Question

The reaction 2A → A2​​​​​ was experimentally determined to be second order with a rate constant, k, equal to 0.0265 M–1min–1. If the initial concentration of A was 4.00 M, what was the concentration of A (in M) after 180.0 min?

Answer 1

Answer: 0.2 M

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$2A\rightarrow A_2$

$rate=k[A]^2$

Integrated rate law for second order kinetics is given by:

$\frac{1}{a}=kt+\frac{1}{a_0}$

t= time taken for the reaction = 180 min

k = rate constant = $0.0265M^{-1}min^{-1}$

$a_0$ = initial concentration = 4 M

a= concentration left after time t = ?

Putting in the values we get:

$\frac{1}{a}=0.0265\times 180+\frac{1}{4}$

$a_0=0.2M$

Thus the concentration of A after 180.0 min is 0.2M