Conisder a gaseous reaction, the rate of which is given by `k[A][B]`. The volume of the reaction vessel containing these gases is suddenly reduced to `1//4th` of the initial volume. The rate of the reaction as compared with original rate is

Options

A. `1//16` times
B. `16` times
C. `1//8` times
D. `8` times

Explanation

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

### Step-by-Step Solution:

1. **Understand the Rate Law**:
The rate of the reaction is given by the equation:
\[ \text{Rate} = k[A][B] \]
where \( k \) is the rate constant, and \([A]\) and \([B]\) are the concentrations of the reactants A and B.

2. **Relate Concentration to Volume**:
Concentration is defined as:
\[ [A] = \frac{n_A}{V} \quad \text{and} \quad [B] = \frac{n_B}{V} \]
where \( n_A \) and \( n_B \) are the number of moles of A and B, and \( V \) is the volume of the reaction vessel.

3. **Effect of Volume Change**:
The problem states that the volume of the reaction vessel is suddenly reduced to \( \frac{1}{4} \) of the initial volume \( V \). Therefore, the new volume \( V' \) is:
\[ V' = \frac{1}{4} V \]

4. **Calculate New Concentrations**:
When the volume is reduced, the concentrations of A and B will change as follows:
\[ [A'] = \frac{n_A}{V'} = \frac{n_A}{\frac{1}{4} V} = 4 \cdot \frac{n_A}{V} = 4[A] \]
\[ [B'] = \frac{n_B}{V'} = \frac{n_B}{\frac{1}{4} V} = 4 \cdot \frac{n_B}{V} = 4[B] \]

5. **Substitute New Concentrations into the Rate Law**:
Now, substituting the new concentrations into the rate law:
\[ \text{Rate}' = k[A'][B'] = k(4[A])(4[B]) = 16k[A][B] \]

6. **Compare New Rate with Original Rate**:
The original rate is:
\[ \text{Rate} = k[A][B] \]
Therefore, the new rate is:
\[ \text{Rate}' = 16 \cdot \text{Rate} \]

### Conclusion:
The rate of the reaction after the volume is reduced to one-fourth of its original value is **16 times** the original rate.