The half-life period of a \( 1^{s t} \) order reaction is \( 60 \) minutes. What percentage will be left over after
\( 240 \) minutes?

For first order reaction, half-life is,
\( t_{1 / 2}=\frac{0.693}{k} \)
\( 60=\frac{0.693}{k} \)
\( k=\frac{0.693}{60} \)
\( t=\frac{2.303}{k} \log \frac{[A]_{0}}{[A]} \)
\( 240=\frac{2.303}{0.693} \times 60 \log \frac{[A]_{0}}{[A]} \)
\( \log \frac{[A]_{0}}{[A]}=\frac{240 \times 0.693}{2.303 \times 60} \)
\( \log \frac{[A]_{0}}{[A]}=1.20 \)
\( \frac{a}{a-\chi}=\frac{[A]_{0}}{[A]}= \) Antilog \( 1.20 \)
\( \frac{a}{a-x}=15.98 \)
\( a=15.98 a-15.98 x \)
\( 15.98 x=15.98 a-a \)
\( x=\frac{14.98 a}{15.98} \)
\( =0.9375 a \)
\( \% \) convertion \( =\frac{0.9375 a \times 100}{a} \)
\( =93.75 \)
The percentage left over \( =100-93.75 \)