In a first order reaction, the concentration of reactant decrease from 400 mol `L^(-1)` to 25 mol `L^(-1)` in 200 seconds. The rate constant for the reaction is

Options

A. `1.01386 s^(-1)`
B. `2 xx 10^(-4) s^(-1)`
C. `1.386 xx 10^(-2)s^(-1)`
D. `3.4 xx 10^(-4) s^(-1)`

Explanation

To find the rate constant \( k \) for a first-order reaction where the concentration of the reactant decreases from 400 mol L\(^{-1}\) to 25 mol L\(^{-1}\) in 200 seconds, we can use the first-order rate equation:

### Step 1: Write the first-order rate equation
The integrated rate law for a first-order reaction is given by:

\[ \ln \left( \frac{[A_0]}{[A]} \right) = kt \]

Where:
- \([A_0]\) is the initial concentration
- \([A]\) is the final concentration
- \(k\) is the rate constant
- \(t\) is the time

### Step 2: Substitute the values
In this case:
- \([A_0] = 400 \, \text{mol L}^{-1}\)
- \([A] = 25 \, \text{mol L}^{-1}\)
- \(t = 200 \, \text{s}\)

Substituting these values into the equation:

\[ \ln \left( \frac{400}{25} \right) = k \cdot 200 \]

### Step 3: Calculate the ratio
Calculate the ratio:

\[ \frac{400}{25} = 16 \]

### Step 4: Take the natural logarithm
Now, take the natural logarithm:

\[ \ln(16) = k \cdot 200 \]

### Step 5: Calculate \(\ln(16)\)
Using the property of logarithms:

\[ \ln(16) = \ln(2^4) = 4 \ln(2) \]

Using \(\ln(2) \approx 0.693\):

\[ \ln(16) \approx 4 \times 0.693 = 2.772 \]

### Step 6: Solve for \(k\)
Now substitute \(\ln(16)\) back into the equation:

\[ 2.772 = k \cdot 200 \]

To find \(k\):

\[ k = \frac{2.772}{200} \approx 0.01386 \, \text{s}^{-1} \]

### Step 7: Convert to scientific notation
Expressing \(k\) in scientific notation:

\[ k \approx 1.386 \times 10^{-2} \, \text{s}^{-1} \]

### Final Answer
The rate constant \(k\) for the reaction is:

\[ \boxed{1.386 \times 10^{-2} \, \text{s}^{-1}} \]