1.

The time taken for 10% completion of a first order reaction is 20 mins. Then for 19% completion, the reaction will take

Answer»

40 mins
60 mins
30 mins
50 mins

Solution :`t = (2*303)/(k) LOG (a)/(a-x) therefore 20 = (2*303)/(k) log ""(a)/(a-0*10a)`
or `20 = (2*303)/(k)log""(100)/(90)`
In 2ND case, `t= (2*303)/(k)log""(a)/(a-0*19a)`
`= (2*303)/(k) log ""(100)/(81)=(2*303)/(k)log((10)/(9))^(2)""…(ii)`
Eqn. (ii) Eqn. (i) gives `(t)/(20) =2` or `t = 40 min.`


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