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 $2.2\times 10^2s$

Explanation :

The expression used for second order kinetics is:

$kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}$

where,

k = rate constant = $9.4\times 10^{-3}M^{-1}s^{-1}$

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:

$(9.4\times 10^{-3})\times t=\frac{1}{0.40}-\frac{1}{2.16}$

$t=216.706s\aprrox 2.2\times 10^2s$

Therefore, the time taken by the reaction is $2.2\times 10^2s$