DPP-27 to 28 Physical chemistry with Answer

FACULTY COPY DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 27 Class : XIII Course : DPP No.1 Total Marks : 30 Max. Time : 30 min. Single choice Objective ('–1' negative marking) Q.1 to Q.5 (3 marks 3 min.) [15, 15] True or False (no negative marking) Q.6 (2 marks 2 min.) [2, 2] Assertion and Reason (no negative marking) Q.7 (3 marks 3 min.) [3, 3] BooSt YoUr PreViouS ConCept Single choice Objective ('–1' negative marking) Q.8 to Q.9 (3 marks 3 min.) [6, 6] Subjective Questions ('–1' negative marking) Q.10 (4 marks 4 min.) [4, 4] 1. For a reversible reaction the rate constants for the forward and backward reactions are 2.38 × 10–4 and 8.15 × 10–5 respectively. The equilibrium constant for the reaction is (1) 0.342 (2*) 2.92 (3) 0.292 (4) 3.42 2. Select the reaction for which the equilibrium constant is written as [MX ]2 = K[MX ]2 [X ] 3 2 2 (1) MX (g) MX (g) + 1 X (g) (2) 2MX (g) 2MX (g) + X (g) 3 2 2 2 3 2 2 (3*) 2MX (g) + X (g) 2MX (g) (4) MX (g) + 1 X (g) MX (g) 2 2 3 2 2 2 3 3. For a reaction H2 (g) + I2 (g) 2HI (g) at 721 K, the value of equilibrium constant is 50. If 0.5 moles each of H2 and I2 is added to the system the value of equilibrium constant will be - (1) 0.02 (2) 0.2 (3*) 50 (4) 25. 4. In a reaction A + 2B 2C, 2.0 mole of ‘A’, 3.0 mole of ‘B’ and 1.0 mole of ‘C’ are placed in a 2.0 L flask and the equilibrium concentration of ‘C’ is 1.0 mole/L. The equilibrium constant (K) for the reaction is (1) 0.33 (2*) 1.33 (3) 1.66 (4) 0.66. 5. The correct order of boiling point is : (1) F2 > Cl2 > Br2 > I2 (3) HI > HBr > HF > HCl (2) NH3 > SbH3 > AsH3 > PH3 (4*) H2O > H2Te > H2Se > H2S 6. Give the correct order of initials T or F for following statements. Use T if statement is true and F if it is false. S1 : HF boils at a higher temperature than HCl S2 : HBr boils at lower temperature than HI S3 : Bond length of N2 is less than N2+ (1) T F T (2*) T T F (3) T T T (4) F T T Sol. S1 : Due to intermolecular H-bonding in HF it boils at higher temperature than HCl S2 : Mol. wt. of HBr < Mol. wt. of HI S3 : Bond order of N2 is more than N2+ . 7. STATEMENT - 1 : Boron halides involves p-p back bonding whereas Al halides do not. STATEMENT - 2 : Al has larger size than boron. (1*) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (2) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (3) Statement-1 is True, Statement-2 is False (4) Statement-1 is False, Statement-2 is True BooSt YoUr PreViouS ConCept 8. If a piece of iron gains 10% of it's weight due to partial rusting into Fe2O3, according to reaction : Fe + O  Fe O . The percentage of total iron that has rusted is : (1) 23 (2) 13 (3*) 23.3 (4) 25.67 9. The order of metallic bond strength in Ca, K, Fe, Rb (1) Ca > Fe > K > Rb (2*) Fe > Ca > K > Rb (3) Ca > K > Fe > Rb (4) Rb > K > Fe > Ca 10. Calculate the number of hours of service that can be derived from an acetylene lamp containing 1 kg CaC2 if 60 liter acetylene gas at 1 atm & 27°C is needed for 1 hour. Sol. no. of hours = 1000  .0821 300 64  1 60 = 6.414 hrs. Ans. FACULTY COPY DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 28 Class : XIII Course : DPP No.2 Total Marks : 36 Max. Time : 36 min. Single choice Objective ('–1' negative marking) Q.1 to Q.6 (3 marks 3 min.) [18, 18] Subjective Questions ('–1' negative marking) Q.7 to Q.8 (4 marks 4 min.) [8, 8] BooSt YoUr PreViouS ConCept Single choice Objective ('–1' negative marking) Q.9 to Q.10 (3 marks 3 min.) [6, 6] Match the Following (no negative marking) Q.11 (4 marks 4 min.) [4, 4] 1. 2 2. 3 3. ANSWER KEY DPP No.-28 4 4. 3 5. 1 6. 2 (2n  y)y 7. KC = 4.0. 8. (n  y)2.P 9. 3 10. 3 11. (1  p, q) ; (2  p, q, r) ; (3  p, q, r, s) ; (4  p, q, r, s) 1. For the following equilibrium : (omitting charges) I : M + CI  MCI Keq = 1 II : MCI + CI  MCI2 Keq = 2 III : MCI2 + CI  MCI3 Keq = 3 IV : M + 3CI  MCI3 Keq = K then relation between K,  ,  and  is : (1) K =  +  +  (2*) K =    1 2 3 1 2 3 (3) log K = log  [MCl] × log  × log  (4) all of the above Sol.  = 2 = 3 = [M][Cl] [MCl2 ] [MCl][Cl] [MCl3 ] [MCl2 ][Cl] [MCl3 ] K = [M][Cl]3 K =  .  .  or log k = log  + log  + log  Total Pressure Pk = P1 + P2 + P3 2. For the equilibrium in a closed vessel PCI5 (g) PCI3 (g) + CI2(g) Kp is found to be double of Kc. This is attained when: (1) T = 2 K (2) T = 12.18 K (3*) T = 24.36 K (4) T = 27.3 K Sol. Kp = KC (RT)n 2KC = KC (RT) = K 2–1 (RT)1 2 = RT 2 T = 0.821 = 24.36 K Ans. 3. For the equilibrium 2H2O(g) 2H2 (g) + O2(g) equilibrium constant is K1. For the equilibrium 2CO2 2CO(g) + O2 (g), equilibrium constant is K2. The equilibrium constant for CO2(g) + H2(g) CO(s) + H2O(g) is : (1) K1 K2 K1 (2) K2 (3) (3) (4*) Sol. K1 = K = [H2 ]2[O2 ] [H2O]2 [CO2 ][HO2 ] [CO2 ][H2 ] = Ans.  K = [CO2 ][O2 ] [CO2 ]2 4. For the reaction A(g) + B(g) C(g) at equilibrium the partial pressure of the species are PA = 0.15 atm, PC = PB = 0.30 atm. If the capacity of reaction vessel is reduced, the equilibrium is reestablished. In the new situation partial pressure A and B become twice. What is the partial pressure of C? (1) 0.30 (2) 0.60 (3*) 1.20 (4) 1.80 Sol. A(g) + B (g) C(g) (PC ) 0.30 100 20 KP = (PA )(PB ) = 0.15  0.30 = 15 = 3 PA = 0.30 atm PB = 0.60 atm 5. For the reaction N2O4 2NO2, at 350 K, the value of Kc = 0.4. The value of Kp for the reaction at the same temperature would be : (1*) 11.49 atm (2) 1.148 atm (3) 1.4 × 10–2 atm (4) 1.4 × 10–3 atm Sol. KP = K (RT)1 = 0.4  0.0821  350 = 11.49 atm 6. The equilibrium N2 + O2 estabilished in a reaction vessel of 2.5 L capacity. The amounts of N2 and O2 taken at the start were respectively 2 moles and 4 moles. Half a mole of nitrogen has been used up at equilibrium. The molar concentration of nitric oxide is (1) 0.2 (2*) 0.4 (3) 0.6 (4) 0.1 Sol. N2 + O2 2NO Initial 2 moles 4moles 0 mole At Eqm, 2 – 1 4 – 1 2 × 1 = 1 mol 2 Molar conc. of NO at eqm. = 2 2 1 = 0.4 2.5 7. For an equilibrium A + 2B 2C + D; A and B are mixed in a reaction vesel at 300 K. The initial concentration of B was 1.5 times the initial concentration of A. After the equilibrium, the equilibrium concentration of A and D are same. Calculate KC. Ans. KC = 4.0. A + 2B 2C + D a 1.5a 0 0 a – x (1.5a – 2x) 2x x [A] = [D] a – x = x a = 2x x = a 2 a 0.5a a a 2 2 [C]2 [D] a2  a 2 KC = [A] [B]2 = a  a 2 = 4    2  2  8. n mole of PCl3 and n mole of Cl2 are allowed to react at constant temperature T under a total pressure P, as Ans. PCl3 (g) + Cl2 (g) PCl5 (g) If Y mole of PCl5 are formed at equilibrium, find KP (2n  y)y (n  y)2.P PCl3 (g) + Cl2 (g) PCl5 (g) n n 0 n – y n – y y Total mole equilibrium = 2n – y PPCl3  n  y  =  2n  y   P PCl2 n  y = 2n  y  P    y  PPCl5 =  2n  y   P   PPCl5 yP 2n  y (2n  y)y KP = (PPCl )(PCl ) =  n  y   n  y   = (n  y)2 P 3 2   P  P 2n  y 2n  y     BooSt YoUr PreViouS ConCept 9. Which one of the following is the correct set with respect to molecule, hybridization and shape? (1) BeCl , sp2, linear (2) BeCl , sp2, triangular planar 2 2 (3*) BCl , sp2, triangular planar (4) BCl , sp3, tetrahedral 3 3 10. Which of the following statements are correct ? (I) structure is not allowed because octet around 'O' can not be expanded. (II) H2O2 is ionic compound (III) In B2 molecule, the H.O.M.O. is  molecular orbital. (IV) The lp–bp repulsion is stronger than bp–bp repulsion. (1) (I) and (III) (2) (II) and (III) (3*) (I) and (IV) (4) (III) and (IV) Sol. (I) The electronic structure which is allowed is as central oxygen has complete octet whereas in the central atom has 10 electrons. (II) H2O2 is covalent compound. (III) (1s)2 (*1s)2 (2s)2 (*2s)2 (2p1 = 2p1 ) (p )0 x y z The highest occupied molecular orbital is  molecular orbital. (IV) The repulsive interaction of electron pairs decreases in the order : lone pair (𝑙p) - lone pair (𝑙p) > lone pair (𝑙p) - bond pair (bp) > bond pair (bp) -bond pair (bp) 11. Match the pair of species given in column-I with the identical characteristic(s)/type of hybridisation given in column-II : Column - Ι Column - ΙΙ (Pair of species) (Identical Property in pairs of species) (1) PCl3F2 and PCl2F3 (p) Hybridisation of central atom (2) BF3 and BCl3 (q) Shape of molecule/ion (3) CO2 and CN (r)  (dipole moment) –2 (4) C6H6 and B3N3H6 (s) Total number of electrons Ans. (1  p, q) ; (2  p, q, r) ; (3  p, q, r, s) ; (4  p, q, r, s)