`Ararr` Product, `[A]_(0) = 2M`. After `10 min` reaction is `10%` completed. If `(d[A])/(dt) = k[A]`, then `t_(1//2)` is approximately

Options

A. `0.693 min`
B. `69.3 min`
C. `66.0 min`
D. `0.0693 min`

Explanation

To solve the problem step-by-step, we will follow the information provided in the question and the video transcript.

### Step 1: Understand the Reaction and Given Data
We are given a first-order reaction where the initial concentration of A, \([A]_0\), is 2 M. After 10 minutes, the reaction is 10% completed.

**Hint:** Identify the type of reaction and the initial conditions.

### Step 2: Calculate Remaining Concentration
Since the reaction is 10% completed, this means that 10% of the initial concentration has reacted. Therefore, the remaining concentration of A after 10 minutes is:
\[ [A] = [A]_0 - 0.10 \times [A]_0 = 2M - 0.2M = 1.8M \]

**Hint:** Calculate the remaining concentration after the specified completion percentage.

### Step 3: Use the First-Order Rate Equation
For a first-order reaction, the rate constant \(k\) can be calculated using the formula:
\[ k = \frac{2.303}{t} \log\left(\frac{[A]_0}{[A]}\right) \]
Where:
- \([A]_0 = 2M\)
- \([A] = 1.8M\)
- \(t = 10 \text{ minutes}\)

**Hint:** Recall the formula for the rate constant of a first-order reaction.

### Step 4: Substitute Values into the Rate Constant Formula
Now, substituting the values into the equation:
\[ k = \frac{2.303}{10} \log\left(\frac{2}{1.8}\right) \]
Calculating the logarithm:
\[ \frac{2}{1.8} = \frac{10}{9} \approx 1.111 \]
Thus,
\[ k = \frac{2.303}{10} \log(1.111) \]

**Hint:** Simplify the fraction and calculate the logarithm.

### Step 5: Calculate the Logarithm
Using a calculator, we find:
\[ \log(1.111) \approx 0.0458 \]
Now substituting this back into the equation for \(k\):
\[ k = \frac{2.303}{10} \times 0.0458 \approx 0.0105 \text{ min}^{-1} \]

**Hint:** Use a calculator to find the logarithm and multiply accordingly.

### Step 6: Calculate the Half-Life
The half-life \(t_{1/2}\) for a first-order reaction is given by:
\[ t_{1/2} = \frac{0.693}{k} \]
Substituting the value of \(k\):
\[ t_{1/2} = \frac{0.693}{0.0105} \approx 66 \text{ minutes} \]

**Hint:** Use the half-life formula for first-order reactions.

### Final Answer
The half-life \(t_{1/2}\) is approximately **66 minutes**.