Question

A gaseous reaction : ${A_2}\left( g \right) \rightleftarrows B\left( g \right) + \dfrac{1}{2}C\left( g \right)$ shows increase in the pressure from $100mm$ to $120mm$ in $5\min $ . The rate of disappearance of ${A_2}$ is:

Answer 1

Hint:
The rate of reaction is defined as the speed at which the products are formed from the reactants. In order to solve this we need to know the rate law equation first. Rate of disappearance is different for reactants and products.

Complete step by step answer:
The rate of reaction is given by the formula as follows:
$R = \dfrac{P}{t}$
Where, $R = $ rate of reaction, $P = $ change in pressure, $t = $ time
The rate of a reaction is the change in concentration of reactant and product in a given time.
The given statement is expressed in two ways:
The rate of decrease in concentration of any of the reactants or rate of increase in any one of the products.
Time taken in the concentration change.
Thus the rate of disappearance is given by the formula as follows:
$K = \dfrac{{\Delta \left( R \right)}}{{\Delta t}}$
Where, $K = $ rate of disappearance, $\Delta R = $ reactants, $\Delta t = $ time
now using this formula we will find the rate of disappearance of ${A_2}$ in the reaction given below:
${A_2}\left( g \right) \rightleftarrows B\left( g \right) + \dfrac{1}{2}C\left( g \right)$
Initial pressure of the reactant ${A_2}$ will be $'P'$ and that of products will be zero.
Pressure at equilibrium of the reactant will be $'P - x'$ and that of product $B$ and $\dfrac{1}{2}C$ will be $'x'$ and $'\dfrac{x}{2}'$ respectively.
The final pressure will be given as:
 ${P_t} = P - x + x + \dfrac{x}{2}$
Therefore, ${P_t} = P + \dfrac{x}{2}$
Where, ${P_t} = $ final pressure, $P = $ Initial pressure, $\dfrac{x}{2} = $ pressure at equilibrium.
Given data:
Initial pressure $ = 100mm$
Final pressure $ = 120mm$
Formula to be used: ${P_t} = P + \dfrac{x}{2}$
Now substituting the values of initial and final pressure in the above formula we get,
$120 = 100 + \dfrac{x}{2}$
$120 - 100 = \dfrac{x}{2}$
$20 = \dfrac{x}{2}$
$x = 40$
Therefore, $x = \Delta {A_2}$
Now using the value of $x = 40$ and $\Delta t = 5\min $ we will find the rate of disappearance of ${A_2}$.
$K = \dfrac{{\Delta \left( R \right)}}{{\Delta t}}$
$K = \dfrac{{\Delta \left( {{A_2}} \right)}}{{\Delta t}}$
Substituting the value we get,
$K = \dfrac{{40}}{5}$
$K = 8mm/\min $

Note: There are various factors that affect the rate of the reaction. It completely depends on the nature of the reaction. The other factors that affect the rate of a chemical reaction are: number of reactants.