At $30^{\circ} \mathrm{C}$, the half life for the decomposition of $\mathrm{AB}_{2}$ is $200 \mathrm{~s}$ and is independent of the initial concentration of $\mathrm{AB}_{2}$. The time required for $80 \%$ of the $\mathrm{AB}_{2}$ to decompose is
Given: $\log 2=0.30$ $\quad \log 3=0.48$
Solution
Since, half-life is independent of the initial concentration of $A B_{2}$. Hence, the reaction is "First Order".
<br/><br/>
$k=\frac{2.303 \log 2}{t_{1 / 2}}$
<br/><br/>
$\frac{2.303 \log 2}{t_{1 / 2}}=\frac{2.303}{t} \log \frac{100}{(100-80)}$
<br/><br/>
$\frac{2.303 \times 0.3}{200}=\frac{2.303}{t} \log 5$
<br/><br/>
$t=467 \mathrm{~s}$
About this question
Subject: Chemistry · Chapter: Chemical Kinetics · Topic: Rate of Reaction
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