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.

Answer 2

1. Understand the effect of volume change on concentration
If the volume of the vessel is reduced to one-half of the initial volume, the concentrations of the reactants will double. This is because concentration is inversely proportional to volume (Concentration = moles/volume).

Let's denote the initial concentrations of A and B as $[A]_1$ and $[B]_1$ respectively, and the initial rate as $R_1$. The new concentrations after the volume is halved are $[A]_2 = 2[A]_1$ and $[B]_2 = 2[B]_1$.

2. Write the initial and final rate expressions
The initial rate expression is given as:

$$ R_1 = k[A]_1[B]_1^2 $$

The new rate expression ($R_2$) after the volume is halved is:

$$ R_2 = k[A]_2[B]_2^2 = k(2[A]_1)(2[B]_1)^2 = k(2[A]_1)(4[B]_1^2) = 8k[A]_1[B]_1^2 $$

3. Determine the ratio of the new rate to the initial rate
Divide the new rate expression by the initial rate expression:

$$ \frac{R_2}{R_1} = \frac{8k[A]_1[B]_1^2}{k[A]_1[B]_1^2} = 8 $$

This means the new rate ($R_2$) is 8 times the initial rate ($R_1$).

Final Answer:
(C) 8