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:
$nA + yB \rightarrow$ Product
Let the rate of the reaction:-
Rate $_{1}= k [ A ]^{ n }[ B ]^{ y } \rightarrow a$
When the conc. of $B$ is doubled;
Rate $_{2}=\frac{\text { Rate }_{1}}{4}= k [ A ]^{ n }[2 B ]^{ y } \rightarrow b$
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 $-2$