Question:
If the rate constant of a first-order reaction is $4.606\times10^{-3}\,\mathrm{s^{-1}}$, then the time taken (in seconds) for the concentration of the reactant to decrease from $1.0\,\mathrm{M}$ to $0.1\,\mathrm{M}$ is:}
If the rate constant of a first-order reaction is $4.606\times10^{-3}\,\mathrm{s^{-1}}$, then the time taken (in seconds) for the concentration of the reactant to decrease from $1.0\,\mathrm{M}$ to $0.1\,\mathrm{M}$ is:}
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For first-order reactions, reducing concentration to one-tenth gives
\[
t=\frac{2.303}{k}.
\]
Updated On: Jun 17, 2026
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The Correct Option is B
Solution and Explanation
Concept:
For a first-order reaction,
\[
k=\frac{2.303}{t}\log\frac{[A]_0}{[A]}.
\]
Step 1: Substitute the given data. \[ k=4.606\times10^{-3}\,\mathrm{s^{-1}} \] \[ [A]_0=1.0 \] \[ [A]=0.1. \] Hence \[ 4.606\times10^{-3} = \frac{2.303}{t} \log\left(\frac{1}{0.1}\right). \]
Step 2: Evaluate logarithm. \[ \log 10=1. \] Thus \[ 4.606\times10^{-3} = \frac{2.303}{t}. \]
Step 3: Calculate time. \[ t=\frac{2.303}{4.606\times10^{-3}} \] \[ t=500\,\mathrm{s}. \] Therefore, \[ \boxed{500\ \mathrm{s}} \]
Step 1: Substitute the given data. \[ k=4.606\times10^{-3}\,\mathrm{s^{-1}} \] \[ [A]_0=1.0 \] \[ [A]=0.1. \] Hence \[ 4.606\times10^{-3} = \frac{2.303}{t} \log\left(\frac{1}{0.1}\right). \]
Step 2: Evaluate logarithm. \[ \log 10=1. \] Thus \[ 4.606\times10^{-3} = \frac{2.303}{t}. \]
Step 3: Calculate time. \[ t=\frac{2.303}{4.606\times10^{-3}} \] \[ t=500\,\mathrm{s}. \] Therefore, \[ \boxed{500\ \mathrm{s}} \]
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