Question

Define the following terms:
(a) Pseudo first order reactions
(b) Half -life period of reaction (${t_{{\dfrac{1}{2}}}}$)

Answer 1

Hint: The rate of reaction depends on the concentration of reactants, the rate at which the reactants react to give product. Order of a reaction gives the relation between the rate of reaction and concentration of reactant taking part in the reaction.

Complete step by step answer:
(a) Pseudo first order reaction: Some reactions are first order with respect to two different reactants:
$A + B \to \operatorname{P} $
$Rate = k[A][B]$
However if one of the reactants is present in high concentration like solvent then there is a very little change in
concentration. In other words concentration of the reactant remains practically constant during the reaction.
So the rate law now can be written as:
$Rate = k[A]$
Thus the reaction therefore behaves as first order reaction in A. Such reactions are known as pseudo first order reactions.
For example: Hydrolysis of ester:
$C{H_3}COO{C_2}{H_5} + {H_2}O\overset {Acid} \leftrightarrows C{H_3}COOH + {C_2}{H_5}OH$
Here concentration of water can be taken as constant so the rate of reaction depends on the concentration of ethyl acetate and hence the order of the reaction is one.
Hence, ${\text{Rate}} = k[C{H_3}COO{C_2}{H_5}]$ .
So, we can say that the above reaction appears to be first order but it follows second order kinetics hence called as pseudo first order reaction.
(b) Half- life period of reaction: It is defined as the time during which the concentration of the reactant is reduced to half of its initial concentration. In other words we can say that the time in which half of the reaction is completed. The half-life period of the reaction may be calculated as:
Let us consider the reaction:
$X \to $ Products
The equation for the rate of first order reaction is given as:
$
   \Rightarrow kt = 2.303\log \dfrac{{{{[A]}_0}}}{{[A]}} \\
   \Rightarrow t = \dfrac{{2.303}}{k}\log \dfrac{{{{[A]}_0}}}{{[A]}} \\
   \\
$----------------(1)
Where $[{A_0}]$ is the initial concentration of reactant and $[A]$ is the concentration of reactant that is left behind.
Putting $[A] = \dfrac{{[{A_0}]}}{2},t = {t_{{\dfrac{1}{2}}}}$ we get
 $
   \Rightarrow {t_{{\dfrac{1}{2}}}} = \dfrac{{2.303}}{k}\log \dfrac{{[{A_0}]}}{{[{A_0}/2]}} = \dfrac{{2.303}}{k}\log 2 \\
   \Rightarrow {t_{{\dfrac{1}{2}}}} = \dfrac{{2.303x0.3010}}{k} = \dfrac{{0.693}}{k} \\
   \Rightarrow {t_{{\dfrac{1}{2}}}} = \dfrac{{0.693}}{k} \\
$
Thus half -life period is independent of the initial concentration of the reactant. It is also inversely proportional to the rate constant.

Note: For pseudo first order reactions the molecularity is equal to the two because the reaction is bimolecular which means two reactants are taking part in the reaction but the reaction follows first order kinetics.
From the equation for half-life we can see that half life does not depend on the concentration of radioactive substances.