Question

Which option is valid for zero order reaction?
A.${t_{1/2}} = \dfrac{3}{2}{t_{1/4}}$
B.${t_{1/2}} = \dfrac{4}{3}{t_{1/4}}$
C.\[{t_{1/2}} = 2{t_{1/4}}\] ​
D.\[{t_{1/4}} = {({t_{1/2}})^2}\]

Answer 1

Hint: The half-life of a zero-order reaction is the time needed for half of the reactant(s) to be consumed, which is the same as the half-life involved in nuclear decay, a first-order reaction. For a zero-order reaction, the half-life is given by, \[A = \;{A_0} - kt\]t

Complete step by step answer:
A reaction that has a rate that is independent of the concentration of the reactant(s) is known as zero-order reaction.
The rate law for a zero-order reaction is rate = k, where ‘k’ is the rate constant. In the case of a zero-order reaction, the rate constant ‘k’ has units of concentration/time, like M/s.
The half-life of a reaction describes the time needed for half of the reactant(s) to be depleted, which is the same as the half-life involved in nuclear decay, for a first-order reaction. For a zero-order reaction, the half-life is given by:
For zero order,
\[A = \;{A_0} - kt\]
Where, [A]0 represents the initial concentration and ‘k’ is the zero-order rate constant.
\[{t_{1/2}} = \dfrac{{{A_0}}}{{2K}}\]
${t_{1/4}} = \dfrac{{{A_0}}}{{4K}} = \dfrac{{{t_{1/2}}}}{{{t_{1/4}}\:}} = \dfrac{2}{1}$
∴ $t_{1/2} = 2t_{1/4​}$

Therefore, the correct answer is option (C).

Note: For a zero-order reaction, when the concentration of the reacting species is increased, it won’t speed up the rate of the reaction. Zero-order reactions are typically found when a material that is required for the reaction to proceed, such as a surface or a catalyst, is saturated by the reactants. A reaction is zero-order if concentration data is plotted versus time and the result is a straight line. The Haber process is a well-known process used to manufacture ammonia from hydrogen and nitrogen gas. The reverse of this is known as the reverse Haber process which is an example of a zero-order reaction because its rate is independent of the concentration of ammonia.