If time taken for a first order reaction to get \(90\%\) complete is \(24\, \ min\), its \(t_{99.9\%}\) will be:

72 min

The correct answer is option 4. 72 min.

We know, for a first order reaction,

\(k =  \frac{2.303}{t}log\frac{a}{a - x}\) ------(1)

For \(90\%\) completion of the reaction

Given, \(t_{90\%} = 24\, \ min\)

Let, \(a = 100\) then \(a - x = 100 - 90 =10\)

Applying these values in the equation (1), we get

\(k =  \frac{2.303}{24}log\frac{100}{10}\)

\(⇒ k = \frac{2.303}{24}log (10)\)

\(⇒ k = \frac{2.303}{24}\)

Now, for \(99.9\%\) of the reaction,

\(a - x = 100 - 99.9 = 0.1\)

Thus, equation (1) can be written as

\(k =  \frac{2.303}{t_{99.9\%}}log\frac{100}{0.1}\)

\(⇒ \frac{2.303}{24} = \frac{2.303}{t_{99.9\%}} log (1000)\)

\(⇒ \frac{2.303}{24} = \frac{2.303}{t_{99.9\%}} \times 3\)

\(⇒ t_{99.9\%} = \frac{2.303 \times 24 \times 3}{2.303}\)

\(⇒ t_{99.9\%} = 72\, \ min\)