Question

The decomposition of hi has a rate constant of 1.80 × 10-3 m-1s-1. how much hi remains after 8.50 min if the initial concentration of hi is 4.78 m?

Answer 1

by use second order integrated integrated law
1/(A)t =Kt + 1/(A)o
k= constant
t= time in second= 8.50 x60= 510 sec
( A)o= initial concentration
(A)t =final concentration
1/(A)t =(1.80 x10^-3) (510) +1 /4.78
1(A)t= 1.127 multiply both side by (A)t
1=1.127 (A)t divide both side by 1.127
(A)t= 0.887 M