Question

For the \[{1^{st}}\]order reaction, $A(g)\xrightarrow{{}}2B(g) + C(s)$, ${t_{1/2}} = 24\,\min $. The reaction is carried out taking a certain mass of $'A'$ enclosed in a vessel in which it exerts a pressure of $400\,mm\,Hg.$The pressure of the reaction mixture after expiry of $48\,\min $ will be:
A. $700\,mm$
B. $600\,mm$
C. $500\,mm$
D. $1000\,mm$

Answer 1

Hint:A first-order reaction is a reaction that proceeds at a rate that depends linearly on only one reactant concentration.The square brackets in the formula are used to express molar concentrations.

Complete step by step answer:
Order of reaction: ${1^{st}}$;
${t_{1/2}} = 24\,\min $; ${t_{1/2}}$ is known as half-life of the reaction. It is the time in which the concentration of a reactant is reduced to one half of its initial concentration.
For \[{1^{st}}\]order reaction - ${t_{1/2}} = \dfrac{{0.693}}{k}$. Initial pressure of $A = 400\,mm\,Hg.$
Final pressure after $48\,\min $ is to be calculated.
Formula used= $[A] = [{A_0}]{e^{ - kt}}$;
Where , $[A]$= Final pressure
$[{A_0}]$= Initial pressure
$k = $Rate constant;
$t = $time
 In \[{1^{st}}\]order reaction - ${t_{1/2}} = \dfrac{{o.693}}{k}$
 $k = \dfrac{{0.693}}{{{t_{1/2}}}}\,$
 $A(g)\xrightarrow{{}}2B(g) + C(s)$
      $[{A_0}]$ $0$ $0$
   $[{A_0}]$$ - a$ $2$$a$ $a$
Initial pressure= $[{A_0}]$=$400\,mm\,Hg.$;
Final pressure=$[{A_0}] - a + 2a = [{A_0}]\, + a$;
Put the values in equation:
$
  ([{A_0}] + a) = [{A_0}]{e^{ - kt}}\, \\
  (400 + a) = 400{e^{ - \dfrac{{0.693}}{{24}}*48}}\, \\
$
calculated $k = \dfrac{{0.693}}{{{t_{1/2}}}}\,$$ \Rightarrow k = \dfrac{{0.693}}{{24}}\,$
$a = 300$;
The final pressure of the reaction mixture after the expiry of $48\,\min $ will be:
$[{A_0}] + a$
Put the value of $[{A_0}]$ and $a$ in the above equation; we get
$700\,mm\,of\,Hg$ because $[{A_0}] = 400$ and $a = 300$;

Note:Rate constant is the proportionality factor in the rate law. It can be determined from rate law or its integrated rate equation. It is unique for each reaction.