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 4.50 M, what was the concentration of A (in M) after 180.0 min?

The concentration of A is 0.2 mol/L.

Explanation:

The chemical equation for your reaction is

                                2A  →  A 2

The integrated rate law for a second-order reaction is

                          1 [ A ] t  =  1 [ A ] 0 +  k t

where

  1 [ A ] t  =  the concentration of  A  at time  t  

  1 [ A ] 0  =  the concentration of  A  at time  t  =  0

(i.e., at the beginning of the reaction)

k m l l  =  the rate constant for the reaction

In this problem,

[ A ] 0  =  4.50 mol/L

k m l  =  0.0265 L⋅mol -1  min - 1

t m l l  =  180.0 min

∴  1 [ A ] t  =  (1  / 4.50 mol/L ) +  0.0265 L⋅mol -1  min - 1  ×  180.0  min

          =  0.2222 L⋅mol -1  +  4.770 L⋅mol -1

          =  4.992 L⋅mol -1

    [ A ] t =  1 /  4.992 L⋅mol -1

           = 0.2 mol/L

The concentration of A is 0.2 mol/L.