33dx/dt=k(A0-2x)
dx/(A0-2x)=kdt
integrating we get...
(A0-2x)=A0e-2kt
[A]=A0e-2kt
Now....theres 2 in power...
For a first order reaction 2A→Products what is [A] as a function of time? Does that coefficient 2 makes a difference?
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19 Answers
3333Yippie........
i got it........
IN RC mukherjee and Arihant also this is given wrong and has generalised nA->B
as k=2.303/nt*log(.......).......
But at last i got it........
Bhagwaan tumhe kabhi maaf nahi karega RCM and arihant...
33ok so for first order rkn, rate of dissosiation of A is proportional to [A] not rate of rkn is proportional to [A}.????
1357you wrote 1/2(d[A]/dt) as the rate of dissociation of A
which is not true because it is -d[A]/dt
33:(
But by differeenciating earlier by taking that 1/2d[A]/dt....
i mean
(-1/2)d(A0-2x)dt.............
gave that...
i'm not getting any mistake in earlier steps.....
i think all steps are mathematically correct.........right from the def of rate of rkn....
but may be.........
my chemistry "god knows"....
1357rate of dissociation of A is -d[A]/dt
rate of dissociation is k[A]
1357The expression you are getting
dx/dt=k[A]
here RHS is the rate of disintegration of A
while LHS is the rate of disintegration of 2A
1no....
coefficient changes molecularity... not order...
n order depends upon reaction rate.. i mean it has nothing to do wid balanced reaction...
(correct me if wrong..)
33Ok tell me where i'm wrong...
2A → B
(A0-2x) (x)
rate=-d[A]/2dt .................................(i)
rate=k[A]1.......................rate law expression
rate=k(A0-2x) ..........................(ii)
from (i) and (ii)
(-1/2)d[A]/dt=k(A0-2x)
(-1/2)d(A0-2x)/dt=(A0-2x)
contd........
1yup...
have got another point,,
take an eqn 2A ->P
here at time t=0, [A]=a, [P]=0
at time t=t, [A]= a-x and [P]=x/2
but rate only depends upon reactants....
so it depends only on (a-x) n not coefficient....
62Dude NCERT is always correct. I have not known it to have even one sngle mistake ever!
and anyway sky gave a perfect explanation to the thing...
33some book say:
k=2.303/2t*log(......)
but in NCERT 1 question is solved as
k=2.303/t*log(........)..........