For the reaction $A + B ightarrow ext{products}$, it is observed that:
(i) On doubling the initial concent

Question

For the reaction $A + B ightarrow 	ext{products}$, it is observed that:
(i) On doubling the initial concentration of A only, the rate of reaction is also doubled.
(ii) On doubling the initial concentrations of both A and B, there is a change by a factor of 8 in the rate of the reaction.

The rate of this reaction is given by:

Accepted Answer

Let the rate law be $	ext{Rate} = k[A]^x[B]^y$.
When the initial concentration of A is doubled while keeping B constant, the rate doubles:
$\frac{	ext{Rate}_2}{	ext{Rate}_1} = \frac{k(2[A])^x[B]^y}{k[A]^x[B]^y} = 2^x = 2 \implies x = 1$
When the initial concentrations of both A and B are doubled, the rate increases by a factor of 8:
$\frac{	ext{Rate}_3}{	ext{Rate}_1} = \frac{k(2[A])^x(2[B])^y}{k[A]^x[B]^y} = 2^x \cdot 2^y = 8$
Substituting $x = 1$, we get:
$2^1 \cdot 2^y = 8 \implies 2 \cdot 2^y = 8 \implies 2^y = 4 \implies y = 2$
Therefore, the overall rate law for the reaction is $	ext{Rate} = k[A][B]^2$.

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