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:

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

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:

$0.20\times 14.0=\frac{1}{[A_t]}-\frac{1}{1.50}$

$[A_t]=0.29M$

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