10
$\begingroup$

Rearranging the Eyring equation leads to the following:

$$\Delta^\ddagger S^\circ = R \ln{\frac{k \times h}{{k_\text{B}}{T}}}+\frac{\Delta^\ddagger H^\circ}{T}$$

where $k$ is the rate constant, $h$ is the Planck constant, $R$ is the universal gas constant, $k_\text{B}$ is the Boltzmann constant, $T$ is temperature, $H$ is enthalpy, and $S$ is entropy. The entropy term should ultimately be in units of $\mathrm{\frac{J}{mol\ K}}$. The right term of the equation will clearly yield the correct units. For the left term, the universal gas constant can be expressed in $\mathrm{\frac{J}{mol\ K}}$, which means that the log term should be dimensionless. For the bottom part of the fraction, the units evaluate to $\mathrm{J}$. For the top part, the Planck constant has units of $\mathrm{J\cdot s}$. To get the top part of the fraction to evaluate to $\mathrm{J}$, the rate constant would have to be in units of $\mathrm{s^{-1}}$. For first order reactions, this is indeed the case. For reactions of higher order, this will never work out to the correct units.

This makes me think that the Eyring equation is simply using the unitless values of these variables. This still leads to some ambiguity about which value of the rate constant to use. Take for instance a second-order reaction in which the rate constant is in units of $\mathrm{M^{-1}\ s^{-1}}$. If we rearrange the units to SI units ($\mathrm{m^3\ mol^{-1}\ s^{-1}}$), the value of the rate constant changes. So is the rate constant $k$ in the Eyring equation simply the value of the rate constant when that rate constant is expressed in certain units? And if so, are the concentration units always expressed in molarity? I kept missing a homework problem because I was converting $k$ from molarity to SI units, which makes me believe this is indeed the case.

$\endgroup$
8
$\begingroup$

Using Transition State Theory (TST) one has to be aware of at least some of the many flaws of it. And may also often lead to wrong assumptions. However, it is a wonderful tool to describe basic kinetics in a theoretical manner.

The most important/ fundamental assumptions are

  • separation of electron and nuclear movement (comparable to Born-Oppenheimer approximation)
  • (all possible) transition states (TS) are in equilibrium with the reactants $$\ce{A + B <=> [AB]^{\ddagger}}\tag1$$
  • combined Boltzmann statistic for reactants and TS
  • All molecules that reach the transition state from the reaction coordinate of the reactants will leave the TS to the reaction coordinate of the product(s) - and vice versa \begin{align} \ce{A + B <=> &[AB]^{\ddagger}->P}\tag{2a}\\ \ce{P <=> &[AB]^{\ddagger}-> A + B}\tag{2b} \end{align}
  • at the TS the transition vector can be separated from all other motions and hence may be treated classical as translation

In other words TST reduces a very complicated multidimensional problem to a one-dimensional translation. That implies that all treated molecules must be connected on a potential energy hypersurface with one vector. (Borrowing from the amazing wikipedia)

reaction diagramm

Hence reactions that are described with TST are of first order by default. One would have to make various adjustments to treat higher orders (and I would assume that not matching units would be of minor concern).

$\endgroup$
3
$\begingroup$

You are correct in your analysis that the rate constant is first order with units $\ce s^{-1}$. The problem you have is that the Eyring equation you use is not complete as I try to show below after some introduction. (The derivation I use can be found in several texts, but I outline that in Steinfeld, Fransisco & Hase, 'Chemical Kinetics & Dynamics')

The idea behind Transition State Theory (TST) is that the reactants are assumed to be in thermal equilibrium with the transition state. This is a small region at the top of the potential energy barrier that separates reactants and products. A contour plot of an A + BC reaction ($\ce{F + H2}$ ) is shown in the first figure (together with a single reaction trajectory).

F+H2 contour plot

The transition state is labelled with a dot and lies at a saddle point, which is to say that the potential is convex along the reaction path and concave in a perpendicular direction to this, see second figure.
Following the reaction path from right to left (contour first figure) we see that on collision the reactants climb the (rather small) barrier and cross to products. The product is formed with some considerable vibrational energy. The profile along the reaction path (or reaction coordinate) is the familiar potential energy hill between reactants and products.

saddle point

There are a number of assumptions in deriving TST. The two basic ones are
(a) that nuclear and electron motion can each be separated (similarly to the Born-Oppenheimer approximation use to solve the Schroedinger equation), and it is further assumed that the energy for translation, vibration and rotational motion can be treated additively) and
(b) the reactant molecule’s energy is distributed among its states, i.e. translation, vibration, rotation according to the Maxwell-Boltzmann distribution.

Other assumptions are necessary for the theory and are that
(c) the transition state is crossed only once, i.e. it is a point of no return,
(d) The motion along the reaction coordinate at the transition state can be separated from other motions and treated classically as a translation.
(e) even in the absence of equilibrium between reactants and product molecules the energy in the modes of the transition state is distributed according to Maxwell-Boltzmann. This assumption is sometimes called the ‘quasi-equilibrium hypothesis’.

To calculate the rate constant k the speed of crossing the transition state and the number of molecules doing so has to be calculated. Let the reaction be a general second order one $\ce{A + B \rightarrow C }$ with rate equation $\ce{d[A]/dt=-$k$[A][B]}$.

In the reaction $\ce{A + B \rightarrow X^\ddagger \rightarrow C}$, where $\ce{X^\ddagger}$ is the transition state, suppose that there is a small region of length $\delta$ at the top of the barrier that the reactants must all pass through. Imagine this as a very thin dividing surface, reactants on one side and products on the other. At equilibrium the rate of forwards and backwards motion through this region must be the same, consequently if the cooncentration is $N^\ddagger$ then
$$N^\ddagger = K^\ddagger\ce{[A][B]}$$ Now suppose that all the products are removed from equilibrium and as these are half of the total $N^\ddagger$ the number going forwards to products is $N_f^\ddagger = K^\ddagger\ce{[A][B]}/2$ which is the quasi equilibrium hypothesis.

The rate of reaction to reactants to products is $dN/dt = \delta N^\ddagger / \delta t $ where the number of transition states with velocity v to $v+\delta v$ in one direction is $\delta N^\ddagger = N^\ddagger/2$. The average time to cross the dividing surface at average velocity $v_a$ is $\delta t = \delta/v_a$ , where $\delta$ is the thickness of the dividing surface and by substituting gives $$\frac{dN}{dt} = \frac{ N^\ddagger }{2} \frac{ v_a }{ \delta }$$

The average velocity in one direction can be calculated in the standard way$^1$ assuming equilibrium distribution of velocities and is $$v_a = ( \frac{2k_BT}{ \pi\mu } )^{1/2}$$ where $\mu$ is the reduced mass. Substituting again gives $$\frac{dN}{dt}=\frac{N^\ddagger}{2\delta}(\frac{2k_BT}{\pi\mu})^{1/2} $$ The equilibrium constant is given by $$K^\ddagger= \frac{N^\ddagger}{\ce{[A][B]}} = \frac{Z_{tot}^\ddagger}{Z_AZ_B}e^{-E_0/k_BT} $$ where $Z_{tot}^\ddagger$, $Z_A$ and $Z_B$ are the respective partition functions per unit volume and are derived using statistical mechanics.

The $Z_{tot}^\ddagger$ partition function is that of the combined species AB at the transition state which by the condition (d) above, is separated into two terms, $Z^\ddagger = Z_rZ^\ddagger$, one term for the reaction coordinate, $Z_r$ and $Z^\ddagger $ for all other $3N-1 $ degrees of freedom. The translational partition function in one dimension is $$Z_r=(2\pi\mu k_BT)^{1/2}\delta/h$$ and thus substituting and simplifying produces $$\frac{dN}{dt}= \frac{k_BT}{h}\frac{Q^\ddagger}{Q_aQ_B}e^{-E_0/k_BT}\ce{[A][B]}$$ from which the second order rate constant is $$k= \frac{k_BT}{h}\frac{Q^\ddagger}{Q_aQ_B}e^{-E_0/k_BT}$$ and the constants $\delta$ and $\mu$, which were used in the derivation, cancel out. The constants $k_BT/h = 6.25~10^{12} \ce{s}^{-1}$ and is often called the frequency factor. The rate constant will have units of $\pu {m^3s^{-1}}$

All that now remains is to calculate the partition functions. The parameters needed have to be guessed for the transition state as they are generally unknown, no transition state ever having being identified spectroscopically. The values needed are vibrational frequencies, moments of inertia reactants and shape, linear or bent. The general partition function is split into a product of terms, $Z_{ trans} Z_{rot}Z_{vib}Z_{elec}$. This is assumption (a) above, separating electronic and nuclear motions.

In the case of atom plus diatom reaction $\ce{A + BC \rightarrow AB + C}$ of which many have been measured, $\ce{O + H2, Cl + H2, H + H2, F + H2 etc.}$, the rate constant equation becomes the monster equation $$k=\frac{k_BT}{h}[\frac{Z^\ddagger}{Z_AZ_{BC}}]_{vib}[\frac{Z^\ddagger}{Z_AZ_{BC}}]_{rot}[\frac{Z^\ddagger/V}{Z_A/V.Z_{BC}/V}]_{trans}e^{-E_0/k_BT}$$

Finally, the whole rate constant may need to be multiplied by a symmetry or statistical factor, e.g. in $\ce{F + H2}$ the rate constant needs an addition factor of 2.

The rate constant has units $\pu{m^3s^{-1}}$ which when multiplied by the Avogadro number becomes $\pu{dm^3mol^{-1}s^{-1}}$, normal $\ce{2^{nd}}$ order units.

Notes:

(A) The volume appears because we need to convert the thermodynamic equilibrium constant k (ratio of partition functions) into $k_c$ the chemical one which uses concentrations.

(B) The equations for partition functions are well known, are listed on many phys. chem. textbooks, e.g. McQuarrie & Simon ‘Physical Chemistry’ chapter 1, but the Wikipedia page is rather technical so they are listed below$^{2}$.

(C) It is important to note that partition functions have widely differing values, translation $Z^\ddagger/V \approx 10^{11} \ce{m^{-3}}$, rotation $Z^\ddagger \approx 10^{4} $, vibration $Z^\ddagger \approx 10 $ and electronic $Z^\ddagger \approx 1$ and all are dimensionless, thus the translational partition function divided by V has dimensions of $\pu{m^{-3}}$).

Limitations:

(A)As the transition state properties are not known, reactions between complicated molecules are unlikely to give accurate values.
(B) Molecules in solution are limited by diffusion to a maximum value determined by diffusion constants and hence solution viscosity.
(c) More recent treatments by Kramers discuss how, in solution, the rate constant depends on friction of the reaction coordinate (see Daune ‘Molecular Biophysics).

(1) The average of a quantity x with distribution $p(x)$ is calculated with the integral ratio $\int xp(x)dx~/\int p(x)dx$. (The denominator is the normalising term, equivalent to the total number when making a simple average of any set of objects, i.e. $\Sigma x/N)$.

In the case in point, the average velocity distribution is $p(v) = exp(-\mu v^2/(2k_BT))$ and the integration limits are 0 to $\infty$.

(2) The partition function is defined as the ‘sum over states’, $$Z=\Sigma _i g_i exp(-E_i/(k_BT))$$ where $g_i$ is the degeneracy of the energy levels $E_i$. When the energy levels are close (as in translation) the summation is replaced with an integral.

The partition functions are
$$Z_{trans} = \frac{(2\pi \mu k_BT)^{3/2}V}{h^3} $$
For a generally shaped molecule, with moments of inertia $I_a$ etc. about the three principal axes, and symmetry factor $\sigma$ (which is the number of ways the molecule can be rotated back to the original) $$Z_{rot} = (\frac{8\pi ^2k_BT}{h^2})^{3/2}\frac{(\pi I_aI_bI_c)^{1/2}}{\sigma}$$

For each vibrational mode. $$Z_{vib} = \frac{1}{1-e^{-h\nu/k_BT} }$$ and these partition functions are multiplied together for many vibrational modes; $\nu$ is the vibrational frequency.

The electronic partition function is the multiplicity of the state, most commonly 1.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.