Why do rates of reaction change with pH?

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Why do rates of reaction change with pH?

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Answer 1 · 2018-05-25T23:38:28

Do they really?

A counterexample is:

$N_{2} O_{4} (g) ⇌ 2 {NO}_{2} (g)$ $"N"_2"O"_4(g) rightleftharpoons 2"NO"_2(g)$

The forward reaction has a rate constant of $6.49 \times 10^{5}$ $6.49 xx 10^5$ $s^{- 1}$ $"s"^(-1)$ at $273 K$ $"273 K"$ , and the reverse reaction has a rate constant of $8.85 \times 10^{8} M^{- 1} \cdot s^{- 1}$ $8.85 xx 10^8 "M"^(-1)cdot"s"^(-1)$ at $273 K$ $"273 K"$ . $^{[1]}$ $""^([1])$

The forward reaction is first-order, with a rate law of:

$r_{f w d} (t) = k_{f w d} [N_{2} O_{4}]$ $r_(fwd)(t) = k_(fwd)["N"_2"O"_4]$

The reverse reaction is second-order, with a rate law of:

$r_{r e v} (t) = k_{r e v} {[{NO}_{2}]}_{2}^{2}$ $r_(rev)(t) = k_(rev)["NO"_2]^2$

Clearly, no $[H^{+}]$ $["H"^(+)]$ and no $[{OH}^{-}]$ $["OH"^(-)]$ appears in either rate law.

Thus, the reaction is completely $pH$ $"pH"$ -independent.

$^{[1]}$ $""^([1])$ Markwalder, B.; Gozel, P.; van den Bergh, H., Temperature-jump measurements on the kinetics of association and dissociation in weakly bound systems, J. Chem. Phys., 1992, 97, 5472 − 5479

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Why do rates of reaction change with pH?

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