Question

The half-life of a zero order reaction is 30 minutes. What is the concentration of the reactant left after 60 minutes?

Answer 1

Hint : Half-life of a chemical reaction is defined as the time taken for the reactant concentration to reach half of its initial concentration. For a zero order reaction, the mathematical expression that can be used is:
${t_{\dfrac{1}{2}}} = \dfrac{{[{A_0}]}}{{2k}}$
Where,
${t_{\dfrac{1}{2}}}$ $=$ Half-life of the reaction
$[{A_0}] =$ Initial reactant concentration
$k =$ Rate constant of the reaction.

Complete Step By Step Answer:
For zero order reaction the mathematical expression is:
$\begin{gathered} {\text{ }}{t_{\dfrac{1}{2}}} = \dfrac{{[{A_0}]}}{{2k}} \\\ \therefore 30 = \dfrac{{[{A_0}]}}{{2k}} \\\ \therefore k = \dfrac{{[{A_0}]}}{{60}} \\\ \end{gathered}$ (Substituting the half-life, ${t_{1/2}}$ value)
The expression of zero-order rate constant is:
$k = \dfrac{{[{A_0}] - [A]}}{t}$
(Here, $[A]$ is the current reactant concentration)
Substituting the value of $k$ , we get
$\begin{gathered} {\text{ }}[{A_0}] = [{A_0}] - [A] \\\ \therefore [A] = 0 \\\ \end{gathered}$
Therefore, the concentration of the reactant left after $60$ minutes is 0.

Note :
Apart from zero-order reaction, there are two more reactions- first-order reaction and second-order reaction.
The mathematical expression to find the half-life of first-order reaction is:
${t_{\dfrac{1}{2}}} = \dfrac{{\ln 2}}{k} \approx \dfrac{{0.693}}{k}$
From this expression we can clearly see that the half-life of first-order reaction depends on the reaction rate constant, $k$ .
The mathematical expression for second-order reaction is:
${t_{\dfrac{1}{2}}} = \dfrac{1}{{k[{A_0}]}}$
The mathematical expression for second-order shows the half-life of second order reaction on the initial concentration and the rate constant.