For a hypothetical reaction `A+B rarr 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 these

Explanation

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.

### Given:
The rate of the reaction is given by:
\[ \text{Rate} = k[A]^{-\frac{1}{2}}[B]^{\frac{3}{2}} \]

### Step 1: Write the initial rate expression
Let the initial concentrations of A and B be \( [A] \) and \( [B] \). The initial rate (let's call it \( r_1 \)) can be expressed as:
\[ r_1 = k[A]^{-\frac{1}{2}}[B]^{\frac{3}{2}} \]

### Step 2: Write the new concentrations after doubling
When the concentrations of A and B are doubled, the new concentrations will be:
\[ [A]_{new} = 2[A] \]
\[ [B]_{new} = 2[B] \]

### Step 3: Write the new rate expression
Now, we can write the new rate (let's call it \( r_2 \)):
\[ r_2 = k[2A]^{-\frac{1}{2}}[2B]^{\frac{3}{2}} \]

### Step 4: Substitute the new concentrations into the rate expression
Substituting the new concentrations into the rate expression gives:
\[ r_2 = k(2[A])^{-\frac{1}{2}}(2[B])^{\frac{3}{2}} \]

### Step 5: Simplify the expression
Now we simplify:
\[ r_2 = k \cdot 2^{-\frac{1}{2}}[A]^{-\frac{1}{2}} \cdot 2^{\frac{3}{2}}[B]^{\frac{3}{2}} \]
\[ r_2 = k \cdot 2^{-\frac{1}{2} + \frac{3}{2}}[A]^{-\frac{1}{2}}[B]^{\frac{3}{2}} \]
\[ r_2 = k \cdot 2^{1}[A]^{-\frac{1}{2}}[B]^{\frac{3}{2}} \]
\[ r_2 = 2 \cdot k[A]^{-\frac{1}{2}}[B]^{\frac{3}{2}} \]

### Step 6: Relate the new rate to the initial rate
Notice that:
\[ r_2 = 2 \cdot r_1 \]

### Conclusion
Thus, when the concentrations of A and B are doubled, the rate of the reaction becomes:
\[ r_2 = 2r_1 \]

### Final Answer
The rate will be doubled.

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