Question

rate = k[AB]2 and k = 0.20 L/mol·s. If the initial concentration of AB is 1.50 M, what is [AB] after 14.0 s?

Answer 1

Answer : The concentration of AB after 14.0 s is, 0.29 M

Explanation :

The expression used for second order kinetics is:

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

where,

k = rate constant =
`0.20M^(-1)s^(-1)`

t = time = 14.0 s

`[A_t]` = final concentration = ?

`[A_o]` = initial concentration = 1.50 M

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

`0.20* 14.0=(1)/([A_t])-(1)/(1.50)`

`[A_t]=0.29M`

Therefore, the concentration of AB after 14.0 s is, 0.29 M