NEET Chemistry Notes - Chapter 8: Chemical Equilibrium PDF

60 E3 Chapter 8 Chemical Equilibrium (2) Irreversible reactions : Reaction in which entire amount of the reactants is converted into products is termed as irreversible reaction. ID Whenever we hear the word Equilibrium immediately a picture arises in our mind an object under the influence of two opp...

60 E3 Chapter 8 Chemical Equilibrium (2) Irreversible reactions : Reaction in which entire amount of the reactants is converted into products is termed as irreversible reaction. ID Whenever we hear the word Equilibrium immediately a picture arises in our mind an object under the influence of two opposing forces. For chemical reactions also this is true. A reaction also can exist in a state of equilibrium balancing forward and backward reactions. (a) These reactions proceed only in one direction (forward direction), U Reversible and Irreversible reactions (i) Characteristics of irreversible reactions D YG A chemical reaction is said to have taken place when the concentration of reactants decreases, and the concentration of the products increases with time. The chemical reactions are classified on the basis of the extent to which they proceed, into the following two classes; (1) Reversible reactions : Reaction in which entire amount of the reactants is not converted into products is termed as reversible reaction. (b) These reactions can proceed to completion, (c) In an irreversible reaction, G < 0, (d) The arrow () is placed between reactants and products, (ii) Examples of irreversible reactions (i) Characteristics of reversible reactions (a) Neutralisation between strong acid and strong base e.g., (a) These reactions can be started from either side, NaOH  HCl NaCl  H 2 O 13.7 kcal (b) These reactions are never complete, (c) These reactions have a tendency to attain a state of equilibrium, in which Free energy change is zero (G = 0), U (d) This sign (⇌) represents the reversibility of the reaction, (ii) Examples of reversible reactions (a) Neutralisation between an acid and a base either of which or both are weak e.g., ST CH 3 COOH  Na OH ⇌ CH 3 COONa  H 2 O (b) Salt hydrolysis, e.g., Fe Cl 3  3 H 2 O ⇌ Fe OH 3  3 HCl (c) Thermal decomposition, e.g., PCl5 (g ) ⇌ PCl3 (g)  Cl 2(g) Q (b) Double decomposition reactions or precipitation reactions e.g., BaCl2(aq)  H 2 SO 4 (aq) BaSO4 (s)  2 HCl(aq) (c) Thermal decomposition, e.g., MnO 2 ,  2 KCl (s)  3O 2  2 KClO3 (s)  (d) Redox reactions, e.g., SnCl 2(aq)  2 FeCl3 (aq) SnCl 4 (aq)  2 FeCl2(aq) Equilibrium and Its dynamic nature “Equilibrium is the state at which the concentration of reactants and products do not change with time. i.e. concentrations of reactants and products become constant.” Products CH 3 COOH  C2 H 5 OH ⇌ CH 3 COOC 2 H 5  H 2 O (e) Evaporation of water in a closed vessel, e.g., H 2 O(l) ⇌ H 2 O(g) Q Concentration (d) Esterification, e.g., Reactants Time Equilibrium state The important characteristics of equilibrium state are, (2) Equilibrium state can only be achieved in close vessel. (3) Equilibrium state is reversible in nature. (4) Equilibrium state is also dynamic in nature. (5) At equilibrium state, Rate of forward reaction = Rate of backward reaction n = number of moles of gaseous products – number of moles of gaseous reactants in chemical equation. As a general rule, the concentration of pure solids and pure liquids are not included when writing an equilibrium equation. Value of n 0 >0 Relation between Kp and Kc Units of Kp Kp = Kc Kp > Kc No unit No unit (atm)n (mole l–1)n K. K   K or (K)1 / 2 (i) Similarly, if a particular equation is multiplied by 2, the equilibrium constant for the new reaction (K) will be the square of the equilibrium constant (K) for the original reaction i.e., K   K 2 (ii) If the chemical equation for a particular reaction is written in two steps having equilibrium constants K1 and K 2 , then the equilibrium constants are related as K  K1  K 2 Reactants Products QK (Reverse reaction) Fig. 8.1 Types of equilibria The equilibrium between different chemical species present in the same or different phases is called chemical equilibrium. There are two types of chemical equilibrium. E3 Applications of equilibrium constant (1) Homogeneous equilibrium : The equilibrium reactions in which all the reactants and the products are in the same phase are called homogeneous equilibrium reactions. C 2 H 5 OH (l)  CH 3 COOH (l) ⇌ CH 3 COOC2 H5 (l)  H 2O(l) ID Reaction proceeds hardly at all Q Kc Q Reaction proceeds to completion 10 10 3 –3 Both reactants and products are present at equilibrium N 2 (g)  3 H 2 (g) ⇌ 2 NH 3 (g) U 2SO 2 (g)  O 2 (g) ⇌ 2SO 3 (g) D YG (2) Predicting the direction of reaction : The concentration ratio, i.e., ratio of the product of concentrations of products to that of reactants is also known as concentration quotient and is denoted by Q. (2) Heterogeneous equilibrium : The equilibrium reactions in which the reactants and the products are present in different phases are called heterogeneous equilibrium reactions. [ X ][Y ] Concentration quotient, Q . [ A][B] It may be noted that Q becomes equal to equilibrium constant (K) when the reaction is at the equilibrium state. At equilibrium, Q  K  K c  K p. Thus, 2 NaHCO 3 (s) ⇌ Na 2 CO 3 (s)  CO 2 (g)  H 2 O (g) Ca(OH )2 (s)  H 2 O (l) ⇌ Ca 2 (aq)  2OH  (aq) CaCO 3 (s) ⇌ CaO (s)  CO 2 (g) H 2 O (l) ⇌ H 2 O (g) (i) If Q > K, the reaction will proceed in the direction of reactants (reverse reaction). U (ii) If Q < K, the reaction will proceed in the direction of the products (forward reaction). Table : 8.1 Homogeneous equilibria and equations for equilibrium constant (Equilibrium pressure is P atm in a V L flask) ⇌ ST H2 + I2 2 HI (g) n  0; K p  Kc n  0 ; K p  Kc n  0 ; K p  Kc N 2  3 H 2 ⇌ 2 NH 3 (g) Initial mole Mole at Equilibrium Total mole at equilibrium Active masses 1 (1–x) (1– x) 1  x  1  x       V   V  2x V  1  x   1  x   2x    3     V   V   V   2  2x   1  x       V   V  Mole fraction 1  x  1  x       2   2  2x 2 x 1 x 31 x    2 2  x  2  2  x  (2  x )  2  2x     3x  1 0 2x (g) 1 (1–x) (g) (g) 3 (3–3x) 0 2x 2 SO 2  O2 ⇌ 2 SO 3 (g) (4 – 2x) 2 (g) 2 (2–2x) (g) (g) 1 (1–x) 0 2x PCl 5 ⇌ PCl3  Cl 2 (g) 1 (g) (g) 0 (1–x) (3 – x) 0 x x (1 + x)  2x     V   1  x   2x      3  x 3  x 1  x     V  x   V x   V 1 x   x   x        1  x  1  x  1  x  Partial pressure  2x  1  x  1  x   p  p  p  2   2   2   1  x   3(1  x )  Px  P   P    2(2  x ) _   2(2  x )  (2  x )  2  2x  P   3x  Kc 4x2 1  x  2 4 x 2V 2 27 1  x  4 x 2V 1  x  3 Kp 4x2 1  x  2 16 x 2 2  x  2 27 1  x  P 4  2x  P  3  x 1 x  P  3 x  x   1 x  P  x   1 x  P x2 1  x  V x 2 3  x  P 1  x  2 1 x   1 x  P Px 2 1  x2  3  Table : 8.2 Heterogeneous equilibria and equation for equilibrium constant (Equilibrium pressure is P atm) 1 (1–x) 0 0 x 1 1 (1–x) (1–x) (1+x) x 2x x 1  2x 2 P 2 1  x  P  1  x  ID P 2 1 (1–x) 1  x    1  x  1 2  2x    1  x  Partial pressure 0 2x E3 Initial mole Mole at equilibrium Total moles at equilibrium (solid not included) Mole fraction NH 2CO 2 NH 4 (s) ⇌ 2 NH 3 (g)  CO 2 (g) 60 C(s)  CO 2 (g) ⇌ 2CO (g) NH 4 HS (s) ⇌ NH 3 (g) + H 2 S (g) 0 2x 3x x 2 3 1 3 2P 3 0 P 3  2x  P  1  x  P2 4 4P x2 (1  x 2 ) U Kp D YG Relationship between equilibrium constant and G° G for a reaction under any condition is related with G° by the relation, G  G  2.303 RT log Q Standard free energy change of a reaction and its equilibrium constant are related to each other at temperature T by the relation, Go  2.303 RT log K For a general reaction aA  bB ⇌ cC  dD K (aC )c (aD )d (a A )a (aB )b ST U Where a represent the activity of the reactants and products. It is unit less. For pure solids and liquids: a  1. For gases: a  pressure of gas in atm. 4 P3 27 For components in solution: a  molar concentration. Le-Chatelier's principle Le-Chatelier and Braun (1884), French chemists, made certain generalizations to explain the effect of changes in concentration, temperature or pressure on the state of system in equilibrium. When a system is subjected to a change in one of these factors, the equilibrium gets disturbed and the system readjusts itself until it returns to equilibrium. The generalization is known as Le-Chatelier's principle. It may be stated as : “Change in any of the factors that determine the equilibrium conditions of a system will shift the equilibrium in such a manner to reduce or to counteract the effect of the change.” The principle is very helpful in predicting qualitatively the effect of change in concentration, pressure or temperature on a system in equilibrium. Table : 8.3 The effect of varying conditions on the equilibrium a A + b B ⇌ c C + d D, n = (c + d) – (a + b) Change imposed on the system in equilibrium Equilibrium position moves Equilibrium constant Any other points Conc. of A and/or B increased Conc. of C and /or D increased Pressure increased To right To left No change No change No change No change No change No change No change Very little effect, if any, on reactions in liquid solution. Value decreased Value increased Equilibrium achieved faster No change Equilibrium achieved faster To right if (c  d )  (a  b) , i.e. n  ve To left if (c  d )  (a  b) , i.e. n  ve No change if (c  d )  (a  b) , i.e. n  0 Temperature increased To left if H  ve (exothermic) To right if H  ve (endothermic) Addition of catalyst No change Application of Le-Chatelier's principle The Le-Chateliers principle has a great significance for the chemical, physical systems and in every day life in a state of equilibrium. (1) Applications to the chemical equilibrium (i) Synthesis of ammonia (Haber’s process) N 2  3 H 2 ⇌ 2 NH 3  23kcal (exothermic) 3 vol 2 vol (b) Low temperature (d) Removal of NH 3 favours forward reaction. (ii) Formation of sulphur trioxide 2 SO 2  O 2 ⇌ 2 SO 3  45 kcal (exothermic) On the other hand if it is accompanied by evolution of heat, solubility decreases with increase in temperature; e.g., CaCl 2 , (b) Low temperature ⇌ 2 NO  43.2 kcal (endothermic ) (a) High temperature (c) Since reaction takes place without change in volume n  0 , pressure has no effect on equilibrium. (iv) Formation of nitrogen dioxide D YG i.e., 2 vol (a) High pressure (b) Low temperature (c) Excess of NO and O 2 favours the reaction in forward direction. (v) Dissociation of phosphours pentachloride PCl5 ⇌ PCl3  Cl 2  15 kcal 1 vol 1 vol 1 vol U (a) Low pressure or high volume of the container, n  0 (b) High temperature (c) Excess of PCl5. ST (2) Applications to the physical equilibrium (i) Melting of ice (Ice – water system) Ice (Greater Volume) ⇌ Water In the following reversible chemical equation. A ⇌ yB Initial mol 1 0 At equilibrium (1–x) yx x = degree of dissociation Number of moles of A and B at equilibrium  1  x  yx  1  x (y  1) If initial volume of 1 mole of A is V, then volume of equilibrium mixture of A and B is,  [1  x (y  1)]V Molar density before dissociation, molecular weight m D  volume V Molar density after dissociation, D m Dd ;  [1  x (y  1)] ; x  d [1  x (y  1)]V d d (y  1) y is the number of moles of products from one mole of reactant. U (b) Excess of N 2 and O 2 1 vol Ca(OH )2( s)  (aq)  Ca(OH )2 (aq)  x kcal Relation between vapour density and Degree of dissociation 2 vol 2 NO  O 2 ⇌ 2 NO 2  27.8 Kcal Ca(OH )2 , NaOH, KOH etc. ID (c) Excess of SO 2 and O 2 , favours the reaction in forward direction. (iii) Synthesis of nitric oxide 2 vol (Higher volume) KNO 3(s)  (aq)  KNO 3(aq )  x kcal 2 vol (a) High pressure (n  0) N 2  O2 1 vol 1 vol Water Vapours  x kcal 60 (c) Excess of N 2 and H 2 1 vol Water ⇌ (Low volume) (It is accompanied by absorption of heat and increase in volume.) (a) At high temperature more vapours are formed. (b) At higher pressure, vapours will be converted to liquid as it decreases volume. (c) At higher pressure, boiling point of water is increased (principle of pressure cooker). (iv) Solubility of salts : If solubility of a salt is accompanied by absorption of heat, its solubility increases with rise in temperature; e.g., NH 4 Cl, K2 SO 4 , KNO 3 etc. (a) High pressure (n  0) 2 vol (iii) Boiling of water (water- water vapour system) E3 1 vol (b) At higher pressure, less sulphur will melt as melting increases volume. (c) At higher pressure, melting point of sulphur is increased.  x kcal (Lesser Volume) (In this reaction volume is decreased from 1.09 c.c. to 1.01 c.c. per gm.) (a) At high temperature more water is formed as it absorbs heat. (b) At high pressure more water is formed as it is accompanied by decrease in volume. (c) At higher pressure, melting point of ice is lowered, while boiling point of water is increased. (ii) Melting of sulphur : S (s) ⇌ S (l)  x kcal (This reaction accompanies increase in volume.) (a) At high temperature, more liquid sulphur is formed. D is also called Van’t Hoff factor. d M m In terms of molecular mass, x  (y  1) m Where M  Initial molecular mass, m  molecular mass at equilibrium Thus for the equilibria (I) PCl5(g) ⇌ PCl3(g)  Cl 2(g) , y  2 (II) N 2 O 4 (g) ⇌ 2 NO 2(g) , y  2 (III) 2NO 2 ⇌ N 2 O 4 , y  1 2 Dd 2(d  D) (for I and II) and x  (for III) d d Also D  2  Molecular weight (theoretical value) d  2  Molecular weight (abnormal value) of the mixture.  x    by increasing the pressure. As the water so obtained on melting is below 0°C, it refreezes when pressure is reduced. It is called regelation of ice. Increase in external pressure always increases the boiling point and viceversa. If the reaction is multipled by 2, the equilibrium constant is squared. When a bottle of coca or beer is opened, the pressure is decreased and dissolved CO gas escapes out with a fizze. Increase in pressure favours melting of ice into water Flash evaporation is a technique generally used for concentrating certain aqueous solutions which cannot be concentrated by normal boiling. Freeze drying is a technique where water is made to sublime off at a temperature below 0°C. 60  Pure ice can be made to melt at a temperature slightly below 0°C ST U D YG U ID  E3 2  