DPP-35 to 36 Faculty Copy

DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 35 to 36 Class : XIII Course : DPP No.1 Total Marks : 38 Max. Time : 38 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.11 (3 marks 3 min.) [9, 9] Multiple choice objective ('–1' negative marking) Q.12 (4 marks 4 min.) [4, 4] 1. The decompostion NH3 gas on a heated tungsten surface gave the following results : Initial pressure (mm) 65 105 y 185 Half-life (sec) 290 x 670 820 Calculate approximately the values of x and y. xeZVxaLVulrgijNH3xSldsfo?kVulsfuEuifj.k eçkIrgksrsgSaA izkjfEHkdnkc(mm) 65 105 y 185 v)Z&vk;q¼lSd.M½ 290 x 670 820 x rFkk ydk yxHkx eku ifjdfyr dhft,A (1) x = 410 sec ¼lSd.M½ (2*) x = 467 sec ¼lSd.M½ (3) x = 490 sec ¼lSd.M½ (4) x = 430 sec ¼lSd.M½ y = 115 mm y = 150 mm y = 120 mm y = 105 mm Sol. Initial pressure 65 105 y 185 Half life 290 x 670 820 Initial pressure of gas  Initial moles of gas in above anestion Half life  Initial pressure So, it must be zero order reaction t 1 = 2 CO = PO 2k 2k 290 = 65 2 k = k = 65 2  290 mm of Hg/sec 105  2  290 x = 2  65 = 468 sec 670 = y  2  290 2  64 = y = 150 mm of Hg 2. In a first order reaction, the concentrations of the reactant, 30 minutes and 40 minutes after the starts are C1 and C2 (in moles/litre) respectively. What was C0, intial concentration ? ,dizFkedksfVvfHkfØ;kesaizkjEHkgksusds30fefuVrFk 40fefuVi'pkrlkUnzrk,¡Øe'k%C1rFk C2(eksy/yhVj)gSarks C0,izkjfEHkdlkUnzrkD;kgksxh\ C3   C 4  C 3 C4  (1) C0   1    (2) C0   1  (3) C 0   1  (4*) C0   1     2   CO   2   2   2  Sol. K t = In    Ct   CO  K × 30 = In  C  ....(1)  1   CO  K × 40 = In  CO     C2  .....(2) In   3 C  C   C  =  1  4  CO  = 3 In  O   C2  = 4 In  O   C1  In    C2   C 3  C 4 C4  O  =  O  = C = 1  C   C  O C3  2   1  2 3. A certain reaction obeys the three-half power law for its order. What function of the concentration may be plotted on the Y-axis against, t on the X-axis to get a linear plot ? 1 (1) C 2 vst (2) 1 vs t C  1 (3) C–2 vs t (4*) C 2 vst dksbZ,dvfHkfØ;kdksfVdsfy,3/2?k rfu;edkikyudjrhgSA,djs[kh;vkjs[kizkIrdjusdsfy,Xv{kijtrFk Y-v{kijlkUnzrkdsfdlQyudkvkjs[kcukukpkfg,A 1 1 – 1 (1) C 2 rFkktds e/; (2) C rFkkt ds e/; (3) C–2rFkktds e/; (4*) C 2 rFkkt ds e/; dx Sol. dt = k [C]3/2 For the order  (n – 1) kt =  1 n1 1  n1  k t =  1 1 t = 2 [Ct ] [C ] 1  [C [CO ]  ] 1  2 t =   [Ct ]2 1 [CO ]2  2 k  t  1 2 O  [C ] 2 So, For Ct 2 Vs t will give linear plot 4. For the reaction A  C + D, the initial concentration of A is 0.01 M. After 100 sec, the concentration of A is 0.001 M. The rate constant of the reaction has the numerical value of 9.0. What is the unit of the reaction rate constant ? vfHkfØ;kA  C+Dds fy,] Adh izkjfEHkdlkUnzrk0.01M gS]100sec i'pkr~ AdhlkUnzrk0.001 M gSaA vfHkfØ;kdsfy,njfu;rkaddkeku9.0gSaAvfHkfØ;kdsnjfu;rkaddhbdkbZD;kgSa\ (1*) M–1s–1 (2) Ms–1 (3) s–1 (4) M–1.5 s–1 5. The half lives of decomposition of gaseous CH3CHO at constant temperature but at initial pressure of 364 mm and 170 mm Hg were 410 second and 880 second respectively. Hence order of reaction is fu;rrkiysfdu364mmrFk 170mmHgizkjfEHkdnkcksaijxSlh;CH3CHOdsfo?kVudhv)Zvk;qØe'k%410lSd.M rFk 880lSd.MgSAvr%vfHkfØ;kdhdksfVfuEugSA (1*) 2 (2) 3 (3) 1.5 (4) 1 Sol. t1/2  1 an1 880 [Here a  P] 410  364 n1 = 170   n = 2 6. The hydrolysis of cane sugar was studied using an optical polarimeter and the following readings were taken: time (min.) : 0 84 min  observed rotation : (degrees) 50 20 –10 When was the mixture optically inactive? (log 2 = 0.3, log 3 = 0.48) xUusdh'kdZjkdsty&vi?kVudsv/; uizdk'kh;iksysjhehVjdsmi;ksx}kjkfd;ktkrkgS]vkSjfuEuikB~;kadfy;s x;sgSA le; (feuV) : 0 84 feuV  izsf{kr?kw.kZu : 50 20 –10 (fMxzh) feJ.k izdk'kh; vlfØ; dc gksrk gSA (log 2 = 0.3, log 3 = 0.48) (1) 118 min (2*) 218 min (3) 318 min (4) 418 min  r – ro   – 10 – 50  Sol. Kt = ln  r    – r – 10 – 20 2   t    K × 84 = ln2  K = ln2 min–1 84 optically in active  v = oº  r – ro   – 10 – 50  K × t = ln  r – r  = ln   – 10 – 0 = ln 6.   t    t = ln 6 K ln6 = ln2 × 84 min = 218 min. 7. Consider the plots for the types of reaction nA  B + C – d [A] dt [A] 1 [A] These plots respectively correspond to the reaction orders : (1) 0, 2, 1 (2) 0, 1, 2 (3) 1, 1, 2, (4*) 1, 0, 2 nA B+CvfHkfØ;kdsfofHkUuçdkjksadsfy,fuEuvkjs[k sijfopkjdhft,µ – d [A] dt [A] 1 [A] ;gvkjs[kØe'k%fdldksfVdhvfHkfØ;klslEcaf/krgSA (1) 0, 2, 1 (2) 0, 1, 2 (3) 1, 1, 2, (4*) 1, 0, 2 Sol. nA d[A]  B + C dt [A]t 1 [A]t  [A] (Ist order)  t (zero order)  t (2nd order) 8. Decomposition of N O follows first-order kinetics, 2N O (g) K  4NO (g) + O (g) If pressure of the system 2 5 2 5 2 2 at time, t and  are Pt and P respectively, find the expression of rate contant (k). N O dk fo;kstu izFke dksfV cyxfrdh dk vuqlj.k djrk gS,2N O (g) K  4NO (g)+O (g);fn le; trFkk  2 5 2 5 2 2 ijfudk;dsnkcØe'k%Pt rFkk P gksrksnjfu;rkad(k)dkO;atdKkrdhft,A 1  3P  Ans. ln 2K 5(P   P )   t  BooSt YoUr PreViouS ConCept 9. In which of the following arrangements, the sequence is not strictly according to the property written against it? (1) HF < HCl < HBr < HI : increasing acid strength (2*) NH3 < PH3 < AsH3 < SbH3 : increasing basic strength (3) B < C < O < N : increasing first ionization enthalpy (4) CO2 < SiO2 < SnO2 < PbO2 : increasing oxidising power fuEuO;oLFk vksaesalsfdlesabudslkeusfn,x;sxq.k/keZdsvuqlkjØeughagSa? (1)HF Li+ > Mg2+ > Be2+ (2) Li+ > Na+ > Mg2+ > Be2+ (3) Mg2+ > Be2+ > Li+ > Na+ (4) Li+ > Be2+ > Na+ > Mg2+ Sol. Down the group ionic radii increases with increasing atomic number because of the increase in the number of atomic shells but across the period the ionic radii decreases due to increase in effective nuclear charge as electrons are added in the same shell. Li+ and Mg2+ are diagonally related but Mg2+ having higher charge is smaller than Li+, so correct order is Na+ > Li+ > Mg2+ > Be2+. Be2+ = 0.31 Å ; Mg2+ = 0.72 Å Li+ = 0.76 Å ; Na+ = 1.02 Å gy. lewgesaÅijlsuhpstkusijijek.kqØekadc<+usdslkFkvk;fudf=kT;kc<+rhgStksfdijek.kqdks'ksadhla[;kc<+us dsdkj.kgksrhgSAysfduvkorZdsvuqfn'kizHkohukfHksdh;vkos'kc<+usdsdkj.kvk;fudf=kT;k?kVrhgSD;ksafdbysDVªkWu lekudks'kesatqM+rsgSALi+rFk Mg2+esafod.kZlEcU/kgksrkgSAysfduMg2+ijvf/kdvkos'kgksusdsdkj.k;gLi+ls NksVk gksrk gSA vr% lgh Øe Na+ >Li+ >Mg2+ >Be2+ gSA Be2+ = 0.31 Å ; Mg2+ = 0.72 Å Li+ = 0.76 Å ; Na+ = 1.02 Å 12. Which of the following statement is/are correct ? (1*) The first ionisation energies (in kJ mol–1) of carbon, silicon, germanium, tin, and lead are 1086, 786, 761, 708 and 715 respectively : (2) Down the group, electronegativity decreases from B to Tl in boron family. (3*) Among oxides of the elements of carbon family, CO is neutral, GeO is acidic and SnO is amphoteric. (4*) The 4f-and 5f-inner transition elements are placed separately at the bottom of the periodic table to maintain its structure. fuEuesalsdkSulk@dkSulsdFkulghgS\ (1*)dkcZu]flfydkWu]tesZfu;e]fVurFk ysMdhizFkevk;u ÅtkZ,¡(kJmol–1 esa)Øe'k%1086,786,761,708rFk 715gSA (2)cksjksuifjokjesaoxZesaBlsTlrduhpstkusijfo|qr_.krk?kVrhgSA (3*)dkcZuifjokjdsrRoksadsvkWDlkbMksaesaCOmnklhu]GeOvEyh;rFk SnOmHk;/kehZgksrkgSA (4*)4f- o5f-vUrjlaØe.krRovkorZlkj.khdsryij]bldhlajpukcuk;sj[kusdsfy,i`Fkd:ilsfLFkrgksrsgSA Sol. (1), (3) and (4) are correct statements (2) First decreases from B to Al and then increases marginally owing to discrepancies in atomic size of the element. (1) ,(3) rFkk (4)lgh dFku gSA (2) BlsAlrdigys?kVrhgSarFk fQjrRodsijek.oh;f=kT;kesaFk sM+hdehgksuslsoS|qr_.krkesavYio`f)gksrhgSA DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 36 Class : XIII Course : DPP No.2 Total Marks : 39 Max. Time : 39 min. Single choice Objective ('–1' negative marking) Q.1 (3 marks 3 min.) [3, 3] Multiple choice objective ('–1' negative marking) Q.2 (4 marks 4 min.) [4, 4] Subjective Questions ('–1' negative marking) Q.3 to Q.7 (4 marks 4 min.) [20, 20] BooSt YoUr PreViouS ConCept Single choice Objective ('–1' negative marking) Q.8 to Q.11 (3 marks 3 min.) [12, 12] 1. 2 2. 1, 2 3. k = t ln  P  k = t ln    5. 2P – P k = t ln  P – P    t  6. 4 hr–1 7. 3 8. 2 9. 4 10. 2 11. 2 1. Rate constant k = 2.303 min–1 for a particular reaction. The initial concentration of the reaction is 1 mol/litre then rate of reaction after 1 minute is : (1) 2.303 M min–1 (2*) 0.2303 M min–1 (3) 0.1 M min–1 (4) none of these ,dfuf'prvfHkfØ;kdsfy,njfu;rkadk=2.303feuV–1gSA;fnvfHkfØ;kdhçkjfEHkdlkanzrk1eksy@yhVjgSrks1 feuVi'pkr~vfHkfØ;k dhnjD;kgksxh% (1) 2.303 M feuV–1 (2*) 0.2303 M feuV–1 (3) 0.1 M feuV–1 (4)buesalsdksbZugha Sol. Rate = K Ct = K CO e kt – = 2.303 × 1 × e– 2.303 × 1 M min–1 2.303 = 2.303 × e– in 10 = 10 = 0.2303 2. For a certain reaction A  2 Which of the following is true : products, the t½ as a function of [A]0 is given as below : (1*) The order is 1 (2*) t 2 ½ would be 100 10 min for [A]0 = 1 M (3) The order is 1 (4) t½ would be 100 min for [A]0 = 1 M ,d fuf'prvfHkfØ;k A  mRikn] dsfy;s t½ dks[A]0 dsQyu ds:iesafuEuçdkjlsfn;kx;k gSA [A]0 (M) : 0.1 0.025 t 1 (min.) : 100 50 2 fuEuesalsdkSulklR;gSA (1*) 1 dksfVgSA (2*) [A] = 1 M ds fy;s t , 100 10 min gksxkA 2 0 ½ (3)çFkedksfVgSA (4) [A]0 = 1 M ijt½, 100 min gksxkA Sol. A  Products. t1/2 = [conc.]1 – n n  order of reaction. 100  0.1 1–n 50 =  0.025    2 = 2 – 2n  2n = 1.  n = 1 . 2 order of reaction is 1 . 2 100  0.11/ 2 t1/ 2 =    1   t1/2 = 100 10 min for [A0] = 1M. 3. Let there be as first-order reaction of the type, A  B + C. Let us assume that all the three species are gases. We are required to calculate the value of rate constant based on the following data. Time 0 T  Partial pressure of A P0 Pt – ekukfdA  B+C,dçFkedksfVçdkjvfHkfØ;kgSa]geekursgSafdrhuksaçtkfrxSlh;gSaAgesafuEuvkWdMksij vk/kkfjrnjfu;rkaddkekuifjdfyrdjukvko';dgSaA 1  P0  Ans. k = t ln  P   t  4. Let there be a first order reaction, A  B + C. Let us assume all there are gases. We are required to calculate the value of rate constant based on the following data Time 0 t  Total pressure P0 Pt – Calculate the expression of rate constant. ekukfdA  B+CçFkedksVhdhvfHkfØ;kgSaAvc;gekursgSfdrhuksaçtkfrxSlh;gSa]gesafuEuvkWdMksaij vk/kkfjrnjfu;rkaddkekuifjdfyrdjukvko';dgSaA nj fu;arkddk O;tadKkr dhft,A 1  P0  Ans. k = t ln  2P – P   0 t  5. A(g)  B(g) + C(g) Calculate the expression of rate constant. A(g)  B(g) + C(g) nj fu;arkddk O;tadKkr dhft,A 1  P  Ans. k = t ln  P – P    t  6. The plot of log (V – V) versus t (where V is the volume of nitrogen collected under constant temperature and pressure conditions) for the decomposition of C6H5N2Cl is given at 50°C with an amount of C6H5N2Cl equivalent to 58.3 cc N2. log (V – V) dk t ds lkis{k vkjs[k (tgk¡ fu;r rki o nkc ifjfLFkrh esa ukbVªkstu dk vk;ru V fy;k x;k gS½ C6H5N2Cl ds fo;kstu ds fy, 50°C rki fn;k x;k gSA ftl ij C6H5N2Cl dh ek=kk 58.3 cc N2 ds leku gSA 1.75 log(V – V) 1.0 5 10 15 20 25 30 time(min.) Calculate the rate constant for the reaction in hr–1 expressing your answer in a single significant digit. njfu;rkaddhx.kuk?k.Vs–1 esadjksomÙkjdsoy,dvadesanksA Ans. 4 Sol. k = – 2.303 (slope) min–1 = – 2.303 (– 0.03) min–1 = 0.06909 min–1 = 4.14 hr–1  4 hr–1 7. For the reaction, 2NO + H  N O + H O the value of -dp/dt was found to be 1.50 Torr s–1 for a pressure 2 2 2 of 359 Torr of NO and 0.25 Torr s-1 for a pressure of 152 Torr, the pressure of H being constant. On the other hand, when the pressure of NO was kept constant, –dp/dt was 1.60 Torr s–1 for a hydrogen pressure of 289 Torr and 0.79 Torr s–1 for a pressure of 147 Torr. Determine the order of the reaction. vfHkfØ;k 2NO+H2  N2O+H2Ods fy, H2 ds fu;r nkc ij -dp/dtds eku NOds 359Vksj nkc ij 1.50Vksj lSd.M–1 ,oa NOds 152Vksj nkcij 0.25VksjlSd.M–1 ik;s x;stcfd ;fn NOdkfu;r nkc fy;k tkrkrks –dp/dt dsekuH ds289Vksjij1.60VksjlSd.M–1 ,oaH ds147Vksjij0.79Vksjlsd.M–1 ik;sx;sblvfHkfØ;kdhdksfV 2 KkrdjksA Ans. 3 2 BooSt YoUr PreViouS ConCept 8. Which is not true about B2H6 (1) Both ‘B’ atoms are sp3 hybridised (2*) Boron atom is in ground state (3) Two hydrogens occupy special positions (4) There are two, three centre two electron bonds B2H6dsfy,fuEuesalsdkSulkdFkulR;ughagS& (1)nksuksa‘B’¼cksjksu½ijek.kqsp3ladfjrgSA (2*)cksjksuijek.kqvk|¼ewy½voLFk esagSA (3)nksgkbMªkstu]fof'k"VfLFkfrizkIrdjrsgSA (4);gk¡nks]rhudsfUnz;ijek.kq&nksbysDVªkWuca/kgSA 9. In P4 O10 molecule (1) There are 4 P–P bond (2) There are 8 P–O bond ^ (3) The P OP bond angle is 180° (4*) The phosphorus atom is in excited state P4O10v.kqesa (1) blesa4P–P cU/kgS (2) 8P–O cU/kgS (3) P OP dkca/kdks.k180°dscjkcjgSA (4*)QkLQksjlijek.kqmRrsftrvoLFk esagSA 10. Consider the following statements In    v CH2  CH C  C H 1. There are 6  and ‘3’  and 2. Carbon  &  are sp2 hybridised 3. Carbon  & V are sp hybridised, The above statements 1, 2, 3 respectively are (T = True, F = False) (1) T T T (2*) F T T (3) F T F (4) T F T fuEu dFkuksa dk voyksdu dhft,&    v esa CH2  CH C  C H 1. 6rFk ‘3’ca/kgksrsgS 2. dkcZurFk sp2ladfjrgksrsgS 3. dkcZurFk V,spladfjrgksrsgS mijksDr 1, 2, 3oDrO; Øe'k% gS (T= lR;, F= vlR;) (1) T T T (2*) F T T (3) F T F (4) T F T 11. Consider the following statements S1 : Steric number ‘7’ gives ‘sp3d3 hybridisation. S2 : In C𝑙F3 at least one bond angle is exactly 180° S3 : Lone pair does not cause any distortion in the bond angle. S4 : Correct order of dipole moment NF3 > NH3 > CO2 fuEudFkuksadkvoyksdudhft,& S1 : f=kfoela[;k‘7’,‘sp3d3ladj.knsrkgSA S2 : C𝑙F3esadelsde,dca/kdks.kBhd180°gksrkgSA S3 : ca/kdks.kesa,dkdh;qXedsdkj.kfdlhizdkjdhfod`fÙkughagksrhgSA S4 : f}/kzqovk?kw.kZdklghØegSNF3>NH3>CO2 (1) T F F T (2*) T F F F (3) T T F F (4) T T F T