 Reversible catalytic relaxation

Here is an explicitly stipulated elementary kinetic model, not experimental data. A well-mixed ideal closed system has constant temperature and volume. Alongside an uncatalysed elementary reaction, allow an illustrative catalytic encounter that returns C immediately:Both channels obey the law of mass action. In the second, the forward reaction rate is ${\alpha }_f[\mathrm{A}][\mathrm{C}]$ and the reverse reaction rate is ${\alpha }_r[\mathrm{B}][\mathrm{C}]$. C cancels in each encounter, so its free concentration $c_C$ stays constant without an external supply. This toy elementary reaction encodes regeneration without binding; it is not a claim that real catalytic cycles universally consist of one encounter. A binding cycle can accumulate a reaction intermediate, in which case constant total C does not imply constant free $c_C$.

Adding the parallel reaction rates gives effective first-order reaction rate constantsBecause $k_f^u/k_r^u={\alpha }_f/{\alpha }_r=3$, each channel balances at the same $[\mathrm{B}]/[\mathrm{A}]=3$, and so does their sum. At $c_C=0$ the pair is $(0.030,0.010){\mathrm{s}}^{-1}$; at $c_C=0.001\mathrm{mol}{\mathrm{L}}^{-1}$ it is $(0.300,0.100){\mathrm{s}}^{-1}$. Both give $K=3$ with ideal thermodynamic activities and a common concentration standard state. At $T=300\mathrm{K}$ their compatible standard reaction change in Gibbs free energy is ${\Delta }_r{G}^{\circ }=-RT\ln 3\approx -2.7403\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Choosing an unrelated reverse rate constant would break this consistency.

Let $S=[\mathrm{A}]+[\mathrm{B}]$ be the conserved concentration, and define the dimensionless quantity $x=[\mathrm{B}]/S$. ThenSolving this first-order linear differential equation givesThe chemical relaxation time is $25\mathrm{s}$ without C and $2.5\mathrm{s}$ with the chosen C. From either $x_0=0$ or $x_0=1$, the exponential decay of the deviation leads to the same chemical equilibrium $x=3/4$. Reducing that deviation to one percent takes $\ln (100)\tau$: about $115.13\mathrm{s}$ versus $11.51\mathrm{s}$.

At this chemical equilibrium, conversion continues in both directions. For $S=1\mathrm{mol}{\mathrm{L}}^{-1}$, the uncatalysed channel has equal forward and reverse reaction rates $0.0075\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$. Adding C gives total directional reaction rates $0.075\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$ each, yet zero net reaction rate. Faster molecular exchange and unchanged composition coexist. This exact calculation explains how catalysis can matter enormously within an observation window while leaving the final equilibrium constant intact.