Equilibrium presentation grade 11 chemis

PPt on Equilibrium

1. Unit-7 Equilibrium

2. Portion to be done Equilibrium in physical and chemical processes, dynamic nature of equilibrium law of mass action, equilibrium constant, factors affecting equilibrium - Le Chatelier's principle. Ionic equilibrium- ionization of acids and bases, strong and weak electrolytes, degree of ionization, ionization of poly basic acids, acid strength, concept of pH, buffer solution, solubility product, common ion effect (with illustrative examples). Log and antilog calculations

3. Deleted portion hydrolysis of salts (elementary idea) Henderson Equation

4. Chemical Equilibrium In a chemical reaction chemical equilibrium is defined as the state at which there is no further change in concentration of reactants and products. For example, At equilibrium the rate of forward reaction is equal to the rate of backward reaction. Equilibrium  mixture:  The mixture of reactants and products in the equilibrium state is called an equilibrium mixtures.

5. Based on the extent to which the reactions proceed to reach the state of equilibrium , these may be classified in three groups: (i) The reactions which proceed almost to completion and the concentrations of the reactants left are negligible. (ii) The reactions in which most of the reactants remains unchanged, i.e. only small amounts of products are formed. (iii) The reactions in which the concentrations of both the reactants and products are comparable i.e., when the system is in equilibrium.

6. Equilibrium in Physical Processes (i) Solid-Liquid Equilibrium : The equilibrium is represented as Rate of melting of ice = Rate of freezing of water. The system here is in dynamic equilibrium s and following can be inferred. (a) Both the opposing processes occur simultaneously (b) Both the processes occur at the same rate so that the amount of ice and water – remains constant.

7. (ii) Liquid- Vapour Equilibrium The equilibrium can be represented as Rate of evaporation = Rate of condensation When there is an equilibrium between liquid and vapours , it is called liquid- vapour equilibrium.

8. (iii) Solid- Vapour Equilibrium This type of equilibrium is attained where solids sublime to vapour phase. For example, when solid iodine is placed in a closed vessel, violet vapours start appearing in the vessel whose intensity increases with time and ultimately, it becomes constant.

9. Equilibrium involving Dissolution of Solid in Liquid Solution:  When a limited amount of salt or sugar or any solute dissolves in a given amount of water solution is formed. At a given temperature, a state is reached when no more solute can be dissolved then the solution is called saturated solution. The equilibrium between a solid and its solution is indicated by the saturated solution and may be represented as

10. Here dissolution and precipitation takes place with the same speed. On adding a small amount of radioactive sugar to the saturated solution it will be found that the sugar present in the solution as well as in the solid state is radioactive.

11. Equilibrium between a Gas and its Solution in Liquid This type of equilibrium can be seen by the following example: Let us consider a sealed soda water bottle in which C02 gas is dissolved under high pressure. A state of equilibrium is attained between CO2 present in the solution and vapours of the gas.

12.  DYNAMIC Equilibrium in Chemical Processes Like equilibria in physical systems it can also be achieved in chemical process involving reversible chemical reactions carried in closed container. Synthesis of ammonia is called Habers Process Forward reaction: N2 and H2  NH3 Backward reaction: NH3 N2 + H2

13. The dynamic nature of chemical equilibrium can be demonstrated in the synthesis of ammonia by Haber’s process. Haber started his experiment with the known amounts of N2 and H2 at high temperature and pressure. At regular intervals of time he determined the amount of ammonia present. He also found out concentration of unreacted N2 and H2.

14. After a certain time he found that the composition of mixture remains the same even though some of the reactants are still present. This constancy indicates the attainment of equilibrium. In general, for a reversible reaction the chemical equilibria can be shown by

15. Law of Chemical Equilibrium At a constant temperature, the rate of a chemical reaction is directly proportional to the product of the molar concentrations of the reactants each raised to a power equal to the corresponding stoichiometric coefficients as represented by the balanced chemical equation. Let us consider a hypothetical equation:

16. The subscript ‘c’ indicates that Kc is expressed in concentrations of mol L–1 . At a given temperature, the product of concentrations of the products raised to the respective stoichiometric coefficient in the balanced chemical equation divided by the product of concentrations of the reactants raised to their individual stoichiometric coefficients has a constant value. This is known as the Equilibrium Law or Law of Chemical Equilibrium.

17. At particular instant of time: The equilibrium constant is called as reaction quotient (Q). That is: Q=[C] c [D] d  /  [A] a [B] b At equilibrium Q= K c

18. Iddeal gas equation is , PV= nRT P= (n/v) RT We know, n/v is nothing but molarity,M /C P= CRT Here, C is the concentration of reactant in his gaseous state

19. Equilibrium constant in gaseous system Ideal gas equation:

20. HETEROGENOUS EQUILIBRIA The reactants and products are not in same state of matter.

21. Numerical-1 What is Kc for the following equilibrium when the equilibrium concentration of each substance is: [SO 2 ]= 0.60 M, [O 2 ] = 0.82 M and [SO 3 ] = 1.90 M ?

22. The following concentrations were obtained for the formation of NH3 from N2 and H2 at equilibrium at 500K. [N2 ] = 1.5 × 10 –2 M. [H2 ] = 3.0 ×10 –2 M and [NH3 ] = 1.2 ×10 –2 M. Calculate equilibrium constant( K c ) N 2 + 3H 2 2NH 3 At a certain temperature and total pressure of 10 5 Pascal, iodine vapour contains 40% by volume of I atoms I 2 (g)  

23. At a certain temperature and total pressure of 10 5 Pascal , iodine vapour contains 40% by volume of I atoms. Calculate Kp for the equilibrium. I 2 (g) K p =(P I ) 2 /P I2  

24. Important formula Relationship between Kp and Kc Delta n= no of moles of products- no of moles of reactants

25. Find out the value of Kc for each of the following equilibria from the value of Kp : R= 0.083 bar L/ mol /K Delta n=?

26. Characteristics of chemical equilibrium The Rate of forward reaction = Rate of backward reaction. Catalyst does not affect the equilibrium .It just helps to achieve equilibrium faster. The Equilibrium is attained only is closed vessel. Equilibrium can be achieved from either direction. At equilibrium concentration of reactants and products becomes equal.

27. Effect of temperature on K c We know that Kc = K f /K b Now if temperature is increased, then the rate of reaction also increases. The extent of reaction depends upon activation energy of reactants i.e R f  and R b  will be different and also the value of K f  and K b  .So, the value of K will also change. A+B C+D+heat For endothermic reactions: K f >K b . Therefore, Kc increases. For exothermic reactions: K f <k b  .Therefore, Kc decreases.

28. Applications of equilibrium constant to: predict the extent of a reaction on the basis of its magnitude, predict the direction of the reaction, and calculate equilibrium concentrations

29. Predicting the extent of reaction If Kc > 10 3  Products predominate If Kc < 10 3  reactants predominate If Kc =1 reaction is at equilibrium

30. Predicting the direction of the reaction Difference between Kc and Qc(reaction quotient) Qc  and  Kc  are calculate the same way, but  Qc  is used to determine which direction a reaction will proceed, while  Kc  is the  equilibrium  constant (the ratio of the concentrations of products and reactants when the reaction is at  equilibrium ). If Qc > Kc , the reaction will proceed in the direction of reactants (reverse reaction). If Qc < Kc , the reaction will proceed in the direction of the products (forward reaction). If Qc = Kc , the reaction mixture is already at equilibrium

31. numerical Q.1:

32. Qc sis reaction quotinet Kc is equilibrium constant Qc can be measured at any instant in the reaction Kc can be measured only at Equilibrium point A B [B]= 1.2 [A]= 1.4 Kc = 2.89 Qc= [B]/[A]1.2/1.4=_____

33. CALCULATING EQUILIBRIUM CONCENTRATION Step 1. Write the balanced equation for the reaction. Step 2. Under the balanced equation, make a table that lists for each substance involved in the reaction: (a) the initial concentration, (b) the change in concentration on going to equilibrium, and (c) the equilibrium concentration

34. Step 3. Substitute the equilibrium concentrations into the equilibrium equation for the reaction and solve for x. If you are to solve a quadratic equation choose the mathematical solution that makes chemical sense. Step 4. Calculate the equilibrium concentrations from the calculated value of x. Step 5. Check your results by substituting them into the equilibrium equation.

35. Relationship between K and Reaction Quotient and Gibbs Free Energy We know ,Qc=[product]/[reactant] At equilibrium ,we know : Q=K

36. Factors affecting state of equilibrium “Le Chartlier’s principle Le chartlier’s principle : We know that at equilibrium, the state variables become constant. If we disturb any state variable, the reaction gets disturbed and the reaction diverts itself in a direction where disturbance is reduced. So, according to this “ if any kind of reaction is subjected to any kind of stress, then the reaction proceeds in a direction where the stress gets reduced “.

37. Effect of change in concentration : If concentrations of reactants are increased, then the reaction favors products.(forward reaction) If concentration of products is increased, then the reaction favors backward reaction.

38. Effect of temperature : We have two types of reactions: Endothermic and Exothermic. If we increase the temperature, the reaction proceeds in such a direction in which the heat is absorbed i.e., endothermic side is favoured . If we decrease the temperature, the reaction proceeds in such a direction in which the heat is released i.e., exothermic side is favoured . Exothermic reactions are favored by low temperatures Endo reactions are favored by high temperatures

39. Effect of change in Volume According to Boyle’s law, we know pressure is inversely proportional to volume .So, if we increase the volume, moles per litre decreases. Therefore, the reactions move in that reaction in which moles per litre is increased. If we decrease the volume, moles per litre increase .Therefore, the reaction move in that reaction in which moles per litre is decreased. P If volume increases, P decrease, the rxn moves towards greater no of moles If volume decreases, P increases, The rxn move towards lesser no of moles  

40. Introduction of Inert gas Whenever inert gas is introduced in any reaction, it does not effect the equilibrium of the reaction. Effect of catalyst : It just increases the rate of reaction without getting consumed. Catalyst will help in attaining the equilibrium faster from either sides. Concn Pressure Volume Temp Inert gas Catalyst

41. Numerical PCl 5 , PCl 3 and Cl 2 are at equilibrium at 500 K and having concentration 1.59M PCl 3 , 1.59M Cl 2 and 1.41 M PCl 5 . Calculate Kc for the reaction, PCl5 ⇌ PCl3 + Cl2

42. Ionic equilibrium in solutions Electrolytes:  Substances which conduct electricity in their aqueous solution/molten state Strong Electrolytes:  COMPLETE DISSOCIATION.Those electrolytes which on dissolution in water are ionized almost completely are called strong electrolytes .  Weak electrolyte:  PARTIAL DISSOCIATION. Those electrolytes which on dissolution in water partially dissociated are called weak electrolyte. Ionic Equilibrium:  The equilibrium formed between ions and unionised substance is called ionic equilibrium, e.g.,  

43. Acids:  Acids are the substances which turn blue litmus paper to red and liberate dihydrogen on reacting with some metals. Bases:  Bases are the substances which turn red litmus paper blue. It is bitter in taste. Common Example: NaOH , Na 2 C0 3 . Three acid-base theories Arrhenius theory Bronsted and Lowry Lewis acid-base theory

44. Arrhenius theory Acids:  According to Arrhenius theory, acids are substances that dissociates in water to give hydrogen ions H + ( aq ). Bases:  Bases are substances that produce OH – ( aq ) after dissociation in water.

45. limitation • Limitations of the Arrhenius Concept (i) According to the Arrhenius concept, an acid gives H +  ions in water but the H+ ions does not exist independently because of its very small size (~H -18  m radius) and intense electric field. (ii) It does not account for the basicity of substances like, ammonia which does not possess a hydroxyl group. : NH 3

46. • The Bronsted -Lowry Acids and Bases According to Bronsted -Lowry, an acid is a substance which is capable of donating a hydrogen ion H +  and bases are substances capable of accepting a hydrogen ion H + . In other words, acids are proton donors and bases are proton acceptors. This can be explained by the following example.

47. Conjugate acid & conjugate base base+H + conjugate acid Acid-H + conjugate base

48. • Acid and Base as Conjugate Pairs The acid-base pair that differs only by one proton is called a conjugate acid-base pair. base+H + conjugate acid Acid-H + conjugate base

49. limitations It cannot explain the acidic character of CO2, SO2 and SO3 It cannot explain the basic character of CaO , MgO, BaO Substances like AlCl3, BF3 are to known to behave as acids though they do not have protons.

50. Lewis acid-base theory According to Lewis, acid is a substance which accepts electron pair and base is a substance with donates an electron pair. Electron deficient species like AlCl 3 , BH 3 , H +  etc. can act as Lewis acids while species like H 2 O, NH 3  etc. can donate a pair of electrons, can act as Lewis bases. BH 3 + :NH 3  H3B:NH3 BH3 Lewis acid NH3 lewis base H 2 O

51. Ionisation of acids and bases The strength of an acid is determined by its tendency to lose proton whereas the strength of the base depends on its tendency to lose OH- ions. Ionisation constant of water and its ionic product: pH

52. pOH= - log [OH-]

53. Ionisation constant or dissociation constant of weak acids

54. Degree of dissociation or ionisation of weak acids

55. Calculate the degree of ionisation of 0.05 M ammonia solution having the ionization constant (Kb) is1.7 x 10 -5 .Also calculate the ionisation constant of the conjugate acid of this base.

56. Factors affecting acid strength Acidity tendency to lose H + ions

57. Deleted pprtion

58. Buffer solutions They are those solutions which has a constant value of pH , no matter what we add in  to them . Example: The pH of Blood is 7.4-7.6. Blood act as a buffer. As u all know, in spite of eating so many types of food items, the pH of blood remains same. The buffer solution is defined as:  A solution which resists the change in hydrogen ion concentration on addition of a small amount of acid or a base in to it. Buffer action : It is the ability to resist the change in pH on addition of acid or base.

59. There are two types of buffers Acidic buffer Basic buffer Acidic buffer : It contains equimolar concentration of weak acid and its salt with a strong base. For example: We take any weak acid say, acetic acid. To it, if we add equimolar concentration of sodium acetate, then the concentration of acetate ion increases. This acetate ion will neutralize the effect of hydrogen ion and the pH remains same.

60. Basic buffer Basic buffer : It is equimolar concentration of solution of weak base and its salt with a strong acid. Let’s say, ammonium hydroxide dissociates to ammonium ion and hydroxide ion. It has ammonium chloride which also gives ammonium ion. So, this ammonium ion combines with hydroxide ion and neutralizes its effect. Therefore, pH remains same.

61. Deleted portion

62. Solubility product It is the product of molar concentrations of the ions produced raised to power of the stoichiometric coefficients.

63. Calculate the solubility of A2X3 in pure water, assuming that neither kind of ion reacts with water. The solubility product of A2 X3 , Ksp = 1.1 × 10 -23

64. Q.no 1

65. Q.no 2:

66. Q.no 3:

67. Q.no 4: