For a hypothetical reaction, `A+B to C+D`, the rate `=k[A]^(-1//2)[B]^(3//2)`. On doubling the concentration of A and B the rate will be

Options

A. `4` times
B. `2` times
C. `3` times
D. none of the above

Answer

To solve the problem, we need to analyze how the rate of the reaction changes when the concentrations of reactants A and B are doubled.

### Step-by-Step Solution:

1. **Write the initial rate expression**:
The rate of the reaction is given by:
\[ \text{Rate} = k[A]^{-1/2}[B]^{3/2} \]
where \( k \) is the rate constant, and \([A]\) and \([B]\) are the concentrations of A and B, respectively.

2. **Define the initial concentrations**:
Let the initial concentrations of A and B be \([A] = a\) and \([B] = b\). Therefore, the initial rate \( R_1 \) can be expressed as:
\[ R_1 = k[a]^{-1/2}[b]^{3/2} \]

3. **Determine the new concentrations after doubling**:
When the concentrations of A and B are doubled, we have:
\[ [A] = 2a \quad \text{and} \quad [B] = 2b \]

4. **Substitute the new concentrations into the rate expression**:
The new rate \( R_1' \) when the concentrations are doubled becomes:
\[ R_1' = k[2a]^{-1/2}[2b]^{3/2} \]

5. **Simplify the new rate expression**:
\[ R_1' = k(2a)^{-1/2}(2b)^{3/2} \]
This can be simplified as:
\[ R_1' = k \cdot \frac{1}{(2^{1/2} \cdot a^{1/2})} \cdot (2^{3/2} \cdot b^{3/2}) \]
\[ R_1' = k \cdot \frac{2^{3/2}}{2^{1/2}} \cdot \frac{b^{3/2}}{a^{1/2}} = k \cdot 2^{3/2 - 1/2} \cdot [a]^{-1/2}[b]^{3/2} \]
\[ R_1' = k \cdot 2^{2/2} \cdot [a]^{-1/2}[b]^{3/2} = k \cdot 2 \cdot [a]^{-1/2}[b]^{3/2} \]

6. **Relate the new rate to the initial rate**:
Thus, we can express the new rate as:
\[ R_1' = 2 \cdot R_1 \]

7. **Conclusion**:
Therefore, when the concentrations of A and B are doubled, the rate of the reaction becomes:
\[ R_1' = 2R_1 \]
This means the rate will be **2 times** the initial rate.

### Final Answer:
The rate will be **2 times** the initial rate.