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

Step-by-step 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:

`(1)/(a)=kt+(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:

`(1)/(a)=0.0265* 180+(1)/(4)`

`a_0=0.2M`

Thus the concentration of A after 180.0 min is 0.2M