Why do rates of reaction change with pH?

1 Answer
May 25, 2018

Do they really?


A counterexample is:

"N"_2"O"_4(g) rightleftharpoons 2"NO"_2(g)

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

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

r_(fwd)(t) = k_(fwd)["N"_2"O"_4]

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

r_(rev)(t) = k_(rev)["NO"_2]^2

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

Thus, the reaction is completely "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