CHAPTER 8 CHEMICAL EQUILIBRIUM TEXT BOOK EXERCISE Q1. Multiple Choice questions (i) For which system does the equilibrium constant, K c has units of (concentration) -1 (a) N 2 + 3H 2 2NH 3 (b) H 2 + I 2 2HI (c) 2NO 2 N 2 O 4 (d) 2HF H 2 + F 2 (ii) Which statement about the following equilibrium is correct 2SO 2 + O 2 2SO 3 (a) The value of K p falls with a rise in temperature (b) The value of K P falls with increasing Pressure (c) Adding V 2 O 5 catalyst increases the equilibrium yield of SO 3 (d) The value of K p is equal to K c (iii) The ph of 10-3 mol/dm 3 of an equous solution of H 2 SO 4 is (a) 3 (b) 2.7 (c) 2 (d) 1.5 (iv) The solubility product of AgCI is 2 x 10 10 mol 2 dm -6. The maximum concentration of Ag+ ions in the solution is (a) 2 x 10-10 mol dm -3 (b) 1.41 x 10-10 mol dm -3 (c) 1 x 10-10 mol dm -3 (d) 4 x 10-20 mol dm -3 (v) An ecxcess of aqueous silver nitrate is added to aqueous barium chloride and precipitate is removed by filtration. What are the main ions in the filtrate? (a) Ag + and NO -1 3 only (b) Ag + and Ba 2+ -1 and NO 3 (c) Ba 2+ and NO -1 3 only 1

(d) Ba 2+ and NO -1 3 and CI -1 Ans. i)c ii)a iii)b iv)b v)b Q2. Fill in the blanks (i) Law of mass action states that the rate at which a reaction proceeds is directly proportional to the product of the active masses of the. (ii) In an exothermic reversible reaction, in temperature will shift the equilibrium towards the forward direction. (iii) The equilibrium constant for the reaction 2O 3 3O 2 is 10 55 at 25 o C, it tells that ozone is at room temperature. (iv) In a gas phase reaction, if the number of moles of reactants are equal to the number of moles of the products, K C of the reaction is to the K p. (i) Buffer solution is prepared by by mixing together a weak base and its salt with or a weak acid and its salt with. Ans. i)reactants ii)decrease iii)unstable iv)equal v)strong acid, strong base Q3. Label the sentences True or False (i) When a reversible reaction attains equilibrium both reactants and products are present in a reaction mixture. (ii) The K c of the reaction A+B C+D is given by (iii) (iv) (v) K c = Therefore it is assumed that [A] = [B]=[C]=[D] A catalyst is a compound, which increases the speed of the reaction and consequently increases the yield of the product. Ionic product K w of pure water at 25 o C is 10-14 dm -6 and is represented by an expression K w =[H + ][OH - ]=10-14 mol 2 dm -6 AgCI is a sparingly soluble ionic solid in water. Its solution produces excess of Ag + and CI - ions. 2

Ans. i)true ii)false iii)false iv)true v)false Q4. (a) Explain the terms reversible reaction and state of equilibrium See Section 8.1 and 8.1.2 (b) Define and explain the law of mass action and drive the expression for the equilibrium constant K C. See Section 8.1.3 (c) Write K C for the following reactions (i) Sn 2+ (aq) + 2Fe 3+ (aq) Sn 4+ (aq) + 2Fe 2+ (aq) (ii) Ag+ (aq) +Fe 2+ (aq) Ag (s) + Fe 3+ (aq) (iii) N 2(g) + O 2(g) 2NO (g) (iv) 4NH 3(g) 5O 2(g) 4NO (g) +6H 2 O (g) (v) PCI 5(g) PCI 3(g) +CI 2(g) (i) K c = (ii) K c = (iii) Kc = (iv) K c = (v) Kc = Q5. (a) Reversible reactions attain the position of equilibrium, which is dynamic in nature and not static. Explain it. See Section 8.1.1 (b) Why do the rates of forward reactions slow down when a reversible in nature and not static. Explain it. See Section 8.1.1 Q6. When a graph is plotted between time on X-axis and the concentration of reactants and products on Y-axis for a reversible reaction, the curve becomes parallel to time axis at a certain stage. (a) At what stage the curves become parallel? 3

(b) Before the curves become parallel, the steepness of curves falls? Give reasons. (c) The rate of decrease of concentrations of the reactants and rate of increase of concentrations of any of products may or may not be equal for various types of reactions before the equilibrium time. Explain it. See section 8.1.1 Q7. (a) Write down the relationship of different types of equilibrium constants. i.e. K c and K p for the following general reactions. aa + bb cc +dd See section 8.1.2 (b) Decide the comparative magnitudes of K c, K p for the following reactions. Synthesis of NH 3 N 2(g) +3H 2(g) 2NH 3(g) K C is given by K c = K p is given by K p = For this reaction, change in number of moles is given by n =number of mole of product-number of moles of reactants =2 (1+3)=-2 Hence K p =K c x (RT) -2 Or K p = K c x Thus Dissociation of PCI 5 if T is such that RT > 1, then K p < K c If T is such that RT< 1, then K p > K c PCI 5(g) PCI 3(g) + CI 2(g) 4

K p is given by K c is given by according to law of Mass of Action K c = K p = For this reaction, change in number of moles is given by n =number of mole of product-number of moles of reactants =2 1 =1 K p =K c x (RT) Thus If T is such that RT > 1, then K p > Kc If T is such that RT< 1, then K p <K c Q8. (a) Write down K C for the following reversible reactions. Suppose that the volume of reaction mixture in all the cases is V dm 3 at equilibrium stage. (i) CH 3 COOH+ CH 3 CH 2 OH CH 3 COOC 2 H 5 + H 2 O (ii) H 2 +1 2 2HI (iii) 2HI H 2 +I 2 (iv) PCI 5 PCI 3 + CI 2 (v) N 2 + 3H 2 2NH 3 See section 8.1.3 (b) How do you explain that some of the reactions mentioned above are affected by change of volume at equilibrium stage? Q9. Explain the following two applications fo equilibrium constant. Give examples. (i) Direction of reaction (ii) Extent of reaction See section 8.1.5 Q10. Explain the following with reasons. (a) The change of volume disturbs the equilibrium position for some of the gaseous phase reactions, but not the equilibrium constant. 5

(b) (c) The change of temperature disturbs both the equilibrium position and the equilibrium constant of a reaction. The solubility of glucose in water is increased by increasing the temperature. Q11. (a) What is ionic product of water? How does this value vary with the change in T? Is it true that this value is 75 times when the T of water increased form 0 o C to 100 o C. Ionic Product of water is given by the equation K w [H + ][OH - ] Value of K w increases with increase in temperature. It is because increase in temperature increase the ionization of H 2 O. Thus, more H + or OH - ions are produced. Hence value of K w increase. e.g. At 25 o C K w =1 x 10-14 (b) (c) and At 100 o C K w =7.5 x 10-14 Further At 0 o C (K w ) o =0.1 x 10-14 (1) At 100 o C (K w ) 100 =7.5 x 10-14 (2) Divide eq (2) by eq (1) = =75 or (K W ) 100 =75 x (K w ) 0 Hence, K W at 100 o C is 75 times more than at 0 o C What is the Justification for the increase of ionic product with temperature? Value of K w increase with increase in temperature. It is because increase in temperature increase the ionization of H 2 O. Thus, more H + or OH - ions are produced. Hence value K w increases. How do you prove that at 25 O C in 1 dm 3 of water, there are 10-7 moles of H 3 O +? 6

At 25 O C K w =[H 3 O + ][OH -14 (1) Since ionization of water gives equal no. of H 3 O + and OH - ions therefore [H 3 O + ]=[OH - ] Hence, eq (1) can be written as K w =[H 3 O + ][H 3 O + ]=10-14 Or [H 3 O + ] 2 =10-14 Taking square root on both sides [H 3 O + ] 2 =10-7 mol/dm 3 Hence, at 25 o C, water has 10-7 mole/dm 3 of H 3 O + ions. Q12. (a) Define ph and poh. How are they related with pk w. (b) What happens to the acidic and basic properties of aqueous solutions when ph varies form 0 to 14. (c) Is it true that the sum of ph a and pk b is always equal to 14 to all temperatures for any acid? If not why? Q13. (a) What is Lowry bronsted idea of acids and bases? Explain conjugate acids and bases. (b) Acetic acid dissolves in water and gives proton to water. But when dissolves in H 2 SO 4, it accepts proton. Discuss the role of acetic acid in both cases. CH 3 COOH + H 2 O CH 3 COO - + H 3 O + However, H 2 SO 4 is a stronger acid than acetic acid, therefore, H 2 SO 4 donates proton and acts as an acid while acetic acid accepts proton and acts as a base. H 2 SO 4 + CH 3 COOH - + HSO 4 + CH 3 COOH 2 Q14. In the equilibrium PCI 5(g) PCI 3(g) + CI 2(g) H=90 kj/ mol (a) The position of equilibrium (b) Equilibrium constant? If (i) Temperature is increased (ii) Volume of the container is decrease. 7

(iii) (iv) Catalyst is added CI 2 is added See section 8.2 Explain your answer. Q15. Synthesis of NH 3 by Haber s process is an exothermic reaction. N 2 + 3H 2 2NH 3 H=92.46 kj/ mol (a) What should be the possible effect of change of temperature at equilibrium stage? (b) How does the change of pressure or volume shifts the equilibrium position of this reaction? (c) What is the role of the catalyst in this reaction? (d) What happens to equilibrium position of this reaction if NH 3 is removed form the reaction vessel form time to time. See Section 8.2.1 Q16. Sulphuric acid is the king of chemicals. It is produced by the burning of SO 2 to SO 3 through an exothermic reversible process. (a) Write the balanced reversible reaction (b) What is the effect of pressure change on this reaction? (c) Reaction is exothermic but still the temperature of 400-500 o C is required to increase the yield of SO 3. Give reasons. See Section 8.2.2 Q17. (a) What are buffer solutions? Why do we need them in daily life? (b) (c) See Section 8.2 How does the mixture of sodium acetate and acetic acid give us the acidic buffer? See Section 8.7 Explain that a mixture of NH 4 OH and NH 4 CI gives us the basic buffer? See Section 8.7 8

(d) How do you justify that the greater quantity of CH 3 COONa in acetic acid decrease the dissociating power of acetic acid and so the ph increases. CH 3 COOH is a weak acid and ionizes very small, while CH 3 COONa is a strong electrolyte and it ionizes in water to greater extent and provides acetate ions. CH 3 COOH +H 2 O CH 3 COO - + H 3 O + CH 3 COONa CH 3 COO - + Na + Thus CH 3 COONa decreases the ionization of CH 3 COOH due to common CH 3 COO - ion and ph of solution increases. (d) Explain the term buffer capacity. See Section 8.7.1 Q18. (a) What is the solubility product? Derive the solubility product expression for sparingly soluble compounds, AgCI, Ag 2 CrO 4 and PbCI 2. See Section 8.8 (b) How do you determine the solubility product of a substance when its solubility is provided in grams/100 g of water? See Section 8.8 (c) How do you calculate the solubility of a substance from the value of solubility product. See Section 8.8 Q19. K c value for volume for the following reaction is 0.076 at 520 o C 2HI H 2 + I 2 Equilibrium mixture constants [HI] =0.08 M, [H 2 ] =0.01 M. To this mixture more HI is added so that its new concentration is 0.096 M. what will be the concentration of [HI], [H 2 ] and [I 2 ] when equilibrium is re-established. Solution 2HI H 2 + I 2 9

Equilibrium conc. (mol/dm3) Initial conc. After adding more HI (mol/dm3) Equilibrium conc. 0.08 0.01 0.01 0.096 0.01 0.01 0.096 2x 0.01 0.01 When equilibrium is re-established (mol/dm3) According to law of mass action K c = K c = =0.016 = =0.016 Taking square root on both sides = =0.126 0.01 + x=0.126 (0.096 2x) 0.01 + x =0.0121 0.252 x x + 0.252 x=0.0121 0.01 1.252 x=0.0021 x == x=0.00168 mol/dm -3 Thus Concentrations when equilibrium is re-established are [H 2 ] =0.01 + x =0.01 +0.00168 =0.01168 mol dm -3 [I 2 ] =0.01 + x =0.01 +0.00168 =0.01168 mol dm -3 [HI] =0.096-2 x =0.096-2x0.00168 =0.0926 mol dm -3 Q20. The equilibrium constant for the reaction between acetic and ethyl alcohol is 4. A mixture of 3 moles of acetic acid and one 10

mole of C 2 H 5 OH is allowed to come to equilibrium. Calculate the amount of ethyl acetate at equilibrium stage in number of moles and grams. Also, calculate the masses of reactants left behind. Solution CH 3 COOH + C 2 H 5 OH CH 3 COOC 2 H 5 +H 2 O Initial conc. 3 1 0 0 (mol/dm3) Equilibrium conc. (mol/dm3) 3- x 1 x x x According to law of mass action K c = K c = X 2 =4(3 x) (1 x) X 2 =4(3 3x) (1 x) X 2 =4(3 4x + x 2 ) X 2 =12 16x + 4 x 2 ) Or 12 16x + 4x 2 x 2 =0 3x 2 16x + 12=0 It is quadratic equation and can be solved by using quadric formula Here a = 3, b = 12, c = 12 Thus x= x= x= x= 11

x= Either x= or x= x= 4.43 mol dm -3 or x= 0.9 mol dm -3 x= 4.43 is not possible as it is greater than the initial concentrations of reactants, therefore, x= 0.9 mol dm -3 Therefore Moles of ethyl acetate =x=0.9 moles Mass of ethyl acetate = 0.9x88=79.46g. Moles of water =x=0.9 moles Mass of water =0.9 x 18=16.2 g. Moles of acetic acid = 3 x =3 0.9 =2.1 moles (left behind) Moles of ethyl alcohol = 1 x =1 0.9= 0.1 moles (Left behind) Mass of ethyl alcohol =0.1 x 46=4.6g. (left behind) Q21. study the equilibrium H 2 O + CO H 2 + CO 2 (a) Write an expression of K p K p = (b) Solution When 1 mole of steam and 1 mole of CO are allowed to reach equilibrium, 33.3% of the equilibrium mixture is hydrogen. Calculate the value of K p. state the units of K p. Initial conc. (mol/dm3) Equilibrium conc. (mol/dm3) H 2 O + CO H 2 + CO 2 1 1 0 0 1 x 1 x x x Total no. of moles at equilibrium =1 x + 1 x + x + x=2 Hence 12

% of H 2 = 33.3= or no of moles of H 2 =0.67 moles Hence At equilibrium Moles of H 2 =x=0.67 moles Moles of CO 2 =x=0.67 moles Moles of H 2 O =1 x = 1-0.67= 0.33 moles Moles of CO =1 x = 1-0.67= 0.33 moles Hence, K c is given as K c = K c = Since K p =K c (RT) and n =n products -n rectants =0, therefore K p =K c =4 Q22. Calculate the ph of (a) 10-4 moles/ dm 3 of HCI HCI ionizes as HCI H + + CI - Since HCI is a strong acid, and it is 100% dissociated. Hence 10-4 mol/dm 3 of HCI produces 10-4 mol/dm 3 of H + ions. Thus [H + ] = 10-4 mol/dm 3 So ph = - log [H + ] ph = - log [10-4 ] ph = 4 (b) 10-4 moles/dm 3 of Ba(OH) 2 Ba(OH) 2 ionizes as Ba(OH) 2 Ba +2 + 2 OH - 13

Since Ba(OH) 2 is a strong base it is 100% dissociated. Hence 10-4 mol/dm 3 of Ba(OH) 2 produces 2 x 10-4 mol/dm 3 of OH - ions. Thus [OH - ]= 2 x 10-4 mol/dm 3 So poh= - log[oh - ] poh= - log[2 x 10-4 ] poh= 3.699 Since ph + poh=14 Therefore ph= 14 poh =14 3.699 =10.301 (c) 1 mol/dm 3 of H 2 X, which is 50% dissociated H 2 X ionizes as H 2 X 2H + + X - 1mole of H 2 X produces 2 moles of H + ions if 100% dissociated However, since H 2 X is 50% dissociated therefore 1 mole of H 2 X produces 1 mole of H + ion Thus [H + ]=1 mol/dm 3 So ph= -log [H + ] ph= -log [1] ph= 0 (d) ions. Hence 1 mol/dm 3 of NH 4 OH which is 1% dissociated NH 4 OH ionizes as + NH 4 OH NH 4 + OH - It shows that 1 mole of NH 4 OH produces 1 mole of OH - NH 4 OH is only 1% dissociated 14

% dissociation = x 100 1 = x 100 mol of OH - x1=0.01 mol/dm 3 Thus [OH - ] =0.01 mol/dm 3 So poh = - log [OH - ] poh = - log [0.01] poh = 2 Since ph + poh =14 Therefore ph=14 poh =14 2=12 Q23. (a) Benzoic aicd C 6 H 5 COOH is weak mono-basic acid (Ka=6.4 x10-5 mol/dm 3 ). What is the ph of a solution containing 7.2 g of sodium benzoate in one dm 3 of o.02 mol/dm 3 benzoic acid. Mass of sodium benzoate =7.2 g /dm 3 Formula of sodium benzoate is C 6 H 5 COONa Mol. Mass of sodium benzoate =144 g/mol Moles of sodium benzoate = =0.05 mol/dm 3 Moles of benzoic acid =0.02 mol/dm 3 K a of benzoic acid 6.4 x 10-5 mol/dm 3 Thus pk a = - log K a = - log (6.4 x10-5 ) =4.2 Hence, according to Henderson s eq. 15

ph =pk a + log ph =4.2 + log ph = 4.2 + 0.39 ph =4.59 (b) A buffer solution has been prepared by mixing 0.2 M CH 3 COONa and 0.5 M CH 3 COOH in 1 dm 3 of solution. calculate the ph of solution. pk a of acid is 4.74 at 25 o C. How the value of ph will change by adding 0.1 mole 0.1 mole of NaOH and 0.1 mole mol HCI respectively. Solution [CH 3 COOH] =0.5 M [CH 3 COONa] =0.2 M pk a of CH 3 COOH =4.74 ph =? or Since ph =pk a + log ph =pk a + log ph =4.74 + log ph =4.74-0.4 ph =4.34 When 0.1 mole of NaOH is added NaOH is strong base. It dissociates completely. Therefore, it produces 0.1 moles of OH - ions. Thus, 0.1 moles of OH - ions reacts with 0.1 moles of CH 3 COOH. Hence, out of 0.5 moles of CH 3 COOH, 0.4 moles of CH 3 COOH are behind. On the other hand, due to slat formed by the neutralization reaction, conc. of salt (CH 3 COONa) is increased from 0.2 moles to 0.3 moles. Hence, new conc. will be 16

COONa] Thus [CH 3 COOH] =0.4 M [CH 3 =0.3M ph =pk a + log ph =4.74 + log ph =4.74-0.12 ph =4.62 Addition of 0.1 mole of HCI HCI is a strong acid. It dissociates completely. Therefore, it produces 0.1 moles of H + ions. Thus, 0.1 moles of H + ions react with 0.1 moles of CH 3 COO - ions. Hence, out of 0.2 moles of salt (CH 3 COO - Na + ), 0.1 moles of salt are left behind. On the other hand, conc. of acid (CH 3 COOH) is increased form 0.5 moles to 0.6 moles. Hence, new conc. will be [CH 3 COOH] =0.6 M [CH 3 COONa] =0.1M Thus ph =pk a + log ph =4.74 + log ph =4.74-0.78 ph =3.96 (See Section 8.7.1 for complete understanding of this numerical) Q24. Solubility of CaF 2 in water at 25 O C is found to be 2.05 x 10-4 mol dm -3. What is the value of K sp at this T. Solubility of CaF 2 = 2.05 x 10-4 mol dm -3 According to balanced chemical eq. CaF 2(aq) Ca 2+ (aq) + 2F - (aq) At initial stage 2.05 x 10-4 0 0 (mol/dm 3 ) After solubility 0 2.05 x 10-4 2x2.05 x 10-4 Hence K sp =[Ca +2 ][F - ] 17

K sp =[0.05 x 10-4 ][2 x 2.05 x 10-4 ] 2 K sp =3.446 x 10-11 mol 3 dm -9 Q25. The solubility product of Ag 2 CrO 4 is 2.6 x 120-2 at 25 o C. Calculate the solubility of the compound. K sp of Ag 2 CrO 4 =2.6 x 10-2 We know Ag 2 CrO 4(aq) 2Ag + 2- CrO 4 (aq) Initial stage (mol/dm3) Ag 2 CrO 4 0 0 At equilibrium (mol/dm3) Ag 2 CrO 4 2S S Hence K sp =[Ag + ] 2 [CrO 4 2- ] K sp =[2S] 2 [S] =2.6 x 10-2 4[S] 3 =2.6 x 10-2 [S]= or [S]=0.1866 mol/dm 3 Hence at equilibrium [Ag + ] =2 x 0.1866 mol/dm 3 =0.3732 mol/dm 3 and 2- [CrO 4 ] =0.1866 mol/dm 3 Since 1 mole of Ag 2 CrO 4 gives 1 mole of CrO 2-4 ions, hence Solubility of Ag 2 CrO 4 =[CrO 2-4 ]=0.1866 mol/dm 3 18