# Why do rates of reaction change with pH?

May 25, 2018

Do they really?

A counterexample is:

${\text{N"_2"O"_4(g) rightleftharpoons 2"NO}}_{2} \left(g\right)$

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

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

${r}_{f w d} \left(t\right) = {k}_{f w d} \left[{\text{N"_2"O}}_{4}\right]$

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

${r}_{r e v} \left(t\right) = {k}_{r e v} {\left[{\text{NO}}_{2}\right]}^{2}$

Clearly, no $\left[{\text{H}}^{+}\right]$ and no $\left[{\text{OH}}^{-}\right]$ appears in either rate law.

Thus, the reaction is completely $\text{pH}$-independent.

""^([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