SATHEE: Chemical Kinetics - Result Question 37

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) $R a t e_{1} = k [A]^{x} [B]^{y}$

$\frac{{Rate}_{1}}{4} = k [A]^{x} [2 B]^{y}$

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 $(2)^{- 2} = 2^{y}$

$y = - 2$ .

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Chemical Kinetics - Result Question 37

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