Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / 4 / b / iii / Solution

“Low NO” means $k_2[\mathrm{NO}]\ll k_{-1}$, or $[\mathrm{NO}]\ll 6.5\times 10^{-6}\mathrm{M}$. ThenThe effective orders are two in NO and one in oxygen, giving overall order three. For the production-rate convention,If the conventional reaction rate $v=\frac{1}{2}d[\mathrm{N}{\mathrm{O}}_2]/dt$ is used, its coefficient is $2.91\times 10^7{\mathrm{M}}^{-2}{\mathrm{s}}^{-1}$. These are the same kinetics under explicitly different normalizations. At high NO the denominator changes, so the effective NO order approaches one rather than remaining two.