JEE MAIN - Chemistry (2022 - 28th July Morning Shift - No. 18)
For the given first order reaction
$$\mathrm{A} \rightarrow \mathrm{B}$$
the half life of the reaction is $$0.3010 \mathrm{~min}$$. The ratio of the initial concentration of reactant to the concentration of reactant at time $$2.0 \mathrm{~min}$$ will be equal to ___________. (Nearest integer)
Answer
100
Explanation
$\mathrm{t}_{1 / 2}=\frac{0.693}{\mathrm{~K}} \quad\quad \mathrm{t}_{1 / 2}$ given $=0.3010$
$$ \begin{aligned} &K=\frac{0.693}{0.3010} \\\\ &K=2.30 \end{aligned} $$
$\mathrm{K}=\frac{2.303}{\mathrm{t}} \log \frac{\left(\mathrm{A}_{0}\right)}{\left(\mathrm{A}_{\mathrm{t}}\right)}$
$A_{0} \rightarrow$ initial concentration of reactant
$A_{t} \rightarrow$ concentration of reactant at time $t$
$2.303=\frac{2.303}{2} \log \frac{\left(A_{0}\right)}{\left(A_{t}\right)}$
$2=\log \frac{\left(A_{0}\right)}{\left(A_{t}\right)}$
$100=\frac{A_{0}}{A_{t}}$
$$ \begin{aligned} &K=\frac{0.693}{0.3010} \\\\ &K=2.30 \end{aligned} $$
$\mathrm{K}=\frac{2.303}{\mathrm{t}} \log \frac{\left(\mathrm{A}_{0}\right)}{\left(\mathrm{A}_{\mathrm{t}}\right)}$
$A_{0} \rightarrow$ initial concentration of reactant
$A_{t} \rightarrow$ concentration of reactant at time $t$
$2.303=\frac{2.303}{2} \log \frac{\left(A_{0}\right)}{\left(A_{t}\right)}$
$2=\log \frac{\left(A_{0}\right)}{\left(A_{t}\right)}$
$100=\frac{A_{0}}{A_{t}}$
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