Question

For the second-order reaction below, the rate constant of the reaction is 9.4 x 10^-3 M^-1 s^-1. How long (in sec) is required to decrease the concentration of A from 2.16 M to 0.40M?

2A --> B
rate = K[A]^2

Answer 1

Answer : The time taken by the reaction is
`2.2* 10^2s`

Explanation :

The expression used for second order kinetics is:

`kt=(1)/([A_t])-(1)/([A_o])`

where,

k = rate constant =
`9.4* 10^(-3)M^(-1)s^(-1)`

t = time = ?

`[A_t]` = final concentration = 0.40 M

`[A_o]` = initial concentration = 2.16 M

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

`(9.4* 10^(-3))* t=(1)/(0.40)-(1)/(2.16)`

`t=216.706s\aprrox 2.2* 10^2s`

Therefore, the time taken by the reaction is
`2.2* 10^2s`