Question

The rate constant for a first order reaction
is 50s-1, How much time it will taken
to
reduce the concentration of
of the reactant
to 1/10th of ites initial value,
​

Answer 1

Given :

The rate constant of the first order reaction = 50 /s

To find :

Time taken by the reaction to reduce the concentration of the reactant to 1/10th of its initial value .

Solution :

The Integral representation of the first order reaction is given by -

ln ( A$_t$ / A$_o$ ) = -kt

where , A$_t$ is the concentration of the reactant at any time t ,

A$_o$ is  the initial concentration ,

k is rate constant ,

and t is time .

now ,

ln [ ( A$_o$ - $\frac{1}{10}$ A$_o$ ) / A$_o$ ] = - ( 50 ) t

=> ln [ ( 9 / 10 ) A$_o$  / A$_o$ ] = - ( 50 ) t

=> ln ( 9 / 10 ) = - ( 50 ) t

=> t = ( 0.105 ) / 50

=> t = 0.0021 s

Time taken by the reaction to reduce the concentration of the reactant to 1/10th of its initial value is 0.0021 seconds .