naa....how do i relate it with solvolysis....plz explain dada????
A certain reaction A + B → Product is first-order w.r.t. each reactant with k = 5.0 x 10-3 M-1s-1. Calculate the concentration of A remaining after 100 s if the initial concentration of A was 0.1 M and that of B was 6.0 M. State any approximation made in obtaining your result.
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4 Answers
.06 mole
- Bitan Chakraborty No................
Hint:
[B]>>[A], hence, rate is dependent on A,ie, 1st order w.r.t A.Upvote·0· Reply ·2013-04-22 23:04:34
Dada first see whether my approach is correct or not......
-d[A]/dt=k[A] (As the reaction is first order w.r.t A and B is present in excess)
Given,the initial concentration of A is 0.1M.Let the final concentration be xM.
A.T.P
∫d[A]/[A]=-∫k.dt (Integration limit for [A] is from 0.1 to x and integration limit for time is from 0 to 100 secs)
After integrating we get,
log[x]-log(0.1)=-5*10^-1
[x]=10^-3/2........whether this is correct????
- Anurag Ghosh sry I forgot to write the unit of [x] i.e. 10^-3/2M...
Since [B] = 6.0, we can consider that its concentration remains constant throughout.
Thus, rate = k[A][B] = 6k[A] = k /[A]
where, k / ( = 6k ) is the new constant and reaction under this condition is supposed to follow 1st order kinetics.
k / = 2.303t log (aa-x)
Now putting in the values we have,
aa-x = 20
So that (a-x) = 0.005 M
- Anurag Ghosh But when something is present in excess...we neglect its concentration na???
- Bitan Chakraborty Yes true..... But conc of A is given in the sum, so although [B]>[A], still A has some relavance in terms of conc. Also think in terms of solvolysis. Get it??