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)`
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, we will follow these steps:

### Step 1: Understand the relationship between rate and concentration
The rate of a reaction can be expressed using the rate equation:
\[ r = k [A]^n \]
where:
- \( r \) is the rate of the reaction,
- \( k \) is the rate constant,
- \([A]\) is the concentration of the reactant,
- \( n \) is the order of the reaction.

### Step 2: Set up the equations based on the given information
From the problem, we know:
1. The initial rate \( r_1 = 4 \times 10^{-6} \, \text{mol L}^{-1} \text{s}^{-1} \) when the concentration \( [A]_1 = 4 \times 10^{-4} \, \text{mol L}^{-1} \).
2. The rate increases four-fold when the concentration is increased sixteen times.

This gives us:
- \( r_2 = 4 \times r_1 = 4 \times (4 \times 10^{-6}) = 16 \times 10^{-6} \, \text{mol L}^{-1} \text{s}^{-1} \)
- \( [A]_2 = 16 \times [A]_1 = 16 \times (4 \times 10^{-4}) = 64 \times 10^{-4} \, \text{mol L}^{-1} \)

### Step 3: Write the equations for both rates
Using the rate equation for both conditions:
1. For the first condition:
\[ r_1 = k [A]_1^n \]
\[ 4 \times 10^{-6} = k (4 \times 10^{-4})^n \]

2. For the second condition:
\[ r_2 = k [A]_2^n \]
\[ 16 \times 10^{-6} = k (64 \times 10^{-4})^n \]

### Step 4: Set up the ratio of the two equations
Taking the ratio of the two equations:
\[ \frac{r_2}{r_1} = \frac{k [A]_2^n}{k [A]_1^n} \]
This simplifies to:
\[ \frac{16 \times 10^{-6}}{4 \times 10^{-6}} = \frac{(64 \times 10^{-4})^n}{(4 \times 10^{-4})^n} \]
\[ 4 = \left(\frac{64}{4}\right)^n \]
\[ 4 = 16^n \]

### Step 5: Solve for \( n \)
Since \( 16 = 4^2 \), we can rewrite:
\[ 4 = (4^2)^n \]
This implies:
\[ 4 = 4^{2n} \]
Thus, equating the exponents gives:
\[ 1 = 2n \]
\[ n = \frac{1}{2} \]

### Step 6: Calculate the rate constant \( k \)
Now that we have \( n \), we can substitute back into one of the original equations to find \( k \):
Using the first condition:
\[ 4 \times 10^{-6} = k (4 \times 10^{-4})^{1/2} \]
\[ 4 \times 10^{-6} = k (2 \times 10^{-4}) \]
\[ k = \frac{4 \times 10^{-6}}{2 \times 10^{-4}} \]
\[ k = 2 \times 10^{-2} \, \text{mol}^{1/2} \text{L}^{-1/2} \text{s}^{-1} \]

### Final Answer
The rate constant \( k \) is:
\[ k = 2 \times 10^{-2} \, \text{mol}^{1/2} \text{L}^{-1/2} \text{s}^{-1} \]

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