Consider the reaction,
`2A + B rarr C + D` , If the rate expression is `r = K[A]^2[B]^1` and if volume is reduced to `1/3` , the rate of reaction will increase by

Options

A. 27 times
B. 9 times
C. 8 time
D. Rate will not get affected

Answer

To solve the problem, we need to analyze how the change in volume affects the rate of the reaction given by the rate expression \( r = K[A]^2[B]^1 \).

### Step-by-Step Solution:

1. **Understand the Rate Expression**:
The rate of the reaction is given by:
\[ r = K[A]^2[B]^1 \]
where \( K \) is the rate constant, and \( [A] \) and \( [B] \) are the concentrations of reactants A and B, respectively.

2. **Relate Concentration to Volume**:
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.

3. **Effect of Reducing Volume**:
If the volume is reduced to \( \frac{1}{3} \) of its original value, the new volume \( V' \) becomes:
\[ V' = \frac{V}{3} \]
Consequently, the new concentrations will be:
\[ [A]' = \frac{n_A}{V'} = \frac{n_A}{\frac{V}{3}} = 3\frac{n_A}{V} = 3[A] \]
\[ [B]' = \frac{n_B}{V'} = \frac{n_B}{\frac{V}{3}} = 3\frac{n_B}{V} = 3[B] \]

4. **Substituting New Concentrations into the Rate Expression**:
Now, we can substitute the new concentrations into the rate expression:
\[ r' = K[3A]^2[3B]^1 \]
Simplifying this gives:
\[ r' = K(9[A]^2)(3[B]) = 27K[A]^2[B] \]

5. **Relate the New Rate to the Original Rate**:
Since the original rate \( r = K[A]^2[B] \), we can express the new rate as:
\[ r' = 27r \]

6. **Conclusion**:
Therefore, the rate of reaction increases by a factor of 27 when the volume is reduced to one-third.

### Final Answer:
The rate of reaction will increase by **27 times**.