Question

A substance ' $A$ ' decomposes by a first order reaction starting initially with $[A]=2.00M$ and after $200\min$,$[A]$ becomes $0.15M$. What will be the half life $\left(t_{1/2}\right)$ for this reaction in minutes? Write answer upto two decimal place. (Given $\ln \frac{200}{15}=2.58,\ln 2=0.693$ )

Answer 1

Answer: 53.72

Given initial concentration (a) $=2.00M$;
Time taken $(t)=200\min$ and
final concentration $(a-x)=0.15M$.
For a first order reaction rate constant,
$k=\frac{2.303}{t}\log \frac{a}{a-x}$
$=\frac{2.303}{200}\log \frac{2.00}{0.15}$
$=1.29\times 10^{-2}mi{n}^{-1}$
Further
$\left(t_{1/2}\right)=\frac{0.693}{k}=\frac{0.693}{1.29\times 10^{-2}}$
$=53.72\min .$