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-by-Step Solution:

1. **Understand the Reaction and Rate Law**:
The reaction is given as:
\[ 2A + B \rightarrow C + D \]
The rate law for this reaction is:
\[ \text{Rate} = k[A][B] \]
where \( k \) is the rate constant, and \([A]\) and \([B]\) are the concentrations of reactants A and B, respectively.

2. **Determine the Effect of Volume Change**:
It is stated that the volume of the container is reduced to \( \frac{1}{4} \) of its initial volume. If the initial volume is \( V \), the new volume \( V' \) becomes:
\[ V' = \frac{V}{4} \]

3. **Calculate Concentrations**:
Concentration is defined as the number of moles of a substance divided by the volume of the container. If the number of moles of A and B remains constant, the new concentrations can be calculated as follows:
\[ [A]' = \frac{n_A}{V'} = \frac{n_A}{\frac{V}{4}} = \frac{4n_A}{V} = 4[A] \]
\[ [B]' = \frac{n_B}{V'} = \frac{n_B}{\frac{V}{4}} = \frac{4n_B}{V} = 4[B] \]
Thus, both concentrations of A and B increase by a factor of 4.

4. **Substitute New Concentrations into the Rate Law**:
The new rate of reaction \( \text{Rate}' \) can be expressed as:
\[ \text{Rate}' = k[A]'[B]' = k(4[A])(4[B]) = 16k[A][B] \]

5. **Compare the New Rate to the Initial Rate**:
The initial rate \( \text{Rate} \) is:
\[ \text{Rate} = k[A][B] \]
Therefore, we can express the new rate in terms of the initial rate:
\[ \text{Rate}' = 16 \times \text{Rate} \]

### Conclusion:
The rate of the reaction, when the volume is reduced to one-fourth of its initial value, becomes **16 times** the initial rate.