DPP-37 to 38 With Answer Physical Chemistry

DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 37 to 38 Class : XIII Course : DPP No.1 Total Marks : 29 Max. Time : 29 min. Single choice Objective ('–1' negative marking) Q.1 to Q.9 (3 marks 3 min.) [21, 21] Multiple choice objective ('–1' negative marking) Q.10 (4 marks 4 min.) [4, 4] Subjective Questions ('–1' negative marking) Q.11 (4 marks 4 min.) [4, 4] 1. The reaction A(g) + 2B(g)  C(g) is an elementary reaction. In an experiment involving this reaction, the initial partial pressures of A and B are PA = 0.40 atm and PB = 1.0 atm respectively. When pressure of C becomes 0.3 atm in the reaction the rate of the reaction relative to the initial rate is : 1 (1) 12 1 (2) 50 1 (3*) 25 (4) none of these vfHkfØ;kA(g)+2B(g) C(g),dvk/k jHkwr¼çk:fid½vfHkfØ;kgSaftls,dç;ksxesaç;qDrfd;ktkrkgS]ArFk Bdk çkjfEHkd nkc Øe'k%PA =0.40ok;qe.Myh; rFkkPB =1.0 ok;qe.Myh; gSA vfHkfØ;k ds nkSjku Cdk nkc tc0.3 ok;qe.Myh;gkstkrkgSAçkjfEHkdlkanzrkdslkis{kvfHkfØ;kdhnjgksxh\ 1 (1) 12 1 (2) 50 1 (3*) 25 (4)mijksDresalsdksbZugha Sol. A(g) + 2B(g)  C(g) t = 0 0.4 atm 1 atm 0.3 atm t = t (0.4 –0.3)atm (1 – 0.6)atm Since reaction is elementary. So, Rate of reaction w.r.t. A & B will be of order equal to stoichiometric coefficient Rate = K [A] [B]2  Rate = K [0.4] [1]2  Rate = K [0.1][0.4]2 R(tt) R(to) K[0.1] [0.4]2 = K[0.4] [1] 1 = 25 2. A reaction is catalysed by H+ ion;and in the rate law the dependence of rate is of first order with respect to the concentration of H+ ions, in presence of HA rate constant is 2 × 10–3 min–1 and in presence of HB rate constant is 1 × 10–3 min–1. HA and HB (both strong acids) have relative strength as : ,dvfHkfØ;kH+vk;u}kjkmRizsfjrdhtkrhgSrFk blesaosxfu;eH+vk;udhlkUnzrkdslkis{kizFkedksfVdhvfHkfØ;k gSA HAdh mifLFkfresanjfu;rkad 2×10–3 min–1 gSrFkk HBdh mifLFkfresanjfu;rkad 1×10–3 min–1 gSrcHA rFk HB(nksuksaizcyvEygS)dhvkisf{kdlkEFkZ;rkgS: (1) 0.5 (2) 0.002 (3) 0.001 (4*) 2 Sol. We know Rate = k [conc.] Given Rxn catalysed by HA and HB Rate constant kA = k1 [H ]  k = k + [H+] Then relative strength of acids A and B is k A kB = [H ]A [H ]B  2 = 1 [H ]A [H ]B = strength of [Acid A] [AcidB] 3. The gaseous decomposition reaction, A(g)  2B(g) + C(g) is observed to first order over the excess of liquid water at 25ºC. It is found that after 10 minutes the total pressure of system is 188 torr and after very long time it is 388 torr. The rate constant of the reaction (in hr–1) is : [Given : vapour pressure of H O at 25º is 28 torr (ln 2 = 0.7, ln 3 = 1.1, ln 10 = 2.3)] (1) 0.02 (2*) 1.2 (3) 0.2 (4) none of these. 25ºCijnzotydsvkf/kD;dh lrgij xSlh;fo;kstuvfHkfØ;kA(g)  2B(g)+C(g)dsfy,çFkedksfVçsf{kr dh tkrhgSA ;gik;k x;k gSfd 10feuV i'pkr~ra=k dkdqy nkc188torr rFkkdkQh yEcsle; i'pkr~;g388torr gSA rksvfHkfØ;kdsfy, njfu;rkad(hr–1esa)gS %[fn;kx;k gS: 25ºCijH O dk ok"inkc28torrgSA] (ln 2 = 0.7, ln 3 = 1.1, ln 10 = 2.3)] (1) 0.02 (2*) 1.2 (3) 0.2 (4)mijksDresalsdksbZughaA Sol. A(g)  2B(g) + C(g) Let initial pressure P0 0 0 After 10 min. (P0 – x) 2x x After long time (t  ) 0 2P0 P0 as per given (P0 – x) + 2x + x + vaour pressure of H2O = 188 P0 + 2x = 160 and 3P0 + 28 = 388 so, P0 = 120 and x = 20 torr 1  P0  ln  t P  x  0  1 ln  120   1  10   100 10 × (ln 4 + ln 3 – ln 10)   = 0.02 min–1 = 1.2 hr–1 4. For a certain reaction the variation of the rate constant with temperature is given by the equation ln kt = ln k0 + 0.0693 t (t  0°C) The value of the temperature coefficient of the reaction rate is therefore fdlhfuf'prvfHkfØ;kdsfy,rkidslkFknjfu;rkaddsifjorZudksfuEuvfHkfØ;k}kjkiznf'kZrfd;ktkrkgSA ln kt = ln k0 + 0.0693 t (t  0°C) blvfHkfØ;knjdsrkixq.k addkekuD;kgksxkA (1) 0.1 (2) 1.0 (3) 10 (4*) 2 kt Sol. = (TC)t–0/10 0 t Taking log gives loge kt – loge k0 = 10 ln (TC)  ln (TC)  loge (TC)  ln kt = ln k0 +  10  t Comparison indicates 10 = 0.0693  ln (TC) = 0.693  TC = 2 5. What percentage fraction of the molecule will cross over the energy barrier at 1000 K temperature for 18.424 KJ activation energy (R = 8J mol –1 K–1) (1*) 10% (2) 20% (3) 90% (4) 80% 1000Krkiij18.424KJlfØ;.kÅtkZdsfy,ÅtkZ&vojks/kdksfdrusizfr'krv.kqikjdjldsxsaA (R = 8J mol –1 K–1) (1*) 10% (2) 20% (3) 90% (4) 80% 6. For the decomposition of H the following logarithmic plot is shown : [R = 1.98 cal/mol-K] H dsfo;kstudsfy;sfuEuy?kqxq.kdvkjs[kfn;kgS[R = 1.98cal/mol-K] The activation energy of the reaction is about vfHkfØ;kdhlfØ;.kÅtkZgksxhA (1*) 45600 cal (2) 13500 cal (3) 24600 cal (4) 32300 cal Sol. log k = – Ea 2.303 R 1 + constant = – T Ea 103 Ea 2.303 R 4 × 10–3 × 103 T + constant thus slope of graph will be – 2.303 R = – 0.4  Ea = 2.303 × 1.98 × 104 = 45600 cal 7. Decomposition of A follows first order kinetics by the following equation. 4A(g)  B(g) + 2C(g) If initially, total pressure was 800 mm of Hg and after 10 minutes it is found to be 650 mm of Hg. What is half- life of A ? (Assume only A is present initially) (1) 10 mins (2*) 5 mins (3) 7.5 mins (4) 15 mins Adkfo;kstuizFkedksfVdhxfrdhdsvuqlkjfuEuvfHkfØ;k}kjkgksrkgSA 4A(g)  B(g) + 2C(g) izkjEHkesa;fndqynkc800mmHgrFk 10fefuVdsi'pkr;g650mmHgik;kx;kgS]rksAdhv)Z&vk;qD;kgSA ¼;gekudjfdizkjEHkesadsoyAmifLFkrgS½ (1) 10fefuV (2*) 5fefuV (3) 7.5fefuV (4)15fefuV Sol. 4A(g)  B(g) + 2C(g) t = 0 800 – – t = 10 minutes, 800 – 4p p 2p 800 – p = 650  p = 150 Pressure of A = 200, so  2 x t = 10 minutes t = 5 minutes 8. For 2A B + 3C, 2C k2  3D, assuming all reactions to be single step (Elementary) reactions, which of the following is correct: 2A B+3C,2C k2  3Ddsfy,] ;gekudj fdlHkh vfHkfØ;k,a,d inesa gksrhgSA fuEuesalsdkSulklR;gSA (1*) d[C] / dt = 3k1 [A]2 – 3 k–1 [B] [C]3 – 2k2 [C]2 (2) d[B] / dt = k1 [A]2 (3) d[A] / dt = 2k–1 [B] [C]3 – 2k1 [B] [C]3 (4)NonedksbZugha 9. For the follwing parallel chain reaction if the sum of the concentration of B and C at any time is 2M then what will be [B] t and [C] t respectively? lekUrjJ`a[kykvfHkfØ;k dsfy,;fnfdlhle;BvkSjCdhlkUnzrkvksadk;ksx2MgSArc[B]trFk [C]t Øe'k%D;kgksxsa\ 11 13 3 5 4 6 8 18 (1) 12 M 12 M Sol. [B] + [C] = 2 M [B]  2k1  4 (2) 4 M, 4 M (3) 5 M, 5 M (4*) 13 M,13 M [C] 3k 2 9 10. For the reaction CH4 + Br2  CH3Br + HBr the experimental data require the following rate equation : d k1[CH4 ][Br2 ] dt [CH3Br] = 1 k 2 [HBr] /[Br2 ] Which of the following is/are true regarding this ? (1) The reaction is a single step reaction (2*) The reaction is 2nd order in the initial stages {[HBr]  0} (3*) The reaction is 2nd order in the final stages {[Br2]  0} (4) The molecularity of the reaction is two. vfHkfØ;kCH4 + Br2CH3Br+HBr dsizk;ksfxd:ilsvk¡dM+s]fuEunjlehdj.kdkikyudjrsgSA d k1[CH4 ][Br2 ] dt [CH3Br] = 1 k 2 [HBr] /[Br2 ] bldsfy,fuEuesalsdkSulk@dkSulsdFkulR;gS@gSaA (1)vfHkfØ;k,dinvfHkfØ;kgSA (2*)izkjfEHkdvoLFk {[HBr]0}esavfHkfØ;kf}rh;dksfVdhgSA (3*)vfUrevoLFk {[Br2]  0} esavfHkfØ;kf}rh;dksfVdhgSA (4)vfHkfØ;kdhv.kqla[;rk2gSA 11. For the reaction Cl2 + CO  Cl CO the rate law is d [Cl2CO] dt = k [Cl ]3/2 [CO] The mechanism which is accepted is Cl2 + M 2Cl + M (fast) Cl + CO + M ClCO + M (fast) ClCO + Cl2 Cl2CO + Cl (slow) Find the expression relating k with the other constants given. d [Cl2CO] vfHkfØ;k Cl + CO Cl CO ds fy, nj fu;e = k [Cl ]3/2 [CO] gSA 2 2 dt 2 fØ;k&fof/ktksfdvuqekfurgSA Cl2 + M 2Cl + M (rhoz) Cl + CO + M ClCO + M (rhoz) ClCO + Cl2 Cl2CO + Cl (/kheh) Kdksnwljsfu;rkadksdslkFklEcfU/krdjusokykO;atdKkrdjksA Sol. Using the r.d.s. approximation method d [Cl2CO] dt = k3 [ClCO]1 [Cl ]1 but K1 = [Cl]2 [M] [Cl2 ] [M] and K2 = [ClCO] [M] [Cl] [CO] [M]  [ClCO] = K [Cl] [CO] and [Cl] =  d [Cl2CO] = k K dt 3 2 [Cl] [CO] [Cl2] = k3K2 = k3K2 [CO] [Cl2] [Cl ]3/2 [CO] comparision with the rate law given shows that k = k3K2 DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 38 Class : XIII Course : DPP No.2 Total Marks : 43 Max. Time : 43 min. Single choice Objective ('–1' negative marking) Q.1 to Q.6 (3 marks 3 min.) [18, 18] Multiple choice objective ('–1' negative marking) Q.7 (4 marks 4 min.) [4, 4] Subjective Questions ('–1' negative marking) Q.8 to Q.9 (4 marks 4 min.) [8, 8] BooSt YoUr PreViouS ConCept Single choice Objective ('–1' negative marking) Q.10 to Q.12 (3 marks 3 min.) [9, 9] Multiple choice objective ('–1' negative marking) Q.13 (4 marks 4 min.) [4, 4] 1. Select the incorrect statement : (1) Larger the value of Ea , smaller the value of rate constant (k). (2) Larger the value of Ea , greater is the effect on the value of k for a given temperature change. (3) At lower temperature, increase in temperature cause more change in the value of k than at higher temperature. (4*) k is affected by change in concentration. xyrdFkudksigpkfu;s\ (1)Ea dkekuvf/kdgksusij]njfu;rkad(k)dkekudegksrkgSA (2)nhxbZrkiekuijkldsfy,;fnEadkekuvf/kdgSrkskdsekuijvf/kdçHkkogksxkA (3)fuEurkiekuijrkiekuc<+kusij]mPprkiekudhrqyukesakdsekuesavf/kdifjorZugksrkgSA (4*)lkanzrkesaifjorZuds}kjkkçHkforgksrkgSA Sol. K is only dependent on temperature. 2. The IUPAC name for the complex compound Li[AlH4] is : (1) lithium aluminium hydride (2) hydrido - aluminiumlithium () (3*) lithium tetrahydridoaluminate () (4) lithium tetrahydridoaluminate () ladqy ;kSfxd Li[AlH4] dkIUPACukegS % (1) fyfFk;e,Y;qfefu;egkbMªkbM (2)gkbMªkbMks-,Y;qfefu;efyfFk;e() (3*)fyfFk;eVsVªkgkbMªkbMks,Y;qfeusV() (4)fyfFk;eVsVªkgkbMªkbMks,Y;qfeusV() 3. The IUPAC name for K [CrVINH (CN) O (O )] is 2 3 2 2 2 (1*) potassium amminedicyanido-C-dioxidoperoxidochromate(V) (2) potassium amminedicyanatotetraoxochromium() (3) potassium amminedicyanochromate (V) (4) potassium amminocyanodiperoxochromate (V) K [CrVINH (CN) O (O )] dk IUPAC uke gSa % 2 3 2 2 2 (1*)ikSVsf'k;e,EehuMkblk;ukbMks-C-MkbvkWDlkbMksijvkWDlkbMksØksesV(V) (2)ikSVsf'k;e,EehuMkblk;usVksVsVªkvkWDlksØksfe;e() (3) ikSVsf'k;e,EehuMkblk;uksØksesV(V) (4)ikSVsf'k;e,Eehuksalk;uksMkbijvkWDlksØksesV(V) 4. The IUPAC name for K2 [OsCl5N] is (1) potassium pentachloroazidoosmate (VIII) (2) potassium pentachloroazoosmate (V) (3*) potassium pentachloridonitridoosmate (V) (4) potassium nitroosmate () K2 [OsCl5N] dk IUPAC uke gSa % (1)ikSVsf'k;eisUVkDyksjks,tkbMksvkWLesV(VII) (2) ikSVsf'k;eisUVkDyksjks,stksbMksvkWLesV(V) (3*)ikSVsf'k;eisUVkDyksjkbMksukbVªkbMksvkWLesV(V) (4)ikSVsf'k;eukbVªksvkWLesV( ) 5. The formula of the complex sodium hydridotrimethoxoborate () is ladqylksfM;egkbMªkbMksVªkbZesFkWDlkscksjsV( )gS& (1) Na4 [B4H2(OCH3)3] (2) Na2 [BH(OCH3)3] (3) Na [BH2(OCH3)3] (4*) Na [BH (OCH3)3] 6. For the reaction 2NO + Br2  2NOBr, the following mechanism has been given NO + Br2 NOBr2 NOBr + NO slow  2NOBr. Hence, rate law is : 2NO+Br2  2NOBr,vfHkfØ;kdsfy,fuEufØ;kfof/knhxbZgSA NO + Br2 NOBr2 NOBr + NO /khek 2NOBr. vr%njfu;egSa% (1*) k[NO]2[Br ] (2) k[NO][Br ] (3) k[NOBr ][NO] (4) k[NO][Br ]2 2 2 2 2 7. In the following gaseous phase first order reaction A(g)  2B(g) + C(g) initial pressure was found to be 400 mm and it changed to 1000 mm after 20 min. Then : (1*) Half life for A is 10 min (2*) Rate constant is 0.0693 min–1 (3*) Partial pressure of C at 30 min is 350 mm (4) Total pressure after 30 min is 1100 mm fuEuxSlh;izFkedksfVdhvfHkfØ;kesa A(g)  2B(g) + C(g) izkjfEHkdnkc400mmFk rFkk;g20feuVi'pkr~1000mmesaifjofrZrgkstkrkgSrc: (1*)Adkv)ZZvk;qdky 10feuVgSA (2*)osx fu;rkad 0.0693feuV–1 gSA (3*)30feuV ij Cdk vkaf'kd nkc 350mm gSA (4)30feuV i'pkr~ dqy nkc 1100mm gSA Sol. Moderate A(g)  2B(g) + C(g) t = 0 400 0 0 t = 20 min 400 – p 2p p Given 400 – p + 2p + p = 1000 400 + 2p = 1000 1 p = 300 mm ; k = 20 400 ln 400  300 1 = 20 ln 2 ln4 ; k = 10 min–1 T = 10 min ; Value of K = 0.0693 min–1 400 200 100 50 After 30 min Partial Pressure of A is 50 mm, After 30 min Partial Pressure of B is 700 mm After 30 min Partial Pressure of C is 350 mm , After 30 min total pressure become 1100 mm 8. Write IUPAC names of the following fuEu;kSfxdksadsIUPACukefyf[k,& (a) [Co(NH3)6][CuCl5] (b) [V(H2O)6]Cl3 (c) (NH4)3[Co(C2O4)3] (d) Cl4 (e) (f) Na2[SiF6] (g) K2 [CrO4] (h) [(NH3)5Cr – OH – Cr(NH3)5]Cl5 (i) [Fe(en)3] [Fe(CO)4] (j) [TiCl4(Et2O)2] (k) Mn2(CO)10 (l) [VO(acac)2] (m) Fe4 [Fe(CN)6]3 Ans. (a) Hexaamminecobalt(III) pentachloridocuprate(II) (b) Hexaaquavanadium(III) chloride (c) Ammonium tri(oxalato)cobaltate(III) (d) Tetraamminecobalt(III)-di- -hydroxidobis(ethylenediamine)cobalt(III) chloride (e) Bis (ethylenediamine) cobalt (III)--imido--hydroxidobis(ethylenediamine) cobalt (IV) (f) Sodium hexafluoridosilicate(IV) (g) Potassium tetraoxidochromate(VI) (h) -hydroxidobis-(pentaamminechromium(III)) chloride (i) Tris(ethylene diamine) iron (III) tetracarbonyl ferrate (–III) (metal in this complex can also be iron (II)) (j) Tetrachloridobis(diethylether)titanium(IV) (k) Decacarbonyldimanganese(0) (l) Bis(acetylacetonato)oxidovanadium(IV) (m) Iron (III) hexacyanido-C-ferrate(II) ion (also called Prussian blue). 9. Write the structure formula corresponding to each of the following IUPAC names : (a) hexaamminechromium (III) tetrachloridocuprate (II) (b) diamminedichloridoplatinum (II) (c) tetracarbonyl nickel(0) (d) tetraammineplatinum(II) amminetrichloridoplatinate (II) (e) sodium dithiosulphatoargentate(I) (f) potassium tetracyanido-C-nickelate(0) (g) bis( 5  cyclopentadienyl)iron (II) (h) tetrathiocyanato-N-zincate (II) ion (i) potassium tetraoxidomanganate(VII) (j) potassium trioxalatoaluminate (III) (k) tetrapyridineplatinum (II) tetrachloridoplatinate (II) fuEuIUPACukedslaxrizR;sddslajpuklw=kdksfyf[k,& (a)gSDlk,EehuØksfe;e(II)VsVªkDyksjkbMksD;wizsV(I) (b)Mkb,EehuMkbDyksjkbMksIysfVue(I) (c)VsVªkdkcksZfuyfudy(0) (d)VsVªk,EehuIysfVue(I),EehuVªkbDyksjkbMksIysfVusV(I) (e)lksfM;eMkbFk;kslYQsVksvtsZUVsV(I) (f)ikSVsf'k;eVsVªklk;ukbMks-C-fudysV(0) (g)fcl~(5–lkbDyksisUVkMkb Zukby)vk;ju(II) (h)VsVªkFk;kslk;usVks-N-ftUdsV(I)vk;u (i)ikSVsf'k;eVsVªkvkWDlkbMkseSaXusV(VI) (j)ikSVsf'k;eVªkbvkWDlsysVks,Y;qfeusV(II) (k)VsVªkfijhMhuIysfVue(I)VsVªkDyksjkbMksIysfVusV(I) Ans. (a) [CrIII(NH3)6]2[CuIICl4]3 (b) [PtII(NH3)2Cl2] (c) [Ni0(CO)4] (d) [PtII(NH3)4][PtIINH3Cl3]2 (e) Na3[AgI(S2O3)2] (f) K4[Ni0(CN)4] (g) Fe[5 – C5H5]2 5 means that all the five carbon atoms of cyclopentadienyl anion are coordinated to the metal ion (h) [ZnII(NCS)4]2– (i) K[MnVIIO4] (j) K3[AllII(C2O4)3] (k) [PtII(Py)4][PtIICl4]. BooSt YoUr PreViouS ConCept 10. Which of the following are isoelectronic and isostructural ? (1) NO +, CO (2) NO –, BF (3) NH , CH – (4*) all of these. 2 2 3 3 3 3 fuEueslsdkSulebySDVªkWfudolelajpukRedgS\ (1) NO +, CO (2) NO –, BF (3) NH , CH – (4*)mijksDrlHkh 2 2 3 3 3 3 11. There is no S–S bond in : fuEuesalsfdlesaS–Sca/kughagS% (1) S O 2– (2) S O 2– (3) S O 2– (4*) S O 2– 12. What is the correct order from the weakest to the strongest carbon-oxygen bond for the following species ? CO, CO , CO 2– , CH OH. 2 3 3 (1*) CH OH < CO 2– < CO < CO (2) CO < CO < CO 2– < CH OH (3) CH OH < CO 2– < CO < CO (4) CH OH < CO < CO 2– CO fuEuLih'khtdsfy,nqcZylsizcydkcZu&vkWDlhtuca/kdklghØegS\ CO, CO , CO 2– , CH OH. 2 3 3 (1*) CH OH < CO 2– < CO < CO (2) CO < CO < CO 2– < CH OH 3 3 2 2 3 3 (3) CH OH < CO 2– < CO < CO (4) CH OH < CO < CO 2– CO 3 3 2 3 2 3 Sol. C – O = 1.43 Å ; C = O 1.23 Å ; C  O 1.09 Å 13. Which of the following species are correctly matched with their geometries according to the VSEPR theory? (1*) BrF +  octahedral. (2*) SnCl –  trigonal bipyramidal. 6 5 (3*) ClF –  linear. (4*) IF +  see – saw. 2 4 VSEPRfl)kUrdsvuqlkj]fuEuesalsdkSulhLih'khtmudhT;kfefrdslkFklqesfyrgS\ (1*) BrF +  v"VQydh; (2*)SnCl –f=kdks.kh;f}fijsfefM; 6 5 (3*)ClF – js[kh; (4*) IF + lh&lkW 2 4