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For a certain reaction of order 'n' the ...

For a certain reaction of order 'n' the time for half change `t_(1//2)` is given by : `t_(1//2)=(2-sqrt(2))/KxxC_(0)^(1//2)` where `K` is rate constant and `C_(0)` is the initial concentration. The value of `n` is:

A

`1`

B

`2`

C

`0`

D

`0.5`

Text Solution

Verified by Experts

The correct Answer is:
d
This is intergrated rate expression of `(-dC)/(dt)=KC^(1//2)`
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