Question

The relationship between rate constant and half-life period of zero order reaction is given by:
\[
  A.{\text{ }}{t_{\dfrac{1}{2}}} = \left[ {{A_0}} \right]2k \\
  B.{\text{ }}{t_{\dfrac{1}{2}}} = \dfrac{{0.693}}{k} \\
  C.{\text{ }}{t_{\dfrac{1}{2}}} = \dfrac{{\left[ {{A_0}} \right]}}{{2k}} \\
  D.{\text{ }}{t_{\dfrac{1}{2}}} = \dfrac{{2\left[ {{A_0}} \right]}}{{2k}} \\
 \]

Answer 1

Hint: In order to solve the given problem we will first see the meaning of the zero order reaction and the characteristics. Further in order to find the relation between the rate constant of the zero order reaction and the half life of zero order reaction we will use the general rate expression for the reaction and substitute the criterion for zero order further we will solve the rate equation by integrating it and we will find the general equation for zero order and further we will put the constant and half life to finally get the relation between the two.

Complete step by step answer:
A zeroth order reaction has a rate which is independent of the reactant(s) concentration(s). The intensity is evidently independent of the concentration of reactants. With rising or declining concentrations of reactants, the frequencies of these zero-order reactions do not differ. It implies that the rate of the response is equal to the rate constant, k, of that response. Zero-order reactions are usually encountered where the reactants saturate a substrate that is required for the reaction to occur, such as a surface or a catalyst.

Let us consider the general zero order reaction.
$A \to {\text{product}}$
As we know that the rate expression of any reaction is:
${\text{Rate}} = - \dfrac{{d\left[ A \right]}}{{dt}} = K{\left[ A \right]^n}$
Where “n” is the order of reaction and “K” is the rate constant.
As we have n=0 for zero order reaction, so the rate expression changes to:
${\text{Rate}} = - \dfrac{{d\left[ A \right]}}{{dt}} = K{\left[ A \right]^0} = K$
Now let us arrange the above expression for further calculation.
\[
   \Rightarrow - \dfrac{{d\left[ A \right]}}{{dt}} = K \\
   \Rightarrow d\left[ A \right] = - K \cdot dt - - - - - (1) \\
 \]
Let us integrate equation (1) taking suitable limits at T=0 and T=“t”
The initial concentration of the reactant is \[{\left[ A \right]_0}\] at T=0 and the concentration becomes \[\left[ A \right]\] at T=“t”.
\[ \Rightarrow \int\limits_{{{\left[ A \right]}_0}}^{\left[ A \right]} {d\left[ A \right]} = \int\limits_0^t { - K \cdot dt} \]
Let us now integrate with the given limits to get the equation.
\[
   \Rightarrow {\left[ A \right]_0} - \left[ A \right] = - Kt \\
   \Rightarrow \left[ A \right] = {\left[ A \right]_0} - Kt - - - - (2) \\
 \]
Here the equation (2) is the general equation for the second order equation.
Now let us find the relation.
As we know that the half life of the reaction is the time period in which the half of the reactants is consumed in the reaction and half is left.
Let \[{t_{\dfrac{1}{2}}}\] is the half life of the reaction.
Initially the amount of reactant was \[{\left[ A \right]_0}\] . So, after one half life the amount of reactant left is \[\dfrac{{{{\left[ A \right]}_0}}}{2}\] at time \[{t_{\dfrac{1}{2}}}\] .
So, let us substitute these values in equation (2).
\[
  \because \left[ A \right] = {\left[ A \right]_0} - Kt \\
   \Rightarrow \dfrac{{{{\left[ A \right]}_0}}}{2} = {\left[ A \right]_0} - K{t_{\dfrac{1}{2}}} \\
 \]

Let us now solve this equation to find the relation.
\[
   \Rightarrow K{t_{\dfrac{1}{2}}} = {\left[ A \right]_0} - \dfrac{{{{\left[ A \right]}_0}}}{2} \\
   \Rightarrow K{t_{\dfrac{1}{2}}} = \dfrac{{{{\left[ A \right]}_0}}}{2} \\
   \Rightarrow {t_{\dfrac{1}{2}}} = \dfrac{{{{\left[ A \right]}_0}}}{{2K}} \\
 \]
Hence, the relationship between rate constant and half-life period of zero order reaction is given by \[{t_{\dfrac{1}{2}}} = \dfrac{{{{\left[ A \right]}_0}}}{{2K}}\] .
So, the correct answer is “Option C”.

Note: In order to solve these types of problems students must remember the general rate expression as well as the unique characteristics of each type of reaction. Students must try to solve the problem by the basic method but can also solve by directly substituting the final equation for the reaction. The final equation found is very important and has significant use while solving numerical problems.