Question

A certain first-order reaction is 58% complete in 95 s. What are the values of the rate constant and the half-life for this process?

Answer 1

`0.005734 s^(-1)` and 20.86 seconds are the values of the rate constant and the half-life for this process respectively..

Step-by-step explanation:

Expression for rate law for first order kinetics is given by:

`a_o=a* e^(-kt)`

where,

k = rate constant

t = age of sample

`a_o` = let initial amount of the reactant

a = amount left after decay process

We have :

`a_o=x`

`a=58\%* x=0.58x`

t = 95 s

`0.58x=x* e^(-k* 95 s)`

`\k= 0.005734 s^(-1)`

Half life is given by for first order kinetics::

`t_(1/2)=(0.693)/(k)`

`=(0.693)/(0.005734 s^(-1))=120.86 s`

`0.005734 s^(-1)` and 20.86 seconds are the values of the rate constant and the half-life for this process respectively..