Practicing Success
\(t_{99.9\%}\) with respect to \(t_{90\%}\) for a first-order reaction is : |
Four times of \(t_{50\%}\) One and half time It is same Tripled |
Tripled |
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. |