Question

For the reaction $2{N_2}{O_5} \to 4N{O_2} + {O_2}$ rate at a particular time and rate constant are $1.02 \times {10^{ - 4}}M{s^{ - 1}}$ and $3.4 \times {10^{ - 5}}M{s^{ - 1}}$ respectively. Then the concentration of ${N_2}{O_5}$ at that time will be :

Answer 1

Hint :The rate of a reaction is the rate at which a chemical reaction occurs. It is mostly expressed in terms of the concentration (in terms of amount per unit volume) of a substance produced in a unit of time or the concentration of a reactant absorbed in a unit of time.

Complete Step By Step Answer:
The rate of a reaction can be determined using the rate law. The rate law is an expression that describes the relationship between the rate of the reaction and the concentrations of the reactants involved for a chemical reaction.
If the representation of a reaction is given by:
$aA + bB \to cC + dD$
Here $a$ , $b$ , $c$ and $d$represent the stoichiometric coefficients of the reactants and products . $A$ , $B$ are the reactants and $C$ , $D$ are the products. The rate for the reaction is said to be proportional to the product of the reactants to the partial reaction orders of the respective reactants.
$Rate \propto {\left[ A \right]^x}{\left[ B \right]^y}$
$ \Rightarrow Rate = k{\left[ A \right]^x}{\left[ B \right]^y}$
$x$ and $y$ represent the partial reaction orders for the reactants $A$ and $B$ respectively and $k$ is the rate constant. $\left[ A \right]$ and $\left[ B \right]$ are concentrations of the reactants.
According to the question:
The reaction is $2{N_2}{O_5} \to 4N{O_2} + {O_2}$
We must note that the above reaction has one reactant and the above reaction is a first order reaction ( here $x = 1$ ). So, by the rate law equation ,
 $Rate = k{\left[ {{N_2}{O_5}} \right]^1}$
The rate of the reaction at a particular time $ = $ $Rate$ $ = 1.02 \times {10^{ - 4}}M{s^{ - 1}}$
The rate constant $ = k$ $ = 3.4 \times {10^{ - 5}}M{s^{ - 1}}$
Therefore, the concentration of ${N_2}{O_5}$ at that time is given by $\left[ {{N_2}{O_5}} \right]$ ,
$\left[ {{N_2}{O_5}} \right] = \dfrac{{Rate}}{k}$
$\left[ {{N_2}{O_5}} \right] = \dfrac{{1.02 \times {{10}^{ - 4}}}}{{3.4 \times {{10}^{ - 5}}}}$
$\left[ {{N_2}{O_5}} \right] = 3M$
Thus, the concentration of ${N_2}{O_5}$ at that particular time is $3M$ .

Note :
Molarity ( $M$ ) is one of the most general units used to determine the concentration of a solution. The Molarity denotes the number of moles of solute per litre of solution (moles per litre). The volume of solvent or the weight of solute may be calculated using molarity.