A first order reaction was commenced with 0.2 M solution of the reactants. If the molarity of the solution falls to 0.02M after 100 minutes the rate constant of the reaction is

Options

A. `2 xx 10^(-2) min^(-1)`
B. `2.3 xx 10^(-2) min^(-1)`
C. `4.6 xx 10^(-2) min^(-1)`
D. `2.3 xx 10^(-1) min^(-1)`

Explanation

To find the rate constant (k) for the first-order reaction, we can use the first-order rate equation:

\[ k = \frac{2.303}{T} \log \left( \frac{[A_0]}{[A]} \right) \]

Where:
- \( [A_0] \) is the initial concentration,
- \( [A] \) is the final concentration,
- \( T \) is the time in minutes.

### Step 1: Identify the given values
- Initial concentration, \( [A_0] = 0.2 \, \text{M} \)
- Final concentration, \( [A] = 0.02 \, \text{M} \)
- Time, \( T = 100 \, \text{minutes} \)

### Step 2: Substitute the values into the equation
Substituting the values into the equation, we have:

\[ k = \frac{2.303}{100} \log \left( \frac{0.2}{0.02} \right) \]

### Step 3: Calculate the ratio of concentrations
Calculate the ratio of the initial and final concentrations:

\[ \frac{0.2}{0.02} = 10 \]

### Step 4: Calculate the logarithm
Now, calculate the logarithm:

\[ \log(10) = 1 \]

### Step 5: Substitute back into the equation
Now substitute back into the equation for \( k \):

\[ k = \frac{2.303}{100} \cdot 1 \]

### Step 6: Calculate the rate constant
Now calculate \( k \):

\[ k = \frac{2.303}{100} = 0.02303 \, \text{min}^{-1} \]

### Step 7: Express in scientific notation
Expressing this in scientific notation gives:

\[ k = 2.303 \times 10^{-2} \, \text{min}^{-1} \]

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

\[ k = 2.3 \times 10^{-2} \, \text{min}^{-1} \]

---