The time taken for a certain amount of completion in a first order reaction can be found using the relationship between the concentration of reactant remaining and the time. The formula for a first order reaction is:
ln()=ktwhere:
[A]0 is the initial concentration of the reactant,
[A] is the concentration of the reactant at time
t,
k is the rate constant,
t is the time.
For a first order reaction, the half-life
(t1∕2) is related to the rate constant by the relationship:
t1∕2= At
90% completion,
10% of the reactant remains, so
[A]=0.1[A]0. Plugging into the equation, we have:
ln()=ktln10=ktWe know that
t1∕2=,
s0 :
k=Therefore, substituting for
k :
ln10=()tt1∕2⋅ln10=tln2t=t1/2Since
ln10≈2.302585 and
ln2≈0.693147, the equation simplifies to:
t≈t1∕2t≈3.32t1/2Hence, the time taken for the reaction to proceed to
90% completion is approximately 3.32 times the half-life of the reaction. The correct answer is Option D.