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The rate of a certain reaction is given by, rate $=k\left[\mathrm{H}^{+}\right]^{n}$. The rate increases 100 times when the pH changes from 3 to 1 . The order $(n)$ of the reaction is
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Given, rate $=k\left[\mathrm{H}^{*}\right]^{n}$
Initial $\mathrm{pH}=3 \mathrm{so}\left[\mathrm{H}^{+}\right]=1 \times 10^{-3}$, initial rate $=\mathrm{r}_{\mathrm{1}}$
Final $\mathrm{pH}=1$ so $\left[\mathrm{H}^{+}\right]=1 \times 10^{-1}$
Final rate $=r_{2}=100 r_{1}$
On substituting values, we get
$$
\begin{array}{l}
r_{1}=k\left[1 \times 10^{-3}\right]^{n} \\
r_{2}=100 r_{1}=k\left[10^{-1}\right]^{n}
\end{array}
$$
Dividing eqs (i) by (ii).
$$
\frac{r_{1}}{100 r_{1}}=\left[\frac{1 \times 10^{-3}}{1 \times 10^{-1}}\right]^{n}
$$
$\therefore$
$$
\begin{array}{c}
\frac{1}{100}=\left[\frac{10^{-2}}{1}\right]^{n} \Rightarrow\left[\frac{1}{100}\right]^{1}=\left[\frac{1}{100}\right]^{n} \\
n=1
\end{array}
$$
Thus, the reaction is of first order.
Initial $\mathrm{pH}=3 \mathrm{so}\left[\mathrm{H}^{+}\right]=1 \times 10^{-3}$, initial rate $=\mathrm{r}_{\mathrm{1}}$
Final $\mathrm{pH}=1$ so $\left[\mathrm{H}^{+}\right]=1 \times 10^{-1}$
Final rate $=r_{2}=100 r_{1}$
On substituting values, we get
$$
\begin{array}{l}
r_{1}=k\left[1 \times 10^{-3}\right]^{n} \\
r_{2}=100 r_{1}=k\left[10^{-1}\right]^{n}
\end{array}
$$
Dividing eqs (i) by (ii).
$$
\frac{r_{1}}{100 r_{1}}=\left[\frac{1 \times 10^{-3}}{1 \times 10^{-1}}\right]^{n}
$$
$\therefore$
$$
\begin{array}{c}
\frac{1}{100}=\left[\frac{10^{-2}}{1}\right]^{n} \Rightarrow\left[\frac{1}{100}\right]^{1}=\left[\frac{1}{100}\right]^{n} \\
n=1
\end{array}
$$
Thus, the reaction is of first order.
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