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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
$-1$
C
1
D
$-2$
Detailed Solution
Let rate $= k[A]^x[B]^y$.
When [B] is doubled, the rate becomes $\dfrac{1}{4}$ of its original value.
$\dfrac{r_2}{r_1} = \dfrac{k[A]^x(2[B])^y}{k[A]^x[B]^y} = 2^y$
$2^y = \dfrac{1}{4} = 2^{-2}$
$\Rightarrow y = -2$
So the order with respect to B is $-2$ (B inhibits the reaction).
When [B] is doubled, the rate becomes $\dfrac{1}{4}$ of its original value.
$\dfrac{r_2}{r_1} = \dfrac{k[A]^x(2[B])^y}{k[A]^x[B]^y} = 2^y$
$2^y = \dfrac{1}{4} = 2^{-2}$
$\Rightarrow y = -2$
So the order with respect to B is $-2$ (B inhibits the reaction).