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In a first order reaction, molar concentration of a reactant decreases from $0.1$ to $0.01$ in $100 \mathrm{~s}$. The rate constant of the reaction is
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Verified Answer
The correct answer is:
$0.02303$
$\left[A_{0}\right]=0.1$
$$
\begin{aligned}
{\left[A_{t}\right] } &=0.01 \\
t &=100 \mathrm{~s}
\end{aligned}
$$
1st order reaction,
$$
\begin{aligned}
k &=\frac{2303}{t} \log \frac{\left[A_{0}\right]}{\left[A_{t}\right]} \\
&=\frac{2303}{100} \log \frac{0.1}{0.01}=0.02303 \mathrm{~s}^{-1}
\end{aligned}
$$
$$
\begin{aligned}
{\left[A_{t}\right] } &=0.01 \\
t &=100 \mathrm{~s}
\end{aligned}
$$
1st order reaction,
$$
\begin{aligned}
k &=\frac{2303}{t} \log \frac{\left[A_{0}\right]}{\left[A_{t}\right]} \\
&=\frac{2303}{100} \log \frac{0.1}{0.01}=0.02303 \mathrm{~s}^{-1}
\end{aligned}
$$
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