Solved: CHEMISTRY Question for NEET

NEET CHEMISTRYMedium

For the reaction $A + B \rightarrow \text{products}$ , 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:

$\text{Rate} = k[A]^2[B]$

$\text{Rate} = k[A][B]^2$

$\text{Rate} = k[A]^2[B]^2$

$\text{Rate} = k[A][B]$

Step-by-Step Solution

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

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