99% of a 1st order reaction completed in 2.303 minutes. What is the rate constant and half life of the reaction

Options

A. 2.303 and 0.3010
B. 2 and 0.3465
C. 2 and 0.693
D. 0.310 and 0.693

Answer

To solve the problem step by step, we will follow the principles of first-order kinetics.

### Step 1: Understand the Reaction Completion
We are given that 99% of a first-order reaction is completed in 2.303 minutes. This means that only 1% of the reactant remains.

### Step 2: Define Initial and Remaining Concentrations
Let’s denote:
- Initial concentration (A₀) = 100% (or 1 for simplicity)
- Remaining concentration (A) after 99% completion = 100% - 99% = 1% (or 0.01)

### Step 3: Use the First-Order Rate Constant Formula
For a first-order reaction, the rate constant (k) can be calculated using the formula:
\[ k = \frac{2.303}{t} \log\left(\frac{A_0}{A}\right) \]
Where:
- \( t \) = time taken for the reaction (in minutes)
- \( A_0 \) = initial concentration
- \( A \) = concentration at time \( t \)

### Step 4: Substitute Values into the Formula
Given:
- \( t = 2.303 \) minutes
- \( A_0 = 1 \)
- \( A = 0.01 \)

Now substituting these values into the equation:
\[ k = \frac{2.303}{2.303} \log\left(\frac{1}{0.01}\right) \]

### Step 5: Simplify the Equation
The equation simplifies to:
\[ k = 1 \cdot \log(100) \]
Since \( \log(100) = 2 \):
\[ k = 2 \, \text{min}^{-1} \]

### Step 6: Calculate the Half-Life
The half-life (\( t_{1/2} \)) for a first-order reaction is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
Now substituting the value of \( k \):
\[ t_{1/2} = \frac{0.693}{2} \]
\[ t_{1/2} = 0.3465 \, \text{minutes} \]

### Final Answers
- Rate constant \( k = 2 \, \text{min}^{-1} \)
- Half-life \( t_{1/2} = 0.3465 \, \text{minutes} \)