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The rate of first order reaction is 0.04...

The rate of first order reaction is `0.04 mol L^(-1) s^(-1)` at `10 min` and `0.03 mol L^(-1) s^(-1)` at `20 min` after initiation. Find the half life of the reaction.

Text Solution

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`24.14" min "r_(2) =k_(1) C_(1) and r_(2) =k_(2) C_(2)`
Since rate of first-order reaction is directly proportional to the concentration of its reactant,
`therefore (r_(1))/(r_(2))=(C_(1))/(C_(2))=(0.04)/(0.03)`
According to first-order reaction `k=(2.303)/(t_(20)-t_(10))" log "(C_(1))/(C_(2))`
On substituting the various values `j=0.0287" m in"^(-1)`
`t_(1//2) =(0.693)/(k) =(0.693)/(0.0287)=24.14` m in
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