For a reaction between gaseous compounds, `2A + B rarrC + D` , the reaction rate law is rate k[A][B]. If the volume of the container is made `1//4^(th)` of the initial, then what will be the rate of reaction as compared to the initial rate?

Options

A. 16 times
B. 4 times
C. `(1)/(8)` times
D. `(1)/(16)` times

Explanation

To solve the problem, we need to analyze how the change in volume affects the concentrations of the reactants and subsequently the rate of the reaction.

**Step 1: Write the initial rate expression.**
The rate of the reaction is given by the rate law:
\[ \text{Rate} = k[A][B] \]
Where \( k \) is the rate constant, and \( [A] \) and \( [B] \) are the concentrations of reactants A and B, respectively.

**Step 2: Define the initial concentrations.**
Assume the initial volume of the container is \( V \). The initial concentrations of A and B can be expressed as:
\[ [A]_{\text{initial}} = \frac{n_A}{V} \]
\[ [B]_{\text{initial}} = \frac{n_B}{V} \]
Where \( n_A \) and \( n_B \) are the number of moles of A and B.

**Step 3: Calculate the initial rate of reaction.**
Substituting the initial concentrations into the rate expression:
\[ \text{Rate}_{\text{initial}} = k \left(\frac{n_A}{V}\right) \left(\frac{n_B}{V}\right) = k \frac{n_A n_B}{V^2} \]

**Step 4: Determine the new volume.**
If the volume of the container is made \( \frac{1}{4} \) of the initial volume, the new volume \( V_{\text{final}} \) is:
\[ V_{\text{final}} = \frac{V}{4} \]

**Step 5: Calculate the new concentrations.**
The new concentrations of A and B will be:
\[ [A]_{\text{final}} = \frac{n_A}{V_{\text{final}}} = \frac{n_A}{\frac{V}{4}} = \frac{4n_A}{V} \]
\[ [B]_{\text{final}} = \frac{n_B}{V_{\text{final}}} = \frac{n_B}{\frac{V}{4}} = \frac{4n_B}{V} \]

**Step 6: Calculate the final rate of reaction.**
Now, substituting the final concentrations into the rate expression:
\[ \text{Rate}_{\text{final}} = k \left(\frac{4n_A}{V}\right) \left(\frac{4n_B}{V}\right) = k \frac{16n_A n_B}{V^2} \]

**Step 7: Compare the final rate to the initial rate.**
Now, we can relate the final rate to the initial rate:
\[ \text{Rate}_{\text{final}} = 16 \times \text{Rate}_{\text{initial}} \]

Thus, the final rate of reaction is 16 times the initial rate.

**Final Answer:**
The rate of reaction, when the volume is made \( \frac{1}{4} \) of the initial, will be 16 times the initial rate.

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