Chemical Equilibrium Solved Problems

Page # 1 CHEMICAL EQUILIBRIUM CHEMICAL EQUILIBRIUM 1. REACTION The transformation of one substance into another is known as reaction. (b) Reactions proceed in both directions. (c) These type of reactions never get completed. (d) Always carried out in closed vessel. 2. REACTANT The substances used at initial stage of reaction are known as reactant. B. Some examples of reversible reactions are as follows – (a) All thermal dissociation in closed vessel. 3. PRODUCT The substances formed at final stage of reaction are known as product. 4. TYPES OF REACTIONS ON THE BASIS OF DIRECTION On the basis of the direction, reactions are of two types. Reaction Irreversible Reversible 4.1 Irreversible reactions : The chemical reactions in which the products formed do not combine to give back the reactants are known as irreversible reaction. A. Properties of irreversible reactions – (a) Reactions in which reactant react to form product only. (b) Reactions proceed in one single direction. CaCO3 CaO PCl5 PCl3 NH4Cl + H2O Salt of weak base + strong acid (b) All the neutralisation reaction of strong acid and strong base. (c) All precipitation reactions. (d) All gaseous reactions in open vessel. 4.2 Reversible reactions : The reactions in which the products can react with one another to give back the reactants again under suitable conditions called as reversible reaction. A. Properties of reversible reactions – (a) In these type of reactions, reactant react to form product and product further react to form reactant. Cl2 NH4OH + HCl (c) All gaseous reactions in closed vessel. N2 + O2 2NO H 2 + I2 2HI 2 SO3 2SO2 + O2 N2 + 3H2 2NH3 (d) All the reactions of organic hydrolysis. CH3 – C – OC2H5 + H–OH || O CH3 – (d) Always carried out in open vessel. B. Some examples of irreversible reactions are as follows – (a) All thermal decomposition in open vessel. + CO2 (b) All types of salt hydrolysis except strong acid strong base salt. (c) Always proceed to completion. (e) In these type of reactions, If product are gaseous in state, then they can escape from reacting site. If they are solid in state they will precipitate. + C – OH + C2H5 – OH || O Reversible reactions are also of two type- Reversible Reaction Homogenous reversible reaction Heterogeneous reversible reaction 4.2.1 Homogenous reversible reactions : If in the reversible reactions physical state of all reactants and products are same, it is known as called as homogenous reversible reaction. N2(g) + O2 (g) 2 NO(g) Graphite (s) SRh. (s) Diamond (s) Smono. (s) Page # 2 CHEMICAL EQUILIBRIUM If in the reversible reaction, physical state of reactant and product are different, it is known [R]0 Fe(s) + + CO(g) Fe(s) + CO2(g) H2O () Fe (OH)2(s) + H2(g) FeCl3(s) + 3H2O () Fe(OH)3(s) + 3HCl () H2O ( ) [P] [P]0 forward reaction becomes equal to the rate of backward reaction is called chemical equilibrium. (b) State where rate of formation of product is equal to the rate of dissociation of reactant is known as chemical equilibrium. (c) State of reversible reaction in which the concentration of reactant and concentration of product remain unchanged but not necessarily equal,is called chemical equilibrium. (d) State where exchange of energy becomes zero is also called as chemical equilibrium. 5.1 Characteristics of chemical equilibrium : (a) It is a dynamic equilibrium i.e. at this stage of reaction takes place in both the directions of same speed although appears to be stopped. (b) On the stage of equilibrium the reaction proceeds both the side. (c) At the state of equilibrium, both reactants and products are present and their concentration do not change with respect to time. (d) The state of equilibrium is not effected by the presence of catalyst : It only helps to attain the equilibrium state in less or more time. tem pera ture or concentration favours one of the reactions and thus shifts the equilibrium point in one direction. (f) Change in free energy is equal to zero at Time Equilibrium State [R] [P]0 Time 6. LAW OF MASS ACTION (a) This law was given by Guldberg and Waage. (b) Accordi ng to th is l aw, the rate of dissociation of reactant and the rate of formation of product is directly proportional to active mass of itself at constant temperature. A + B  C + D dA  [A] dt dB  [B] dt dC   [C] dt dD   [D] dt While, the rate of a chemical reaction is directly proportional to the product of active masses of reactants. r  [A] [B] r = K [A] [B] 6.1 Active Mass or activity or effective concentration or molar concentration – (a) It is the molecular concentration of a substance i.e. the no. of gram moles per litre. (b) It is represented by enclosing its formula in square brackets ( [ ] ) of a substance. Active mass = Molar concentration No. of moles No. of grams of subs tan ce = = volume (in litres) Mol. wt.  volume (in litre) Note : The active mass of solids is one. eg. What is the active mass of 4 gm H2 in 2 litre volume ? the state of equilibrium (G = 0). [H2] = 1 [P]0 Time [R] = [P] Equilibrium State [P] (a) State of reversible reaction at which rate of in pressure, Equilibrium State [R]0 H2O (g) 5. CHEMICAL EQUILIBRIUM (e) Change [R] Concentration FeO(g) CaO(s) + CO2(g) [R]0 Concentration as heterogeneous reversible reaction. CaCO3(s) 5.2 Graphical representation of chemical equilibrium – R P Concentration 4.2.2 Heterogenous reversible reaction : 4 = 1 22 Page # 3 CHEMICAL EQUILIBRIUM eg. What is the active mass of 14 gm N2 in 2 litre volume ? 14 1 [N2] = = 28  2 4 1 eg. What is the active mass of 14 gm N2 in 2 litre volume ? 14  2 [N2] = = = 1 1 28 28  2 Note: – Law of mass action is applicable only for reversible reaction. Here, V is the volume of container in litre. According to the law of mass action [NO] 2 [N2 ][O 2 ] Substituting the values in the above equation KC = FG 2x IJ H vK FG a  x IJ FG b  x IJ H v KH v K 2 14 6.2 Mathematical Expression of law of mass action : At the constant temperature, let us consider the following reversible reaction. A + B C + D According to law of mass action Rate of forward reaction rf  [A] [B] or rf = K1 [A] [B] where K1 is the rate constant of the forward reaction. Rate of backward reaction rb  [C] [D] or rb = K2 [C] [D] where K2 is the rate constant of the backward reaction. At equilibrium : Rate of forward reaction = Rate of backward reaction K1 [A] [B] = K2 [C] [D] K1 or K 2 [C] [D] = [ A ] [B] or Here, k is equilibrium constant of given reversible reaction. 6.3 Applications of law of mass action : – 6.3.1 Formation of Nitric Oxide : (n = 0) A. Calculation of KC : – Suppose the initial concentration of N2 and O2 is a and b respectively. x is the degree of dissociation. N2 + O2 2NO Initial a b 0 concentration moles at (a–x) (b–x) 2x equilibrium Active mass (mol l–1) (a  x) V (b  x) V 2x V KC = KC = 4 x2 (a – x) (b – x) B. Calculation of KP : All the things being same as above, except pressure. Let P atmosphere is the pressure at equilibrium. N2 + Initial a O2 2 NO b 0 concentration moles at (a–x) (b–x) 2x equilibrium Total no. of moles = (a–x) + (b–x) + 2x = (a+b) Now, the partial pressure of the above three species can be calculated as below(a  x) P PN = 2 (a  b) (b  x) P PO2 = ( a  b) (2x) P PNO = (a  b) According to the law of mass action KP [PNO ] 2 = [P ][P ] N2 O2 substituting the value of PNO , PN2 , PO2 in the above equation of KP - LM (2x)P OP N (a  b) Q LM (a  x)P OP LM (b  x)P OP N (a  b) Q N (a  b) Q 2 KP = KP = 4x2 ( a  x)(b  x) Page # 4 CHEMICAL EQUILIBRIUM 6.3.2 Thermal Dissociation of Phosphorus Penta-chloride- (n > 0) A. Calculation of KC - Suppose one mole of PCl5 is take in a closed container of V litre. Further at equilibrium x mol of it dissociated PCl5 PCl3 + Cl2 Initial 1 0 0 concentration moles at (1–x) x x equilibrium 1 x x x Active mass v v v (mol l–1) According to law of mass action At equilibrium PPCl3 = PCl2 = Substituting the values in the above equation. FG x IJ FG x IJ H vK H vK FG 1  x IJ H vK KC = So, x2 v = KC .V  V KC = x2 x2 x V If we increase the volume, the dissociation x is also increased. B. Calculation of KP PCl5 PCl3 + Cl2 initial 1 0 0 concentration moles at 1–x x x equilibrium Total no. of moles at equilibrium, (1 – x) + x + x = (1 + x) moles According to law of mass action KP = PPCl3  PCl2 PPCl5 (1  x)P (1  x) Substituting the values in the above equation of KP - KP = KP FG x  P IJ FG x  P IJ H 1 x K H 1 x K (1  x)  P (1  x) x 2P = 1  x2 The equation of K P is not independent of pressure. suppose, x << 1 then 1 – x2  1 x2 P KP = P 1  P KP = x2 x2 The formula of KC has V in the denominator, hence the equilibrium will be affected by V of the reaction container. If x << 1 then 1 – x  1 xP (1  x) PPCl5 = [PCl 3 ][Cl 2 ] [PCl 5 ] KC = xP (1  x) x 1 P The degree of dissociation of PCl5 is inversely proportional to the square root of pressure so, decrease of pressure increases dissociation of PCl5. 6.3.3 Formation of Ammonia – (n < 0) A. Calculation of KC : – N2 + 3H2 2NH3 Initial 1 3 0 concentration moles at (1–x) (3–3x) 2x equilibrium 1 x 3  3 x  2x    Active mass v v  v  (mol l–1) According to law of mass action FG H KC = IJ K FG H IJ K [NH 3 ] 2 [N 2 ][H 2 ] 3 Substituting the values in the above equation- FG 2x IJ H vK FG 1  x IJ FG 3  3x IJ H v KH v K 2 KC = 3 Page # 5 CHEMICAL EQUILIBRIUM KC = KC  The equation of K P is not independent of pressure suppose, x << 1 then, (1 –x)4 = 1 (2 – x)2 = 4 4x 2 v 2 (1  x)(3  3 x) 3 4x 2 v 2 27(1 – x )4 KP = The formula of KC has V in the denominator, hence the equilibrium will be affected by V of the reaction container. Dependence If, x << 1 then, (1 – x)4 = 1 KC x2 4x2 v 2 27 = V2 1 x2   x2  P 2 xP If we increase the pressure the degree of dissociation x is also increased. V2 1 V x If we increase the volume of the container the degree of dissociation x is decreased. B. Calculation of KP : N2 + 3H2 2NH3 Initial concentration 1 3 0 moles at equilibrium (1 – x) (3 – 3x) 2x Total number of moles at equilibrium = (1 – x) + (3 – 3x) + 2x = (4 – 2x) According to the law of mass action KP 27P 2 6.3.4 Liquid equilibrium (Ester formation) It is proposed by Bartholate and Gill. In this reaction all the reactants and products are in liquid state. KC = 64 x 2 = (PNH3 ) 2 PN2 = (1  x). P ( 4  2x) PH2 = (3  3 x).P (4  2x) (2x). P (4  2x) KP = (1  x) 3 (3  3 x) 2 P 2 KP  16 x (2 – x ) (b – x) x x bx v x v ax v (mole –1) According to law of mass action KC  2 27(1 – x ) 4 P 2 x v x2 (a – x )(b – x ) Note : The value of KP is equal to unity in liquid and solid equilibrium. 3 7. TYPES OF EQUILIBRIUM CONSTANT Equilibrium constant (K) is also of two types Equilibrium Constant (K) 4 x 2 (4  2x) 2 2 0 FG IJ FG IJ H K H K FG IJ FG IJ H v KH v K 2 KP = 0 Substituting the values in the above equation x x  v v KC = a  x b  x Substituting the values in the above equation of KP. FG 2x .PIJ H 4  2x K FG 1  x .PIJ FG 3  3x .PIJ H 4  2x K H 4  2 x K Active mass b [CH 3 COOC 2H 5 ][H 2 O] KC = [CH COOH][C H OH] 3 2 5 (PN2 )  (PH2 ) 3 At equilibrium PNH3 = Initial a concentration Moles at (a – x) equilibrium KC Equilibrium constant (in concentration terms) KP Equilibrium constant (in Pressure terms) Page # 6 CHEMICAL EQUILIBRIUM 7.1 Equilibrium concentration constant (Kc) : If we use the term concentration in given reaction with the equilibrium constant (K) is called equilibrium concentration constant (Kc). [C] [D] K = Kc = [ A ] [B] p C  pD Kp = p  p A B pA = pB = pC = pD = Partial pressure formula. Partial pressure of A Partial pressure of B Partial pressure of C Partial pressure of D of a gas given by the following p = No. of mole of given gas at equilibrium  P Total no. of moles at equilibrium or p = mole fraction × P. Here, P is the total pressure. 7.3 Factors which do not influence equilibrium constant : (a) Concentration of reactants and products. (b) Pressure and volume. (c) Presence of catalyst. (d) Addition of the inert gas at constant Pressure and volume. 7.4 Factors influence the equilibrium constant : A. Mode of representation of chemical reaction. B. Stoichiometry of reaction. C. Temperature. A. Mode of representation of reaction – If we take reaction2HI H2 + I2 Then, we write the value of equilibrium constant K C1 for the above reaction as following. K C1 = [H2 ][I 2 ] .......(i) [HI] 2 Now, if we take reaction H2 + I2 2HI Then, we write the value of equilibrium constant K C2 for above reaction as following K C2 [HI] 2 1 = [H ][I ]  K 2 2 C1 H 2 + 2 2H The expression of its equilibrium constant is- [H]2 [H2 ][ 2 ] If the equation of above reaction is written by following method – K C1 = 7.2 Equilibrium Pressure constant (Kp) : It we use the term pressure in given reaction with the equilibrium constant (K) is called equilibrium pressure constant (KP). where Now, we consider the formation of HI(g) by the combination of H2(g) and 2(g). ......(ii) B. Stoichiometry of the reaction – Method of writing the equation of the reversible reaction is called as stoichiometry of the reaction. 1 1 H2(g) +  (g) H(g) 2 2 2 The expression for the equilibrium constant is – K C2 = [H ] [H2 ] ½ [ 2 ] ½ on the basis of comparing both the equilibrium constant equation. K C2 = or ( K C 1 )1/2 K C1 Result – When we divide a reaction by a factor 'n' in the equation, the value of new equilibrium constant is equal to the root of n of the previous equilibrium constant. For Example – Suppose, the equilibrium constant for the following reaction. A + B C + D is K1 then for the reaction 1 1 1 1 A + B C + D n n n n the value of the equilibrium constant K2 is equal to n K 1 or (K1)1/n. C Temperature – Increase in temperature favours the endothermic reaction and decrease in temperature favours the exothermic reaction for the forward reaction. Means for exot hermic react ions, t he value of Kc and Kp decrease with rise in temperature while for endothermic reactions, the value of Kc and Kp increases with rise in temperature. This type of variation in equilibrium constant with temperature given by Van't Hoff equation as follows log K2 – log K1 = or K2 log K 1 = LM 1  1 OP NT T Q LM 1  1 OP H T Q 2.303R N T H 2.303R 1 2 1 2 Where, K2 = equilibrium constant at temperature T2 K1 = equilibrium constant at temperature T1 H = Energy of reaction of constant Page # 7 CHEMICAL EQUILIBRIUM temperature R = Molar gas constant According to the temperature, reaction are of three types. (a) Athermic or Non-thermic reaction means H = 0 log K2 – log K1 = 0 log K2 = log K1 Result : – There is no effect of temperature on this type of reaction. (b) Endothermic reaction  H = ( + ) ve log K2 – log K1 = ( + ) ve, means K2 > K1 Result : – On increase of temperature equilibrium constant will also increase for this type of reaction. (c) Exothermic reaction  H = ( – ) ve log K2 – log K1 = ( – ) ve, means K2 < K1 Result– On the increase of temperature equilibrium constant will decreases for exothermic reaction. 7.5 Units of Kp and KC– Partial pressure are measured by the unit of atmospheres and therefore units of Kp will be (Atmospheres)n. In the same way, the concentration is expressed in the term of moles per litre. Therefore, units of K will be (moles litre–1)n. c Here [ A ] n1 [B] n2 PV = n RT [C] m1 [D] m2 (RT) m1 m2 [ A] n1 [B] n2 (RT ) n1 n2 KP = KC ( RT) ( m1  m2 )  ( n1  n 2 ) K = K (RT)n P C n = (m1 + m2) – (n1 + n2) n = number of moles of product – number of moles of reactant. T = Absolute temperature. Note : n only includes gaseous moles. I. on the basis of n, the reactions are of three types. (a) Reaction in which n = 0 N2 + O2 2NO  n = 2 – 2 = 0 By the relation of K = K (RT)n   n = 0 P KP =  C KC (RT) So, (b) Reaction in which the value of n is negative2 CO + O2 2 CO2 n = 2 – 3 = – 1 By the relation of K = K (RT)n KP  (RT)–1 P C KC means KP × RT = KC RT (c) Reaction in which the value of n is positive– 2 SO3 2 SO2 + O2 n = 3 – 2 = + 1 By the relation of KP = KC (RT)n KP = KC (RT) +1 means KP = KC × RT So, .......(i) no. of moles n = = [ ] = Active mass v lit. (p C ) m1 (p D ) m2  KP = (p ) n1 (p ) n2 A B on putting the value of 'p' in the formula of KP by the equation . ........(i) 1 ([ A ] RT) n1 ([B] RT) n2 So, [C] m1 [D] m2 FG IJ H K KP = or 7.6 Relation between Kp and KC: Let us consider the following reaction n1 A + n2 B m1 C + m2 D. The value of Kc for the reaction is, According to gas law n P = RT v KP = KP = KC V alue of  n Units of K p Units of K c 0 No Unit No unit n (atm ) > 0 (m oles l –1) n n (atm ) < 0 (m oles l –1) n Kc = ([C] RT) m1 ([D] RT) m2 II. On the basis of the value of pressure, reactions are also of three types as follows(a) Reaction in which the value of n = 0 N2 + O2 2 NO n = 2 – 2 = 0 K = K (RT)n According to relationship KP = KC (RT)0 means P C Page # 8 CHEMICAL EQUILIBRIUM In this type of reaction, there is no effect of pressure. (b) Reaction in which the value of n is (+)ve. PCl5 PCl 3 + Cl 2 n = 2 – 1 = + 1 In this type of reaction, when pressure is increases then the degree of dissociation decreases. (c) Reaction in which the value of n is ( – )ve. N2 + 3H2 2NH3 n = 2 – 3 = – 1 In this type of reaction, when the pressure is increases then the degree of dissociation is also increases. 8. LE-CHATELIER'S PRINCIPLE (a) According to this principle, if a system in equil ibrium is subjected to a change of temperature, pressure or concentration, then the equilibrium shifts in a direction, as the effect of the changed condition is nullified. (b) It explain s th e effect of chan ges in temperature, pressure and concentrations on various reactions in equilibrium. A. Effect of change in temperature (a) If the temperature is raised, reaction will proceed in that direction in which the effect of raised heat can be destroyed so that the temperature on the system remains constant. (b) Athermic or nonthermic reaction not affected by the change in temperature. (i) Endothermic reaction – A + B + energy or, A + B Here, K = C + D C + D H = (+) ve [C][D] [ A][B] If we increase in temperature, the equilibrium shifts in forward direction of endothermic reaction which proceed with absorption of heat. eg. N2 + O2 + 43.2 k. cal. 2NO or N2 + O 2 2NO H = + 43.2 kcal (ii) Exothermic reaction : – A + B C + D + energy or A + B C + D  H = (–) ve If we increase in temperature, the equilibrium shifts in backward direction of exothermic reaction which proceed with evolution of heat. eg. N2 + 3H2 2NH3 + 23 k cal or N2 + 3H2 2NH3 1 H = –23 k cal B. Effect of change in pressure : According to Le-Chatelier's principle, if the pressure is increased, reaction will take place in that direction, which will bring about lowering of pressure or lowering in number of moles. This implies that (i) Increase of pressure shifts the equilibrium in that direction where the number of moles decreases or n < 0. eg. N2 + 3H2 2NH3 n = 2 – 4 = – 2 < 0 (ii) Decrease of pressure shifts the equilibrium in that direction where the number of moles increase or n > 0. eg. PCl 5 PCl3 + Cl 2  n = 2 – 1 = 1 > 0 (iii) Change of pressure has no effect, if the number of moles of gaseous product remain same as the gaseous reactants i.e.  n = 0 eg. H2 +  2 2H n = 2 – 2 = 0 CO + H2O CO2 + H2 n = 2 – 2 = 0 C. Effect of change in concentration : If in any reversible reaction, concentration of the reactants is increased at equilibrium state, the equilibrium shifts in the forward direction, means more product are formed. It is due to the fact that increase in concentration of reactant increases the number of effective collisions between the reactant molecule. Similarly, if the concentration of products is increased, then to maintain the equilibrium constant, concentration of reactant has to increase and backward reaction is favoured. Thus, it implies as (i) Increase in the concentration of any reactants, the equilibrium shifts in the forward direction. (ii) Increase in the concentration of products, the equilibrium shifts in the backward direction. Note : addition or removal of gaseous species only can affect equilibrium position. D. Effect of addition of catalyst : There is no effect on equilibrium by the addition of catalyst. It is due to the fact that the catalyst increase the rate of forward and backward reaction equally i.e. the catalyst helps in acquiring the equilibrium state quickly. E. Effect of addition of inert gas : Addition of inert gas at constant pressure to the equilibrium system causes the under mentioned effect. Page # 9 CHEMICAL EQUILIBRIUM system and decreases the molar concentration On increase in pressure, the equilibrium will shift towards less volume means solid is not converted into liquid and thus, melting point of sulphur increases. of reactant and product and the dissociation of 9. CALCULATION OF DEGREE OF DISSOCIATION (i) For the equilibrium system where the n > 0, the addition of inert gas at constant pressure increases the total volume of the reactant increases. eg. PCl 5(g) PCl3(g) + Cl2(g) (ii) For the equilibrium system where n < 0, the addition of inert gas at constant pressure will shifts the equilibrium to the backward BY VAPOUR DENSITY MEASUREMENT In those type of reactions in which there is a change in the number of moles after dissociation, the extent of dissociation can be determined by vapour density measurement. Let us, consider the following reaction - direction. eg. N2 + 3H2(g) 2 NH3(g) (iii) For the equilibrium system where n = 0, the addition of inert gas at constant pressure has no effect. eg. H2(s) + I2(g) 2HI(s) (iv) The addition of inert gas at constant volume has no effect on the equilibrium system. 8.1 Application of Le chatelier principle on physical equilibria A. Melting of Ice : Si nce, if we i ncrease the pressure, the equilibrium will be shifted in the direction of less volume. Hence, the rise of pressure, more ice will melt into water i.e. melting point of ice is PCl5 Therefore decreased by rise of pressure. H2O less volume more volume (g) vapour Vaporization of a liquid is endothermic process heat, so the rise of temperature will favour vaporization. On the other hand in this process, on increase of pressure the equilibrium will shift in the direction of less volume means water cannot be converted into vapour and boiling point increases. C. Melting of Sulphur : Sulphur(s) less volume Sulphur () more volume D  or D = d or  in the nature i.e. the evaporation of a liquid into its vapour is completed by absorption of + Cl2 1 V 1 (1   )V 1 V = (1 + ) 1 (1   ) V d  B. Vaporization of liquid H2 O ( ) water PCl 3 Initially 1 0 0 moles at (1 – )   equilibrium (Here '' is the degree of dissociation ) Total number of moles at equilibrium = (1 – ) +  +  = (1 + ) Now, the V is the volume occupied by 1 mol of PCl5(s) which have vapour density is 'D' before dissociation and after dissociation is 'd'. So, under the same conditions, the volume occupied by (1 + ) moles at equilibrium would be (1 + ) V litre. 1 Since Density  Volume = D Dd 1 = d d or Note: When one mole of a reactant on dissociation gives 'n' moles of gaseous products, the above equation takes in the form of