Question

In the first order reaction A → products, the initial concentration of A is 0.1 108M, and 44 s later, 0.0554M. What is the initial rate of this reaction? (Initial rate-k[A]). (t %-0.693/k)

Answer 1

Answer : The initial rate of the reaction is,
`1.739* 10^(-3)s^(-1)`

Explanation :

First we have to calculate the rate constant of the reaction.

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

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

where,

k = rate constant = ?

t = time taken for the process = 44 s

`[A_o]` = initial amount or concentration of the reactant = 0.1108 M

`[A]` = amount or concentration left time 44 s = 0.0554 M

Now put all the given values in above equation, we get:

`k=(2.303)/(44)\log(0.1108)/(0.0554)`

`k=0.0157`

Now we have to calculate the initial rate of the reaction.

Initial rate = K [A]

At t = 0,
`[A]=[A_o]`

Initial rate = 0.0157 × 0.1108 =
`1.739* 10^(-3)s^(-1)`

Therefore, the initial rate of the reaction is,
`1.739* 10^(-3)s^(-1)`