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, we need to understand the characteristics of a first-order reaction and how its kinetics work.

### Step-by-Step Solution:

1. **Understanding the Half-Life of a First-Order Reaction**:
- The half-life (t₁/₂) of a first-order reaction is the time required for the concentration of the reactant to decrease to half of its initial value. It is given that the half-life is \(1.26 \times 10^{14}\) seconds.

2. **Recognizing the Nature of First-Order Reactions**:
- In a first-order reaction, the reactant is never completely consumed. Each half-life results in half of the remaining reactant being converted into products. Therefore, after one half-life, 50% of the reactant remains; after two half-lives, 25% remains; after three half-lives, 12.5% remains, and so on.

3. **Calculating Time for 100% Completion**:
- Since the first-order reaction never reaches 100% completion, we can conclude that the time required for complete consumption of the reactant is theoretically infinite. This is because, as time progresses, the amount of reactant approaches zero but never actually reaches zero.

4. **Conclusion**:
- Therefore, the time taken for 100% completion of a first-order reaction is infinite.

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