Question

2A → B + C

The above reaction is run and found to follow second order kinetics with a rate constant of 1.30 x 10-3 M-1sec-1. If the initial concentration of A is 1.68 M, what is the concentration after 195 seconds

Answer 1

Answer : The concentration after 195 seconds is, 1.18 M

Explanation :

The expression used for second order kinetics is:

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

where,

k = rate constant =
`1.30* 10^(-3)M^(-1)s^(-1)`

t = time = 195 s

`[A_t]` = final concentration = ?

`[A_o]` = initial concentration = 1.68 M

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

`1.30* 10^(-3)* 195=(1)/([A_t])-(1)/(1.68)`

`[A_t]=1.18M`

Therefore, the concentration after 195 seconds is, 1.18 M