# Supplemental Information: Rate Constants

The reversible exchange reaction of the form $B \mathrel{\mathop{\leftrightharpoons}^{\mathrm{\textit{ k }_1}}_{\mathrm{\textit{ k }_{-1}}}} A \mathrel{\mathop{\rightleftharpoons}^{\mathrm{\textit{ k }_2}}_{\mathrm{\textit{ k }_{-2}}}} C$ has rate equations given by the following: $\frac{d[A]}{dt} = k_{-1}[B]-k_1[A]-k_2[A]+k_{-2}[C] \\$ $\frac{d[B]}{dt} = k_1[A]-k_{-1}[B] \\$ $\frac{d[C]}{dt} = k_2[A]-k_{-2}[C]$ Note that, for the purposes of this analysis, the inter-conversion of the major isomers is ignored. This is a valid assumption if the rate of inter-conversion of $$B \longleftrightarrow C$$ is small. We estimate from cursory analysis of the data that the rate constants for this process are at least two orders of magnitude smaller than the pathway that proceeds through the minor isomer, $$A$$. However, a steady-state approximation for the system of rate equations is not valid for the minor isomer because its concentration changes appreciably and is not significantly different from those concentrations of either major isomer. Thus, the differential equations in (2)-(4) are solved explicitly. The exact solutions to these equations are: $A(t) = B_0[\frac{k_{-1}k_{-2}}{mn}-\frac{k_{-1}(k_{-2}-m)}{m(m-n)}\mathrm{e}^{-mt}-\frac{k_{-1}(k_{-2}-n)}{n(m-n)}\mathrm{e}^{-nt}] \\$ $B(t) = B_0[\frac{k_{1}k_{-2}}{mn}+\frac{k_{-1}(m-k_2-k_{-2})}{m(m-n)}\mathrm{e}^{-mt}+\frac{k_{-1}(k_2+k_{-2}-n)}{n(m-n)}\mathrm{e}^{-nt}] \\$ $C(t) = B_0[\frac{k_{-1}k_{2}}{mn}-\frac{k_{-1}k_{2}}{m(m-n)}\mathrm{e}^{-mt}+\frac{k_{-1}k_{2}}{n(m-n)}\mathrm{e}^{-nt}]$ Where, $m = \frac{1}{2}(p+q)\\$ $n = \frac{1}{2}(p-q)\\$ $p = k_1+k_{-1}+k_2+k_{-2}\\$ $q = (p^2-4(k_{-1}k_2+k_{1}k_{-2}+k_{-1}k_{-2}))^\frac{1}{2}$ The values of m and n generally differ by less than an order of magnitude within the range of rate constant values expected for Reaction (1). Thus, fits to time-dependent data in the form of $$y(t) = y_0+\Lambda\mathrm{e}^{-\lambda{}t}+K\mathrm{e}^{-\kappa{}t}$$ tend to converge to the form $$y(t) = y_0+\alpha{}\mathrm{e}^{-\beta{}t}$$. We take advantage of this to approximate $$\textit{k}_{-1}$$ by fitting data for major isomer B with a single exponential such that equation (6) may be rearranged as: $B(t) = B_0[\frac{k_{1}k_{-2}}{k_{-1}k_2+k_{1}k_{-2}+k_{-1}k_{-2}}+[\frac{k_{-1}k_2+k_{-1}k_{-2}}{k_{-1}k_2+k_{1}k_{-2}+k_{-1}k_{-2}}]\mathrm{e}^{-x_1t}]$ Similarly, the integrated rate laws for isomers A and C may be rewritten as: $A(t) = B_0[\frac{k_{-1}k_{-2}}{k_{-1}k_2+k_{1}k_{-2}+k_{-1}k_{-2}}+[\frac{k_{-1}k_{-2}}{k_{-1}k_2+k_{1}k_{-2}+k_{-1}k_{-2}}]\mathrm{e}^{-x_2t}]\\$ $C(t) = B_0[\frac{k_{-1}k_{2}}{k_{-1}k_2+k_{1}k_{-2}+k_{-1}k_{-2}}+[\frac{k_{-1}k_{2}}{k_{-1}k_2+k_{1}k_{-2}+k_{-1}k_{-2}}]\mathrm{e}^{-x_3t}]$ Note that in equations (12) and (14), $$x_1$$ and $$x_3$$ are roughly equivalent to $$k_{-1}$$ and $$k_{-2}$$, respectively. $$x_2$$ does not have an obvious analogue. At equilibrium, for a normalized data set, the following relations also hold: $1 =[A]+[B]+[C]\\$ $[B]_e =\frac{k_{1}}{k_{-1}}[A]_e={K_1}[A]_e\\$ $[C]_e =\frac{k_2}{k_{-2}}[A]_e=K_2[A]_e$ Where $$K_1$$ and $$K_2$$ are the equilibrium constants for the first and second step of Reaction (1), respectively. Substituting Eqns (16) and (17) into (15) and rearranging yields the following relationship: $[A]_e = \frac{k_{-1}k_{-2}B_0}{k_{-1}k_2+k_{1}k_{-2}+k_{-1}k_{-2}}$ Then, in solving for Eqn (3) at equilibrium, $[B]_e = \frac{k_{-1}k_{1}k_{-2}}{k_{-1}k_2+k_{1}k_{-2}+k_{-1}k_{-2}}+\frac{k_{-1}k_2+k_{-1}k_{-2}}{k_{-1}k_2+k_{1}k_{-2}+k_{-1}k_{-2}}\mathrm{e}^{-k_{-1}t}$ Thus, by inspection of Eqns (12) and (19), $$x_1$$ $$\approx$$ $$k_{-1}$$. Equation (4) may be similarly rearranged and integrated to solve assuming equilibrium conditions, where $$x_3$$ $$\approx$$ $$k_{-2}$$. Unfortunately, equations (12)-(14) cannot be expressly used to determine exact values for $$k_1$$, $$k_{-1}$$, $$k_2$$, and $$k_{-2}$$ explicitly because those variables are coupled due to the simplifying assumption that a three-component exponential fit to the experimental data converges to a two-component exponential fit when values of $$m$$ and $$n$$ are similar. It is, however, relatively straightforward to determine $$K_1$$ and $$K_2$$ by fitting the experimental data with a function of the form $$y(t) = y_0+\alpha{}\mathrm{e}^{-\beta{}t}$$ to obtain $$y_0$$ and $$\beta$$ for $$A$$, $$B$$, and $$C$$ at each temperature. It then follows that $K_1=\frac{y_0(B)}{y_0(A)}$ and $K_2=\frac{y_0(C)}{y_0(A)}$ Finally, $$K_1$$ and $$K_2$$ may be used to estimate the remaining two unknowns, $$k_{1}$$, $$k_2$$, from the estimated values of $$k_{-1}$$, and $$k_{-2}$$.