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The first order integrated rate equation...

The first order integrated rate equation is

A

`k = (x)/(t)`

B

`k = - ""(2.303)/(t) "log" (a)/(a-x)`

C

`k = (1)/(t) "ln" (a)/(a-x)`

D

`k =(1)/(t) (x)/(a(a-x))`

Text Solution

Verified by Experts

The correct Answer is:
C
For first order ,
rate = `(d[R])/(dt) = k[R]`
or `(d[R])/([R]) = kdt " " … (i)`
On integration Eq. (i)
`int (d[R])/([R]) = k int dt`
In [R] = `-kt + C " " … (ii)`
At t = 0 , [R] = `[R_(0)]`
[where , R = final concentration , ie , a - x and `R_(0)` is the initial concentration ie,a ]
In `[R_(0)] = C`
On putting the value of C in Eq. (ii) we get
ln[R] = -kt + ln `[R_(0)]`
`-kt ` = ln [R] - ln `[R_(0)]`
kt = ln `[R_(0)] - `ln [R]
or `k = (1)/(t)` ln `([R_(0)])/([R])`
or `k = (1)/(t) "ln" (a)/(a-x)`
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