What is the time taken to complete 75 percent of the reaction if the rate of the first-order reaction is 0.023 min-1?

Options

A. 60.28 minutes
B. 69.28 minutes
C. 50.37 minutes
D. 65.97 minutes

Answer

To solve the problem of finding the time taken to complete 75% of a first-order reaction with a rate constant of 0.023 min⁻¹, we can follow these steps:

### Step 1: Understand the Reaction Completion
In a first-order reaction, if 75% of the reaction is 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 remaining concentration \( A \) is:
\[ A = A_0 - 0.75 A_0 = 0.25 A_0 \]

### Step 2: Set Up the First-Order Reaction Equation
The integrated rate law for a first-order reaction is given by:
\[ \ln \left( \frac{A_0}{A} \right) = k t \]
where:
- \( A_0 \) is the initial concentration,
- \( A \) is the concentration at time \( t \),
- \( k \) is the rate constant,
- \( t \) is the time.

### Step 3: Substitute Values into the Equation
We know:
- \( A_0 = 100 \) (for simplicity, we can assume this value),
- \( A = 25 \) (since 25% remains),
- \( k = 0.023 \, \text{min}^{-1} \).

Substituting these values into the equation gives:
\[ \ln \left( \frac{100}{25} \right) = k t \]

### Step 4: Calculate the Natural Logarithm
Calculate the left side:
\[ \ln \left( \frac{100}{25} \right) = \ln(4) \]

### Step 5: Solve for Time \( t \)
Now we can rearrange the equation to solve for \( t \):
\[ t = \frac{\ln(4)}{k} \]
Substituting \( k = 0.023 \):
\[ t = \frac{\ln(4)}{0.023} \]

### Step 6: Calculate \( \ln(4) \)
Using the value of \( \ln(4) \):
\[ \ln(4) \approx 1.386 \]

### Step 7: Final Calculation
Now plug this value back into the equation for \( t \):
\[ t = \frac{1.386}{0.023} \approx 60.26 \, \text{minutes} \]

Thus, the time taken to complete 75% of the reaction is approximately:
\[ t \approx 60.28 \, \text{minutes} \]

### Summary of the Solution
The time taken to complete 75% of the reaction is approximately **60.28 minutes**.