`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 1: Understand the Initial and Final Concentrations
Assume the initial concentration of the reactant \( A_0 \) is 1 (for simplicity). Since 90% of the reaction is completed, the remaining concentration \( A_t \) after 100 minutes is:
\[ A_t = A_0 - 0.9 \times A_0 = 1 - 0.9 = 0.1 \]

### Step 2: Use the First-Order Rate Equation
The first-order rate equation is given by:
\[ k = \frac{2.303}{t} \log \left( \frac{A_0}{A_t} \right) \]
Substituting the known values:
- \( A_0 = 1 \)
- \( A_t = 0.1 \)
- \( t = 100 \) minutes

We can substitute these values into the equation:
\[ k = \frac{2.303}{100} \log \left( \frac{1}{0.1} \right) \]

### Step 3: Calculate the Logarithm
Calculate the logarithm:
\[ \log \left( \frac{1}{0.1} \right) = \log(10) = 1 \]
Now substitute this back into the equation for \( k \):
\[ k = \frac{2.303}{100} \times 1 = 0.02303 \, \text{min}^{-1} \]

### Step 4: Calculate 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 \( k \):
\[ t_{1/2} = \frac{0.693}{0.02303} \]

### Step 5: Perform the Calculation
Calculating the above expression:
\[ t_{1/2} \approx 30 \, \text{minutes} \]

### Final Answer
The half-life of the reaction is approximately **30 minutes**.

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