The rate of a gaseous reaction is given by the expression `k[A][B]`. If The volume of reaction vessel is suddenly reduced to one-fourth of the initial volume, the reaction rate relative to the original rate will be :

Options

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

Answer

To solve the problem, we need to analyze how the change in volume affects the rate of the reaction given by the expression \( \text{Rate} = k[A][B] \).

### Step-by-Step Solution:

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

**Hint**: Recall that the rate of a reaction depends on the concentrations of the reactants.

2. **Effect of Volume Change on Concentration**:
When the volume of the reaction vessel is reduced to one-fourth of its initial volume, the concentration of the gases will change. Concentration is defined as:
\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume}} \]
Therefore, if the volume decreases, the concentration increases. Specifically, if the volume is reduced to one-fourth, the concentration increases by a factor of four.

**Hint**: Remember that concentration is inversely proportional to volume.

3. **Calculate New Concentrations**:
If the original concentrations of A and B are \( [A] \) and \( [B] \), after reducing the volume to one-fourth, the new concentrations will be:
\[ [A]_{\text{new}} = 4[A] \]
\[ [B]_{\text{new}} = 4[B] \]

**Hint**: Think about how reducing the volume affects the number of moles per unit volume.

4. **Substitute New Concentrations into the Rate Expression**:
Now, substitute the new concentrations back into the rate expression:
\[ \text{New Rate} = k[4A][4B] \]
This simplifies to:
\[ \text{New Rate} = k \cdot 4A \cdot 4B = 16k[A][B] \]

**Hint**: Factor out the constants to see how they affect the rate.

5. **Relate New Rate to Original Rate**:
The original rate was:
\[ \text{Original Rate} = k[A][B] \]
Therefore, we can express the new rate in terms of the original rate:
\[ \text{New Rate} = 16 \times \text{Original Rate} \]

**Hint**: Compare the new rate with the original rate to find the factor by which it has changed.

6. **Conclusion**:
The reaction rate relative to the original rate will be:
\[ \text{New Rate} = 16 \times \text{Original Rate} \]

Thus, the answer is that the reaction rate will be **16 times the original rate**.

### Final Answer:
The reaction rate relative to the original rate will be **16 times**.