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 $0.2006=1.587$)
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Answer (63)
Solution
Half life of bromine $-82=36$ hours
$t _{1 / 2}=\frac{0.693}{K}$
$K=\frac{0.693}{36}=0.01925 hr^{-1}$
$1^{\text {st }}$ order rxn kinetic equation
$t=\frac{2.303}{K} \log \frac{a}{a-x}$
$\log \frac{a}{a-x}=\frac{t \times K}{2.303} \quad(t=1$ day $=24 hr)$
$\log \frac{a}{a-x}=\frac{24 hr \times 0.01925 hr^{-1}}{2.303}$
$\log \frac{a t h}{a-x}=0.2006$
$\frac{a}{a-x}=\operatorname{anti} \log (0.2006)$
$\frac{a}{a-x}=1.587$
If $a=1$
$\frac{1}{1-x}=1.587 \Rightarrow 1-x=0.6301=$ Fraction remain after one day
