The rate of a reaction increases four-fold when the concentration of reactant is increased `16` times. If the rate of reaction is `4 xx 10^(-6) mol L^(-1) s^(-1)` when the concentration of the reactant is `4 xx 10^(-4) mol L^(-1)`. The rate constant of the reaction will be

Options

A. `2 xx 10^(-4) mol^(1//2) L^(1//2) s^(1//2)`
B. `1 xx 10^(-2) s^(-1)`
C. `2 xx 10^(-4) mol^(-1//2) L^(1//2) s^(-1)`
D. `25 mol^(-1) L min^(-1)`

Explanation

To solve the problem step by step, we will follow the principles of chemical kinetics and the relationship between rate, concentration, and the rate constant.

### Step 1: Understand the relationship between rate and concentration
The problem states that the rate of reaction increases four-fold when the concentration of the reactant is increased sixteen times. This indicates a relationship between the rate (R) and the concentration ([A]) of the reactant.

### Step 2: Establish the rate law
From the information given, we can express the rate law as:
\[ R = k [A]^n \]
where \( k \) is the rate constant and \( n \) is the order of the reaction.

### Step 3: Determine the order of the reaction
Given that increasing the concentration by a factor of 16 increases the rate by a factor of 4, we can write:
\[ \frac{R_2}{R_1} = \left(\frac{[A_2]}{[A_1]}\right)^n \]
Substituting the known values:
- \( R_2/R_1 = 4 \)
- \( [A_2]/[A_1] = 16 \)

Thus, we have:
\[ 4 = 16^n \]
Taking logarithm on both sides:
\[ \log(4) = n \log(16) \]
Since \( 16 = 4^2 \), we can express \( \log(16) \) as \( 2 \log(4) \):
\[ \log(4) = n \cdot 2 \log(4) \]
Dividing both sides by \( \log(4) \) (assuming \( \log(4) \neq 0 \)):
\[ 1 = 2n \]
Thus, \( n = \frac{1}{2} \).

### Step 4: Write the rate equation with the determined order
Now we can write the rate equation as:
\[ R = k [A]^{1/2} \]

### Step 5: Substitute the known values to find the rate constant \( k \)
We know:
- The rate \( R = 4 \times 10^{-6} \, \text{mol L}^{-1} \text{s}^{-1} \)
- The concentration \( [A] = 4 \times 10^{-4} \, \text{mol L}^{-1} \)

Substituting these values into the rate equation:
\[ 4 \times 10^{-6} = k (4 \times 10^{-4})^{1/2} \]

### Step 6: Calculate the square root of the concentration
Calculating \( (4 \times 10^{-4})^{1/2} \):
\[ (4 \times 10^{-4})^{1/2} = 2 \times 10^{-2} \, \text{mol L}^{-1} \]

### Step 7: Substitute back to find \( k \)
Now substituting back:
\[ 4 \times 10^{-6} = k (2 \times 10^{-2}) \]
To find \( k \), rearranging gives:
\[ k = \frac{4 \times 10^{-6}}{2 \times 10^{-2}} \]
\[ k = 2 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

### Conclusion
Thus, the rate constant \( k \) of the reaction is:
\[ k = 2 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

---