DPP-31 to 32 With Answer

FACULTY COPY DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 31 Class : XIII Course : DPP No.1 Total Marks : 27 Max. Time : 27 min. Single choice Objective ('–1' negative marking) Q.1 to Q.7 (3 marks 3 min.) [21, 21] Comprehension ('–1' negative marking) Q.8 to Q.9 (3 marks 3 min.) [6, 6] ANSWER KEY DPP No.-31 1. 3 2. 1 3. 1 4. 2 5. 3 6. 2 7. 1 8. 1 9. 2 1. The reaction, SO2 + Cl2 SO2Cl2 is exothermic and reversible. A mixture of SO2(g), Cl2(g) and SO2Cl2(𝑙) is at equilibrium in a closed container. Now a certain quantity of extra SO2 is introduced into the container, the volume remaining the same. Which of the following are true? (1) The pressure inside the container will not change (2) The temperature will not change (3*) The temperature will increase (4) The temperature will decrease Sol. addition of extra SO2, reaction more forward direction and release energy so temperature will increases. 2. What are the most favourable conditions for the reaction ; SO2(g) + 1 2 2 (g) SO3(g) ; H = – ve to occur ? (1*) low temperature, high pressure (2) low temperature, low pressure (3) high temperature, low pressure (4) high temperature, high pressure Sol. H (–) exothermic low T High P When pressure increases, reaction moves in such direction where No. of moles decreases so reaction shift in forward direction. 3. For the reaction 2NO2(g) 2NO(g) + O2(g) Kc = 1.8 × 10–6 at 184ºC Given : R = 0.083 kJ K–1 mol–1 when Kp and Kc are compared at 184ºC it is found that : (1*) Kp is greater than Kc (2) Kp is less than Kc (3) Kp = Kc (4) whether Kp is greater than, less than or equal to Kc depends upon the total gas pressure Sol. Kp = Kc (RT)n n = 3 – 2 = 1 Kp = Kc (RT)1  Kp > Kc 4. A 10L container at 300K contains CO2 gas at pressure of 0.2 atm and an excess solid CaO (neglect the volume of solid CaO). The volume of container is now decreased by moving the movable piston fitted in the container. What will be the maximum volume of container when pressure of CO2 attains its maximum value given that CaCO3 (s) CaO(s) + CO2(g) Kp = 0.800 atm (1) 5 L (2*) 2.5 L (3) 1 L (4) The information is insufficient. Sol. K = 0.800 atm = PCO = maximum pressure of CO2 in the container to calculate maximum volume of container the PCO = 0.8 atm and none of CO2 should get converted into CaCO3(s). so V(0.800 atm) = (10 L) (0.2 atm) so v = 2.5 L 5. At 1400 K, K = 2.5×10–3 for the reaction CH (g) + 2H S(g) CS (g) + 4H (g) . A 10.0L reaction vessel at 1400 K contains 2.00 mole of CH4, 3.0 mol of CS2, 3.0 mole of H2 and 4.0 mole of H2S. Then (1) This reaction is at equilibrium with above concentrations. (2) The reaction will proceed in forward direction to reach equilibrium (3*) The reaction will proceed in backward direction to reach equilibrium (4) The information is insufficient to decide the direction of progress of reaction  3 4  3  Sol. QC =          2  4 2 243 = 32  10–2 = 7.59 × 10–2 > K     10  10  so, reaction will proceed in backward direction. 6. The graph which will be representing all the equilibrium concentrations for the reaction N2O4(g) 2NO2 (g) will be : (the concentrations of N2O4 (g) and NO2 (g) for which the following reaction will be at equilibrium will lie on which of the following graphs) (1) [NO2] [N2O4] (2*) [NO2] [N2O4] (3) [NO2] [N2O4] (4) [NO2] [N2O4] Sol. All the equilibrium concentrations will satisfy the equation [NO2 ]2 K = [N2O4 ]  y2 = kx the required parabolic graph 7. The unit of equilibrium constant KC of a reaction is mol–2 l2. For this reaction, the product concentration increases by - (1*) increasing the pressure (2) lowering the temperature (3) lowering the pressure (4) both B and C Sol. KC = (mole/litre)n where n = no of moles on product side – no. of moles on reactant side Hence n = – 2 so moles on reactant side > moles on product side so on increasing pressure reaction will get shifted in forward direction. Comprehension : (Q.8 to Q. 9) Read the following passage carefully and answer the questions. Hetrogeneous system : If more than one phase is present in the reversible reaction then it is said to be hetrogeneous system. Ex. CaCO3(s) CaO(s) + CO2(g) Expression of equilibrium constant for the above reaction can be taken as [CaO(s)] [CO2(g)] K = [CaCO3 (s)] .... (i) moles of CaO Now concentration of CaO(s) = [CaO(s)] = Volume of CaO as density of CaO   CaO(s)) and molar mass of CaO (MCaO(s)) are a fixed quantity therefore concentration of pure solid and liquid term is unchanged with respect to time. Hence equilibrium constant for the equation (i) can be written as Kc = [CO2(g)] or Kp = PCO2 As Kp and Kc is not containing solid terms therefore addition or removal of pure solid and pure liquid has no. effect on the equilibrium process. 8. 200 gm of CaCO3(s) taken in 4 lit container at a certain temperature. Kc for the dissociation of CaCO3 at this temperature is found to be 1/4 mole lit–1 then calculate the remaining amount of CaCO3. (1*) 100 gm (2) 50 gm (3) 150 gm (4) 25 gm 9. CaCO3(s) CaO(s) + CO2(g) At equilibrium in the above case ‘a’ moles of CaCO3, ‘b’ moles of CaO and ‘c’ moles of CO2 are found then identify the wrong statement : (1) ‘a’ will decrease with the addition of inert gas at constant pressure (2*) ‘a’ will remain constant with the increase in volume (3) It volume of the vessel is halved then ‘a’ increases (4) ‘b’ decreases with the increase in pressure Sol. CaCO3(s) CaO(s) + CO2(g) a mole b mole c mole (1) On constant pressure (inert gas add) V  P  , reaction shift in forward direction, a decrease (2) V  P  , reaction shifts forward direction, a decrease (3) If volume is halved then pressure increase, reaction shift in backward direction so a increase. (4) By increasing pressure, reaction shift in backward direction so mole of b decrease. FACULTY COPY DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 32 Class : XIII Course : DPP No.2 Total Marks : 34 Max. Time : 34 min. Single choice Objective ('–1' negative marking) Q.1 to Q.6 (3 marks 3 min.) [18, 18] Subjectve Questions ('–1' negative marking) Q.7 to Q.10 (4 marks 4 min.) [16, 16] 1. Attainment of the equilibrium A(g) 2C(g) + B(g) gave the following graph. Find the correct option. (% dissociation = fraction dissociated × 100) (1) At t = 5 sec equilibrium has been reached and Kc = 128 (mol/litre)2 (2) At t = 5 sec equilibrium has been reached and % dissociation of A is 60% (3*) At t = 5 sec equilibrium has been reached and % dissociation of A is 30% (4) None of these Sol. t = 5 sec. eq. has reached and kc = (8)2 x 6 7 = 384 7 10  7 = 54.8 (mole/litre)2 %dissociation of = 10 x 100% = 30%. 2. What concentration of CO be in equilibrium with 2.5 × 10–2 mol litre–1 of CO at 100ºC for the reaction : FeO (s) + CO (g) Fe (s) + CO2 (g) , KC = 5.0. (1*) 12.5 × 10–2 (2) 2.5 × 10–2 (3) 0.5 × 10–2 (4) 5 × 10–2 Sol. FeO(s) + CO (g) Fe (s) + CO2 (g) KC = 5 KC = [CO2 ]equ. [CO]equ. [CO2 ]equ. = 2.5 10– 2 = 5 [CO2]equ. = 1.25 × 10–1 mol 3. When alcohol (C2H5OH) and acetic acid are mixed together in equimolecular proportions at 25ºC, 66.5% is converted into ester. Calculate how much ester will be formed if 1 mole of acetic acid is treated with 0.5 mole of alcohol at the same temperature. (1*) 0.423 mole (2) 0.523 mole (3) 0.2 mole (4) 0.15 mole Sol. 1 1 0 0 Initial moles C2H5OH ( 𝑙 ) + CH3COOH( 𝑙 ) CH3COOC2H5 ( 𝑙 ) + H2O ( 𝑙 ) (1 – 0.665) (1 – 0.665) (0.665) (0.665) Moles at equilibrium where V is the volume in litres of the reaction mixture. 0.665  0.665 K  [CH3COOC2H5 ].[H2O]  V V  4  C [C H OH].[CH COOH] 0.335 0.335 2 3 3  V V Now, for the second reaction 0.5 1 0 0 Initial moles C2H5OH + CH3COOH CH3COOC2H5 + H2O (0.5 – x) (1 – x) x x Moles at equilibrium x is the moles of ester produced at equilibrium. x . x KC  V V  0.5  x .1 x = x2 (0.5  x)(1 x) = 4 V V (Kc = has the same value as temperature remains the same) x = 0.423 mole. 4. In an evacuated closed isolated chamber at 2500 C, 0.02 mole PCl and .01 mole Cl are mixed 2 (PCl5 PCl3 + Cl2) . At equilibrium density of mixture was 2.48 g/L and pressure was 1 atm. The number of total moles at equilibrium will be approximately (1) 0.012 (2) 0.022 (3) 0.032 (4*) 0.045 Sol. PCl5 (g) PCl3 (g) + Cl2 (g) t = 0 0.02 mole 0.01 mole t = teq 0.02 – x x (0.01 + x) Total number at eq m. = (0.03 + x) moles. (0.02 MPCI5 ) (0.01 MCI2 ) density at equilibrium = V = 1.96 L 0.03  0.0821 523 Volumeinlibe  V = 0.48 g/L (0.03  x) 0.0821 523 Pinitial = 1.96  1 atm = 1.96 Total mole = (0.03 + x) = 0.045 mole 5. For the gas phase reaction C H + H C H H = – 136.8 kJ mol–1 2 4 2 2 6 carried out in a vessel, the equilibrium concentration of C2H4 can be decreased by (1) increasing the temperature. (2*) increasing the pressure. (3) removing some H2 (4) adding some C2H6 6. In an aqueous solution of volume 500 ml, when the reaction of 2Ag+ (aq) + Cu (s) Cu2+ (aq) + 2Ag (s) reached equilibrium the [Cu2+] was x M. When 500 ml of water is further added, at the equilibrium [Cu2+] will be (1) 2 x M (2) x M (3) between x M and x/2M (4*) less then x/2 M. 7. The degree of dissociation is 0.39 at 400 K and 1.0 atm for the gaseous reaction PCl5 PCl3 + Cl2. Assuming ideal behaviour of all gases, Calculate the density of equilibrium mixture at 400 K at 1.0 atm. [P = 31, Cl = 35.5, R = 0.08 lt-atm/mol-K] Ans. 4.6875 gm/L. 208.5 Sol. M = 1 + (n – 1)  .  = 1 + 0.39. 1 150 M = 150.  d = 0.082  400 . 8. The density of an equilibrium mixture of N2O4 and NO2 at 1 atm and 625 K is 1.15 g/L. Calculate KP for the 0 a (1 – ) 2a PV = n RT d = PMmix = 1Mmix = 1.15 g/L equ. RT 0.08  625 Mmix = 1.15 × 625 × 0.08 = 57.5  M obs = Mth 1 (n – 1) 92 1   1 + a = 92 = 57.5 92 – 57.5 57.5   = 0.6 PN2O4 = a(1– ) a(1 ) 0.4 + 1 atm PNO2 = 2a  a(1 2) 1.2 = 1.6 PNO2 =  3 2 9 1.6 P 2  4  NO2   16 9  4 9 KP = PN2O4 =  1  = 1   4 4 = 16 = 4 = 2.25 atm.   9. 0.96 gm of H were heated to attain equilibrium 2H H +  . The reaction mixture on titration requires 15 ml of N/10 Hypo solution. Calculate the degree of dissociation of H. [H = 1,  = 127] Ans. 0.2. Sol. 2H H +  0.96 t = 0 128 = 7.5 × 10–3 0 0 t = ta a – 2x x x Meq. of hypo = meq. I2 1 15 × 10 = 2 × millimole of I2 millimole of I2 = 0.75 2  0.75 % dissociation = 7.5 × 100 = 20 % = 0.2 10. Gº for 1 2 N2 (g) + 3 2 H2 (g) NH3 (g) , is – 16.5 kJ mol–1 . Find out K for the reaction at 300 K. Also report KP & Gº for N2 (g) + 3H2 (g) 2NH3 (g) , at 300 K. [antilog (2.87) = 745.6]. Ans. KP = 745.6 (atm–1) ; KP = 5.56 × 105 atm–2 ; Gº = – 33 kJ mol–1.