Question

The rate of a gaseous reaction is given by the expression \(k[A][B]\). If the volume of the reaction vessel is suddenly reduced to \( \frac{1}{4} \) of the initial volume, the reaction rate relative to original rate will be:

• A. 1/10
• B. 1/8
• C. 8
• D. 16

Hint:

If volume decreases by factor \(n\), concentration increases by \(n\). Raise \(n\) to the power of overall order.

Answer 1

The Correct Option is D

Concept: \[ \text{Rate} = k[A][B], \quad [\text{Concentration}] \propto \frac{1}{V} \]

Step 1: Effect of volume change. \[ V \rightarrow \frac{V}{4} \Rightarrow \text{Concentration increases 4 times} \] \[ [A] \rightarrow 4[A], \quad [B] \rightarrow 4[B] \]

Step 2: Apply rate law. \[ \text{New rate} = k(4[A])(4[B]) = 16k[A][B] \]

Step 3: Compare with original rate. \[ \frac{\text{New rate}}{\text{Original rate}} = 16 \]