Question

For the reaction 2A₂(g) + ¼X(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:

• A. Decrease by eight times of its value
• B. Increase by eight times of its value
• C. Increase by four times of its value
• D. Remain unchanged

Answer 1

The Correct Option is A

For the reaction given by \(2A_2(g) + \frac{1}{4}X(g) \rightarrow 2A_2X(g)\), the rate of reaction \(R\) is determined by the rate law: \(R = k[A_2]^2[X]\). Here, \(k\) is the rate constant, \([A_2]\) is the concentration of \(A_2\), and \([X]\) is the concentration of \(X\). 

If the volume is doubled, according to the ideal gas law (\(PV=nRT\)), the pressure is reduced by half. Because concentrations are directly proportional to pressure in gases, each concentration is halved: \([A_2]_{new} = \frac{1}{2}[A_2]_{original}\) and \([X]_{new} = \frac{1}{2}[X]_{original}\).

Substituting these into the rate equation:

\(R_{new} = k\left(\frac{1}{2}[A_2]_{original}\right)^2\left(\frac{1}{2}[X]_{original}\right)\)

\( = k\left(\frac{1}{4}[A_2]_{original}^2\right)\left(\frac{1}{2}[X]_{original}\right)\)

\( = \frac{1}{8}k[A_2]_{original}^2[X]_{original} = \frac{1}{8}R_{original}\)

Thus, the reaction rate decreases to \(\frac{1}{8}\) of its original value, meaning the rate has decreased by eight times.