Question

For a first order reaction, the time required for $99.9\%$ of the reaction to take place is $n$ times the time required for half of the reaction to take place.
What is the value of $n$ to the nearest integer?
(Take $lo{g}_{10}2=0.3$)

• A. $2$
• B. $10$
• C. $0.5$
• D. $100$

Answer 1

The correct option is B $10$

For a first order reaction, $t=\frac{2.303}{k}lo{g}_{10}\frac{[A{]}_a}{[A{]}_t}=\frac{2.303}{k}lo{g}_{10}\frac{a}{a-x}=\frac{2.303}{k}lo{g}_{10}\frac{1}{1-f}$
Here f is the fraction of a reaction completed in time, t.
$t_{50\%}=\frac{2.303}{k}lo{g}_{10}\frac{1}{1-0.5}=\frac{2.303}{k}lo{g}_{10}2...\left(i\right)t_{99.9\%}=\frac{2.303}{k}lo{g}_{10}\frac{1}{1-0.999}=\frac{2.303}{k}lo{g}_{10}\frac{1}{0.001}=\frac{2.303}{k}lo{g}_{10}\frac{1}{{10}^{-3}}=\frac{2.303}{k}lo{g}_{10}{10}^3...\left(ii\right)$
On dividing equation (ii) by (i) we get
$\frac{t_{99.9\%}}{t_{50\%}}=\frac{2.303}{k}\times \frac{k}{2.303}\times log\frac{{10}^3}{log2}=\frac{3.0}{0.3}=10$
Hence, (b) is the correct option.