Question

The rate of reaction between two reactants $A$ and $B$ decreases by a factor of $4$ if the concentration of reactant is doubled . The order of this reaction with respect to reactant $B$ is

Answer 1

Hint: The rate of reaction is defined as the speed at which a chemical reaction will occur. We can also define it as the speed of reaction in which the reactants are converted into products . The rate is equal to the concentration of the reactant in units of time or we can also find it by the concentration of the product which is formed in a unit of time.

Complete step by step answer:
Rate of reaction can be find out by following way: $rate = k{\left[ A \right]^x}{\left[ B \right]^y}$
In the above formula the k is rate constant , A is the concentration of reactant A , B I the concentration of reactant B , x is the order of reaction with respect to A and y is the order of reaction with respect to reactant B.
Order of reaction is the relation between the rate and concentration of the reactant. It is the sum of the exponents of the reactants in a rate expression.
$ \Rightarrow $ $rate{}_1 = k{\left[ A \right]^x}{\left[ B \right]^y}.........\left( 1 \right)$
Now when the reaction rate is decreased by factor 4 and the concentration of the reactant B is doubled.
$ \Rightarrow $ $rat{e_2} = k{\left[ A \right]^x}{\left[ {2B} \right]^Y}$ Here $rat{e_2} = \dfrac{{rat{e_1}}}{4}$
$ \Rightarrow $ $\dfrac{{rat{e_1}}}{4} = k{\left[ A \right]^x}{\left[ {2B} \right]^y}............\left( 2 \right)$
   Let's divide the equations $\left( 1 \right)$ by $\left( 2 \right)$
$\dfrac{{rat{e_1}}}{{\dfrac{{rat{e_1}}}{4}}} = \dfrac{{k{{\left[ A \right]}^x}{{\left[ B \right]}^y}}}{{k{{\left[ A \right]}^x}{{\left[ {2B} \right]}^y}}}$
$ \Rightarrow $ $4 = \dfrac{1}{{{2^y}}}$
$ \Rightarrow $ ${2^2} = \dfrac{1}{{{2^y}}}$
Reciprocal the above equation
$ \Rightarrow {2^y} = \dfrac{1}{{{2^2}}}$
$ \Rightarrow {2^y} = {2^{ - 2}}$
So $y = - 2$


Additional information: Rate of reaction is influenced by pressure, concentration , reaction order , catalyst , isotopes , surface area ,solvent ,electromagnetic radiation , stirring and diffusion limit The rate of reaction depends directly proportional to concentration , which means if the concentration increases the rate of reaction will also increase due to greater number of collisions. Rate of reaction is always positive.
Hence, here the Order of the reaction is $ - 2$ with respect to reactant $B$.

Note: Rate of reaction is $rate = k{\left[ A \right]^x}{\left[ B \right]^y}$ , here $order = x + y$ if asked for complete reaction . Another formula of rate of reaction is $rate = \dfrac{{\Delta C}}{{\Delta T}}$ ,where $\Delta C$ is change of concentration in a time period.