Read the passage carefully and answer the questions.

The speed at which a chemical reaction takes place is called the rate of reaction. The rate of reaction depends on various factors like concentration of the reactants, temperature, etc. The relation between the rate of reaction and the concentration of reacting species is represented by the equation $r = k[A]^x[B]^y$, where $x$ and $y$ are the order of the reaction with respect to the reactants A and B, respectively. The overall order of the reaction is $x + y$. The rate of reaction can also be increased by the use of a catalyst which provides an alternate pathway of lower activation energy. It increases the rate of forward and backward reaction to an equal extent. It does not alter the Gibbs energy of the reaction.

The rate of a gaseous reaction is given by $r = k[A][B]$. If the volume of the reaction vessel is suddenly reduced to (1/4)th of its initial value, the reaction rate relating to the initial rate will become

16 times

The correct answer is Option (3) → 16 times

Given rate law for the gaseous reaction:

$r = k[A][B]$

This is first order in A and first order in B, so overall order = 2.

When the volume is reduced to $\frac{1}{4}$​ of its initial value:

• Concentration of each gas increases 4 times (since concentration ∝ 1/volume).

New rate:

$r' = k(4[A])(4[B]) = 16k[A][B]$

So, the reaction rate becomes 16 times the initial rate.