t_99.9% with respect to t_90% for a first-order reaction is :

The correct answer is option 4. Tripled.

We know, that for a first-order reaction,

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

where,

\(k\) is the rate constant,

\(t\) is time

\(a\) is the initial concentration of the reactant

\((a − x)\) is the final concentration of the reactant

\(x\) is the amount of reactant consumed during the course of the reaction

Let \(a =100\), then

For, \(t_{99.9\%}\), \(x = 99.9\), so, the equation becomes

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

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

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

For, \(t_{90\%}\), \(x = 90\), so, the equation becomes

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

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

\(⇒ t_{90\%} = \frac{2.303}{k} × 1 -----(2)\)

Dividing equation (2) by equation (1), we get

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

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

Hence, the rate of reaction will be doubled. So, the correct answer is (4) Tripled.