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the reaction A->B follows a first order kinetics. the time taken for 0.8 mole of A to produce 0.6 mole of B is 1 hour . what is the time taken for conversion of 0.9 mole of A to produce 0.675 mole of B? plzz tell how did u find the answer

10 years ago

Dear student,

The integrated rate law for a 1st order reaction is:

ln([A]0) - ln([A]) = kt

One interesting thing about first order reactions is that the time required for 1/2 of the beginning reagent to react is independant of the conentration of the reagent. This can easily be seen by combining the two logs in the above integrated rate expression

ln([A]0/[A]) = kt

All the best.

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Sagar Singh

B.Tech, IIT Delhi

10 years ago

for first order reaction ,

t = 2.303log(a/a-x) / k                    ....................1

k is rate constant ..

when .8 mol of A produces .6mol of B then .8-.6 = .2 mol of A is left ...

final moles A = a-x = .2

initial moles = a = .8

time taken = 1hr

by plugging these values in eq 1 , we get

k = 2.303log4 ...............2

in second case , .9mol of A produces .675mols of B so remaining moles

of A are .9-.675 = .225

final moles of A = a-x = .225

initial mole = .9

now againg using eq 1

t = 2.303log(.9/.225)/k

=2.303log4/k

k = 2.303log4 , so

t = 1hr

therefore time in second part will be same 1 hr...

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one year ago
or first order reaction ,
t = 2.303log(a/a-x) / k ....................1
k is rate constant ..
when .8 mol of A produces .6mol of B then .8-.6 = .2 mol of A is left ...
final moles A = a-x = .2
initial moles = a = .8
time taken = 1hr
by plugging these values in eq 1 , we get
k = 2.303log4 ...............2
in second case , .9mol of A produces .675mols of B so remaining moles
of A are .9-.675 = .225
final moles of A = a-x = .225
initial mole = .9
now againg using eq 1
t = 2.303log(.9/.225)/k
=2.303log4/k
k = 2.303log4 , so
t = 1hr
therefore time in second part will be same 1 hr..