too many kinetics !

(2) ur second one is wrong mani.. its 1M.

actually this was askd in one of the CBSE sample papers....

and the given was 1

Pl. post full soltn to Q3 in Sky's 1st post................

mere ko abhi bhi nahi samajh aa raha hai [2]

@ skygirl previously it has been said that molecularity of a complex rxn is the molecularity of its rds but now for complex reaction molecularity is not defined ( refer N.C.E.R.T) look molecularity is something which is not required for solving a problem it is just a way for naming the colliding probablity nothing else and as probablity theory is not included in IIT course they wont be asking question on molecularity but for other exams i can not say anything

no..no the first question asked by lady sky....

Pl. reply here

solution to (1)

its molecularity is not defined as it is not an elementary reaction.....
and check the states of all three substances(C)

(2)
k=1/t2.303log[A]₀/[A]
R=K[A]₀

the rate constant for a first order reaction is equal to the initial rate of reaction

=>[A]₀=1

SO

1=1/t ln1/0.1 (if time is 1sec)

so[A]=0.1 =(D)

(3) (B)

use the definition of rate law and H₂SO₄ follows a 2nd order reaction
but HCl follows 1st order

ok sry i think i read it for its oxidising nature...

sky if i am wrong

THEN I NEED EXPLAINATION OR [342]

[2] [2] [2] [2] [2] [2] [2] [2] [2] [2] [2] [2] [2] [2]

chemsandeep: ok.........let me see the 3rd question
me: k
16:35 chemsandeep: what does K stands for here
16:36 is it equlibrium const of ethyl acetate
hydrolysis
me: kuchh ni bataye
par wahi hona chaiye
chemsandeep: ok..............let me understand first
16:39 if it will be equilibrium const. then .........as we know that it doesnot depend on concentration of reactant
16:40 me: so ?
chemsandeep: ok
it is rake constant of hydrolusis of rthyl acetate
16:41 so we can write the basic hydrolysis reaction as
16:42 me: Etcooet + hcl -> etcocl + etoh
aur ek me heso4
16:43 chemsandeep: thabove reaction which u have written wont be the hydrolysis product
hydrolysis product will be acetic acid and EtOH
16:44 here H+ is the catalyst
me: sry..
haan
chemsandeep: so here catalyst concentartion is different
16:45 me: hmm..
16:46 chemsandeep: 0.075 M HCl =0.075 M H+ but 0.075 M H2SO= 0.150 M H+
me: haan
16:47 par k ke upar iska koi asar kyun hoga ?
iska asar rate par hona chaiye
16:48 chemsandeep: so what will be the effect of a catalyst on the rate of reaction
16:49 rate constant of reaction depends on catalyst
me: rate inc ofcrs.....
rate constant how ?
achha
Ea will change
i mean decrease
chemsandeep: yaa
16:50 yer Ea will decrease
me: ok ok got it
chemsandeep: so if there will be linear relationship
me: kh2so4 <k hcl
haan
16:52 chemsandeep: kh2so4 >k hcl
higher rate const means faster reaction
me: kaise ?
16:53 par h2so4 better catalyst ?
isnt it ?
chemsandeep: rate of reaction(R)=k[conc]..................
as h2so4 is better catalyst then
16:54 its Ea will be low
me: so it eill lower Ea more
chemsandeep: and rate will be more faster
means rate const will be more
me: par k=e^-Ea/RT
16:55 if we compare the k's of both of them.... and Ea's
chemsandeep: as the exponent has -ve sign then so lesser the Ea means higher k
16:56 did u understood the above line
16:57 me: par lnk1/lnk1 = Ea1/Ea2
lnk1/lnk2 = Ea1/Ea2

if we compare
chemsandeep: lnk1/lnk2 = Ea2/Ea1
16:58 its inversely related
me: minus gets cancelled
how?
16:59 k1= e^-Ea1/RT => lnk1 = -Ea1/RT
lnk2 =-Ea2/RT
lnk1/lnk2 = Ea1/Ea2
chemsandeep: k=e^-Ea/RT ...................u see if Ea will be more then k will be less
me: hmm fine
17:00 chemsandeep: so fo H2SO4 case Ea is less ...........this implies k is more
17:01 me: ok
agreeing...
par woh upar jo relation aya .. usse kyun ni aya/
?
17:02 chemsandeep: k=e^-Ea/RT === so Ea is directky proportional to ln(1/k)
17:03 there is something wrong with taking the ration and cancelling the -ve sign
me: hmm...
17:04 chemsandeep: so option a is right
me: actually ans given was C... par wrong bhi ho sakta hai
this book has got mny mistakes
17:06 chemsandeep: sorry i missed the point what u wrote earlier..................please write it again as gtalk had crashed
write it again
me: k
17:07 actually ans given was C... par wrong bhi ho sakta hai
this book has got mny mistakes
17:10 chemsandeep: in H2SO4 case the concentarion of H+ is more.............. i.e .. catalyst concentration is more so rate should be more ...implies rate constant should be higher
me: hmm got it :)
17:11 chemsandeep: any way ........JEE wont ask u catalyst concentratin dependency problem
me: okie.....

who is this

CHEMSANDEEP[7][7][7]

[2] [12] ..........

he is our chem sir in coaching...........

SKY TELL ME
IS HE'S UR TEACHER OR R U HIS TEACHER[7][7][7][3]