The half-life for a zero-order chemical reaction is X/T minutes. How much time will be required by the reaction for the concentration of the reactant to be decreased to 1/3rd of the initial concentration?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 4X/3T minutes
For a zero order reaction the rate does not depend on concentration, so the reactant is used up at a constant rate. The integrated rate law and half-life are:
\( \displaystyle [R]=[R]_0-kt\quad\Rightarrow\quad t=\frac{[R]_0-[R]}{k} \)
\( \displaystyle t_{1/2}=\frac{[R]_0}{2k} \)
From the given half-life:
\( \displaystyle \frac{[R]_0}{2k}=\frac{X}{T}\quad\Rightarrow\quad k=\frac{[R]_0\,T}{2X} \)
The concentration falls to one third, so [R] = [R]0/3:
\( \displaystyle t=\frac{[R]_0-\frac{[R]_0}{3}}{k}=\frac{\frac{2}{3}[R]_0}{k} \)
\( \displaystyle t=\frac{2}{3}[R]_0\times\frac{2X}{[R]_0\,T} \)
\( \displaystyle t=\frac{4X}{3T}\ \text{minutes} \)
So the time required is 4X/3T minutes.
Shortcut: In a zero order reaction the time is proportional to the amount reacted. Half reacts in X/T; two thirds react in (2/3)/(1/2) = 4/3 of that time, i.e. 4X/3T.