Question

For a first order reaction, the half-life period is independent of:
(A) Initial concentration
(B) Cube root of initial concentration
(C) First power of final concentration
(D) Square root of final concentration

Answer 1

Hint : To solve this question, we must first understand the basic concepts of First order reaction, its half-life and slightly about chemical kinetics. Then we need to assess the relation between concentration of sample and half-life and then only we can conclude the correct answer.

Complete Step By Step Answer:
Before we move forward with the solution of this given question, let us first understand some basic concepts:
First Order Reaction: It can be defined as a chemical reaction in which the reaction rate is linearly dependent on the concentration of only one reactant. In other words, a first-order reaction is a chemical reaction in which the rate varies based on the changes in the concentration of only one of the reactants. Thus, the order of these reactions is equal to 1.
Half-life: the half-life of a chemical reaction which is denoted by $ {t_{1/2}} $ is the time taken for the initial concentration of the reactant(s) to reach half of its original value.
Step 1: In this step, we will derive a relation between conc. and half-life of reaction:
Let, initial concentration be denoted by $ [A] $
And the final concentration be denoted by $ {[A]_ \circ } $
So, at $ t = {t_{1/2}}$
$[A] = [A]{_{\circ}}/2 $
Now, substituting $ A = {[A]_ \circ }/2\,\,,\,\,t = {t_{1/2}}\,\, $ in the equation: $ [A] = {[A]_ \circ }{e^{ - kt}} $
 $ \Rightarrow \,\,\dfrac{{{{[A]}_ \circ }}}{2} = {[A]_ \circ }{e^{ - k{t_{1/2}}}}\,\, \Rightarrow \,\,\dfrac{1}{2} = {e^{ - k{t_{1/2}}}}\, $
Taking Natural Logarithm both sides:
 $
  \ln (\dfrac{1}{2}) = - k{t_{1/2}}\,\, \\
   \Rightarrow \,{t_{1/2}} = \dfrac{{0.693}}{k} \\
  $
So, from the above relation that, the half-life period of a first order reaction is independent of initial concentration.
So, clearly we can conclude that the correct answer is Option A.

Note :
Elementary (single-step) reactions and reaction steps have reaction orders equal to the stoichiometric coefficients for each reactant. The overall reaction order, i.e. the sum of stoichiometric coefficients of reactants, is always equal to the molecularity of the elementary reaction.