DPP-29 to 30 With Answer

FACULTY COPY DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 29 Class : XIII Course : DPP No.1 Total Marks : 40 Max. Time : 40 min. Single choice Objective ('–1' negative marking) Q.1 to Q.7 (3 marks 3 min.) [21, 21] Subjective Questions ('–1' negative marking) Q.8 (4 marks 4 min.) [4, 4] BooSt YoUr PreViouS ConCept Single choice Objective ('–1' negative marking) Q.9 to Q.13 (3 marks 3 min.) [15, 15] 1. CH3–CO–CH3(g) CH3–CH3(g) + CO(g) Initial pressure of CH3COCH3 is 100 mm. When equilibrium set up and mole fraction of CO(g) is 1/3 hence Kp is : (1) 100 mm (2*) 50 mm (3) 25 mm (4) 0.6 mm Sol. CH3–CO–CH3(g) CH3–CH3(g) + CO(g) 100 0 0 100 – P P P Total pressure at equilibrium = 100 + P P 1 CO = 100  P = 3 3P = 100 + P P = 50 2. PCl5 is 40% dissociated when pressure is 2 atm. It will be 80% dissociated when pressure is approximately (1*) 0.2 atm (2) 0.5 atm (3) 0.3 atm (4) 0.6 atm 2P Sol. KP = 2P 1 2 2P 1 1 1 2 2 2 = 1 2 0.4  0.4  2  1 0.4  0.4 0.8  0.8  P2 = 1 0.8  0.8 1 1 0.16 0.36 2P2 = 1 0.64 0.84  2 = P2 P2 = 0.214 atm 3. In the reaction C(s) + CO2(g) 2CO(g), the equilibrium pressure is 12 atm. If 50% of CO2 reacts then Kp will be : (1) 12 atm (2*) 16 atm (3) 20 atm (4) 24 atm Sol. C(s) + CO2(g) 2CO(g) P–P/2 P = 3P 2 = 12 so KP = P2 (P / 2) = 2P = 2 × 8 = 16 atm. 4. For the reaction CaCO3(s) CaO(s) + CO2(g) What is the minimum mass of CaCO3 (s), below which it decomposes completely, required to establish equilibrium in a 6.50 litre container for the reaction : [Kc = 0.05 mole/litre] (1*) 32.5 g (2) 24.6 g (3) 40.9 g (4) 8.0 gm Sol. KC = [CO2] = 0.05 mole/litre so moles of CO2 = 6.50 × 0.05 moles = 0.3250 moles CaCO3 CaO + CO2 1 mole of CO2 = 1 mole of CaCO3 0.3250 moles of CO2 = 0.3250 moles of CaCO3 = 0.3250 × 100 gm of CaCO3 = 32.5 gm of CaCO3 . 5. In a homogeneous gaseous reaction, A + 2B 2C, 2.0 mole of ‘A’, 3.0 mole of ‘B’ and 2.0 mole of ‘C’ are placed in a 2.0 L flask and the equilibrium concentration of ‘C’ is 0.5 mole/L. The equilibrium constant (KC) for the reaction is : (1*) 0.05 (2) 0.147 (3) 0.073 (4) 0.026 Sol. A + 2B 2C Initial moles 2 3 2 At eqm. 2.5 4 1 Molar conc. 2.5/2 = 1.25 4/2 = 2 1/2 = 0.5 (0.5)2 K = 1.25  (2)2 = 0.05 Note that 1 mole of C has reacted to form 1 mole of B and 0.5 mole of A. 6. 1 N (g) + O (g) NO (g) ...K 2 2 2 2 1 2NO2 (g) N2O4 (g) ...K2 Given that above reactions have equilibrium constants K1 and K2 respectively. What would be the expression for the equilibrium constant K for the following reaction, in terms of K1 and K2 ? N2O4 (g) N2 (g) + 2O2 (g) 1 N (g) + O (g) NO (g) ...K 2 2 2 2 1 2NO2 (g) N2O4 (g) ...K2 1 (1) K1 K2 (2) K1(K 2 )2 1 1 (3*) K 2 (K1 )2 1 (4) K1 K 2 1 Sol. N2O4 2NO2 k2  2NO N + 2O 2 . 1 7. NH4COONH2 (s) 2NH3(g) + CO2(g). If equilibrium pressure is 6 atm for the above reaction ; Kp will be : (1*) 32 (2) 27 (3) 4/27 (4) 1/27 Sol. PNH : PCO = 2 : 1  PNH = 4 atm ; PCO = 2 atm  Kp = [ PNH ]2 [ PCO ] = 42 × 2 = 32 8. Consider the equilibrium: P(g) + 2Q(g) R(g). When the reaction is carried out at a certain temperature, the equilibrium concentration of P and Q are 3M and 4M respectively. When the volume of the vessel is doubled and the equilibrium is allowed to be reestablished, the concentration of Q is found to be 3M. Find (a) Kc (b) concentration of R at two equilibrium stages. 1 Ans. (a) KC = litre2 mol–2 12 (b) 4M , 1.5 M P(g) + 2Q(g) R(g) eq. 3 4 x KC = x (3)(4)2 x = 48 If volume become double then new con. 3 4 x 2 2 2 x 2 Q =  3 22 2 x = 12   Q > KC backward direction P + 2Q R(g) 3  y 2 4  2y 2 = 3 4  2y 2 x  y 2 4 + 2y = 6 y = 1 put the value of y x  1 2 3 2 x  1 KC = [R(g)] [P][Q]2 x  48 = 2 (2) (3)2  x = 4 Put the value of x x 4 1 K = = =  [R] = x = 4M  [R] = x  1 = 4  1 = 1.5 M. C 48 48 12 i f 2 2 BooSt YoUr PreViouS ConCept 9. I, II, III are three isotherms respectively at T1, T2 and T3. Temperature will be in order (1) T1 = T2 = T3 (2) T1 < T2 < T3 (3*) T1 > T2 > T3 (4) T1 > T2 = T3 Sol. PV = nRT  (PV)I > (PV)II > (PV)III  (nRT ) > (nRT ) > (nRT ) 1  T > T 2 3 > T3 10. The ratio of rates of diffusion of SO2, O2 and CH4 is : (1*) 1 : : 2 (2) 1 : 2 : 4 (3) 2 : : 1 (4) 1 : 2 : Sol. Rate of diffusion ×  rSO : r : rCH = 1 1 1 : : = 1 : : 2 11. The compressibility factor of He is (1) Linearly increasing with volume (2*) Linearly increasing with pressure (3) Equal to 1 (4) Linearly decreasing with pressure Sol. For He  P (Vm–b) = RT or PVm = 1 + RT Pb = Z RT  Z is linearly increasing with pressure 12. Compressibility factor for H2 behaving as real gas is : 1 a  1  Pb  RTV (1) 1 (2)    RTV  (3*)    RT  (4) (1 a) Sol. Compressibility factor for H2 gas is always > 1 for ordinary conditions at all pressures.  a  P  V2  (V – b) = RT   a is negligible P (V – b) = RT PV – Pb = RT PV  RT – Pb RT = 1 PV Pb  RT = Z = 1 + RT 13. The critical density of the gas CO2 is 0.44 g cm–3 at a certain temperature. If r is the radius of the molecule, r3 in cm3 is approximately. (N is Avogadro number) 25 (1) N (2) (2) 100 N (3*) 6.25 N 25 (4)  Sol. VC = Mass d = 44 = 100 cm3 (Molar volume) 0.44 3 × 4 × 4 r3 N = 100 r3 = 3 100 16 NA A 6.25 = NA . FACULTY COPY DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 30 Class : XIII Course : DPP No.2 Total Marks : 44 Max. Time : 44 min. Single choice Objective ('–1' negative marking) Q.1 to Q.4 (3 marks 3 min.) [12, 12] Subjective Questions ('–1' negative marking) Q.5 to Q.8 (4 marks 4 min.) [16, 16] BooSt YoUr PreViouS ConCept Single choice Objective ('–1' negative marking) Q.9 to Q.12 (3 marks 3 min.) [12, 12] Multiple choice objective ('–1' negative marking) Q.13 (4 marks 4 min.) [4, 4] 1. Two moles of HI were heated in a sealed tube at 440°C till the equilibrium was reached. HI was found to be 22% decomposed. The equilibrium constant for dissociation is - (1) 0.282 (2) 0.0796 (3*) 0.0199 (4) 1.99 Sol. 2HI H2 + I2 Initial 2moles 22 At eqm. 2 – 100 × 2 0.22 0.22 = 2 – 0.44 = 1.56 [H2 ] [I2 ] K = [HI]2 0.22  0.22 = (1.56)2 = 0.0199 2. For A  2 B equilibrium constant at total pressure p1 is kp1 & for C  D + E equilibrium constant at total pressure p2 is kp2 . If both reaction are gaseous homogeneous reactions and degree of dissociation of A & C are same , then the ratio of p1/p2 if kp1 = 2 kp2 is : (1*) 1/2 (2) 1/3 (3) 1/4 (4) 2 Sol. A  2 B a (a–a) 0 2a Total mole = (a + a) a  a  1   P =  P =   P A PB = a  a 1 2a a  a  P  1   1 2P1 = 1   2P1 2   [PB ]2  1   42P1 KP1 = [P ] =  1  P  = 2 A  1  1  1  C  D + E C 0 0 C–C C C Total mole = (C + C) C  C  1   P =  P =    P C PD = PE = C  C C C  C C C  C 2  P =  P =  1   2 P2 1  P2 1  kP2 2P2 = 1 2 kP1 kP2 4P1 = P2 2 4P1 1 = P2 P1 = 1 P2 2 3. For the reaction : CaCO3 (s) CaO(s) + CO2 (g), Kp = 1.16 atm at 800°C. If 20g of CaCO3 were kept in a 10 litre vessel at 800°C, the amount of CaCO3 remained at equilibrium is - (1*) 35% (2) 64% (3) 46% (4) none Sol. CaCO3(s) CaO(s) + CO2 (g) KP [PCO2 ] 1.16 = [PCO2 ] PV = nRT 1.16  10 = nCO2  0.0821  1073  nCO2 = 0.13 mole initial nCaCO3 20 = 100 = 0.2  use mole of CaCO3 = 0.13 mole remaining mole of CaCO3 = 0.2 – 0.13 = 0.07 remaining mass of CaCO3 = 0.07  100 = 7 g 7 % remaining mass of CaCO3 = 20  100 = 35% 4. For the following equilibrium N2O4 2NO2 in gaseous phase, NO2 is 50% of the total volume when equilibrium is set up. Hence percent of dissoicaiton of N2O4 is : (1) 50% (2) 25% (3) 66.66% (4*) 33.33% Sol. N2O4 2NO2 1 1 -  0 2 % NO2 2  = 1  1  100 = 50 = 1  = 2 = 4 = 1 +  3 = 1 1  = 3 %  = 33.33 % 5. For NH4HS(s) NH3(g) + H2S(g),for a reaction started only with NH4HS(s) the observed pressure for reaction mixture in equilibrium is 1.12 atm at 106°C. What is the value of Kp for the reaction . Ans. 0.3136 atm2 Sol. NH4HS(s) NH3(g) + H2S(g) at eq. P P Total pressure = 2P = 1.12  P = 1.12 2 = 0.56 K = PNH  PH S  = 0.56 0.56 KP = 0.3136 atm2 6. For the gaseous reaction of XO with O2 to form XO2, the equilibrium constant at 398 K is 1.0 × 10–4 lit/mole. If 1.0 mole of XO and 2.0 mole of O are placed in a 1.0 L vessel and allowed to come to equilibrium, what will be the equilibrium concentration of each of the species? Sol. The equilibrium reaction is 2XO (g) + O2(g) 2XO2 (g) since the unit of Kc given is lit/mole. 2XO(g) + O2 (g) 2XO2 (g) Initial conc. 1 2 0 Conc. at equilib. 1 2x 2 – x 2x  K = [XO2 ]2 [XO]2 [O2 ] 2x2 = 1 2x2 2  x = 4x2 (1 2x)2 (2  x) 4x2 = 2 Since, the value of equilibrium constant is very small (1 × 10–4), so 2x can be ignored with respect to 1 and x can be ignored with respect to 2. 4x2  1 × 10–4 = 2 x = 7.07 × 10–3 we can see that the value of x is very small, so the assumtion made was correct as it is within 1.4% of the actual value. Thus, the assumption made is correct and acceptable.  [XO] = 1 – 0.01414 = 0.985 M [O2] = 2 – 0.00707 = 1.992 M [XO2] = 0.0141 M 7. At a certain temperature the equilibrium constant (Kc) is 16 for the reaction SO2(g) + NO2 (g) SO3 (g) + NO (g) It we take one mole of each of the four gases in a one-litre container, what would be the equilibrium concentration of NO and NO2 ? Sol. We have, SO2 + NO2 SO3 + NO 1 mole 1 mole 1 mole 1 mole Initial moles 1 – x 1 – x 1 + x 1 + x Moles at eqb. or concentration at eqb. where x is the number of moles of each reactant changed to the products at equilibrium. (1 x) (1 x)  (1 x)2 K = (1 x) (1 x) or 1 x = 4 ; x = 0.6 1 x (1 x)2 = 16 (given) [NO] = 1 + x = 1 + 0.6 = 1.6 moles/litre [NO2] = 1 – x = 1 – 0.6 = 0.4 moles/litre. 8. 1 mole of H2, 2 moles of I2 and 3 moles of HI were taken in a 1-litre flask. If the value of KC for the equation H2(g) + I2 (g) 2HI (g) is 50 at 440ºC, what will be the concentration of each species at equilibrium ? Sol. Suppose that x moles each of H2 and I2 are converted to HI at equilibrium. H2 + I2 2HI 1 2 3 Initial moles (1 – x) (2 – x) (3 + 2x) Moles at equilibrium [HI]2  (3  2x)2  K = 50 c [H2 ] [I2 ] (1 x)(2  x) x = 0.7 (approx). [H ] = (1 – x) = 1 – 0.7 = 0.3  [I ] = 2 – x = 2 – 0.7 = 0.3  [HI] = 3 + 2x = 3 + 1.4 = 4.4. 2 2 BooSt YoUr PreViouS ConCept 9. In P4S3 how many P–P bonds are present. (1*) 3 (2) 4 (3) 5 (4) 2 Sol. The structure of P4S3 is ; P4S3fdlajpuk gSA 10. Hybridisation of Iodine atoms in Cl (in its stable form, found in solid state) and  Cl Br are : 3 2 4 2 (1) sp3 & sp3d2 (2) sp3d2 & sp3d3 (3) both sp3d (4*) both sp3d2 Sol. ICI3 exist as dimer as I2Cl6 11. Which among the following molecules have sp3d hybridisation with one lone pair of electrons on the central atom ? (i) SF (ii) [PCl ]+ (iii) XeO F (iv) ClOF 4 4 2 2 3 (1) (i), (ii) and (iii) only (2*) (i), (iii) and (iv) only (3) (i) and (iii) only (4) (iii) and (iv) only. 12. Among the following molecules (i) XeO3 (ii) XeOF4 (iii) XeO2F2 (iv) XeF6 those having different molecular geometry(SHAPE) but same number of lone pairs on Xe are. (1) (i),(ii) and (iii) only (2) (i) ,(ii) and (iv) only (3) (ii), (iii) and (iv) only (4*) (i), (ii) (iii) and (iv) 13. Which of the following statement(s) is / are correct ? (1) Hybridisation of carbon in C3O2 is sp . 2 (2*) In Cr O 2 –, six Cr – O bonds are identical. 2 7 (3) Three centre two electron bonds exist in B2H6 and Al2Cl6 . (4*) In AgI, the colour is attributed to polarisation of – anion. Sol. (1) O  C  C  C  O | | | sp sp sp (2) Resonance takes place Resonance ; Therefore, all Cr – O bond lengths are not equal. takes place (3) This is an example of 3-centre 2-e– bond which is also known as Banana bond. But Al2Cl6 have covalent bond only and there is no electron deficient bonding as depicted in the given structure. Cl Cl Cl Al Al Cl Cl Cl (4) AgI is bright yellow coloured compound due to the polarisation of anion, I–.The bigger anions are more polarised and hence their electrons get excited by partial absorption of visible light.