Question

For the first order reaction $ {{\text{A}}_{\left( {\text{g}} \right)}} \to {\text{2}}{{\text{B}}_{\left( {\text{g}} \right)}}{\text{ + }}{{\text{C}}_{\left( {\text{g}} \right)}} $ , the initial pressure is $ {{\text{P}}_{{\text{A}}}} $ =90 mm Hg. Then pressure after 10 minutes is found to be 180 mm Hg. The half-life period of the reaction is:
(A) ${\text{1}}{\text{.15 x 1}}{{\text{0}}^{{\text{ - 3}}}}{\sec ^{ - 1}}$
(B) 600 sec
(C) $ {\text{3}}{\text{.45 x 1}}{{\text{0}}^{{\text{ - 3}}}}{\sec ^{ - 1}}$
(D) 200 sec

Answer 1

Hint: The first order reactions are reactions in which rate of the reaction is directly proportional to the concentration of the reactants. The half life period of a first order reaction is independent of the concentration of the reactants.

Complete step by step solution:
The half-life period of a reaction is the time taken for the concentration of the reactant to reach half of its initial concentration. It is denoted as $ {{\text{t}}_{{\text{1/2}}}} $ and expressed in the unit of time generally in seconds(sec).
For a first order reaction, the half-life of a reaction is given by the following expression,
 $ {{\text{t}}_{{\text{1/2}}}}{\text{ = }}\dfrac{{0.693}}{{\text{k}}} $
where k is the rate constant of the reaction which is expressed as $ {{\text{M}}^{\left( {{\text{1 - n}}} \right)}}{{\text{s}}^{{\text{ - 1}}}} $ (n = order of the reaction, M is the unit of $ {\text{mol }}{{\text{L}}^{{\text{ - 1}}}} $ ).
For the given first order reaction,
$ {{\text{A}}_{\left( {\text{g}} \right)}} \to {\text{2}}{{\text{B}}_{\left( {\text{g}} \right)}}{\text{ + }}{{\text{C}}_{\left( {\text{g}} \right)}} $

Initially, at time t =0	p	0	0
At equilibrium	p-2x	2x	x

Total pressure = $ {\text{P - x + 2x + x = P + 2x}} $
It is given that initial pressure = 90 mm Hg.
And after 10 minutes, Pressure = 180 mm Hg.
${\text{P + 2x = 180 }} $
$\Rightarrow 90 + {\text{ 2x = 180}} $
$\Rightarrow {\text{2x = }}90 $
$\Rightarrow {\text{x = }}45 $
The rate constant of the first order reaction is given by the following expression:
${\text{k = }}\dfrac{{{\text{2}}{\text{.303}}}}{{\text{t}}}{\text{log}}\dfrac{{\text{P}}}{{{\text{P - x}}}}$
$\Rightarrow {\text{k = }}\dfrac{{{\text{2}}{\text{.303}}}}{{600}}{\text{log}}\dfrac{{90}}{{90 - 45}}$
$\Rightarrow {\text{k = }}\dfrac{{{\text{2}}{\text{.303}}}}{{600}}{\text{log(2)}} $
$\Rightarrow {\text{k = }}1.156{\text{ x 1}}{{\text{0}}^{{\text{ - 3}}}}{\text{ se}}{{\text{c}}^{{\text{ - 1}}}} $

Now substituting the k value in the half-life formula, we get,
  $
  {{\text{t}}_{{\text{1/2}}}}{\text{ = }}\dfrac{{0.693}}{{\text{k}}} \\
   \Rightarrow {{\text{t}}_{{\text{1/2}}}}{\text{ = }}\dfrac{{0.693}}{{1.156{\text{ x 1}}{{\text{0}}^{{\text{ - 3}}}}{\text{ }}}} \\
   \Rightarrow {{\text{t}}_{{\text{1/2}}}}{\text{ = 600 sec}} \\
 $
The half-life period of the reaction is 600 seconds. So, the correct option is B.

Note:
Half life of a reaction varies depending upon the order of reaction and therefore may or may not depend upon the concentration.
Here is the formula of half life for common order of reactions:
Zero order reaction
 $ {{\text{t}}_{{\text{1/2}}}}{\text{ = }}\dfrac{{{\text{[}}{{\text{R}}_{\text{0}}}{\text{]}}}}{{{\text{2k}}}} $
First order reaction
 $ {{\text{t}}_{{\text{1/2}}}}{\text{ = }}\dfrac{{0.693}}{{\text{k}}} $
Second order reaction
 $ {{\text{t}}_{{\text{1/2}}}}{\text{ = }}\dfrac{1}{{{\text{[}}{{\text{R}}_{\text{0}}}{\text{]k}}}} $