The time of completion of 90% of a first order reaction is approximately

Options

A. 1.1 times that of half-life
B. 2.2 times that of half-life
C. 3.3 times that of half-life
D. 4.4 times that of half-life

Explanation

To find the time of completion of 90% of a first-order reaction in terms of the half-life (t_half), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Reaction**:
For a first-order reaction, we denote the reactant as A converting to product P. At time \( t = 0 \), the concentration of A is \( [A]_0 \) and at time \( t \), it is \( [A] = [A]_0 - x \), where \( x \) is the amount reacted.

2. **Using the First-Order Rate Constant**:
The rate constant \( k \) for a first-order reaction is given by:
\[ k = \frac{\ln(2)}{t_{1/2}} \]
where \( t_{1/2} \) is the half-life of the reaction.

3. **Finding the Time for 90% Completion**:
For 90% completion, 10% of the reactant A remains. Therefore, if we assume \( [A]_0 = 100 \) (for simplicity), then:
\[ [A] = 100 - 90 = 10 \]
Thus, we can express this in the first-order kinetics equation:
\[ t = \frac{1}{k} \ln\left(\frac{[A]_0}{[A]}\right) \]
Substituting the values:
\[ t_{90\%} = \frac{1}{k} \ln\left(\frac{100}{10}\right) = \frac{1}{k} \ln(10) \]

4. **Substituting the Value of k**:
Now substituting \( k = \frac{\ln(2)}{t_{1/2}} \):
\[ t_{90\%} = \frac{t_{1/2}}{\ln(2)} \ln(10) \]

5. **Converting Natural Logarithm to Logarithm Base 10**:
We can convert the natural logarithm to base 10 using the relation:
\[ \ln(x) = 2.303 \log(x) \]
Therefore,
\[ t_{90\%} = \frac{t_{1/2}}{\ln(2)} \cdot 2.303 \log(10) \]
Since \( \log(10) = 1 \):
\[ t_{90\%} = \frac{2.303 \cdot t_{1/2}}{\ln(2)} \]

6. **Calculating the Final Value**:
The value of \( \ln(2) \) is approximately 0.693. Thus:
\[ t_{90\%} = \frac{2.303}{0.693} t_{1/2} \approx 3.33 t_{1/2} \]

### Conclusion:
The time for the completion of 90% of a first-order reaction is approximately:
\[ t_{90\%} \approx 3.33 \, t_{1/2} \]