A first-order reaction which is `30%` complete in `30` minutes has a half-life period of

Options

A. `24.2 min`
B. `58.2 min`
C. `102.2min`
D. `120.2 min`

Explanation

To solve the problem of finding the half-life of a first-order reaction that is 30% complete in 30 minutes, we can follow these steps:

### Step 1: Understand the Reaction Completion
Given that the reaction is 30% complete, it means that 30% of the initial concentration (A) has reacted. Therefore, the remaining concentration (A - x) is 70% of the initial concentration.

### Step 2: Set Up the First-Order Reaction Equation
For a first-order reaction, the relationship between time (t), the rate constant (k), and the concentrations can be expressed using the formula:
\[ t = \frac{1}{k} \ln \left( \frac{A}{A - x} \right) \]
Where:
- \( A \) = initial concentration
- \( x \) = amount reacted (30% of A)

### Step 3: Substitute Values into the Equation
In this case, since 30% of A has reacted, we have:
- \( x = 0.3A \)
- Remaining concentration \( = A - x = A - 0.3A = 0.7A \)

Substituting these values into the equation:
\[ 30 \text{ minutes} = \frac{1}{k} \ln \left( \frac{A}{0.7A} \right) \]
This simplifies to:
\[ 30 = \frac{1}{k} \ln \left( \frac{1}{0.7} \right) \]

### Step 4: Calculate the Rate Constant (k)
Rearranging the equation to solve for \( k \):
\[ k = \frac{1}{30} \ln \left( \frac{1}{0.7} \right) \]
Calculating \( \ln(1/0.7) \):
\[ \ln(1/0.7) = \ln(100/70) \approx 0.3567 \]
Thus,
\[ k = \frac{0.3567}{30} \approx 0.01189 \text{ min}^{-1} \]

### Step 5: Calculate the Half-Life (T_half)
The half-life for a first-order reaction is given by the formula:
\[ T_{1/2} = \frac{0.693}{k} \]
Substituting the value of \( k \):
\[ T_{1/2} = \frac{0.693}{0.01189} \approx 58.28 \text{ minutes} \]

### Final Answer
The half-life period of the reaction is approximately **58.28 minutes**.

---