The time taken for 90% of a first order reaction to be completed 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 taken for 90% completion of a first-order reaction, we can follow these steps:

### Step 1: Understand the relationship of first-order reactions
For a first-order reaction, the time taken to complete a certain percentage of the reaction can be related to its half-life. The half-life (\(t_{1/2}\)) for a first-order reaction is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
where \(k\) is the rate constant.

### Step 2: Determine the fraction of the reaction completed
If 90% of the reaction is completed, then 10% of the reactant remains. This can be represented as:
\[ \text{Remaining concentration} = A - x = A - 0.9A = 0.1A \]
where \(A\) is the initial concentration and \(x\) is the amount reacted.

### Step 3: Use the first-order reaction formula
The integrated rate law for a first-order reaction is:
\[ \ln\left(\frac{A}{A - x}\right) = kt \]
Substituting \(A = 100\) and \(x = 90\):
\[ \ln\left(\frac{100}{10}\right) = kt \]
This simplifies to:
\[ \ln(10) = kt \]

### Step 4: Relate \(k\) to \(t_{1/2}\)
From the half-life expression, we can express \(k\) in terms of \(t_{1/2}\):
\[ k = \frac{0.693}{t_{1/2}} \]
Substituting this into our equation gives:
\[ \ln(10) = \frac{0.693}{t_{1/2}} \cdot t \]

### Step 5: Solve for time \(t\)
Rearranging the equation to solve for \(t\):
\[ t = \frac{t_{1/2} \cdot \ln(10)}{0.693} \]
Using the value of \(\ln(10) \approx 2.303\):
\[ t = \frac{t_{1/2} \cdot 2.303}{0.693} \]
Calculating this gives:
\[ t \approx 3.32 \cdot t_{1/2} \]

### Conclusion
The time taken for 90% of a first-order reaction to be completed is approximately:
\[ t \approx 3.3 \cdot t_{1/2} \]

### Final Answer
Thus, the answer is \(3.3\) times the half-life of the reaction.

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