Jp Xii Physical&inorganic Chemistry (15) - Prev Chaps + Inorg. Chem.pdf

PHY./INO. CHEMISTRY

TARGET : JEE (Main + Advanced) 2016 NO. 24

Course : VIJETA (JP)

This DPP is to be discussed in the week (06.07.2015 to 11.07.2015)

DPP No. # 24 (JEE-MAIN) REVISION DPP Total Marks : 102

Max. Time : 66 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.30 ChemINFO : 2 Questions ('–1' negative marking) Q.31 to Q.33

1.

(3 marks, 2 min.) (4 marks, 2 min.)

[90, 60] [12, 06]

The enthalpy of neutralisation of acetic acid and sodium hydroxide is – 55.4 kJ. What is the enthalpy of ionisation of acetic acid ? ,lhfVd vEy rFkk lksfM;e gkbMªksDlkbM ds mnklhuhdj.k fd ,UFkSYih – 55.4 kJ gSA ,lhfVd vEy ds vk;uu fd ,UFkSYih

D;k gS \s Sol. 2.

(A) –1.9 kJ (B*) + 1.9 kJ (C) + 5.54 kJ enthalpy of ionisation vk;uu fd ,UFkSYih = 57.3 – 55.4 = 1.9 kJ

(D) – 5.54 kJ

For the reaction A B at equilibrium. The partial pressure of B is found to be 16 times the partial pressure of A. The value of G° of the reaction is : vfHkfØ;k A B ds fy, lkE; ij B dk vkaf'kd nkc A ds vkaf'kd nkc fd rqyuk esa 16 xquk ik;k x;kA vfHkfØ;k fd

Sol.

G° dk eku gS : (A) RTn 4 G° = – RTn 16

3.

Which of the following titrations may be represented by this titration curve?

(B) RTn 16

(C) – RTn 4

(D*) – RTn 16

fuEu vuqekiu oØ }kjk dkSulk vuqekiu iznf'kZr fd;k tkrk gS \

Sol.

(A*) Na2CO3 vs HCl (B) H2CO3 vs NaOH (C) H3PO4 vs NaOH (D) NH3 vs HCl (E) H4edta vs NaOH A two stage curve reflects a diprotic species. Final pH is lesser, hence acid is added ,d nks inh; oØ ,d f}izksfVd iztkfr dks crkrk gSA vfUre pH de gksrh gS] vr% vEy feyk;k 

tkrk gSA



H

H  H2CO3 (in-2) CO32–   HCO3 (in-1) 

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4.

The burning of solid Caffeine {C8H10N4O2} in gaseous oxygen to produce gaseous CO 2, N2O3 & H2O takes place. { No other oxide is formed }. The maximum mass of N2O3 that can be produced from burning of 9.7 g of caffeine is : xSlh; vkWDlhtu esa Bksl {C8H10N4O2} ds tyus ls xSlh; CO2, N2O3 rFkk H2O mRikn curs gSa { dksbZ vU; vkWDlkbM ugha

curh gS }. 9.7 g dSfQhu ds tyus ls N2O3 dk og vf/kdre nzO;eku tks cu ldrk gSA fuEu gS % Sol.

(A) 3.6 g (B) 5.6 g (C*) 7.6 g (D) 9.6 g Apply POAC on nitrogen (moles of caffeine) × No. of atoms of N in caffeine = (moles of N2O3) × No. of atoms of nitrogen in N2O3  9.7   mass of N 2O 3   ×2   ×4=  194 76     mass of N2O3 =

9. 7 4 × × 76 = 7.6 Ans. 194 2

ukbVªkstu ij POAC yxkus ij (dsQhu ds eksy) × dsfQu esa N ds ijek.kqvksa dh la[;k = (N2O3 ds eksy) × N2O3 esa ukbVªkstu ds ijek.kqvksa dh la[;k 9.7  N O   ×4=  2 3 76  194   N2O3 dk 5.

nzO;eku =

  ×2 

9.7 4 × × 76 = 7.6 Ans. 194 2

Equal amount (mass) of methane and ethane have their total translational kinetic energy energy in the ratio 3 : 1 then their temperatures are in the ratio. leku ek=kk ¼nzO;eku½ ds esFksu rFkk ,sFksu ds fy, dqy LFkkukUrfjr xfrt ÅtkZ dk vuqikr 3 : 1 gS rc muds rkiekuksa dk

vuqikr fuEu gS& (A) 5 : 8 (E) 8 : 15 Sol.

(B) 45 : 8

E1 =

3 M  RT1 2 16

E2 =

3 M  RT2 2 30

(C) 15 : 8

E1 30 T1  E 2 16 T2





(D*) 8 : 5

3 30 T1 T1 8  .  or 1 16 T2 T2 5

6.

A solution saturated in lime water has a pH of 12.4. Then the Ksp for Ca(OH) 2 is pwus ds ikuh esa lar`Ir foy;u dh pH 12.4 gS rc Ca(OH)2 ds fy, Ksp fuEu gksxk (A) 3.2 × 10–3 (B*) 7.8 × 10–6 (C) 7.8 × 10–28 (D) 3.2 × 10 –4

Sol.

pH = 12.4

– log (H+) = 12.4





Ca (OH)2

 log [H+] = 13.6

2 OH + Ca2+

[H+] = 4 × 10–13

10 –14 1 [OH ] = = × 10–1 = 2.5 × 10–2 4  10 –13 4 –

ksp = [Ca2+] [OH–] 1  1    10 –1    10 –1  8 4    

[Ca2+] = 2

=

1 [OH–] 2

1 1 × × 10–3 = 7.8 × 10–6 Ans. 8 16

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7.

A sample of an ideal gas occupies 30 L at 27ºC, 40 L at 127ºC and 60 L at 327ºC. The pressure of the gas in the given situations are : (A*) same (B) different (C) can not be determined

vkn'kZ xSl dk ,d uewuk 27ºC ij 30 L, 127ºC ij 40 L rFkk 327ºC ij 60 L ?ksjrk gSA bu ifjfLFkfr;ksa esa vkn'kZ xSl dk nkc gksxk % (A*) leku (B) fHkUu (C) Kkr ugha fd;k tk ldrk Sol. 8.

V is same. Hence P is constant. T Standard molar enthalpy of formation of CO 2 is equal to (A) the standard molar enthalpy of combustion of C (diamond, s) (B) the standard molar enthalpy of combustion of gaseous carbon (C) the sum of standard molar enthalpies of CO & O 2 (D*) the standard molar enthalpy of combustion of C (graphite, s) CO2 dh ekud eksyj ,UFkSYih fdlds cjkcj gS (A) C (ghjk] s) ds

ngu dh ekud eksyj ,UFksYih ds cjkcj gS

(A) 'kwU; (B) xSlh;

dkcZu ds ngu dh ekud eksyj ,UFkSYih ds cjkcj gS ekud eksyj ,UFkSYih ds ;ksx ds cjkcj gS (D*) C (xzsQkbV] s) ds ngu dh ekud eksyj ,UFksYih ds cjkcj gS (C) CO vkSj & O2 dh Ans.

(D)

Sol.

C (graphite, s) + O 2 (g)  CO2 (g) H0rxn = H0c (graphite, s) = H0F (CO2, g)

9.

One mole of an ideal monoatomic gas expands isothermally against constant external pressure of 1 atm from initial volume of 1L to a state where its final pressure becomes equal to external pressure. If initial temperature of gas is 300 K then total entropy change of system in the above process is : [R = 0.082 L atm mol–1 K–1 = 8.3 J mol–1K–1]. ,d eksy vkn'kZ xSl 1 atm ds fLFkj ckg~; nkc ds fo:) rc rd QSyrh gS tc rd bldk vafre nkc ckg~; nkc

ds cjkcj u gks tk;sA xSl dk izkjfEHkd vk;ru 1L o rkieku 300 K gSA bl izØe esa ra=k dh dqy ,UVªkSi h ifjorZu dk eku fuEu gksxkA [R = 0.082 L atm mol–1 K–1 = 8.3 J mol–1K–1]. (A) 0

Sol.

10.

Ans.

(B*) Rn (24.6)

 Vf S = nR n   Vi

 P  = Rn  i   Pf

  

(C) R n (2490)

(D)

3 Rn (24.6) 2

 300R   = Rn (24.6) = R n   1 L  1 atm 

H2 gas diffuses 4 times as rapidly as a mixture of C2H4 and CO2. The molar ratio of C2H4 to CO2 in the mixture is H2 xSl dk folj.k C2H4 vkSj CO2 ds feJ.k ls 4 xquk rsth ls gksrk gSA rks C2H4 vkSj CO2 ds eksyj feJ.k dk vuqikr gksxk (A) 1 : 1 (B) 2 : 1 (C) 3 : 1 (D) 3 : 2 rH 2  (C) r mix

Mm ix MH 2

or,

4=

Mmix 2

 Mmix = 32

Mmix = XC2H4 + XCO2 .MCO2 32 = (1 – X CO ) MC H + XCO .MCO = 28 + (44 – 27) X CO 2 2 2 4 2 2 

X CO2 = 1/4



X C2H4

3 X CO2 = 1

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11.

12.

Which of the following is path function as will as extensive property? fuEu es ls dkSulk iFk Qyu o ek=kkRed xq.k/keZ (extensive property) Hkh gS? (A) Molar heat capacity (B) Internal energy (C) Temperature (A) eksyj Å"ek /kkfjrk (B) vkUrfjd ÅtkZ (C) rki

(D*) None of these (D*) mijksDr esa ls dksbZ

In the graph of Z vs P for a given gas, which of the following is INCORRECT ? (A*) At T > TC, Z>1 at all pressures . (B) At T = TC, Z<1 at low pressures. (C) At T = TB, the gas follows Boyle's law at low pressures. (D) At T > TB, Z>1 at high pressures. ,d nh xbZ xSl ds fy, Z rFkk P ds e/; vkjs[k ds lUnHkZ esa] fuEu esa ls dkSulk dFku

ugha

xyr gS \

(A*) T > TC

ij] lHkh nkc ij Z > 1 gksrk gSA (B) T = TC ij] lHkh nkc ij Z < 1 gksrk gSA (C) T = TB ij] fuEu nkc ij xSl ckW;y fu;e dk ikyu djrh gSA (D) T > TB ij] mPp nkc ij Z > 1 gksrk gSA 13.

Sol.

Hydrolysis constants of two salts KA & KB of weak acids HA and HB are 10–8 and 10–6. If the dissociation constant of third acid HC is 10-2. The order of acidic strength of three acids will be nks nqcZy vEy HA rFkk HB ds nks yo.k KA rFkk KB ds ty vi?kVu fu;arkd Øe'k% 10–8 vkSj 10–6 gS ;fn rhljs vEy HC dk fo;kstu fLFkjkad 10-2 gS rks bu rhu vEyks (A) HA > HB > HC (B) HB > HA > HC For weak acids nqcZy vEyksa ds fy,

dh vEy lkeF;Z dk Øe gksxkA (C*) HC> HA > HB

(D) HA = HB = HC

Kw Kh = K A HA

Kw 10 14 Ka1 = K = = 10–6 h 10 8

For

HB

Kw 10 14 Ka2 = K = = 10–8 10  6 h

For

HC

Ka3 = 10–2

For

14.

An acid HA (Ka = 10–5) reacts with NaOH at 298 K. What would be the value of the rate constant of the reverse reaction at the same temperature if the rate constant of the forward reaction is 10 –11 mol–1 L sec–1 ? ,d vEy HA (Ka = 10–5) 298 K ij NaOH ls fØ;k djrk gSA izrhi vfHkfØ;k ds osx fLFkjkad dk eku leku rki ij

D;k gksxk ;fn vxz vfHkfØ;k dk osx fLFkjkad 10–11 mol–1 L sec–1 (A) 10–9 Sol.

(B) 109

(C) 10–5

(D*) 10–20

CH3COOH + NaOH CH3COONa + H2O Equilibrium constant of the backward reaction

Kw Kh = K a Equlibrium constant of the forward reaction



Ka Kequilibrium = Kh –1 = K = 109 w Kf 10 11 10 = K = Kb b 9

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15.

H and S for a reaction are +30.558 kJ mol–1 and 0.666 kJ mol–1 at 1 atm pressure. The temperature at which free energy is equal to zero and the nature of reaction below this temperature are (A) 483 K, spontaneous (B) 443 K, non-spontaneous (C) 443 K, spontaneous (D*) 463 K, non-spontaneous

nkc ij ,d vfHkfØ;k ds fy, H rFkk S Øe'k% +30.558 kJ mol–1 rFkk 0.666 kJ mol–1 gSA og rki ftl ij eqDr ÅtkZ 'kwU; ds cjkcj gks tkrh gS rFkk bl rki ds uhps vfHkfØ;k dh izd`fr fuEu gS& (A) 483 K, Lor% izØe (B) 443 K, vLor% izØe (C) 443 K, Lor% izØe (D*) 463 K, vLor% izØe 1 ok;qe.Myh;

Ans.

G = H – TS = 0 

T=

H 30 .558 = = 463 K S 0.066

G = H – TS = 30.558 – T × 0.066 at T < 463 K, 0.066 T < 463 × 0.066 or 0 < 30.558 – 0.066 T or 0 < G Thus at T < 463 K, G > 0 i.e. process is non spontaneous.

bl izdkj T < 463 K, G > 0 vr% izØe vLor% gksrk gSA 16.

Consider the modes of transformation of a gas from state ‘A’ to state ‘B’ as shown in following P–V diagram. Which one of the following is true

voLFkk ‘A’ ls voLFkk ‘B’ esa xSl ds LFkkukarj.k esa rjhds@izØe dks nsf[k;s] tSlk dh P–V vkjs[k esa n'kkZ;k x;k gSA fuEu esa dkSulk ,d dFku lR; gSA

(A) H = q along A  C (B*) S is same along both A  B and A  C  B (C) W is same along A  B and A  C  B (D) W > 0 along both A  B and A  C (A) A  C ds

fy, H = q (B*) A  B rFkk A  C  B nksuksa ds fy, S leku gSA (C) A  B rFkk A  C  B ds fy, W leku gSA (D) A  B rFkk A  C nksuksa ds fy, W > 0 gSA Sol.

Entropy is a state function i.e. the change in entropy depends upon the initial and final states of the system, & not on how that change is brought.

Sol.

,UVªksih ,d voLFkk Qyu gS vFkkZr~ ,UVªksih esa ifjorZu] ra=k ¼fudk;½ dh izkjfEHkd rFkk vfUre voLFkkvksa ij fuHkZj djrk gSA rFkk bl ij ugha dh ;g ifjorZu fdl izdkj izkIr fd;k x;kA

17.

Which of the following diatomic molecules would be stabilised by the removal of an electron ?

fuEu esa ls dkSulk f}ijek.kqd v.kq ,d bysDVªkWu ds fu”dklu ij LFkk;hd‘r gksrk gS \ (B) CN– (C) N2 From O2 the electron is removed from anti bonding orbital. O2 es bysDVªkWu vcaf/kr d{kd ls fu”dkf”kr fd;k tkrk gSA (A*) O2 Sol.

(D) C2

2 2 1 2 KK * 2s2 ,  * 2s2 , 2pz , ( 2p y  2p x ) * 2p y  * 2px .

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18.

Diatomic molecule has a dipole moment of 1.2 D. If is bond length is 1.0 Å, what fraction of an electronic charge exists on each atom ? (A) 11% (B) 20% (C*) 25% (D) None of these ,d f}ijek.kqd v.kq dk f}/kzqo vk?kw.kZ 1.2 D gSA ;fn bldh ca/k yEckbZ 1.0 Å gS rks fo|qfr; vkos’k dk fdruk izfr’kr Hkkx

izR;sd ijek.kq ij fo|eku gksrk gS \ (A) 11% 19.

(B) 20%

(C*) 25%

(D) buesa

ls dksbZ ugha

The correct order of dipole moment is

f}/kzqo vk?kw.kZ dk lgh Øe gS& Sol.

(A*) CH4 < NF3 < NH3 < H2O (B) NF3 < CH4 < NH3 < H 2O (C) NH3 < NF3 < CH4 < H2O (D) H2O < NH3 < NF3 < CH4 CH4  net dipole moment is zero. NF3  resultant dipole moment towards nucleus of nitrogen. NH3  resultant dipole moment towards lone pair. H2O  resultant dipole moment 2P cos 104.5°. CH4  dqy f}/kzqo vk|q.kZ ‘kwU; gksrk gSA NF3  ifj.kkeh

f}/kzqo vk?kw.kZ ukbVªkstu ds ukfHkd dh vksj gksrk gSA NH3  ifj.kkeh f}/kzqo vk?kw.kZ ,dkdh ;qXe dh vksj gksrk gSA H2O  ifj.kkeh f}/kzqo vk?kw.kZ 2P cos 104.5° gksrk gSA 20.

Sol.

Two types FXF angles are present in which of the following molecule (X = S, Xe, C)? fuEu esa ls fdl v.kq esa nks fHkUu izdkj ds FXF dks.k mifLFkr gksrk gS (X = S, Xe, C)? (A*) SF4 (B) XeF4 (C) SF6 (D) CF4 SF4 is sp3d hybridised and with one lp of e– and its shape is see-saw. SF4 ,d lp e– ds lkFk sp3d ladfjr gksrk gS rFkk bldh vkd‘rh ‘khlkW gksrh gSA

F

90°

F

S

F

120°

F

Therefore, two different types of FSF angles are present. vr% nks fHkUu izdkj ds FSF dks.k mifLFkr gksrs gSA 21.

Which one of the following combination is not allowed in the LCAO method for the formation of a molecular orbital (consider Z-axis as the molecular axis) ? fuEu esa ls fdl la;kstu es LCAO fof/k ds vuqlkj vkf.od d{kd dk fuekZ.k ugh gksrk gSA ¼Z-v{k dks vkf.od v{k ekurs

gSA) ? Sol.

(A*) s + px (B) s + pz (C) px + px (D) pz + pz s and px orbitals do not have proper orientation to overlap and hence MO is not formed. s rFkk px d{kd mfpr leferh ds lkFk vfrO;kiu ugh djrs gS vr% MO ugh curk gSA

+

+ +

+

z (No overlap)

s

s

–

–

x px

x px

z ¼vfrO;kiu ugh gksrk gS ½

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22.

Which one of the following compounds is a peroxide ?

fuEu esa ls dkSulk ;kSfxd ijkWDlkbM gS \ (A) KO2 Sol.

(B*) BaO2

(C) MnO2

(D) NO2

KO2 is K  O2– i.e. superoxide BaO2 is Ba2  O2– 2 i.e. peroxide MnO2 is Mn4+ O2– i.e. monoxide NO2 is N4– O2– i.e. monoxide

gy -

KO2, K  O2–

gS] vFkkZr lwijvkWDlkbM gSA

BaO2, Ba2 O2– 2

gS] vFkkZr ijkWDlkbM gSA

MnO2, Mn4+, O2– gS]

vFkkZr eksuksvkWDlkbM gSA NO2, N O gS] vFkkZr eksuksvkWDlkbM gSA 4–

23.

2–

Dihydrogen gas may be prepared by heating caustic soda on

fuEu esa ls fdldks dkWfLVd lksMk ds lkFk xeZ djus ij gkbMªkstu xSl cu ldrh gSA (A) Cu

(B*) Zn

(C) Na

(D) Ag



Sol.

Zn  2NaOH  Na2 ZnO2  H2

24.

Potassium is _________, __________ and _________ than sodium. (A*) lighter, softer and more reactive (B) heavier, softer and less reactive (C) lighter, harder and more reactive (D) None of the above iksVsf'’k;e] lksfM;e dh rqyuk esa _________, __________ rFkk _________ gSA (A*) gYdk]

e‘nq rFkk vf/kd fØ;k'’khy (C) gYdk] dBksj rFkk vf/kd fØ;k'’khy

(B) Hkkjh]

e‘nq rFkk de fØ;k'’khy (D) buesa ls dksbZ ugh

Sol.

Density of K < density of Na; metallic bonds are weaker in K as compared to Na. K dk ?kuRo] < Na dk ?kuRo( Na dh rqyuk esa K esa /kkfRod ca/k detksj gkrs gSA

25.

Which of the following on thermal decomposition yields a basic as well as an acidic oxide ?

fuEu esa ls fdlds rkih; fo?kVu ij {kkjh; rFkk vEyh; vkWDlkbM izkIr gksrk gS \ (A) KCIO3

(B) Na2CO3 

CaO Basic oxide



CO2 Acidic oxide



CaO



CO2

CaCO3  Sol.

CaCO3 

26.

Borox bead test is not given by (A) Copper salts (B) Cobalt salts

(C) NaNO3

(D*) CaCO3

(C) Nickel salts

(D*) Aluminium salts

cksjsDl eudk ifj{k.k fdlds }kjk ugh fn;k tkrk gSA (A) dkWij yo.k (B) dkWckYV yo.k (C) fufdy yo.k Sol.

(D*)

,Y;qfefu;e yo.k

Borax bead test is not given by colourless salts.

jaxghu yo.kksa }kjk cksjsDl eudk ifj{k.k ugha fn;k tkrk gSA

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27.

Boron does not form B3+ ion, due to (A*) Energy required to form B 3+ ion is very high which will not be compensated by lattice enthalpies or hydration enthalpies of such ion. (B) Boron is a non-metal. (C) Boron is a metalloid. (D) None of the above. cksjksu] B3+ vk;u ugh cukrk gSA ( D;ksafd (A*) B3+ vk;u

fuekZ.k ds fy, vko’;d ÅtkZ cgqr mPp gksrh gS rFkk ;g] bl vk;u dh tkyd ,UFksYih rFkk ty;kstu ,UFksYih ls iwfjr ugh gksrh gSA (B) cksjksu v/kkrq gksrh gSA (C) cksjksu mi/kkrq gksrh gSA (D) mijksDr esa ls dksbZ ugha Sol.

Factual statement.

rF;kRed dFku gSA 28.

Which of these is not a monomer for a high molecular mass silicone polymer ?

mPp v.kqHkkj okys flfydkWu cgqyd ds fy, fuEu esa ls dkSulk ,dyd ugh gS \ (A*) Me3SiCl Sol.

(B) PhSiCl3

(C) MeSiCl3

(D) Me2SiCl2



Me3 SiCl  Me3 Si(OH) It will form only dimer as shown.

;g fuEu izdkj f}yd cukrk gSA Me Me

Si

Me Me

Si

Me

Me OH H O

Me

Si

Me



Me

O

Si

Me

Me Me (Dimer) ¼f}yd½

whereas others will give high molecular mass silicon polymer.

tcfd vU; mPp v.kqHkkj okys flfydkWu cgqyd cukrs gSA 29.

What is not true about borax ? (A) Molecular formula is Na2B4O7.10H2O. (B) Crystallic borax contains tetrahedral unit of [B4 O5 (OH)4 ]2– . (C*) It hydrolyses to give an acidic solution. (D) White crystalline solid.

ckWjsDl ds fy, fuEu esa ls dkSulk lgh ugha gS \ (A) v.kqlw=k Na2B4O7.10H2O gSA ckWjsDl esa [B4 O5 (OH)4 ]2– dh prq”Qydh; bdkbZ;k gksrh gSa (C*) ;g ty vi?kfVr gksdj vEyh; foy;u nsrk gSA (D) ‘osr fØLVyh; Bksl gSA (B) fØLVyh;

Na2B4 O7  7H2O  Sol.

2NaOH

 4B(OH)3

Strong base

Weak acid

Na2B4O7  7H2O  2NaOH  4B(OH)3

On hydrolysis, an alkaline solution is formed.

ty vi?kVu ij ,d {kkjh; foy;u curk gSA Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PTC024029

PAGE NO.- 8

30.

AlCl3 is an electron-deficient compound but AlF3 is not, due to (A) Atomic size of F is smaller than Cl, which makes AIF3 more covalent. (B*) AlCl3 is a covalent compound while AlF3 is an ionic compound. (C) Al in AlCl3 is sp3 hybridised but in AIF3, Al is sp2 hybridised. (D) AlCl3 exists as a dimer but AIF3 does not. AlCl3 ,d bysDVªkWu U;wu ;kSfxd gS] ijUrq AlF3 ugh] D;ksafd (A) F dk

ijek.kq vkdkj Cl ls NksVk gksrk gS tks fd AIF3 dks vf/kd lgla;kstd cukrk gSA (B*) AlCl3 lgla;kstd ;kSfxd gksrk gS tcfd AlF3 vk;fud ;kSfxd gksrk gSA (C) AlCl3 esa Al sp3 ladfjr tcfd AIF3 es Al sp2 ladfjr gksrk gSA (D) AlCl3 f}yd ds :i vfLrRo esa gksrk gS ijUrq AIF3 ugh Sol.

Factual statement.

rF;kRed dFku gSA

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PAGE NO.- 9

ChemINFO-1.17

ELECTROCHEMISTRY

Daily Self-Study Dosage for mastering Chemistry

Fuel Cell

Conventionally energy is obtained by the combustion of the fossil fuel. The conversion into electrical energy involves a number of steps and there is loss of energy at every step. The efficiency of the process is around 40%. There is also a viable way of converting the chemical energy of fuel directly into elecrtical energy through catalytically activated redox reactions. Such devices are called fuel cells. In other words, fuel cells are galvanic cells in which the electrical energy is directly derived from the redox reactions of the fuel. The representation of the fuel cell is Fuel | Electrode | Electrolyte | Electrode | Oxidant Overall cell reaction : Fuel + Oxidant  Oxidation product + Reduction product Hydrogen–Oxygen Fuel Cell This fuel cell combines hydrogen and oxygen to produce electricity, heat and water. It converts chemical energy to electrical energy and thus can be compared to a battery. the only difference being that as long as hydrogen is upplied, it continues to produce electricity without being discharged.

Fig. A block diagram of hydrogen-oxygen fuel cell Design and working The whole setup consists of chambers (Fig.). The first chamber serves as an inlet for hydrogen, while the second one serves as an inlet for oxygen. There are two carbon electrodes. Hydrogen passes over the anode and oxygen passes over the cathode. In the middle is a proton exchange membrane separting the two electrodes from one another, where these two gases can interact with each other. The electrolyte separating the two electrodes is an ion-conducting material, such as concentrated aqueous sodium hydroxide. The following reactions take place at the two electrodes : At the anode :

2H2  4H+ + 4e–

At the cathode :

O2  4H+ + 4e–  2H2O

Overall cell reaction: 2H2  O2  2H2O The water is released in the fourth chamber. The anodic reaction gives protons and electrons. These are separated and hydrogen ions are allowed to pass through the electrolyte to the cathode while the electrons travel via an external circuit as direct current to power electrical devices. In the cathode, hydrogen ions combine with oxygen to form water along with heat, thus completing the circuit. Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice.

Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PTC024029

PAGE NO.- 10

ChemINFO-1.17

fo|q r jlk;u

Daily Self-Study Dosage for mastering Chemistry

b±èku lSy

ijaijkxr (Conventionally) ÅtkZ thok'eh; bZa/ku ds ngu }kjk izkIr gksrh gSA oS|qr ÅtkZ es :ikUrj.k vusd inks esa gksrk gS rFkk ;gk¡ izR;sd in ij ÅtkZ dh gkfu gksrh gSA izØe dh n{krk yxHkx 40% gksrh gSA ;gk¡ bZa/ku dh jklk;fud ÅtkZ dks lh/ks mRizsjdh; lfØ;r jsMkWDl vfHkfØ;kvksa }kjk oS|qr ÅtkZ ess :ikUrfjr djus ds fy, Hkh vusd rjhds gksrs gSA bl izdkj dh ;qfDRk;k¡ bZa/ku lSy dgykrh gSA vU; 'kCnks esa bZa/ku lsy XksYosfud lSy gksrs gS] ftuesa oS|qr ÅtkZ bZa/ku dh jsMkWDl vfHkfØ;kvksa ls lh/ks O;qRiUu gksrh gSA b±èku lSy dk izn'kZu gS % b±èku | bysDVªkWM | oS|qr vi?kV~; | bysDVªkWM | vkWDlhdkjh lEiw.kZ lSy vfHkfØ;k : b±èku + vkWDlhdkjh  vkWDlhdj.k mRikn + vip;u mRikn gkbMªkstu&vkWDlhtu bZa/ku lSy ;g bZ/kau lSy gkbMªkstu rFkk vkWDlhtu dks tksM+dj fo/kqr] Å"ek rFkk ty mRiUu djrs gSA ;g jklk;fud ÅtkZ dks fo/kqr ÅtkZ esa cnyrs gS rFkk bl izdkj cSVjh ds rqY; gks ldrs gSA dsoy vUrj ;g gS fd vf/kd yEcs le; rd gkbMªkstu izokfgr djus ij ;g fujUrj fujkosf'kr gq, fcuk fo/kqr mRiUu djrk gSA

fp=k % gkbMªkstu&vkWDlhtu b±èku lSy cukoV rFkk dkf;Zdh lEiw.kZ O;oLFkk esa dks"B (fp=k) gksrs gSA izFke dks"B gkbMªkstu ds fy, izosf'kdk ds :i esa dke djrk gSA tcfd f}fr; vkWDlhtu ds fy, izosf'kdk ds :i es dke djrk gSA ;gk¡ nks dkcZu bysDVªkWM gSA gkbMªkstu dks ,uksM ij rFkk vkWDlhtu dks dSFkksM+ ij izokfgr djrs gSA e/; esa ,d izksVksu fofu;e f>Yyh gksrh gS tks ,d nwljs ls nks bysDVªkWMks dks i`Fkd djrh gS ] tgk¡ nks xSls ijLij vU;ksU; fØ;k dj ldrh gSA nks bysDVªkWMks dks i`Fkd djus okyk oS/kqr vi?kV~; ,d vk;u pkyd inkFkZ] lkfUnzr tyh; lksfM;e gkbMªkWDlkbM ds leku gksrk gSA fuEu vfHkfØ;k;sa nks bysDVªkWMks ij gksrh gSA ,uksM+ ij :

2H2  4H+ + 4e–

dSFkksM+ ij :

O2  4H+ + 4e–  2H2O

lEiw.kZ lSy vfHkfØ;k : 2H2  O2  2H2O ty pkSFks dks"B es fu"dkflr gksrk gSA ,uksfMd vfHkfØ;k izksVkWu rFkk bysDVªkWu nsrh gSA budks i`Fkd djrs gS rFkk gkbMªkstu vk;uks dks dSFkksM+ rd fo/kqr vi?kV~; ds }kjk xqtkjrs gS tcfd bysDVªkWu 'kfDr fo/kqr ;qfDr;ksa rd izR;{k /kkjk ds :i es ckg~;re ifjiFk ds }kjk xfr djrs gSA dSFkksM+ esa gkbMªkt s u vk;u vkWDlhtu ds lkFk tqMdj Å"ek ds lkis{k ty cukrs gS bl izdkj ifjiFk iw.kZ gksrk gSA Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PTC024029

PAGE NO.- 11

31.

32.

In H2–O2 fuel cell, the reaction occuring at cathode is (B) H  OH  H2O

(C*) O2 + 2H2O + 4e–  4OH–

(D) He–  1/2 H2

The thermodynamic efficiency of fuel cell is given by (A)

33.

31.

32.

H G

(B)

nFE G

The cell reaction of a fuel cell is (A) 2H2O(l)2H2 (g) + O2(g) (C) CH4(g) + O2(g) CO2(g) + H2O (l)

H2–O2 bZ/ku

(C*)

nFE H

(D) nEF–

(B*) 2H2O(g) + O2(g)2H 2O(l) (D) C(s) + O2(g) CO2(g)

lsy es] dSFkksM ij vfHkfØ;k gksrh gS%& 

(A) 2H2 + O2  2H 2O (l)

(B) H  OH  H2O

(C*) O2 + 2H2O + 4e–  4OH–

(D) He–  1/2 H2

bZ/ku lsy dh Å"ekxfrfd; {kerk fuEu }kjk nh tkrh gSA (A)

33.



(A) 2H2 + O2  2H 2O (l)

H G

(B)

nFE G

(C*)

nFE H

(D) nEF–

bZ/ku lsy dh lsy vfHkfØ;k gS %& (A) 2H2O(l)2H2 (g) + O2(g) (C) CH4(g) + O2(g) CO2(g) + H2O (l)

(B*) 2H2O(g) + O2(g)2H 2O(l) (D) C(s) + O2(g) CO2(g)

Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PTC024029

PAGE NO.- 12