1.

Half life period of a first order reaction is 10min. Starting with initial concentration 12 M. The rate after 20 min is

Answer»

`0.0693 xx3 MIN^(-1)`
`0.0693 M min^(-1)`
`0.0693 xx4 "M min"^(-1)`
`0.693xx3 "M min"^(-1)`

Solution :(A) `K=(0.693)/(t_(1//2))= (0.693)/(10)`
=0.0693 `min^(-1)`
`t= (2.303)/(k) "LOG" ([R]_(0))/([R])`
`20 = (2.303)/(0.0693)"log" (12)/([R])`
or `"log"(12)/([R]) = 0.601`
`(12)/([R])` = Antilog (0.601)=4 .
or `[R] =(12)/(4) =3`
RATE after 20 min
Rate =`k[R]^(1)`
`=0.0693xx3` M `min^(-1)`


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