For the following reaction:

2A₂(g) + 1/4X(g) → 2A₂X(g)

If the 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 order with respect to A₂, the rate of reaction will:

The correct option is: Decrease by eight times of its initial value

Let's understand why this is the correct answer step-by-step.

Step 1: Write the rate law expression for the reaction.
The reaction is given as:
2A₂(g) + 1/4X(g) → 2A₂X(g)
We are given that the reaction is first order with respect to X and second order with respect to A₂. Therefore, the rate law for the reaction is:

$$\mathrm{Rate}=k{\left[A_2\right]}^2{\left[X\right]}^1$$

where:
- k is the rate constant,
- [A₂] is the concentration of A₂,
- [X] is the concentration of X.

Step 2: Understand the effect of doubling the volume.
Concentration is defined as the number of moles (n) divided by the volume (V):

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

When the volume (V) is doubled to 2V, the concentration of each gaseous reactant becomes half of its initial concentration because concentration is inversely proportional to volume. Thus:
- New concentration of A₂, [A₂]' = [A₂]/2
- New concentration of X, [X]' = [X]/2

Step 3: Calculate the new rate of the reaction.
Substitute the new concentrations into the rate law equation to find the new rate (Rate'):

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

Simplifying the mathematical expression:

$${\mathrm{Rate}}^{\prime }=k\cdot \frac{{\left[A_2\right]}^2}{4}$$

$${\mathrm{Rate}}^{\prime }=\frac{1}{8}\cdot \left(k{\left[A_2\right]}^2\right[X\left]\right)$$

$${\mathrm{Rate}}^{\prime }=\frac{1}{8}\cdot \mathrm{Rate}$$

Therefore, the new rate of the reaction is 1/8th of its initial value, which means the rate of the reaction will decrease by eight times.