Why do rates of reaction change with pH?

1 Answer
Truong-Son N.
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

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