Question

A first order reactions has half life of 14.5 hrs. What percentage of the reactant will remain after 24 hrs?
use antilog(0.498)=3.147

• A. 18.3%
• B. 31.8%
• C. 45.5%
• D. 68.2%

Answer 1

The correct option is B 31.8%

Given, $n=1,t_{1/2}=14.5\text{hrs},t=24hrs$
Let initial amout be a=100
Applying first order formula
$k=\frac{2.303}{t}log\left(\frac{a}{a-x}\right)$
$\frac{0.693}{14.5}=\frac{2.303}{24}log\frac{100}{(a-x)}$
$log\left(\frac{100}{a-x}\right)=0.498$
$\frac{100}{a-x}=3.147$
$\frac{100}{3.147}=a-x$
On solving, (a - x ) = 31.77%
Hence option (b) is correct.