Chemical Kinetics - Result Question 37
####37. 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:
[2005]
(a) 2
(b) -2
(c) 1
(d) -1
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Solution:
- (b) $Rate_1=k[A]^{x}[B]^{y}$
$\frac{\text{ Rate }_1}{4}=k[A]^{x}[2 B]^{y}$
or Rate $_1=4 k[A]^{x}[2 B]^{y}$
From (1) and (2) we get
$\frac{k[A]^{x}[B]^{y}}{4}=k[A]^{x}[2 B]^{y}$
$\frac{[B]^{y}}{4}=[2 B]^{y}$
or $\frac{1}{4}=(\frac{2 B}{B})^{y} \Rightarrow \frac{1}{4}=2^{y}$ or $(2)^{-2}=2^{y}$
$y=-2$.
