Question

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

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: $1^{st}$;
$t_{1/2}=24min$; $t_{1/2}$ 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 - $t_{1/2}={\displaystyle \frac{0.693}{k}}$. Initial pressure of $A=400mmHg.$
Final pressure after $48min$ 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 - $t_{1/2}={\displaystyle \frac{o.693}{k}}$
 $k={\displaystyle \frac{0.693}{t_{1/2}}}$
 $A(g)\stackrel{}{\to }2B(g)+C(s)$
      $[A_0]$ $0$ $0$
   $[A_0]$$-a$ $2$$a$ $a$
Initial pressure= $[A_0]$=$400mmHg.$;
Final pressure=$[A_0]-a+2a=[A_0]+a$;
Put the values in equation:
$$\begin{gathered} ([A_0]+a)=[A_0]e^{-kt} \\ (400+a)=400e^{-{\displaystyle \frac{0.693}{24}}\ast 48} \end{gathered}$$
calculated $k={\displaystyle \frac{0.693}{t_{1/2}}}$$\Rightarrow k={\displaystyle \frac{0.693}{24}}$
$a=300$;
The final pressure of the reaction mixture after the expiry of $48min$ will be:
$[A_0]+a$
Put the value of $[A_0]$ and $a$ in the above equation; we get
$700mmofHg$ 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.