Practicing Success
The time taken for the completion of 90% of a first-order reaction is ‘t’ min. What is the time (in seconds) taken for the completion of 99% of the reaction? (Consider t = 1 min or 60 sec) |
2sec \(\frac{t}{30}\) 120 sec 60 sec |
120 sec |
The correct answer is option 3. 120 sec. Case I: Given, the time taken for the completion of 90% of a first-order reaction is ‘t’ min As the reaction is 90% complete, \(a = 100, \text{ and }a − x = 100 −90 = 10\) So, \(t_{90} = \frac{2.303}{k}log\frac{100}{10}\) or, \(t = \frac{2.303}{k}log(10)\) [since t90 = t min] or, \(t = \frac{2.303}{k} × 1\) ------(i) Case II: When the reaction is 99% complete, then \(a = 100, \text{ and }a − x = 100 − 99 = 1\) So, \(t_{99} = \frac{2.303}{k}log\frac{100}{1}\) or, \(t_{99} = \frac{2.303}{k}log(100)\) or, \(t_{99} = \frac{2.303}{k} × 2\) ------(ii) Dividing equation (ii) by (i), we get \(\frac{t_{99}}{t} = \frac{2}{1}\) or, \(\frac{t_{99}}{t} =2min\) or, \(t_{99} = 2t\text{ min}\) Since \(t = 60 sec\) \(∴ t_{99} = 120 \text{ sec}\) |