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.

Q) 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

The correct option is 16 times.

To understand why this is correct, we can analyze the relationship between the volume of the reaction vessel and the concentrations of the reactants.

For a gaseous reaction, the concentration of any reactant (represented by molarity) is defined as the number of moles of that substance divided by the volume of the vessel:

$$\text{Concentration}=\frac{n}{V}$$

Here, $n$ represents the number of moles and $V$ represents the volume of the vessel. This means that concentration is inversely proportional to the volume. If the volume of the reaction vessel is reduced to $\frac{1}{4}$ of its initial value ($V^{\prime }=\frac{V}{4}$), the concentration of each reactant will increase by a factor of 4:

$${\left[A\right]}^{\prime }=4\left[A\right]$$

$${\left[B\right]}^{\prime }=4\left[B\right]$$

Let the initial rate of the reaction be $r_1$, which is given by:

$$r_1=k\left[A\right]\left[B\right]$$

When the volume is reduced, the new rate $r_2$ becomes:

$$r_2=k{\left[A\right]}^{\prime }{\left[B\right]}^{\prime }$$

Substitute the new concentration values into the rate equation:

$$r_2=k\left(4\right[A\left]\right)\left(4\right[B\left]\right)$$

$$r_2=16\cdot k\left[A\right]\left[B\right]$$

$$r_2=16r_1$$

Therefore, the new reaction rate is 16 times the initial rate.