Which of the following is the correct relationship between the time required for completion of \(99.9\%\) of a first-order reaction and its half-life?

\(t_{99.9\%} = 10 × t_{1/2}\)

The correct answer is option 2. \(t_{99.9\%} = 10 × t_{1/2}\).

For a first-order reaction,

\(t= \frac{2.303}{k}log\left(\frac{a_0}{a_0 - x}\right)\)

Let us consider the initial concentration, \(a_0 = 100\)

the reaction is \(99.9\%\) is complete, so \(x = 99.9\)

So, \(a_0 - x = 100-99.9\)

Then

\(t_{99.9\%}= \frac{2.303}{k}log\left(\frac{100}{100 - 99.9}\right)\)

or, \(t_{99.9\%} = \frac{2.303}{k} × 3\)

or, \(t_{99.9\%} = \frac{6.93}{k}\) ------(i)

Since,

\(t_{1/2} = \frac{0.693}{k}\) ---------(ii)

From equation (i) and (ii) we can write

\(t_{99.9\%} = 10 × t_{1/2}\)