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

Answer 1

The concentration of A is 0.2 mol/L.

Step-by-step 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.