A first orde reaction is 75% completed after 32min. When was 50% of the reaction completed?

Options

A. 16 min
B. 8 min
C. 4 min
D. 32 min

Explanation

To solve the problem, we need to determine when 50% of a first-order reaction is completed, given that it is 75% completed after 32 minutes. We'll follow these steps:

### Step 1: Understand the Reaction Completion
For a first-order reaction, when it is 75% completed, it means that 25% of the reactant remains. If we denote the initial concentration of the reactant as \( A_0 \), then after 75% completion, the concentration of the reactant \( A \) is:
\[ A = A_0 - 0.75 A_0 = 0.25 A_0 \]

### Step 2: Use the First-Order Kinetics Equation
The first-order kinetics equation is given by:
\[ k = \frac{2.303}{t} \log \left( \frac{A_0}{A} \right) \]
Substituting the values for when the reaction is 75% completed (i.e., \( t = 32 \) minutes and \( A = 0.25 A_0 \)):
\[ k = \frac{2.303}{32} \log \left( \frac{A_0}{0.25 A_0} \right) = \frac{2.303}{32} \log(4) \]

### Step 3: Calculate the Rate Constant \( k \)
Now, we need to calculate \( k \):
\[ k = \frac{2.303}{32} \log(4) \]
Using \( \log(4) \approx 0.602 \):
\[ k = \frac{2.303}{32} \times 0.602 \approx 0.0433 \text{ min}^{-1} \]

### Step 4: Calculate the Half-Life of the Reaction
For a first-order reaction, the half-life \( t_{1/2} \) is given by:
\[ t_{1/2} = \frac{0.693}{k} \]
Substituting the value of \( k \):
\[ t_{1/2} = \frac{0.693}{0.0433} \approx 16 \text{ minutes} \]

### Conclusion
Thus, the time when 50% of the reaction is completed is 16 minutes.

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