A first order reaction is 50% completed in $1.26 × 10^{14} s$. How much time would it take for 100% completion?

infinite

The correct answer is Option (3) → infinite

For a first-order reaction, the time taken for a certain percentage completion is given by:

t = (2.303/k) log(100 / (100 – % completion))

Given:

50% completion → t₅₀% = 1.26 × 10¹⁴ s

At 50% completion:

t = (2.303/k) log(100/50) = (2.303/k) log(2)

⇒ 1.26 × 10¹⁴ = (2.303/k) × 0.693

⇒ k = (2.303 × 0.693) / (1.26 × 10¹⁴) ≈ 1.26 × 10⁻¹⁴ s⁻¹

Now, for 100% completion:

100% completion means reactant concentration → 0

⇒ log(100 / 0) → ∞

⇒ t = ∞

A first-order reaction never reaches 100% completion in finite time.