`90%` of a first order reaction was completed in `100 min`.What is the half life of the reaction ?

Options

A. `63.3 min`
B. `53.3 min`
C. `43.3 min`
D. `30 min`

Explanation

To find the half-life of a first-order reaction where 90% of the reaction is completed in 100 minutes, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Reaction Completion**:
- Given that 90% of the reaction is completed in 100 minutes, this means that 10% of the reactant (let's denote it as A) remains.
- If we start with an initial concentration \( [A_0] \), after 100 minutes, the concentration of A is \( [A] = 0.1 [A_0] \).

2. **Use the First-Order Kinetics Equation**:
- The first-order reaction can be described by the equation:
\[ [A] = [A_0] e^{-kt} \]
- Rearranging this equation in terms of the rate constant \( k \):
\[ k = \frac{1}{t} \ln\left(\frac{[A_0]}{[A]}\right) \]

3. **Substituting Known Values**:
- We know \( [A] = 0.1 [A_0] \) and \( t = 100 \) minutes.
- Substituting these values into the equation gives:
\[ k = \frac{1}{100} \ln\left(\frac{[A_0]}{0.1 [A_0]}\right) = \frac{1}{100} \ln(10) \]

4. **Calculating the Rate Constant \( k \)**:
- We can use the natural logarithm value \( \ln(10) \approx 2.303 \):
\[ k = \frac{1}{100} \times 2.303 \approx 0.02303 \, \text{min}^{-1} \]

5. **Finding the Half-Life**:
- The half-life \( t_{1/2} \) for a first-order reaction is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
- Substituting the value of \( k \):
\[ t_{1/2} = \frac{0.693}{0.02303} \approx 30.1 \, \text{minutes} \]

6. **Final Answer**:
- Therefore, the half-life of the reaction is approximately **30 minutes**.