1.

The rate constant for a first order reaction is 60 s^(-1) .How much time will it take to reduce the initial concentration of the reactant to its (1)/(16)^(th) value?

Answer»

SOLUTION :Order of REACTANT =1
rate constant of reaction k=`60s^(-1)`
Suppose ,initial concentration =`[R]_(0) MOL L^(-1)`
Concentration after t time `=(1)/(6)` initial concentration
`therefore[R]_(t)=(1)/(6) mol L^(-1)`
For first order reaction ,
`t=(2.303)/(k)` log `([R]_(0)6)/([R]_(0))`
`=(2.303)/(60 s^(-1))xxlog6(2.303)/(60 s^(-1))xx0.7782`
`=2.9868xx10^(-2)s`
Note:If the concentration is `(1)/(16)` after t time,
`t=(2.303)/(k)` log `([R]_(0)xx16)/([R]_(0))`
`=(2.303)/(k)` log 16 =`(2.303)/(60 s^(-1))xx1.2041`
Note ::Here ,the CALCULATION of `(1)/(16)` is according to english textbook .


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