Question

For a first order reaction, if $t_{3/4}$ and $t_{1/2}$ are time required for completion of 3/4 decay and 1/4 decay, then, $t_{3/4}=t_{1/2}\times n$. What is the value of $n$?

Answer 1

If $t_{3/4}$ and $t_{1/2}$ are time required for completion of 3/4 decay and 1/4 decay then $t_{3/4}=t_{1/2}\times n$, then $n$ is 2.
For 3/4 decay, $t_{3/4}=\frac{2.303}{k}log\frac{100}{25}=\frac{1.39}{k}$
For 1/2 decay, $t_{1/2}=\frac{2.303}{k}log\frac{100}{50}=\frac{0.693}{k}$
$n=\frac{t_{3/4}}{t_{1/2}}=\frac{1.39}{0.693}=2$