50% of a first order reaction gets compl...

50% of a first order reaction gets completed in 10 min. What fraction of reactant of the above reaction would remain after 20min ?

Text Solution

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For a first order reaction , `k=(2.303)/(t)log.(a)/(a-x)`
Now , when `t=10"min",x=axx(50)/(100)=0.5a`
`therefore(a-x)=a-0.5a=0.5a`
`thereforek=(2.303)/(10)log.(a)/(0.5a)=0.0693`
When , `t=20 min , 0.0693=(2.303)/(20)log.(a)/(a-x)`
or , `(x)/(a)=(3)/(4)`
So, `(1-(3)/(4))=(1)/(4)` th fraction of reactant would remain after 20 min.

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