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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

```
9 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

```
9 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...

appr0ve if u like this ans
```
9 years ago
```							or first order reaction ,t = 2.303log(a/a-x) / k ....................1k 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 = .2initial moles = a = .8time taken = 1hrby plugging these values in eq 1 , we getk = 2.303log4 ...............2in second case , .9mol of A produces .675mols of B so remaining molesof A are .9-.675 = .225final moles of A = a-x = .225initial mole = .9now againg using eq 1t = 2.303log(.9/.225)/k=2.303log4/kk = 2.303log4 , sot = 1hrtherefore time in second part will be same 1 hr..Hope it will help you...Pls mark me brainliest...
```
3 months ago
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