The rate of a gaseous reaction is equal to k[A][B]. The volume of the reaction vessel containing these gases is reduced by one fourth of the initial volume. The rate of the reaction would be

Options

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

Answer

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

### Step-by-Step Solution:

1. **Understanding 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. **Initial Concentrations**:
Let's denote the initial concentrations of A and B as \([A]_0\) and \([B]_0\).

3. **Effect of Volume Reduction**:
The volume of the reaction vessel is reduced by one fourth. This means the new volume \( V' \) is:
\[ V' = V - \frac{1}{4}V = \frac{3}{4}V \]
where \( V \) is the initial volume.

4. **Calculating New Concentrations**:
Concentration is defined as:
\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume}} \]
When the volume is reduced, the concentration increases. The new concentrations \([A]'\) and \([B]'\) can be calculated as:
\[ [A]' = \frac{n_A}{V'} = \frac{n_A}{\frac{3}{4}V} = \frac{4}{3} \cdot \frac{n_A}{V} = \frac{4}{3}[A]_0 \]
\[ [B]' = \frac{n_B}{V'} = \frac{n_B}{\frac{3}{4}V} = \frac{4}{3} \cdot \frac{n_B}{V} = \frac{4}{3}[B]_0 \]

5. **Substituting New Concentrations into the Rate Law**:
Now substituting the new concentrations into the rate equation:
\[ \text{Rate}' = k[A]'[B]' = k\left(\frac{4}{3}[A]_0\right)\left(\frac{4}{3}[B]_0\right) \]
\[ = k \cdot \frac{16}{9}[A]_0[B]_0 \]

6. **Comparing New Rate with Initial Rate**:
The initial rate was:
\[ \text{Rate}_0 = k[A]_0[B]_0 \]
Therefore, the new rate can be expressed in terms of the initial rate:
\[ \text{Rate}' = \frac{16}{9} \cdot \text{Rate}_0 \]

7. **Conclusion**:
The rate of the reaction increases by a factor of \(\frac{16}{9}\) when the volume is reduced by one fourth.

### Final Answer:
The rate of the reaction would be \(\frac{16}{9}\) times the initial rate.