2s) + (aq) Ca (aq) + 2OH (aq) a = 3.24, b = – 0.848, c = 0.0424 is a heterogeneous equilibrium. 2 (for quadratic equation ax + bx + c = 0, Heterogeneous equilibria often involve pure solids or liquids. We can simplify ▯▯b ▯ b ▯ ac4 ▯ x ▯ equilibrium expressions for the heterogeneous 2a equilibria involving a pure liquid or a pure 2 solid, as the molar concentration of a pure x = 0.848±√(0.848) – 4(3.24)(0.0424)/ (3.24×2) solid or liquid is constant (i.e., independent of the amount present). In other words if a x = (0.848 ± 0.4118)/ 6.48 substance ‘X’ is involved, then [X(s)] and [X(l)] x = (0.848 – 0.4118)/6.48 = 0.067 1 are constant, whatever the amount of ‘X’ is x2 = (0.848 + 0.4118)/6.48 = 0.194 taken. Contrary to this, [X(g)] and [X(aq)] will Reprint 2026-27