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The half-life of two samples are $0.1$ and $0.8 \mathrm{~s}$. Their respective concentration are 400 and 50 respectively. The order of the reaction is
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It is known,
$\frac{\left(\mathrm{t}_{1 / 2}\right)_{1}}{\left(\mathrm{t}_{1 / 2}\right)_{2}}=\left[\frac{\mathrm{a}_{2}}{\mathrm{a}_{1}}\right]^{(\mathrm{n}-1)}$
Here, $\mathrm{n}=$ order of the reaction
Given, $\left(\mathrm{t}_{1 / 2}\right)_{1}=0.1 \mathrm{~s}, \mathrm{a}_{1}=400$ $\left(\mathrm{t}_{1 / 2}\right)_{2}=0.8 \mathrm{~s}, \mathrm{a}_{2}=50$
On putting the values,
$\frac{0.1}{0.8}=\left[\frac{50}{400}\right]^{(\mathrm{n}-1)}$
Taking log on both sides
$\begin{array}{l}
\log \frac{0.1}{0.8}=(\mathrm{n}-1) \log \frac{50}{400} \\
\log \frac{1}{8}=(\mathrm{n}-1) \log \frac{1}{8} \\
\mathrm{n}-1=1 \Rightarrow \mathrm{n}=2
\end{array}$
$\frac{\left(\mathrm{t}_{1 / 2}\right)_{1}}{\left(\mathrm{t}_{1 / 2}\right)_{2}}=\left[\frac{\mathrm{a}_{2}}{\mathrm{a}_{1}}\right]^{(\mathrm{n}-1)}$
Here, $\mathrm{n}=$ order of the reaction
Given, $\left(\mathrm{t}_{1 / 2}\right)_{1}=0.1 \mathrm{~s}, \mathrm{a}_{1}=400$ $\left(\mathrm{t}_{1 / 2}\right)_{2}=0.8 \mathrm{~s}, \mathrm{a}_{2}=50$
On putting the values,
$\frac{0.1}{0.8}=\left[\frac{50}{400}\right]^{(\mathrm{n}-1)}$
Taking log on both sides
$\begin{array}{l}
\log \frac{0.1}{0.8}=(\mathrm{n}-1) \log \frac{50}{400} \\
\log \frac{1}{8}=(\mathrm{n}-1) \log \frac{1}{8} \\
\mathrm{n}-1=1 \Rightarrow \mathrm{n}=2
\end{array}$
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