Question

Following reaction takes place in one step :
\( 2A + B \rightarrow 2C \)
How will the rate of above reaction change if the volume of the reaction vessel is decreased to one third of its original volume? Will there be any change in the order of reaction with the reduced volume?

Hint:

For gas-phase reactions, remember: Decrease in volume \(\rightarrow\) Increase in Pressure \(\rightarrow\) Increase in Concentration \(\rightarrow\) Increase in Rate.

Answer 1

Step 1: Understanding the Concept:
Since the reaction occurs in one step (elementary reaction), the rate law can be written directly from its stoichiometry. Concentration is inversely proportional to volume.
Step 2: Key Formula or Approach:
1. Rate Law: \( \text{Rate} = k[A]^2[B]^1 \).
2. Concentration (\( C \)) = \( \frac{n}{V} \).
Step 2: Detailed Explanation:
Let initial volume be \( V \). Let initial concentrations be \( [A] \) and \( [B] \).
Initial Rate \( R_1 = k[A]^2[B] \).
If volume \( V' = \frac{1}{3}V \), then the new concentrations become:
\( [A]' = \frac{n_A}{V/3} = 3[A] \) and \( [B]' = 3[B] \).
New Rate \( R_2 = k[3A]^2[3B] = k \cdot 9[A]^2 \cdot 3[B] = 27 \cdot k[A]^2[B] \).
\( R_2 = 27 R_1 \).
The order of a reaction is a characteristic of the reaction mechanism and the nature of reactants. It does not change with changes in physical parameters like volume or pressure.
Step 3: Final Answer:
The rate increases 27 times. There is no change in the order of the reaction.