13actually this was askd in one of the CBSE sample papers....
and the given was 1
1.) the molecularity of the reaction
2 H2(g) + O2(g) --> 2 H2O (l)
is: a)2 b) 3 c)undefined d)1
2) the rate consant for a first order reaction is equal to the initial rate of reaction wheb the initial concentration of the reactant is:
a) 100 M b) 1x 10^-2 C) 1 M d) 0.1 M
3) the hydrolysis of ethyl acetate was carried out separately with 0.075M HCl and 0.075 M H2SO4. Which of the following relations is correct?
a) k(h2so4) >> k(hcl)
b) k(h2so4) = 2 k(hcl)
c) k(h2so4) < k(hcl)
d) k(h2so4)= 8 k(hcl)
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64 Answers
1313actually this was askd in one of the CBSE sample papers....
and the given ans was 1 and even my skul chem teacher said its 1
but m styl not gettin it
33oh 1 is the answer...
well partly i think yes... because which two molecules collide..
i think thers only 1..
and multiplying a eqn doesnot change its molecularity... because.. molecularity can never be 1000.. :D if u multiply eq by 1000 :D
21Pl. post full soltn to Q3 in Sky's 1st post................
mere ko abhi bhi nahi samajh aa raha hai [2]
33:D
crux of Q.3
tukka method...
all other options are same... also anyone will think that H2SO4 will give 2 H+
so unique option... :D
_________
reason according to me...
always we use HCl for hydrolysis...always...
H2SO4 is dehydreating agent... so will convert to ester a litle
13hey this very ques came in my preboard paper .... and molecularity 3 was markd correct...
priyam wat do say nao......
1@ 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
21SUM1 who is sure PL. POST DA DETAILED solution of questions asked by SKY.........
I m really not gettin wats da ans.....
Pl. help!!! ............ AND SKY : can u pl. post da answers to all the 3
11solution 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
11sky provide the sol to 2
and sacchi i didnt get the depth of 3
PLEASE EXPLAIN IN DETAIL....................
1chemsandeep: 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.....
33nahi... maine kahi padha tha
rate for HCl>H2SO4
isliye har jagah HCl likha rahta hai hydrolysis ke liye...
21THIS IS DA BOTTOM LINE :: (literally)
17:11 chemsandeep: any way ........JEE wont ask u catalyst concentratin dependency problem
me: okie.....
