For `A_((s))+B_((s))toC_((s))`,rate=`k[A]^(1//2)[B]^(2)`, if initial concentration of A and B are increased by factors 4 and 2 respectively, then the initial rate is changed by the factor:

Options

A. 4
B. 6
C. 8
D. None of these

Answer

To solve the problem, we need to determine how the initial rate of the reaction changes when the concentrations of reactants A and B are increased by specified factors. The given rate law for the reaction is:

\[ \text{Rate} = k[A]^{1/2}[B]^2 \]

### Step 1: Write the initial rate expression
The initial rate of the reaction can be expressed as:

\[ \text{Rate}_0 = k[A]^{1/2}[B]^2 \]

### Step 2: Determine the new concentrations
According to the problem, the initial concentration of A is increased by a factor of 4, and the initial concentration of B is increased by a factor of 2. Therefore, the new concentrations will be:

- New concentration of A: \( [A]' = 4[A] \)
- New concentration of B: \( [B]' = 2[B] \)

### Step 3: Write the new rate expression
Now, we can write the expression for the new rate (\( \text{Rate}' \)) using the new concentrations:

\[ \text{Rate}' = k[4A]^{1/2}[2B]^2 \]

### Step 4: Simplify the new rate expression
Now, we simplify the new rate expression:

\[ \text{Rate}' = k(4[A])^{1/2}(2[B])^2 \]

Calculating each part:

1. \( (4[A])^{1/2} = 4^{1/2}[A]^{1/2} = 2[A]^{1/2} \)
2. \( (2[B])^2 = 2^2[B]^2 = 4[B]^2 \)

Substituting these back into the rate expression gives:

\[ \text{Rate}' = k(2[A]^{1/2})(4[B]^2) \]

### Step 5: Combine the terms
Now, we can combine the terms:

\[ \text{Rate}' = k \cdot 2 \cdot 4 \cdot [A]^{1/2}[B]^2 = 8k[A]^{1/2}[B]^2 \]

### Step 6: Relate the new rate to the initial rate
We can relate the new rate to the initial rate:

\[ \text{Rate}' = 8 \cdot \text{Rate}_0 \]

### Conclusion
Thus, the initial rate is changed by a factor of 8 when the concentrations of A and B are increased by factors of 4 and 2, respectively.

\[ \text{Factor of change} = 8 \]