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For a First Order Reaction, Show that Time Required for 99% Completion is Twice the Time Required for the Completion of 90% of Reaction. - Chemistry

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For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.

 

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Solution

(a) For a first order reaction,

 

`A->P`

t=0 a 0
t=t (a-x) x

Case 1: If ‘t’ is the time required for 99% completion then x = 99% of a

`(a-x)=1% of a=1/100xxa=a/100`

`t=2.303/klog(a/(a-x))=2.303/k log ((axx100)/a)=2.303/k log10^2`

`therefore t=2[2.303/k]`

 

 

 

Case 2: If ‘t’ is the time required for 90% of completion then x = 90% of a

`(a-x)=1% of a=1/100xxa=a/100`

`t=2.303/klog(a/(a-x))=2.303/k log ((axx10)/a)`

`therefore t=[2.303/k]`

 

Therefore, the time required for 99% completion of 1st order reaction is twice the time required for 90% completion

 

 

Concept: Integrated Rate Equations - Half-life of a Reaction
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