\(t_{99.9\%}\) with respect to \(t_{90\%}\) for a first order reaction is:

Three

The correct answer is option 3. Three.

We know, for a first order reaction

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

For \(t_{99.9\%}\)

Let, \(a =100\) and \(x = 99.9\)

\(∴ a - x  = 100- 99.9 = 0.1\)

Therefore, equation (i) becomes

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

or, \(t_{99.9\%} = \frac{2.303}{k}log (1000)\)

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

For \(t_{90\%}\)

Let, \(a =100\) and \(x = 90\)

\(∴ a - x  = 100- 90 = 10\)

Therefore, equation (i) becomes

\(t_{90\%} = \frac{2.303}{k}log \frac{100}{10}\)

or, \(t_{90\%} = \frac{2.303}{k}log (10)\)

or, \(t_{90\%} = \frac{2.303}{k} × 1\, \  ------(iii)\)

Dividing equation (ii) by equation (iii)

\(\frac{t_{99.9\%}}{t_{90\%}} = \frac{\frac{2.303}{k} × 3}{\frac{2.303}{k} × 1}\)

or, \(\frac{t_{99.9\%}}{t_{90\%}} = \frac{3}{1}\)

or, \(t_{99.9\%} = 3 t_{90\%}\)