Chemical Kinetics Question 6
Question 6 - 2024 (29 Jan Shift 2)
The half-life of radioisotopic bromine - 82 is 36 hours. The fraction which remains after one day is ________ $\times 10^{-2}$
(Given antilog $0.2006=1.587$)
Show Answer
Answer (63)
Solution
Half life of bromine $-82=36$ hours
$\mathrm{t}_{1 / 2}=\frac{0.693}{\mathrm{~K}}$
$\mathrm{K}=\frac{0.693}{36}=0.01925 \mathrm{hr}^{-1}$
$1^{\text {st }}$ order rxn kinetic equation
$t=\frac{2.303}{K} \log \frac{a}{a-x}$
$\log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}=\frac{\mathrm{t} \times \mathrm{K}}{2.303} \quad(\mathrm{t}=1$ day $=24 \mathrm{hr})$
$\log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}=\frac{24 \mathrm{hr} \times 0.01925 \mathrm{hr}^{-1}}{2.303}$
$\log \frac{\mathrm{a} t h}{\mathrm{a}-\mathrm{x}}=0.2006$
$\frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}=\operatorname{anti} \log (0.2006)$
$\frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}=1.587$
If $\mathrm{a}=1$
$\frac{1}{1-\mathrm{x}}=1.587 \Rightarrow 1-\mathrm{x}=0.6301=$ Fraction remain after one day
