The time for 90% of a first order reaction to complete 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.

Answer

To find the time required for 90% completion of a first-order reaction, we can use the first-order kinetics equation. Here’s a step-by-step solution:

### Step 1: Understand the first-order reaction equation
For a first-order reaction, the rate constant \( k \) can be expressed using the following formula:
\[ k = \frac{2.303}{T} \log \left( \frac{A}{A - X} \right) \]
where:
- \( T \) is the time,
- \( A \) is the initial concentration,
- \( X \) is the amount reacted.

### Step 2: Set up the equation for 90% completion
For 90% completion, if we assume the initial concentration \( A \) is 100, then the amount reacted \( X \) will be 90. Thus, the remaining concentration \( A - X \) will be:
\[ A - X = 100 - 90 = 10 \]

### Step 3: Substitute values into the equation
Substituting \( A = 100 \) and \( A - X = 10 \) into the equation gives:
\[ T = \frac{2.303}{k} \log \left( \frac{100}{10} \right) \]

### Step 4: Simplify the logarithm
The logarithm simplifies as follows:
\[ \log \left( \frac{100}{10} \right) = \log(10) = 1 \]
Thus, the equation for time becomes:
\[ T = \frac{2.303}{k} \cdot 1 = \frac{2.303}{k} \]

### Step 5: Relate \( k \) to half-life
The half-life \( t_{1/2} \) for a first-order reaction is given by:
\[ t_{1/2} = \frac{0.693}{k} \]
From this, we can express \( k \) as:
\[ k = \frac{0.693}{t_{1/2}} \]

### Step 6: Substitute \( k \) back into the time equation
Substituting \( k \) back into the time equation gives:
\[ T = \frac{2.303}{\frac{0.693}{t_{1/2}}} = \frac{2.303 \cdot t_{1/2}}{0.693} \]

### Step 7: Calculate the ratio
Calculating the ratio:
\[ T \approx 3.32 \cdot t_{1/2} \]
This can be approximated as:
\[ T \approx 3.3 \cdot t_{1/2} \]

### Conclusion
Thus, the time for 90% completion of a first-order reaction is approximately:
\[ T \approx 3.3 \cdot t_{1/2} \]