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The rate of a first order reaction is $1.5 \times$ $10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}$ at $0.5 \mathrm{M}$ concentration of the reactant. The half life of the reaction is:
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Verified Answer
The correct answer is:
$23.1 \mathrm{~min}$
Let the reaction be
$$
\begin{aligned}
& \mathrm{A} \stackrel{k}{\longrightarrow} \text { Product } \\
& \text { Rate }=k[\mathrm{~A}] \\
& 1.5 \times 10^{-2} \mathrm{~mol} \mathrm{\textrm {L } ^ { - 1 } \mathrm { min } ^ { - 1 } = k [ 0 . 5 ]} \\
& k=\frac{1.5 \times 10^{-2}}{0.5} \\
&=3 \times 10^{-2} \mathrm{~min}^{-1} \\
& t_{1 / 2}=\frac{0.693}{k} \\
& \Rightarrow \quad t_{1 / 2}=\frac{0.693}{3 \times 10^{-2}} \\
&=0.231 \times 10^2 \mathrm{~min} \\
& \Rightarrow \quad t_{1 / 2}=23.1 \mathrm{~min}
\end{aligned}
$$
Related Theory
Decay profiles for first-order reactions with large and small rate constants
$$
\begin{aligned}
& \mathrm{A} \stackrel{k}{\longrightarrow} \text { Product } \\
& \text { Rate }=k[\mathrm{~A}] \\
& 1.5 \times 10^{-2} \mathrm{~mol} \mathrm{\textrm {L } ^ { - 1 } \mathrm { min } ^ { - 1 } = k [ 0 . 5 ]} \\
& k=\frac{1.5 \times 10^{-2}}{0.5} \\
&=3 \times 10^{-2} \mathrm{~min}^{-1} \\
& t_{1 / 2}=\frac{0.693}{k} \\
& \Rightarrow \quad t_{1 / 2}=\frac{0.693}{3 \times 10^{-2}} \\
&=0.231 \times 10^2 \mathrm{~min} \\
& \Rightarrow \quad t_{1 / 2}=23.1 \mathrm{~min}
\end{aligned}
$$
Related Theory
Decay profiles for first-order reactions with large and small rate constants
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