Question

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

• A. 2
• B. -2
• C. 1
• D. -1

Answer 1

Answer: (B)

Let us assume the chemical reaction:
$nA+yB\to$ Product
Let the rate of the reaction:-
Rate ${}_1=k[A{]}^n[B{]}^y\to a$
When the conc. of $B$ is doubled;
Rate ${}_2=\frac{{\text{ Rate }}_1}{4}=k[A{]}^n[2B{]}^y\to b$
Now divide equation a by $b$:-
$4=\frac{1}{2^y}$
$\Rightarrow 2^2=\frac{1}{2^y}$
or $\Rightarrow 2^y=\frac{1}{2^2}$
$\Rightarrow y=-2$ with respect to $B$
Hence the correct answer is option $A=-2$
i.e., the order of the reaction with respect to reactant $B$ is $-2$