Question #a42c0

1 Answer
Mar 3, 2016

Answer:

After #22.45"s"#

Explanation:

This is a 1st order reaction of the form:

#Ararr"products"#

As the reaction is 1st order the following expression applies:

#[A]_t=[A]_0e^(-kt)#

#[A]_0# is the initial concentration of #A#.

#[A]_t# is the concentration of #A# after time #t#.

#k# is the rate constant.

Taking natural logs of both sides gives:

#ln[A]_t=ln[A]_0-kt#

#[A]_0=100%#

#[A]_t=11%#

#:.ln11=ln100-kt#

#kt=ln100-ln11#

#t=(4.6-2.4)/(0.098)#

#t=22.45"s"#