
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
475.2k+ views
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.
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.
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