A second order reaction is 40% completed after one hour. What is the rate constant for this reaction?

1 Answer
Dec 19, 2015

Answer:

#1.1xx10^-2# #"M"^-1# #"min"^-1#

Explanation:

The rate law for a second-order reaction is

#1/([A])=1/([A]_0)+kt#

We can say that the current concentration #[A]# is #0.6# and the original concentration #[A]_0# is #1#. This is true because the reaction is #40%# complete, which means that #60%# still has to be reacted, and #60%# of an original concentration #1# is #0.6#.

#t=60# since #60# minutes have elapsed.

#1/(0.6)=1/1+k(60)#

#1.1xx10^-2=k#

The rate constant is #1.1xx10^-2# #"M"^-1# #"min"^-1#.