`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, let's break it down:

### Step 1: Understand the Given Information
- Initial concentration, \([A]_0 = 2 \, M\)
- The reaction is 10% complete after 10 minutes.
- The rate law is given as \(\frac{d[A]}{dt} = k[A]\), indicating a first-order reaction.

### Step 2: Determine the Concentration After 10 Minutes
Since the reaction is 10% complete, this means that 10% of the initial concentration has reacted. Therefore:
\[ \text{Concentration of A after 10 minutes} = [A] = [A]_0 - 0.1[A]_0 = 2 \, M - 0.2 \, M = 1.8 \, M \]

### Step 3: Use the Integrated Rate Law for First-Order Reactions
For a first-order reaction, the integrated rate law is given by:
\[ \ln \left(\frac{[A]_0}{[A]}\right) = kt \]
Substituting the known values:
- \([A]_0 = 2 \, M\)
- \([A] = 1.8 \, M\)
- \(t = 10 \, \text{minutes}\)

We can write:
\[ \ln \left(\frac{2}{1.8}\right) = k \cdot 10 \]

### Step 4: Calculate \(k\)
First, calculate \(\frac{2}{1.8}\):
\[ \frac{2}{1.8} \approx 1.1111 \]
Now, take the natural logarithm:
\[ \ln(1.1111) \approx 0.1054 \]
Now, substitute back to find \(k\):
\[ 0.1054 = k \cdot 10 \implies k = \frac{0.1054}{10} \approx 0.01054 \, \text{min}^{-1} \]

### Step 5: Calculate the Half-Life \(t_{1/2}\)
For a first-order reaction, the half-life is given by:
\[ t_{1/2} = \frac{0.693}{k} \]
Substituting the value of \(k\):
\[ t_{1/2} = \frac{0.693}{0.01054} \approx 65.7 \, \text{minutes} \]

### Conclusion
Thus, the approximate half-life \(t_{1/2}\) is about **66 minutes**.

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