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

Options

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

Answer

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

### Step 1: Understand the relationship for first-order reactions
For a first-order reaction, the rate constant \( k \) can be expressed using the formula:
\[ k = \frac{2.303}{t} \log \left( \frac{[A_0]}{[A]} \right) \]
where:
- \( [A_0] \) is the initial concentration,
- \( [A] \) is the concentration at time \( t \),
- \( t \) is the time elapsed.

### Step 2: Determine the concentrations
Given that the reaction is 30% complete in 30 minutes:
- Initial concentration, \( [A_0] = 100\% \)
- Concentration after 30 minutes, \( [A] = 100\% - 30\% = 70\% \)

### Step 3: Substitute values into the equation
Now substituting the values into the equation:
\[ k = \frac{2.303}{30} \log \left( \frac{100}{70} \right) \]

### Step 4: Calculate the logarithm
Calculate \( \log \left( \frac{100}{70} \right) \):
\[ \log \left( \frac{100}{70} \right) = \log(1.4286) \approx 0.155 \]

### Step 5: Calculate the rate constant \( k \)
Now substituting this value back into the equation for \( k \):
\[ k = \frac{2.303}{30} \times 0.155 \approx \frac{0.356}{30} \approx 0.01187 \, \text{min}^{-1} \]

### Step 6: Use the half-life formula
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 \( k \):
\[ t_{1/2} = \frac{0.693}{0.01187} \approx 58.2 \, \text{minutes} \]

### Conclusion
The half-life of the reaction is approximately **58.2 minutes**.

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