Question

Consider an elementary reaction: \[ A(g) + B(g) \rightarrow C(g) + D(g) \] If the volume of the reaction mixture is suddenly reduced to \( \frac{1}{3} \) of its initial volume, the reaction rate will become \( x \) times of the original reaction rate. The value of \( x \) is:

• A. \( \frac{1}{9} \)
• B. 9
• C. 3
• D. \( \frac{1}{3} \)

Hint:

In elementary reactions, the rate depends on the concentration of reactants. When the volume is reduced, the concentration increases, which leads to an increase in the rate, depending on the order of the reaction.

Answer 1

The Correct Option is B

For an elementary reaction, the rate of reaction is proportional to the concentrations of the reactants. Specifically, for a reaction where the stoichiometric coefficients are 1 for both A and B, the rate law can be expressed as: \[ \text{Rate} = k[A][B] \] Here, \( k \) is the rate constant, and \( [A] \) and \( [B] \) are the concentrations of reactants A and B. Now, when the volume of the reaction mixture is reduced to \( \frac{1}{3} \) of its original volume, the concentration of the reactants will increase by a factor of 3, as concentration is inversely proportional to volume. Since the rate is directly proportional to the product of the concentrations of A and B, the reaction rate will increase by: \[ \text{New rate} = k(3[A])(3[B]) = 9 \times (\text{Original rate}) \] Therefore, the reaction rate will become 9 times the original rate. The value of \( x \) is 9.  

Answer 2

Step 1: Given reaction.
The elementary reaction is:
\[ A(g) + B(g) \rightarrow C(g) + D(g) \]
Since the reaction is elementary, the rate law is directly determined from the stoichiometric coefficients:
\[ \text{Rate} = k[A][B]. \]

Step 2: Effect of volume reduction.
When the volume of the reaction mixture is reduced to \( \frac{1}{3} \) of its initial volume, the concentrations of all gaseous species will increase because concentration is inversely proportional to volume:
\[ \text{New concentration} = 3 \times \text{Initial concentration}. \]

Step 3: Calculate new rate.
Let the initial concentrations of \( A \) and \( B \) be \([A]\) and \([B]\).
Initial rate: \[ R_1 = k[A][B]. \] After reduction of volume to one-third, new concentrations are \( 3[A] \) and \( 3[B] \).
New rate: \[ R_2 = k(3[A])(3[B]) = 9k[A][B]. \]

Step 4: Ratio of new rate to old rate.
\[ \frac{R_2}{R_1} = \frac{9k[A][B]}{k[A][B]} = 9. \] Hence, the new rate becomes 9 times the original rate.

Final Answer:
\[ \boxed{x = 9} \]