For the following reaction :
\( \ce{2A2 (g)} + \frac{1}{4}\ce{X (g) -> 2A2X (g)} \)
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 order with respect to A2, the rate of reaction will :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Decrease by eight times of its initial value
The rate of a reaction depends on the concentrations of the reactants, each raised to its order. With first order in X and second order in \( \ce{A2} \):
\( \displaystyle \text{Rate}=k[\ce{A2}]^{2}[\ce{X}] \)
For a gas, concentration is the number of moles per unit volume. When the volume is doubled (by lowering the pressure), the amounts stay the same, so every concentration becomes half:
\( \displaystyle \text{Rate}'=k\left(\frac{[\ce{A2}]}{2}\right)^{2}\left(\frac{[\ce{X}]}{2}\right)=\frac{1}{8}\,k[\ce{A2}]^{2}[\ce{X}]=\frac{\text{Rate}}{8} \)
So the rate decreases to one-eighth of its initial value.
Shortcut: The total order is 2 + 1 = 3, and halving every concentration multiplies the rate by (1/2)3 = 1/8.
Note: It seems that there is a mistake in the original NTA paper, which prints the equation as 2A2(g) + ¼X(g) → 2A2X(g); this equation is not balanced.