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Take the logarithm to base ten and write pressure in units of $1\mathrm{bar}$, so its argument is dimensionless. Continuity of the atmospheric pressure-temperature profile givesThusIf the logarithm means $\ln$, the equivalent constant is $C=900/\ln 100\simeq 195.4\mathrm{K}$. A plot of temperature against logarithmic pressure is vertical in each isothermal region and straight between the two endpoints.

The processes can be organized by the supplied pressure ranges, although the exact boundaries require reaction rates, irradiation and mixing information:

The profile identifies plausible chemical regimes, but does not fix their transition pressures by itself. In particular, cloud formation depends on the species-specific condensation curve, and ultraviolet processing depends on shielding.

Assume a hydrogen-helium perfect gas with mean particle mass $2.3m_H$, Jupiter gravity $g\simeq 24.8\mathrm{m}{\mathrm{s}}^{-2}$, and mixing over one atmospheric scale height. At $T=1000\mathrm{K}$,The eddy mixing time is ${\tau }_{\mathrm{mix}}\simeq H^2/K_{zz}$. A chemical quench level at one bar requires ${\tau }_{\mathrm{mix}}\lesssim {\tau }_{\mathrm{chem}}=10^5\mathrm{s}$, hence the vertical eddy diffusion coefficient must be of orderThe number depends quadratically on the assumed mixing length; using a fraction of $H$ reduces it accordingly.

Above the chemical quench level, neglect photochemistry, condensation, and molecular diffusion. The quenched atmospheric mixing ratio $f_A=n_A/n$ is approximately constant, while the number density isThus the abundance fraction is frozen, but the absolute number density falls with pressure and altitude. Examples are carbon monoxide–methane quenching, through $\mathrm{CO}+3{\mathrm{H}}_2\rightleftharpoons {\mathrm{CH}}_4+{\mathrm{H}}_2\mathrm{O}$, and nitrogen–ammonia quenching, through ${\mathrm{N}}_2+3{\mathrm{H}}_2\rightleftharpoons 2{\mathrm{NH}}_3$.