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

Explanation

To solve the question regarding the time required for a first-order reaction to reach 100% completion, we follow these steps:

### Step-by-Step Solution:

1. **Understanding First-Order Reactions**:
- A first-order reaction is characterized by a constant rate that depends on the concentration of one reactant. Importantly, first-order reactions never truly reach 100% completion in a finite amount of time.

2. **Given Information**:
- The reaction is 50% completed in \( t_{1/2} = 1.26 \times 10^{14} \) seconds.

3. **Half-Life of a First-Order Reaction**:
- The half-life (\( t_{1/2} \)) of a first-order reaction is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
- Rearranging this gives us the rate constant \( k \):
\[ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{1.26 \times 10^{14}} \approx 5.49 \times 10^{-15} \, \text{s}^{-1} \]

4. **Calculating Time for 100% Completion**:
- For a first-order reaction, the time required to reach 100% completion theoretically approaches infinity. This is because as the concentration of the reactant approaches zero, the time taken to convert the remaining reactant to product increases indefinitely.

5. **Conclusion**:
- Therefore, while we can calculate the time for a significant percentage (like 99.9%), the time for 100% completion is infinite:
\[ \text{Time for 100% completion} = \infty \]

### Summary:
- A first-order reaction will never reach 100% completion in a finite time. The time required approaches infinity.