For a first order for the reaction `A rarr` products `:` the rate of reaction at `[A]=0.2 M ` is `1.0 xx 10^(-3) mol L^(-1)s^(-1)`. The reaction will occur to `75%` completion in .

Options

A. `138.65`
B. `277.2s`
C. `44.25`
D. `822.5`

Explanation

To solve the problem, we need to determine the time required for a first-order reaction to reach 75% completion. Here's a step-by-step breakdown of the solution:

### Step 1: Understand the Reaction and Given Data
We have a first-order reaction \( A \rightarrow \text{products} \). The initial concentration of \( A \) is given as \( [A]_0 = 0.2 \, \text{M} \) and the rate of reaction at this concentration is \( 1.0 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \).

### Step 2: Calculate the Rate Constant \( k \)
For a first-order reaction, the rate law is given by:
\[ \text{Rate} = k [A] \]
We can rearrange this to find \( k \):
\[ k = \frac{\text{Rate}}{[A]} = \frac{1.0 \times 10^{-3}}{0.2} = 5.0 \times 10^{-3} \, \text{s}^{-1} \]

### Step 3: Determine the Final Concentration After 75% Completion
If the reaction is 75% complete, then 25% of the initial concentration remains. Therefore, the final concentration \( [A] \) after 75% completion is:
\[ [A] = 0.25 \times [A]_0 = 0.25 \times 0.2 = 0.05 \, \text{M} \]

### Step 4: Use the Integrated Rate Equation for First-Order Reactions
The integrated rate equation for a first-order reaction is:
\[ k = \frac{2.303}{t} \log\left(\frac{[A]_0}{[A]}\right) \]
Rearranging this to solve for time \( t \):
\[ t = \frac{2.303}{k} \log\left(\frac{[A]_0}{[A]}\right) \]

### Step 5: Substitute the Values into the Equation
Substituting the known values into the equation:
- \( k = 5.0 \times 10^{-3} \, \text{s}^{-1} \)
- \( [A]_0 = 0.2 \, \text{M} \)
- \( [A] = 0.05 \, \text{M} \)

Calculating the logarithm:
\[ \log\left(\frac{0.2}{0.05}\right) = \log(4) \approx 0.602 \]

Now substituting into the time equation:
\[ t = \frac{2.303}{5.0 \times 10^{-3}} \times 0.602 \]

### Step 6: Calculate the Time
Calculating the above expression:
\[ t = \frac{2.303 \times 0.602}{5.0 \times 10^{-3}} \approx \frac{1.384}{0.005} \approx 276.8 \, \text{s} \]

Rounding this gives approximately:
\[ t \approx 277.2 \, \text{s} \]

### Conclusion
The time required for the reaction to occur to 75% completion is approximately **277.2 seconds**.