Question

The rate of a gaseous reaction is given by the expression k[A][B]². If the volume of vessel is reduced to one half of the initial volume, the reaction rate as compared to original rate is

• A. \(\frac{1}{16}\)
• B. \(\frac{1}{8}\)
• C. 8
• D. 16

Answer 1

The Correct Option is C

We are given the rate expression for a gaseous reaction:
\[ \text{Rate} = k[A][B]^2 \]
When the volume of the vessel is reduced to half, the concentration of gases doubles. Therefore, the new concentrations of \( A \) and \( B \) are \( 2[A]_0 \) and \( 2[B]_0 \), respectively.
Initial rate: \[ \text{Rate}_0 = k[A]_0[B]_0^2 \]
New rate: \[ \text{Rate}_\text{new} = k(2[A]_0)(2[B]_0)^2 = 8k[A]_0[B]_0^2 \]
Therefore, the new rate is 8 times the original rate.

The correct answer is (C) : 8.