The rate of a gaseous reaction is given by the expression k [A] [B]. If the volume of the reaction vessel is suddenly reduced to `1//4th` of the initial volume, the reaction rate relating to original rate will be

Options

A. `1//10`
B. `1//8`
C. 8
D. 16

Explanation

To solve the problem, we need to analyze how the rate of the reaction changes when the volume of the reaction vessel is reduced to one-fourth of its original volume. The rate of the reaction is given by the expression \( R = k [A][B] \).

### Step-by-Step Solution:

1. **Initial Rate Expression**:
The initial rate of the reaction is given by:
\[ R = k [A][B] \]
Here, \( [A] \) and \( [B] \) are the concentrations of reactants A and B, respectively.

2. **Understanding Concentration**:
Concentration is defined as the number of moles of a substance divided by the volume of the solution:
\[ [A] = \frac{n_A}{V} \quad \text{and} \quad [B] = \frac{n_B}{V} \]
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 Reduction**:
When the volume of the reaction vessel is reduced to one-fourth (\( \frac{V}{4} \)), the new concentrations of A and B will be:
\[ [A]' = \frac{n_A}{\frac{V}{4}} = 4 \cdot \frac{n_A}{V} = 4[A] \]
\[ [B]' = \frac{n_B}{\frac{V}{4}} = 4 \cdot \frac{n_B}{V} = 4[B] \]

4. **New Rate Expression**:
The new rate \( R' \) can be expressed as:
\[ R' = k [A]' [B]' = k (4[A])(4[B]) = k \cdot 16[A][B] \]

5. **Relation Between New Rate and Old Rate**:
Since the original rate \( R = k [A][B] \), we can substitute this into the new rate expression:
\[ R' = 16 \cdot R \]

6. **Conclusion**:
Therefore, the new rate \( R' \) is 16 times the original rate \( R \):
\[ R' = 16R \]

### Final Answer:
The reaction rate relating to the original rate will be \( 16R \).

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