JEE Main 2024 — 29 January, Shift 2 — Question 81

Half life of bromine $-82=36$ hours

\begin{array}{*{35}{r}}{} & {{t}_{1/2}}=\frac{0.693}{K} \\{} & K=\frac{0.693}{36}=0.01925\text{h}{{\text{r}}^{-1}} \\{} & {{1}^{\text{st }\!\!~\!\!\text{ }}}\text{ }\!\!~\!\!\text{ order }\!\!~\!\!\text{ rxn }\!\!~\!\!\text{ kinetic }\!\!~\!\!\text{ equation }\!\!~\!\!\text{ } \\{} & t=\frac{2.303}{\text{ }\!\!~\!\!\text{ K}}\text{log}\frac{\text{a}}{\text{a}-\text{x}} \\{} & \text{log}\frac{\text{a}}{\text{a}-\text{x}}=\frac{\text{t}\times \text{K}}{2.303}\left( \text{t}=1\text{ }\!\!~\!\!\text{ day }\!\!~\!\!\text{ }=24\text{hr} \right) \\{} & \text{log}\frac{\text{a}}{\text{a}-\text{x}}=\frac{24\text{hr}\times 0.01925\text{h}{{\text{r}}^{-1}}}{2.303} \\{} & \text{log}\frac{\text{a}}{\text{a}-\text{x}}=0.2006 \\{} & \frac{\text{a}}{\text{a}-\text{x}}=\text{ }\!\!~\!\!\text{ anti }\!\!~\!\!\text{ log }\!\!~\!\!\text{ }\left( 0.2006 \right) \\{} & \frac{\text{a}}{\text{a}-\text{x}}=1.587 \\{} & \text{ }\!\!~\!\!\text{ If }\!\!~\!\!\text{ a}=1 \\{} & \frac{1}{1-\text{x}}=1.587\Rightarrow 1-\text{x}=0.6301=\text{ }\!\!~\!\!\text{Fraction }\!\!~\!\!\text{ remain }\!\!~\!\!\text{ } \\ {} & \text{ }\!\!~\!\!\text{ after }\!\!~\!\!\text{ one}\!\!~\!\!\text{day}\!\!~\!\!\text{} \\\end{array}