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

The intermediate balance isSetting this derivative approximately to zero under the quasi-steady-state approximation givesBoth dissociation and reaction with NO remove the intermediate. Its steady concentration is not zero and does not require the first reversible pair to be at equilibrium.

The last elementary reaction creates two nitrogen dioxide molecules per event, soThe factor of two is required for the production rate requested. The conventional reaction rate for $2NO+{\mathrm{O}}_2\to 2N{\mathrm{O}}_2$ would be one half of this production rate.

“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.