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 (First order formula): For a first order reaction the rate constant is
\[ k = \frac{2.303}{t}\log\frac{[A]_0}{[A]} \]
where \( [A]_0 \) is the initial concentration and \( [A] \) is the concentration left after time \( t \).
Step 2 (Find k from the 20% data): Let \( [A]_0 = 100 \). After 20% decomposition, \( [A] = 80 \), so \( [A]_0/[A] = 100/80 = 1.25 \).
\[ k = \frac{2.303}{10}\log 1.25 \]
\( \log 1.25 = \log\frac{5}{4} = 1 - 3\log 2 = 1 - 3(0.3010) = 1 - 0.9030 = 0.0970 \).
\[ k = \frac{2.303}{10}(0.0970) = 0.02234\ \text{min}^{-1} \]
Step 3 (Time for 80% decomposition): Now \( [A] = 20 \), so \( [A]_0/[A] = 100/20 = 5 \).
\[ t = \frac{2.303}{k}\log 5 \]
\( \log 5 = 1 - \log 2 = 1 - 0.3010 = 0.6990 \).
Step 4 (Substitute):
\[ t = \frac{2.303}{0.02234}(0.6990) = (103.09)(0.6990) = 72.06\ \text{min} \]
\[ \boxed{t \approx 72.1\ \text{minutes}} \]