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(C) Given the rate law: $Rate = k[H^{+}]^n$. At $pH = 3$, the concentration of hydrogen ions is $[H^{+}]_1 = 10^{-3} \ M$. At $pH = 1$, the concentrat

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(C) Given the rate law: $Rate = k[H^{+}]^n$. At $pH = 3$, the concentration of hydrogen ions is $[H^{+}]_1 = 10^{-3} \ M$. At $pH = 1$, the concentration of hydrogen ions is $[H^{+}]_2 = 10^{-1} \ M$. It is given that the rate increases $100$ times, so $Rate_2 = 100 	imes Rate_1$. Using the ratio: $\frac{Rate_2}{Rate_1} = \left(\frac{[H^{+}]_2}{[H^{+}]_1}ight)^n$. Substituting the values: $100 = \left(\frac{10^{-1}}{10^{-3}}ight)^n$. $100 = (10^2)^n$. $10^2 = 10^{2n}$. Comparing the exponents, $2 = 2n$, which gives $n = 1$.

The rate of a certain reaction is given by, $Rate = k[H^{+}]^n$. The rate increases $100$ times when the $pH$ changes from $3$ to $1$. The order $(n)$ of the reaction is:

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