Question

Prove that the time taken to complete 99.9% reaction is ten times the half-life period $(t_{1/2})$ for a first-order reaction.

Hint:

For any first-order decay, the fraction remaining is $f=(a-x)/a=10^{-n}$ $\Rightarrow$ $t=(2.303\,n/k)$. Each factor of $10$ in reduction adds $\dfrac{2.303}{k}$ to the time.

Answer 1

For a first-order reaction, the integrated rate law is
\[ k t \;=\; 2.303\,\log\!\left(\frac{a}{a-x}\right) \] where $a$ is the initial amount and $(a-x)$ is the amount remaining at time $t$.
Half-life $t_{1/2$:} at $x=\dfrac{a}{2}$, so $\dfrac{a}{a-x}=2$.
\[ k t_{1/2} \;=\; 2.303 \log 2 \;=\; 0.693 $\Rightarrow$ t_{1/2}=\frac{0.693}{k} \]
Time for 99.9% completion $t_{99.9$:} 99.9% reacted $\Rightarrow$ 0.1% remains, so $(a-x)=0.001\,a$.
Thus $\dfrac{a}{a-x}=\dfrac{a}{0.001\,a}=1000$.
\[ k t_{99.9} \;=\; 2.303 \log 1000 \;=\; 2.303 \times 3 \;=\; 6.909 \] \[ \Rightarrow\; t_{99.9}=\frac{6.909}{k} \]
Ratio:
\[ \frac{t_{99.9}}{t_{1/2}}=\frac{6.909/k}{0.693/k}=\frac{6.909}{0.693}\approx 9.97 \approx 10 \] \[ \boxed{\,t_{99.9}\ \approx\ 10\,t_{1/2}\,} \]