The rate constant for a first order reac...

The rate constant for a first order reaction is `60s^(-1)` How much time will it take to reduce the initial concentration of the reactant to its `(1)/(10)` th value?

Text Solution

Verified by Experts

For a first order reaction , the integrated rate equation is
`t=(2.303)/(k)log.([A]_(0))/([A])`
We have to calculate t, when `[A]=([A]_(0))/(10)` . putting the values of [A] and k into the above equation , we get
`t=(2.303)/(60)log.([A]_(0))/(([A])/(10))=0.0457s`

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Value of rate constant of a pseudo first order reaction-

depends on the concentration of reactants present in small amount

depends on the concentration of reactants present in excess

is independent of the concentration of reactants

depends only on temperature

The value of rate constant for a first order reaction is 2.303 xx 10^-2 sec^-1 . What will be the time required to reduce the concentration to 1/(10)th of its initial concentration?

100 s

10 s

2303 s

230.3 s

A first order reaction is 50% completed in 1.26xx10^(14)s. How much time would it take for 100% completion-

`1.26xx10^(15)s`

`2.52xx10^(14)`

`2.52xx10^(28)s`

infinite