For the following reaction:

\(2A_2 (g) + \frac{1}{4}X (g) \longrightarrow 2A_2X(g)\)

volume is increased to double its value by decreasing the pressure on it. If the reaction is first order with respect to \(X\) and second with respect to \(A_2\) the rate of reaction will:

Decrease by eight times of its initial value

The correct answer is option 1. Decrease by eight times of its initial value.

To analyze the effect of changing the volume (and consequently the pressure) on the rate of the reaction, we need to understand how the rate of reaction is affected by changes in concentration or pressure.

Given that the reaction is first order with respect to \(X\) and second order with respect to \(A_2\), we can write the rate equation for the reaction as:

\(\text{Rate} = k[A_2]^2[X]\)

Where:

\( k \) is the rate constant.

\( [A_2] \) is the concentration of \( A_2 \).

\( [X] \) is the concentration of \( X \).

Now, let's consider what happens when the volume is increased, leading to a decrease in pressure:

Decreasing the pressure will result in the expansion of the gas volume, which will lead to a decrease in the concentration of all the reactants.

Since the reaction is first order with respect to \(X\) and second order with respect to \(A_2\), the rate of the reaction is determined by the concentration of \(X\) and \(A_2\).

When the volume is doubled, the concentration of \(X\) and \(A_2\) each decrease by a factor of 2.

Using the rate equation, the overall effect on the rate of the reaction can be calculated:

\(\text{Rate} = k \left( \frac{[A_2]}{2} \right)^2 \left( \frac{[X]}{2} \right)\)

\(= k \left( \frac{[A_2]^2}{4} \right) \left( \frac{[X]}{2} \right) \)

\( = \frac{k}{8} [A_2]^2[X] \)

\( = \frac{1}{8} \times \text{Initial Rate} \)

Therefore, the rate of reaction will decrease to half of its initial value.

So, the correct answer is 1. Decrease by eight times of its initial value