A first order reaction is 50% completed in `1.26 xx 10^(14)`s. How much time would it take for 100% completion ?

Options

A. `1.26 xx 10^(15)s`
B. `2.52 xx 10^(14) s`
C. `2.52 xx 10^(28)s`
D. Infinite

Explanation

To solve the problem of how much time it would take for a first-order reaction to reach 100% completion, we can follow these steps:

### Step 1: Understand the concept of half-life in first-order reactions
In a first-order reaction, the half-life (t₁/₂) is the time required for the concentration of a reactant to decrease to half of its initial concentration. The half-life is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
where \( k \) is the rate constant.

### Step 2: Given data
From the problem, we know that the reaction is 50% completed in:
\[ t_{1/2} = 1.26 \times 10^{14} \text{s} \]

### Step 3: Relate half-life to the rate constant
Using the half-life formula, we can find the rate constant \( k \):
\[ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{1.26 \times 10^{14}} \]

### Step 4: Calculate the rate constant \( k \)
Now we can calculate \( k \):
\[ k = \frac{0.693}{1.26 \times 10^{14}} \approx 5.49 \times 10^{-15} \text{s}^{-1} \]

### Step 5: Understanding completion in first-order reactions
For a first-order reaction, it is important to note that it never truly reaches 100% completion in a finite amount of time. The concentration of the reactant approaches zero asymptotically but never actually becomes zero.

### Step 6: Conclusion
Since a first-order reaction can never reach 100% completion in a finite time, we conclude that the time required for 100% completion is infinite.

### Final Answer
The time taken for 100% completion of a first-order reaction is infinite.

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