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Chapter three Concept of chemical equilibrium If two reactants A and B are mixed together a chemical reaction can take place to yield the product C&D The chemical reaction can be presented as follows :aA + bB cC + dD …………….(1) a, b, c, d : stoichiometric coefficients in many chemical reaction the reverse reaction i.e reaction between c with d to form A&B the reaction is also possible hence the reaction is presented by: aA + bB cC + dD aA + bB cC + dD when forward and reverse reaction become equal, the system has reacted to the condition know as " Chemical equilibrium " the reaction appear to stop when equilibrium condition is actually reached since there is no change in the concentration of all substances The equilibrium constant the equilibrium condition is often expressrd by a factor or known as equilibrium constant Chemical Equilibrium Calculations Equilibrium constants are useful for calculating the concentration of varies ions- in equilibrium the following examples give an idea for calculations using equilibrium constants Example |(1) the chemical substances (A) and (B) are reacted to produce (C) and (D) according to the reaction A + B C + D if 0.2 mol of (A) and 0.5mol of (B) are dissolved in 1 liter solution and the reaction takes place with equilibrium constant K = 0.3. calculate the concentrations of reactants and products at equilibrium let x mol/l of A and B that is reacted, equilibrium constant of [A] = 0.2-x mol/l let x mol/l of A and B that is reacted, equilibrium constant of [B] = 0.5 –x mol/l [C] = x mol/l [D] = x mol/l [ ][ ] [ ][ ] K= = = 0.3 x2 = 0.3 (0.2-x) (0.5-x) 0.7x2 + 0.21 x – 0.03 = 0 This is a quadratic equation and can be solved by general quadratic formula x= √ x= -0.21± √ = 0.11M [A] = 0.09 , [B] = 0.39 , [C] = 0.11 , [D] = 0.11M Example (2) Calculate the equilibrium concentration of A and B in a 0.10 M solution of weak electrolyte AB with an equilibrium constant of 3× 10-6 AB A+B K= [ ][ ] [ ] Let x = equil. Conc. of A and B , AB = 0.1- x K= quite small compare to the initial conc. Hence [AB]=0.1-x 0.1 3×10-6 = =√ = 5.5× 10-4 M = A = B Note : this simplification will generally hold if the initial conc. is ≥ 100k , other wise , a quadratic equation will have to be solved factors. Affecting on chemical equilibrium 1. 2. 3. 4. Temp. Pressure Conc. Catalysts Example (3) A solution was prepared by dissolving 0.1 mol of trichloroacetic Acid (CCl3COOH) in enough water to make 1.0liter of solution at 25 the dissociation constant Ka for trichloroacetic acid = 0.129 calculate :a) The percent dissociation of trichloroacetic acid b) The percent dissociation of trichloroacetic if 0.1mole of soluble salt of trichloroacelate (CCl3COO-) is added to the solution (neglect any reaction of trichloroacetate with water) The chemical reaction CCl3COOH↔ CCl3COO- + H+ Ka = [ ][ [ ] = 0.129 ] Let x = moles of the acid dissociates during the reaction = (0.1- x) M [CCl3COOH] = =(x) M [CCl3COO-] = [H+] = = 0.129 x2 + 0.129x – 0.0129 = 0 → √ X= = 0.066 mole % dissoiciation of trichloroacetic acid = × 100 = 66% b) if 0.1 moles of trichloroacetate is added to the solution hence [CCl3COOH] = (0.1 – x) M [H+] = x M [CCl3COO-] =( = 0.129 )M → = 0.129 X2 + 0.229 X – 0.0129 = 0 then X= √ X = 0.0465 mole % dissociation of trichloroacetic acid = 46.5 % Hence It is concluded that the addition of one of the products decreases the dissociation of the acid from 66% to 46.5% which shows the effect of concentation