Chemical Equilibrium Study Guide

CHEMICAL EQUILIBRIUM SOLVED PROBLEMS Ex.1 Sol. Ex.2 Sol. Ex.3 The volume of a closed reaction vessel in which the equilibrium : 2SO2(g) + O2(g) 2SO3(g) sets is halved, Now (A) The rates of forward and backward reactions will remain the same. (B) The equilibrium will not shift. (C) The equilibrium will shift to the right. (D) The rate of forward reaction will become double that of reverse reaction and the equilibrium will shift to the right. Ans. [D] In the reaction 2SO2(g) + O2(g) 2SO3(g) In this reaction three moles (or volumes) of reactants are converted into two moles (or volumes) of products i.e. there is a decrease in volume and so if the volume of the reaction vessel is halved the equilibrium will be shifted to the right i.e. more product will be formed and the rate of forward reaction will increase i.e. double that of reverse reaction. H2(g) + I2(g) 2HI(g) When 46g of I2 and 1g of H2 are heated at equilibrium at 450ºC, the equilibrium mixture contained 1.9g of I2. How many moles of I 2 and HI are present at equilibrium – (A) 0.0075 & 0.147 moles (B) 0.0050 & 0.147 moles (C) 0.0075 & 0.347 moles (D) 0.0052 & 0.347 moles Ans. [C] 46 Moles of I2 taken = = 0.181 254 1 Moles of H2 taken = = 0.5 2 1.9 Moles of I2 remaining = = 0.0075 254 Moles of I2 used = 0.181 – 0.0075 = 0.1735 Moles of H2 used =0.1735 Moles of H2 remaining 0.5 – 0.1735 = 0.3265 Moles of HI formed = 0.1735 × 2 = 0.347 At equilibrium Moles of I2 = 0.0075 moles Moles of HI = 0.347 moles A two litre flask contains 1.4 gm nitrogen and 1.0 gm hydrogen. The ratio of active mass of nitrogen and hydrogen would be (A) 1 : 3 (B) 1 : 5 (C) 1.4 : 1 (D) 1 : 10 Ans. [D] 1 Sol. [N2] = 1 .4 , 28  2 [N2] : [H2] Ex.4 Sol. Sol. Ex.6 1 22 1 1 : 40 4 = 1: 10 = In the reaction : A + B C + D the initial concentration of A is double the initial concentration of B. At equilibrium the concentration of B was found to be one third of the concentration of C. The value of equilibrium constant is – (A) 1.8 (B) 1.008 (C) 0.0028 (D) 0.08 Ans. [A] A + B C +D 1 0.5 0 0 Initially 1-x 0.5-x x x At equilibrium According to question (0.5 – x) = 1/3 x or x = 1.5 – 3x 4x = 1.5 1.5 x = = 0.375 4 KC = Ex.5 [H2] = xx (1  x )(0.5  x ) = 0.375  0.375 (1  0.375)  (0.5  0.375) = 0.375  0.375 = 1.8 0.625  0.125 The value of KC for the reaction : A + 3B 2C at 400°C . Calculate the value of KP (A) 1.64×10–4 (B) 1.64×10–6 –5 (C) 1.64×10 (D) 1.64×10–3 Ans. [A] KP = KC [RT]n n = 2 – 4 = – 2 T = 673K, KC = 0.5, R = 0.082 litre. atm. KP = 0.5 × (0.082×673)–2 = 1.64 × 10– 4 atm. Two moles of ammonia was introduced in an evacuated vessel of 1 litre capacity. At high temperature the gas undergoes partial dissociation according to the equation : 2NH3(g) N2(g) + 3H2(g) At equilibrium the concentration of ammonia was found to be 1 mole. What is the value of ‘K’ ? CHEMICAL EQUILIBRIUM Sol. (A) 3/4 = 0.75 mol2 –2 (B) 3/2 = 1.5 mol2 –2 (C) 27/16 = 1.7 mol2 –2 (D) 27/64 = 0.42 mol2 –2 Ans. [C] Let  be the degree of dissociation : Hence we have : 2NH3(g) N2(g) + 3H2(g) 2 0 0 2–2  3 At equilibrium : [NH3] = 2 – 2 = 1 or  = 1/2 [N2] =  K = 1 , [H2] = 2 Ex.9 2 Sol. In a 2 litre flask, the reaction takes place as : COCl2(g) CO(g) +Cl2(g) The equilibrium conc. of [COCl2] was found to be 0.4. If the excess of COCl2 is added to the system, the equilibrium reestablishes and [COCl2] becomes 1.6. What is the equilibrium conc. of [CO] ? (A) Half of the former value (B) Thrice of the former value (C) Remains unaltered (D) Twice of the former value Ans. [D] COCl2(g) CO(g) + Cl2(g) [CO][Cl2 ] [COCl2 ] = KC Let [CO] = x, then [Cl2] = x x2 x2  = KC or = KC [COCl2 ] 0 .4 [CO][Cl2 ] x2 Again KC = = [COCl2 ] 0 .4 x2 [CO]2 = ; {[CO]= [Cl]} 0 .4 1.6 x2 × 4 = [CO]2 2x = [CO] Ex.8 4.0 gms of hydrogen react with 9.023×1023 molecules of chlorine to form HCl gas. The total pressure after the reaction was found to be 700 mm. The partial pressure of HCl will be – (A) 3900 mm (B) 600 mm (C) 700 mm (D) 350 mm Ans. [B] 2HCl 0 Initially 3 0 Initially 2 3/2 moles of Cl 2 will react with 3/2 moles of 3 H2 to give × 2 = 3.0 moles of HCl. So 2 the total number of moles in the reaction 3  =  2   + 0 + 3 = 3.5 2  3  700 Partial pressure of HCl = = 600 mm 3 .5 1 3 3 3    27 = 2 2 2 2 = = 1.7 mol 2 –2 16 1 1 Ex.7 H2 + Cl 2 23 4 9.023  10 2 6.02  10 23 2 3 , [NH3] = 1 2 [N2 ][H2 ]3 [NH3 ] Sol. Sol.  The equilibrium constant ‘K’ for the reaction N2 + 3H2 2NH3 is 16. What would be value of equilibrium constant ‘K’ for the reaction : 1 3 NH3 N2 + H2 2 2 (A) 0.25 (B) 0.4 (C) 2 (D) 0.625 Ans. [A] We know that : [New reaction = reverse 1 of old reaction × 2 1 1 1 K' = = = = 0.25 K 16 4 Ex.10 1.0 mole of PCl3(g) and 2.0 moles of Cl2 (g) were placed in a 3 litre flask and heated to 400 K. When equilibrium was established , only 0.70 mole of PCl3 (g) remained. What is the value of equilibrium constant for the reaction : PCl3(g) + Cl2(g) PCl5(g) at 400 K ? (A) 0.25 (B) 1.31 (C) 0.76 (D) 2.6 Ans. [C] Sol. PCl3(g) + Cl2(g) PCl5(g) Initial conc. 1.0 2.0 0 At equilibrium 0.7 1.7 0.3 KC 0 .3 / 3 0.3  3 = 0 .7 / 3  1 .7 / 3 0.7  1.7 = 0.76 Ex.11 For the reaction PCl5(g) PCl3(g) + Cl2(g), The moles of each component PCl5, PCl3 and Cl2 at equilibrium were found to be 2. If the total pressure is 3 atm. The KP will be – (A) 1 atm. (B) 2 atm. (C) 3 atm. (D) 1.5 atm. Ans. [A] Sol. Total Moles = 2 + 2+ 2 = 6 PPCl3 = 2 2 2 × 3, PPCl5 = ×3, PCl2 = ×3 6 6 6 CHEMICAL EQUILIBRIUM KP = PPCl3  PCl2 PPCl5 1 1 = = 1 atmosphere. 1 Ex.14 In a reaction vessel of 2 litre capacity 3 moles of N2 reacts with 2 moles of O2 to produce 1 mole of NO. What is the molar Ex.12 For the reaction H2 + 2 Sol. concentration of N2 at equilibrium ? 2H The value of equilibrium constant is 9.0. The degree of dissociation of H will be – (A) 2 (B) 2/5 (C) 5/2 (D) 1/2 Ans. [B] Equilibrium constant of the reaction H2 +  2 So the Sol. (B) 1.50 (C) 0.75 (D) 2.0 N2 + O2 2NO 3 3-x  2H is 9.0 equi lib ri um (A) 1.25 0 2-x 2x 2x = 1 ; x = 0.5 consta nt for the dissociation of Hi.e. 2H 2 Ans. [A] [N2] = H2+ 2will be 3  0. 5 = 1.25 2 1/9. 2H H2 + 2 1 0 x 2 1– x KC = x x × 2 2 Ex.15 H was heated in a sealed tube at 440°C till the equilibrium was established. The 0 x 2 dissociation of H was found to be 22%. The equilibrium constant for dissociation is – 1 1  (1  x ) (1  x ) x2 1 = ; 9 2  2(1  x )2 Sol. 2 – 2x = 3x 5x = 2 x = 2 / 5 Ex.13 For the reaction N2 Sol. 2NH3 , N2 : H2 were taken in the ratio of 1 : 3 . Up to the point of equilibrium 50% each reactant has been reacted. If total pressure at equilibrium is P. The partial pressure of ammonia would be – (A) P/3 (B) P/6 (C) P/4 (D) P/8 Ans. [A] N2 + 3H2 2NH3 Initially 1 3 0 At equilibrium 2 × 0.5 Total moles 1.5) + 1 = 3 (B) 0.0786 (C) 0.0199 (D) 1.99 Ans. [C] x 1 = 2(1  x ) 3 or (A) 0.282 1-0.5 3-1.5 = (1 – 0.5) + (3 – ; PNH3 = 1 1 P 3 The equilibrium of the dissociation of 2H H2 1 0 0 22 100  2 22 100  2 0.11 0.11 1– 22 100 0.78 KC = 0.11 0.11  0.0199 0.78  0.78 + 2