For a first order reaction `A rightarrow B`, half life time is 30 min., calculate the time taken for 75% completion of the reaction is ………………….min.

Answer

To solve the problem, we need to calculate the time taken for 75% completion of a first-order reaction given that the half-life (t₁/₂) is 30 minutes.

### Step-by-Step Solution:

1. **Understand the Concept of Half-Life in First-Order Reactions:**
The half-life of a first-order reaction is constant and is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
where \( k \) is the rate constant.

2. **Calculate the Rate Constant (k):**
Given that the half-life \( t_{1/2} = 30 \) minutes, we can rearrange the formula to find \( k \):
\[ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{30 \text{ min}} \approx 0.0231 \text{ min}^{-1} \]

3. **Determine the Relationship Between Time and Completion Percentage:**
For a first-order reaction, the relationship between the initial concentration \([A_0]\), the concentration at time \( t \) \([A_t]\), and the time \( t \) can be expressed as:
\[ \ln\left(\frac{[A_0]}{[A_t]}\right) = kt \]
If 75% of the reaction is complete, then 25% of the reactant remains. Thus, if we assume \([A_0] = 100\), then \([A_t] = 25\).

4. **Set Up the Equation:**
Substitute the values into the equation:
\[ \ln\left(\frac{100}{25}\right) = kt \]
Simplifying the left side gives:
\[ \ln(4) = kt \]

5. **Substitute k into the Equation:**
Now substitute \( k \) into the equation:
\[ \ln(4) = 0.0231 \cdot t \]

6. **Calculate \( t \):**
We know that \( \ln(4) \approx 1.386 \):
\[ 1.386 = 0.0231 \cdot t \]
Solving for \( t \):
\[ t = \frac{1.386}{0.0231} \approx 60 \text{ minutes} \]

### Final Answer:
The time taken for 75% completion of the reaction is **60 minutes**.

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