= Reversible catalytic relaxation
{title2=$x(t)=x_{\mathrm{eq}}+(x_0-x_{\mathrm{eq}})e^{-t/\tau}$}
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:
$$
\begin{aligned}
\mathrm A&\rightleftharpoons\mathrm B,
& k_f^u&=0.030\,\mathrm{s^{-1}},
& k_r^u&=0.010\,\mathrm{s^{-1}},\\
\mathrm A+\mathrm C&\rightleftharpoons\mathrm B+\mathrm C,
& \alpha_f&=270\,\mathrm{L\,mol^{-1}\,s^{-1}},
& \alpha_r&=90\,\mathrm{L\,mol^{-1}\,s^{-1}}.
\end{aligned}
$$
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 constants>
$$
k_f=k_f^u+\alpha_f c_C,
\qquad k_r=k_r^u+\alpha_r c_C.
$$
Because $k_f^u/k_r^u=\alpha_f/\alpha_r=3$, 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 $c_C=0.001\,\mathrm{mol\,L^{-1}}$ 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 $\Delta_rG^\circ=-RT\ln3\approx-2.7403\,\mathrm{kJ\,mol^{-1}}$. 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> $x=[\mathrm B]/S$. Then
$$
\frac{dx}{dt}=k_f(1-x)-k_rx
=-(k_f+k_r)(x-x_{\mathrm{eq}}),
\qquad
x_{\mathrm{eq}}=\frac{k_f}{k_f+k_r}=\frac34.
$$
Solving this <first-order linear differential equation> gives
$$
x(t)=\frac34+\left(x_0-\frac34\right)e^{-(k_f+k_r)t},
\qquad
\tau=\frac{1}{k_f+k_r}.
$$
The <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 $11.51\,\mathrm s$.
\Image[/catalysis-reversible-equilibrium-relaxation.png]
{title=Two initial compositions and two kinetic speeds approach the same <chemical equilibrium>}
{id=image-catalysis-reversible-equilibrium-relaxation}
{description=Exact curves from the stipulated <reversible catalytic relaxation> model. Solid curves include $0.001\,\mathrm{mol\,L^{-1}}$ of C; dashed curves omit C. Blue starts entirely as A and orange entirely as B. Both panels use the same equations; the right panel resolves the first $15\,\mathrm s$. The common dotted line is $[\mathrm B]/([\mathrm A]+[\mathrm B])=3/4$, not a fitted plateau or an experimental measurement.}
At this <chemical equilibrium>, conversion continues in both directions. For $S=1\,\mathrm{mol\,L^{-1}}$, the uncatalysed channel has equal forward and reverse <reaction rates> $0.0075\,\mathrm{mol\,L^{-1}\,s^{-1}}$. Adding C gives total directional <reaction rates> $0.075\,\mathrm{mol\,L^{-1}\,s^{-1}}$ 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.
Back to article page
