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If $(\frac{d x}{d t}) = k {[H^{+}]}^{n}$ and rate becomes 100 times when pH changes from 2 to 1. Hence, order is :

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answer is B.

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Detailed Solution

$r_{b} = k {[H^{+}]}^{n}$

$pH = 2, [H^{+}] = 10^{- 2} M$

$r_{0} = k {[10^{- 2}]}^{n} \dots 1$

At $pH = 1, [H^{+}] = 10^{- 1} M$

$r_{1} = k {[10^{- 1}]}^{n} \dots . . . 2$

By equation 1 and 2

$\frac{r_{1}}{r_{0}} = 100 = [10]^{n},$

$\Rightarrow n = 2$

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