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.