24arre yaar.. tu to aur bhi diff ans bataa raha hai [3]
The inversion of cane sugar proceeds with half life of 600 min at pH=5 for any concentratiojn of sugar.If pH=6,the half life=60 min.fond rate law
i got the answer Rate=k[H+]1
bt ans is R=k[Sugar][H+]
how is order 1 wrt sugar ??? whwen it si given that any conc of sugar[7][7]
help guys...
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7 Answers
11DUDE, it must be given that the inversion of cane sugar proceeds with constant half life of 600 mins, If it is,
Let the rate law be given by
r = k [sugar] x [H + ] y where k is the rate const
From the above , calculate x and y,
600 = K 1 [Sugar] x [10 -5] y {since pH=5}
60 = K 2 [Sugar] x [10 -6 ] y
Since the half life doesn't change with the conc of sugar, so it must be first order reaction w.r.t sugar
(t 1/2) 1 / (t 1/2) 2 = (a1 / a 2) 1-n
where t 1/2 is the half life time and a is the initial conc.
Therefore,
(600 / 60 ) = (10 -5 / 10 - 6 ) 1-n
That implies n=0
So, it should be zero order w.r.t H + ion
Therefore, r = K [Sugar] [H+] 0
Therefore, x=1, and y=0
1what tushar hs posted is absolutely rite.
eure,the pt. u r hving doubt is :
t1/2 = 500 for all concentrations of sugar.so, t1/2 is directly proportional to [sugar]0.
thus, the reaction is of first order with respect to sugar.
thus t1/2 is proportional to [ H+ ]1-n , where u'll get n=0.
so,R =k [SUGAR] [H+]0
3Since the half life doesn't change with the conc of sugar, so it must be first order reaction w.r.t sugar
y it shud be tushar
11Dude, the half life for a first order reactions independent of concentration., and not only haf life but also the time rquired to complete any definite fraction of the first order reacn is independent of initial conc.
t1/2 = 0.693 / k1