Question

A first order reaction takes 10 min for 20% decomposition. Calculate the time taken for the reaction to go to 80% decomposition. (Given, \( \log_{10}2 = 0.3010 \))

Hint:

Use \( k = \frac{2.303}{t}\log\frac{[A]_0}{[A]} \). Find k from 20% (remaining 80%), then use it to find t for 80% decomposition (remaining 20%).

Answer 1

Step 1: Because \(k\) is the same throughout a first order reaction, the ratio of two times equals the ratio of the two log terms. Take the working formula \(t = \dfrac{2.303}{k}\log\dfrac{[A]_0}{[A]}\) and divide the 80% case by the 20% case to cancel \(\dfrac{2.303}{k}\):
\[ \frac{t_{80}}{t_{20}} = \frac{\log(100/20)}{\log(100/80)} = \frac{\log 5}{\log 1.25} \]
Step 2: Evaluate the logs using \(\log 2 = 0.3010\). \(\log 5 = 1 - \log 2 = 0.6990\). \(\log 1.25 = \log 5 - \log 4 = 0.6990 - 0.6020 = 0.0970\).
Step 3: So \(\dfrac{t_{80}}{t_{20}} = \dfrac{0.6990}{0.0970} = 7.206\).
Step 4: With \(t_{20} = 10\) min,
\[ t_{80} = 10 \times 7.206 = 72.06\ \text{min} \]
\[ \boxed{t_{80} \approx 72\ \text{minutes}} \]
This shortcut avoids computing \(k\) explicitly and gives the same answer.