UNIT # 08 (PART - I)

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. r. d[h d[h.5 7.5 mol L S d[o d[so UNIT # 8 (PRT - I CHEMICL INETICS EXERCISE # 6. d[ x [ x [ x. r [X[C ' [X [[B r '[ [B [C. r [NO [Cl. d[so d[h.5 5 mol L S d[nh d[nh. 5. 6. r [ [B r [x [y r' [x [y r' 7 r 8. r. d[c. log t.5 mol L S. log 5. log 5.66 sc. r [ 5 5 [ 5 5 M s. r [N O [O...(i [N O or [N O [NO r [NO [O compir r [NO [O d[n d[no 5. r / [NO from q. (i d[o [N [NO ' [NO r ' [NO [Cl. 5 ln 5 5 9. ln t ln. q. (ii/(i t ln ln t 8 min. Givn t t n n n... (i... (ii 5. t n.5 n n n 5. E / RT 5. ln ln E RT E. log +. log RT log E.RT + log n E slop.r 5 E 5 8.. 95.7J mol E R E 8 cl 5. E / RT E / RT RT E + E E 6 + J

CHEMICL INETICS EXERCISE #. (g B (g 5 mol mol f.5 s [ f 5 b 7.5 b b.5 L mol s. + B C + D rt [ / [B / dx ( x( x dx ( x. t log x. t log..5 t 6 sc.. X log( log( / + B nd Y.6 5. P(mmHg 5 5 t / (in min. 5 95 t / n n C + D log log log n (t / 5 5 (t / 95 5 ( n n n 6. t t 5 Now rrhnius qution E T T log. R T.T E log ( 8.. E 55. J/mol 7.. s ; E. J mol 6. s E/RT But whn T 6. s. t(n C C t / t / (n (n n n Co qu. ( qu. ( n t / n t / n Co n n n n n Co Co (n n ( t / t / [ n +... (... ( n n ( ( n (. Frction of rctnt consum f C C df for rction : ( f. + B C + D (Rmining mount. rt [ [B t C C C 7C t min, Rt t tht tim C 7C Rt 9C. B 5 /T C D C /T t T ( C 5 C /T /T /T /T /T log /T T log 5. log. E T T R T T R log 8 9 E. log T.. R log 8 9. R 9 8 log log 6. + B products x x.9 d[ [ [B Rctnt r in thir stoichiomtric proportion

d( x ( x ( x dx ( x t /.69 7. +B products d[ [ t whn t l/ /. Rt [ [B [ [ [ {[ }/ r ' [ / [B ordr 6. For t ln 9 For B t ln 5 Eq. ( nd ( 6 ln ln 5.58 7. % B 76.8% %C B%.7% CHEMICL INETICS.( N + H NH [NH Rt mol L s t (b dh dh dnh dn dh. N (g NO (g + O (g d [N / [N d [NO / [N d [O / [N d N O d NO d O 5...(...( 8. N, For mx. no of nucli dn N N / 9. Lt n is th mols of rgnt 'R' whn R is rctd with t tim t B + C t n t t n x x x t t n n n 5n n n 5 n n n + x n x. n log t n x so t ln n 5(n n. Ovrll rt constnt + + 6.9.69 t / sc; 6.9 ftr hlf-lif, P B + P C + P D tm PB P P P B C D 69 P B 69. + B + C D + E dx [ [B [C if incrss conc. tim.5 tm EXERCISE # [ dx [ dx [B [C 8 rt incrs by 8 tim. + B B dx [ [B if V is dcrs to V dx dx dx [ [B 7 rction incrs by 7 tims d O 5. H O H O + O.6 M min ( Rt of formtion of H O d H O d H O.6 7. M min 7. M min

6. B. M s initilly M from th unit of rt constnt w find tht rction is of zro ordr X t rminng. 6 7. [Rmining mol.8 M If. t ddd 6. t. t t.88 min E 8. H O + O HO H 7J Eb T 5 E 77 J/mol H E Eb 7 77 Eb Eb 77 7 Eb 5 J. NO + Br NOBr (i NO + Br NOBr (Est Slow (ii NOBr + NO NOBr Rt is dtrmind by th solwst stp Rt [NOBr [NO But NOBr is visibl of Intrmdit [NOBr [NO[Br Rt [NO [Br rd ordr RxN 5. B E 7 J/mol t / min.69 min log E.R 7. 8. T T T T 5 98 5.88.588.87.87.85 log / x log ( / log / x log( / x.8875 x.85 x.8% % dcomposition 8.8 67.6% 6. (i product (ii log B product E.R log. 8. E E 5.6 J/mol B.69.69 5 min E 6.8 J/mol E log B. 6.8 log B. 8.. log B.5.76.69 5 6.7 min sc 7. H O H O + O ml. q of H O in ml dilutd M q. of MnO titrtd 5.5 5.5 ml. q in ml solu..5 but ml O 68 gm H O ml ml 68 68 ln ml H O 68 in ml H O No. of Mq. in ml of %g ml H O Initily 68 5.7 7. 5.7 log 6.5. hr

CHEMICL INETICS. Initil m mol of cyclobutn ( m Lt ftr min, x m mol cyclobutn isomrizd. (, x m mol m mol of cyclobutn lft x nd m mol of din formd x (m mol x m mol x m mol of Br rquird ftr min x + x + x 6 ( Br m mol x + x + x 6 x 6 ln...(i If y m mol of cyclobutn isomrizd ftr min. ( y m mol ln y...(ii From Eqs. (i nd (ii ((i (ii y 7.7 m mol of Br rquird (Br m mol + y 7.7 Vol. of bromin solution rquird ( 7.7 ml. RT RT T. Solving (, Rt constnt (.6 hr t ln t ln n ( n ( n n (B n (B n n.7 n.7 Totl mols of gss ftr. hr. (..7 +.7. P 5. tm. Lt rt constnt in bsnc of ctlyst is ( Lt rt constnt in prsnc of first ctlyst is ( Lt rt constnt in prsnc of scond ctlyst is ( ln. hr...(i.5 ln 8.86...(ii EXERCISE # [B ln 6 8.7...(iii 8,/RT E / RT E ctivtion nrgy in prsnc of st ctlyst ( st ln E 8, E RT ln.86. 75.8 J 75. J E / RT E ctivtion nrgy in prsnc of nd ctlyst ( nd ln E. N 8, E RT ln 8.7. 797.9 J 7. 975 J NO + NO NO + NO b NO + O + NO c NO + NO NO d[n [N [NO [NO...(i d[no [N [NO [NO b [NO [NO c [NO [NO...(ii d[no b [NO [NO c [NO [NO...(iii From qution (iii ( (iii [NO b [NO c putting this in qution(ii((ii [N [NO { [NO + [NO [N [NO [NO [NO b putting this in qution (i((i d[n [N [N O 5 d[n b[n d[n r 5. t+ b b 5 b [N O d d ordr ( ( b

6. Givn ( : r ' [complx [[ & sinc ( t / t / so ( t / ln ln ' [H b On doubling concntrtion of [H + ion t / gts hlf so b ([H + t / b 7. Lt t C rt constnt( C thn t C rt constnt ( C / ln / E R 76 9 E.5 J Lt t C rt constnt is thn ln.5 8. 9 ( C.5 so tim rquird for juic to gt spoil t C 6.5.7 hr. ( C 8. Givn : 9,. 5 [ [ [C ( t [ 9 ( [ t [C [ ( [ t 9..6,. so /.65 +.95 [B q. 9 t [.57 [ q. [. B + C t t min. x x x t 6( x + x 8x 5 8 on solving (.5, x.5 so t / min. vrg lif /. t / 8.86 min.. t t whn quilibrium is stblishd & (t [P 7. [.. [ [ ( t [. 6.75 [...6 5 sc. 9.7 5 sc. CH OCH CH + CO + H t. t.5hr...9.9.9 t ln P. tm t t P. M 6 P P.78.5 6 ln. P t t.5 hr. P. +.9.98 tm M. 6.9(6 8.98 8.77 givn [B [ & [B [B q. [ ( ( t ( t so ( + t ln t 89.7 sc..8 min.. r P M. 8.77.6 r P M.98 6 B +.69.69 C t / 6 +.69 9 t / 6 min.

6. + B C, C r [ [B [C D d[c [ [B [C [C [[B [C d[d [[B r [ [B [[B r sinc >> nt so nt (E nt E + E E 8. P (g Q (g + R (g + S ( t P t min. P P P P/ t P P / so P P + P + P/ 7.5 i.. P +.5 P 8.5...(i &.5 P 67.5 58.5 so P.8 P.8 t ln.8 t 75 min.5.5 75 ln.8 P P P P 58. P 75.57 P T.5 + P +.5 P 7.55 +.5 P T 79.65 mm Hg (ii.5 t ln 8 t 99.89 min. 9. B n+ B (n++ t so t min. x v.f. of B n+ v.f. of B (n++ 5 Lt normlity of rducing gnt N 5 N ( x + 5x N + x N ln x 7 N.5N ln x 7.5N N. min.. For rction ( E / R T T.79 6 59 T For rction.5 5 5 T For givn condition (.79 8.55 6 59 T.5 5 6 5 T 59 T ln 8.55. 65.9 T T 67.6 97.6 C. B C ( t [ [ [B [C [ 5 5 T ( [ t [ ( [ t sinc V & T r constnt (V T P mols t t, P tm so [ t sc. P. tm so t t, [ + [B + [C....( t P.5 tm...( t t, [B + [C.5 from qution ( ( (.5 +.5.5... (

from qution ( & ( ( t + t sc & ( t ( t + (. so.5.5...5..69.8. 9 Th 8 Pb8 + 6 H + t tim t x 6x givn : 5 x 5 8 6x 7 so.75 9.69.9 t ln x.69 ln.9 t.89 9 yr..69 8.55 9 mol x 5.95.75.55 9 9.8 mol 5. 8 Po8 8 Pb 8 Bi Numbr of nucli of Pb t tim t r N N ( t t (t Pb N For mx. vlu of N (N dn so t ln ( whn.69.68,.69.5, on putting ths vlus ( t.87 ln.5.68. min 6. Lt th mss of smpl in g & initil mss of U 8 is w g th (g U 8 w g t t U 8 Pb 6 w w-x givn w x.5 6x.68 t ln.69 ln 8.77 Totl ctivity is 7.7% of th originl ctivity but only 67.7% found in th thysoid so mss of stbl iodid ion hd migrtd to th thyroid glnd is (7.7% 67.7% 67.7. 7.7.9575 mg so 6x 8 x.6 w.56 t ln.5.9.555 w w x.5 t ln.56.5 t. 8 yr.

CHEMICL INETICS. rt of pprnc of HI d[hi rt of formtion of H d[h rt of formtion of I d[i hnc d[h d[i d[hi d[h d[i d[hi or. Ordr is th sum of th powr of th concntrtions trms in rt lw xprssion. hnc th ordr of rction is +. For zro ordr rction. rt [ i.., rt hnc unit of M.sc For first ordr rction rt [ M.sc / M sc. Rt log C log C t It is clr from th qution tht if w plot grph btwn log C t nd tim, stright lin with slop qul to nd intrcpt qul to. log[ will b obtind. 6. It is zro ordr rction Not : dsorption of gs on mtl surfc is of zro ordr. 7. In qution E /Rt ; Frquncy fctor vlocity constnt, R gs constnt nd E nrgy of ctivtion 8. r [O [NO. Whn th volum is rducd to /, th conc. will doubl. Nw rt [O [NO 8 [O [NO Th nw rt incrss to ight tims of its initil. n m Rt [ [ B 9. [ [. n m Rt [ [B n m n m n m. s th concntrtion of rctnt dcrss from.8 to. in minuts hnc th t / is 5 inuts. To fll th concntrtion from. to.5 w nd two hlf livs i.., minuts.. Th vlocity constnt dpnds on tmprtur only. It is indpndnt of concntrtion of rctnts. n. N t N whr n is numbr of hlf lif priods. totl tim n 6 hlf lif 6 N t.5g. Enthlpy of rction (H E (f E (b for n ndothrmic rction H +v hnc for H to b ngtiv. E (b < E (f. Th molculrity of rction is th numbr of rctnt molculs ting prt in singl stp of th rction. Not : Th rction involving two diffrnt rctnt cn nvr b unimolculr. 5. t /. log. log /.. (log log ( log log..9 (..77 6. Sinc th rction is nd ordr w.r.t. CO. Thus, rt lw is givn s r [CO Lt initil concntrtion of CO is i.. [CO r ( whn concntrtion bcoms doubld,i.., [CO r ( r r So, th rt of rction bcoms tims. 7. In rrhnius qution E/RT, E is th nrgy of ctivtion, which is rquird by th colliding molculs to rct rsulting in th formtion of products. 8. (i NO(g + Br (g NOBr (g (ii NOBr (g + NO(g NOBr(g Rt lw qution [NOBr [NO But NOBr is intrmdit nd must not ppr in th rt lw qution [NOBr from Ist stp C [NO[Br [NOBr C [NO [Br EXERCISE # 5[ Rt lw qution. C [NO [Br hnc ordr of rction is w.r.t. NO.

9. H R E f E b 8 J/mol Th nrst corrct nswr givn in choics my b obtind by nglcting sign.. For first ordr rction t /.69 i.., for first ordr rction t / dos not dpnd up on th concntrtion. From th givn dt, w cn sy tht ordr of rction with rspct to B bcus chng in concntrtion of B dos not chng hlf lif. Ordr of rction with rspct to bcus rt of rction doubls whn concntrtion of B is doubl ping concntrtion of constnt. Ordr of rction + nd units of scond ordr rction r L mol sc.. Th rts of rctions for th rction or B cn b writtn ithr s d [ d [B with rspct to '' with rspct to 'B' from th bov, w hv d d [ [B d or d B. For first ordr rction. log t 99.69. log 69 t.69. 69 t t 6.6 minuts. Sinc th slow stp is th rt dtrmining stp hnc if w considr option ( w find Rt [Cl [H S Now if w considr option ( w find Rt [Cl [HS...(i From qution (i [H [HS [H S or [HS H S H Substituting this vlu in qution (i w find [H S [Cl [H S Rt [Cl ' H [H hnc only, mchnism (i is consistnt with th givn rt qution.. For th rction Product givn t / hour for zro ordr rction [ initil conc. t compltion rt constnt t / [ [ or mol lit hr t / Furthr for zro ordr rction dx chng in concntrtion tim.5.5 tim tim.5 hr. 5. Sinc for vry C ris in tmprtur rt doubls for 5 C ris in tmp incrs in rction rt 5 tims. 6. E / RT E / RT...(i...(ii On dividing q. (i from q. (ii (E E / RT...(iii Givn E E, On substituting this vlu in qn. (iii E /RT 7. For first ordr rction... log log t x.5...6 log.7 R (.7..7 7