A first order reaction requires 30 minutes for 50 % completion . The time required to complete the reaction by 75 % will be __.

Options

A. 45 timutes
B. 15 minutes
C. 60 minutes
D. None of these

Answer

To solve the problem, we need to find the time required for a first-order reaction to reach 75% completion, given that it takes 30 minutes for 50% completion.

### Step-by-Step Solution:

1. **Understanding Half-Life of a First-Order Reaction**:
The half-life (t₁/₂) of a first-order reaction is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
where \( k \) is the rate constant.

2. **Calculate the Rate Constant (k)**:
We know that the half-life for 50% completion is 30 minutes. Thus, we can rearrange the formula to find \( k \):
\[ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{30 \text{ min}} = 0.0231 \text{ min}^{-1} \]

3. **Using the Integrated Rate Law for First-Order Reactions**:
The integrated rate law for a first-order reaction is:
\[ \ln \left( \frac{[A_0]}{[A]} \right) = kt \]
Here, \( [A_0] \) is the initial concentration and \( [A] \) is the concentration at time \( t \).

4. **Determine Concentrations for 75% Completion**:
If the reaction is 75% complete, then 25% of the original concentration remains. Therefore:
\[ [A] = 0.25[A_0] \]
Substituting into the integrated rate law gives:
\[ \ln \left( \frac{[A_0]}{0.25[A_0]} \right) = kt \]
This simplifies to:
\[ \ln(4) = kt \]

5. **Calculate \( \ln(4) \)**:
We know that:
\[ \ln(4) \approx 1.386 \]

6. **Substituting Values to Find Time (t)**:
Now we can substitute \( k \) and \( \ln(4) \) into the equation:
\[ 1.386 = (0.0231)t \]
Rearranging to solve for \( t \):
\[ t = \frac{1.386}{0.0231} \approx 60 \text{ minutes} \]

### Final Answer:
The time required to complete the reaction by 75% is approximately **60 minutes**.

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