Question

A first order reaction complete \(60\)% in \(20\) minutes. How long will the reaction take to complete \(84\)%?

• A. \(54\) min
• B. \(68\) min
• C. \(40\) min
• D. \(76\) min

Hint:

Find $k$ from the 60% data, then use $t=\frac{2.303}{k}\log\frac{100}{100-x}$.

Answer 1

The Correct Option is C

Step 1: Find \(k\)
For first order, \(k=\frac{2.303}{t}\log\frac{a}{a-x}\). After \(60\%\) completion, \(a-x=40\) when \(a=100\).
\[ k=\frac{2.303}{20}\log\frac{100}{40}=\frac{2.303\times0.3979}{20}=0.0458\text{ min}^{-1} \]

Step 2: Time for 84%
Now \(a-x=16\).
\[ t=\frac{2.303}{0.0458}\log\frac{100}{16}=\frac{2.303\times0.7959}{0.0458}=40\text{ min} \]

Step 3: Check
Option (C) is \(40\) min. Others come from assuming linear progress.

Final Answer:
The reaction takes \(40\) min, option (C). \[ \boxed{\text{(C) }40\text{ min}} \]