If time taken for a first order reaction to get \(90\%\) complete is \(24\, \ min\), its \(t_{99.9\%}\) will be:
Answer & explanation
Correct answer: option 4
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
\(k = \frac{2.303}{t_{99.9\%}}log\frac{100}{0.1}\)
\(⇒ \frac{2.303}{24} = \frac{2.303}{t_{99.9\%}} log (1000)\)
\(⇒ \frac{2.303}{24} = \frac{2.303}{t_{99.9\%}} \times 3\)
\(⇒ t_{99.9\%} = \frac{2.303 \times 24 \times 3}{2.303}\)
\(⇒ t_{99.9\%} = 72\, \ min\)