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. 102.2 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 information given
We know that the reaction is 30% complete in 30 minutes. This means that 30% of the reactant has been converted to product, and therefore, 70% of the reactant remains.

### Step 2: Set up the equation for a first-order reaction
For a first-order reaction, the rate constant \( K \) can be calculated using the formula:
\[ K = \frac{2.303}{T} \log \left( \frac{[A_0]}{[A_t]} \right) \]
where:
- \( T \) is the time (30 minutes in this case),
- \( [A_0] \) is the initial concentration (100%),
- \( [A_t] \) is the concentration at time \( t \) (70% remaining).

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

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

### Step 5: Calculate the rate constant \( K \)
Now substitute 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: Calculate the half-life \( T_{1/2} \)
The half-life for a first-order reaction is given by:
\[ T_{1/2} = \frac{0.693}{K} \]
Substituting the value of \( K \):
\[ T_{1/2} = \frac{0.693}{0.01187} \approx 58.34 \text{ minutes} \]

### Final Answer
The half-life period of the reaction is approximately **58.34 minutes**.

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