Question

g The decomposition reaction of A to B is a first-order reaction with a half-life of 2.42×103 seconds: A → 2B If the initial concentration of A is 0.163 M, how many minutes will it take for the concentration of A to be 66.8% of the initial concentration?

Answer 1

In 23.49 minutes the concentration of A to be 66.8% of the initial concentration.

Step-by-step explanation:

The equation used to calculate the constant for first order kinetics:

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

Rate law expression for first order kinetics is given by the equation:

`t=(2.303)/(k)\log([A_o])/([A])` ......(2)

where,

k = rate constant

`t_(1/2)` =Half life of the reaction =
`2.42* 10^3 s`

t = time taken for decay process = ?

`[A_o]` = initial amount of the reactant = 0.163 M

[A] = amount left after time t = 66.8% of
`[A_o]`

[A]=
`(66.8)/(100)* 0.163 M=0.108884 M`

`k=(0.693)/(2.42* 10^3 s)`

`t=(2.303)/((0.693)/(2.42* 10^3 s))\log(0.163 M)/(0.108884 M)`

t = 1,409.19 s

1 minute = 60 sec

`t=(1,409.19 )/(60) min=23.49 min`

In 23.49 minutes the concentration of A to be 66.8% of the initial concentration.