How much time is requred for two - third completion of a first order reaction having, `K=5.48xx10^(-14)S^(-1)`?

Options

A. `2.01xx10^(11) s`
B. `2.01xx10^(13)s`
C. `8.08xx10^(13) s`
D. `16.04xx10^(11)` s

Explanation

To solve the problem of finding the time required for two-thirds completion of a first-order reaction with a rate constant \( K = 5.48 \times 10^{-14} \, \text{s}^{-1} \), we can follow these steps:

### Step 1: Understand the Reaction Completion
For a first-order reaction, if two-thirds of the reactant has reacted, it means that one-third of the reactant remains. If we denote the initial concentration as \( A_0 \), then after two-thirds completion, the remaining concentration \( A \) is:
\[ A = A_0 - \frac{2}{3} A_0 = \frac{1}{3} A_0 \]

### Step 2: Use the First-Order Kinetics Formula
The integrated rate law for a first-order reaction is given by:
\[ t = \frac{1}{k} \ln \left( \frac{A_0}{A} \right) \]
Substituting \( A = \frac{1}{3} A_0 \) into the equation gives:
\[ t = \frac{1}{k} \ln \left( \frac{A_0}{\frac{1}{3} A_0} \right) \]

### Step 3: Simplify the Equation
This simplifies to:
\[ t = \frac{1}{k} \ln \left( 3 \right) \]

### Step 4: Substitute the Value of \( k \)
Now, substituting the given value of \( k \):
\[ t = \frac{1}{5.48 \times 10^{-14}} \ln(3) \]

### Step 5: Calculate \( \ln(3) \)
The natural logarithm of 3 is approximately:
\[ \ln(3) \approx 1.0986 \]

### Step 6: Calculate the Time
Now substituting \( \ln(3) \) back into the equation:
\[ t = \frac{1.0986}{5.48 \times 10^{-14}} \]

### Step 7: Perform the Calculation
Calculating the above expression:
\[ t \approx 2.00 \times 10^{13} \, \text{s} \]

### Final Answer
Thus, the time required for two-thirds completion of the reaction is approximately:
\[ t \approx 2.00 \times 10^{13} \, \text{s} \]