Do catalysts shift equilibrium constant towards 1?

Question

I want to be able to understand shifts in equilibrium from the maxwell boltzmann distribution. One thing I cannot get my head around is the effect of catalysts on the equilibrium position - supposedly it's none at all.

But consider these diagrams of the M.B distribution of both reactants and products of a reversible reaction.

The top diagram shows the distribution of the products of the reaction which equilibrium favours. It has a bigger concentration of particles. The line dividing the shaded region on these diagrams represents activation energy and the $dE$ represents the shift in activation energy due to the catalyst. We can see that the shaded regions, $A_1=A_2$ because forward and backward rates are the same at eqm. But clearly, the rate of the top reaction increases more as a greater fraction of particles have above activation energy. So adding a catalyst should shift equilibrium to the middle?

Basically, catalysts work better on reactants with higher concentration. So shouldn't the side with the lower concentration be favoured?

Answer 1

TL;DR Your Maxwell–Boltzmann diagram up there is not sufficient to describe the variation of rate with $E_\mathrm{a}$. Simply evaluating the shaded area alone does not reproduce the exponential part of the rate constant correctly, and therefore the shaded area should not be taken as a quantitative measure of the rate (only a qualitative one).

There is a subtle issue with the way you've presented your drawing. However, we'll come to that slightly later. First, let's establish that the "proportion of molecules with sufficient energy to react" is given by

$$P(\varepsilon) = \exp \left(-\frac{\varepsilon}{kT}\right) \tag{1}$$

Therefore, for a reaction $\ce{X <=> Y}$ with uncatalysed forward activation energy $E_\mathrm{f}$ and uncatalysed backward activation energy $E_\mathrm{b}$, the rates are given by

$$k_\mathrm{f,uncat} = A_\mathrm{f} \exp \left(-\frac{E_\mathrm{f}}{kT}\right) \tag{2} $$

$$k_\mathrm{b,uncat} = A_\mathrm{b} \exp \left(-\frac{E_\mathrm{b}}{kT}\right) \tag{3} $$

The equilibrium constant of this reaction is given by

$$K_\mathrm{uncat} = \frac{k_\mathrm{f,uncat}}{k_\mathrm{b,uncat}} = \frac{A_\mathrm{f}\exp(-E_\mathrm{f}/kT)}{A_\mathrm{b}\exp(-E_\mathrm{b}/kT)} \tag{4}$$

As you have noted, the change in activation energy due to the catalyst is the same. I would be a bit careful with using "$\mathrm{d}E$" as the notation for this, since $\mathrm{d}$ implies an infinitesimal change, and if the change is infinitesimal, your catalyst isn't much of a catalyst. So, I'm going to use $\Delta E$. We then have

$$k_\mathrm{f,cat} = A_\mathrm{f} \exp \left(-\frac{E_\mathrm{f} - \Delta E}{kT}\right) \tag{5} $$

$$k_\mathrm{b,cat} = A_\mathrm{b} \exp \left(-\frac{E_\mathrm{b} - \Delta E}{kT}\right) \tag{6} $$

and the new equilibrium constant is

$$\begin{align} K_\mathrm{cat} = \frac{k_\mathrm{f,cat}}{k_\mathrm{b,cat}} &= \frac{A_\mathrm{f}\exp[-(E_\mathrm{f} - \Delta E)/kT]}{A_\mathrm{b}\exp[-(E_\mathrm{b} - \Delta E)/kT]} \tag{7} \\[0.2cm] &= \frac{A_\mathrm{f}\exp(-E_\mathrm{f}/kT)}{A_\mathrm{b}\exp(-E_\mathrm{b}/kT)} \frac{\exp(\Delta E/kT)}{\exp(\Delta E/kT)} \tag{8} \\[0.2cm] &= \frac{A_\mathrm{f}\exp(-E_\mathrm{f}/kT)}{A_\mathrm{b}\exp(-E_\mathrm{b}/kT)} \tag{9} \end{align}$$

Equations $(9)$ and $(4)$ are the same, so there is no change in the equilibrium constant.

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The question then arises as to how eq. $(1)$ is obtained. The simplest way is to invoke a Boltzmann distribution, which almost by definition gives the desired form. However, since you have a Maxwell–Boltzmann curve, I guess I should talk about it a bit more. The fraction of molecules with energy $E_\mathrm{a}$ or greater is simply the shaded area under the curve, i.e. one can obtain it by integrating the curve over the desired range.

$$P(\varepsilon) = \int_{E_\mathrm{a}}^\infty f(\varepsilon)\,\mathrm{d}\varepsilon \tag{10}$$

where the Maxwell–Boltzmann distribution of energies is given by (see Wikipedia)

$$f(\varepsilon) = \frac{2}{\sqrt{\pi}}\left(\frac{1}{kT}\right)^{3/2} \sqrt{\varepsilon} \exp\left(-\frac{\varepsilon}{kT}\right) \tag{11}$$

At first glance, we would expect this to be directly proportional to the exponential part of the rate constant, i.e. $\exp(-E_\mathrm{a}/kT)$. Alas, it is not that simple. If you try to work out the integral

$$\int_{E_\mathrm{a}}^{\infty} \frac{2}{\sqrt{\pi}}\left(\frac{1}{kT}\right)^{3/2} \sqrt{\varepsilon} \exp\left(-\frac{\varepsilon}{kT}\right) \,\mathrm{d}\varepsilon \tag{12}$$

you don't get anything close to the form of $\exp(-E_\mathrm{a}/kT)$. Instead, you get some "error function" rubbish, and some nasty square roots and exponentials. (You can use WolframAlpha to verify this.)

Why is this so? Well, it turns out that there are other terms that also depend on $\varepsilon$ and therefore need to go inside that integral (they aren't constants and can't be taken out).

The simplest example is that faster molecules tend to collide more often, so even though the right-hand tail of the diagram seems to contribute very little to the "proportion of molecules with sufficient energy", it actually contributes more significantly to the overall rate because these molecules collide more often. In collision theory this is described using the "relative velocity" of the particles $v_\mathrm{rel}$.

There is also another complication, in that the Maxwell–Boltzmann distribution, the direction of the particles is not accounted for. (For more insight please refer to Levine Physical Chemistry 6th ed., p 467.) Therefore, there has to be yet another term that takes into account the direction of movement of the particles. The idea is that a head-on collision between two molecules is more likely to overcome the activation barrier than is a $90^\circ$ collision. The term that compensates for this is the "collision cross-section" $\sigma$.

If you go through the maths (and I don't really intend to type it out here, it's rather long, but I will give some references) then you will find that at the end you will recover the form $\exp(-\varepsilon/kT)$. Once you have arrived at this, it's very straightforward to see that the increases in rate of both the forward and backward reaction cancel each other out.

Now, as for the promised references, Pilling and Seakins's Reaction Kinetics pp 61-2 have a short outline of the proof. Atkins's Physical Chemistry 10th ed. has a slightly longer proof on pp 883-4.

Answer 2

In a catalysed reaction the activation barrier is not lowered but the catalyst causes the reaction to proceed by a different route which occurs more rapidly.

Thus there is a third party a 'chaperone' if you like that controls the way the two reactive species interact. This can be a surface, or a metal-organic or an enzyme, the details do not matter, but what matters is that the catalyst causes the reactants to undergo different reactions to that which would occur in homogeneous solution to produce the same product.

As the final outcome is the same so is the equilibrium. The kinetic approach is outlined by @orthocresol based on the Arrhenius equation. From a thermodynamic view, which is the most general, at equilibrium $\Delta G^{\mathrm {o}} = -RT\ln(K_e)$ and as $\Delta G^{\mathrm {o}}$ is unchanged in the reaction (the reactants and products are the same) so is the equilibrium constant $K_e$.

Now to your approach using the Maxwell-Boltzmann (MB) distribution. It is not at all clear that Maxwell-Boltzmann distribution applies to a liquid. It is based on ideas around the perfect or ideal gas, and point like particles colliding but have no other interaction. In a liquid molecules do interact with one another since a liquid has a meniscus and is distinct from its vapour.

However, there still has to be a distribution of speeds and hence energy. On collision the energy is converted into increasing vibrational/rotational energy in the colliding partners until by random chance enough energy is reached to cause reaction. As collisions can occur at at least $10^{12} \pu{s^{-1}}$ only very occasionally, perhaps only one in a million collisions, is enough energy present to cause reaction unless the activation energy is very low. In these cases diffusion limits the bimolecular reaction rate constant to $\lt 10^{11} \pu{dm^3 mol^{-1}s^{-1}}$ and often far less but depending on solvent viscosity. Thus in these cases it is the rate of approach that limits reaction not energy, so clearly MB energy distribution does not apply.

In solution it is normal to describe the rate constant of a (non-diffusion controlled) reaction using transition state theory. This involves calculating the properties of the transitions state and reactants using statistical thermodynamics. The weakness of this approach is that the transition state properties have to be 'guestimated' in some way.

Appendix

[ Using the MB distribution an Arrhenius type expression can be obtained for a bimolecular gas phase reaction. The expression for the rate constant at energy E is
$$k(E) \approx \int_0^{\infty} Q(E)\sqrt{E} ~f(E)dE$$ where $f(E)$ is the MB distribution in terms of energy (by substituting $E=mv^2/2$), and $Q(E)$ is the reaction cross-section. Under conditions of thermal equilibrium the rate constant is the product of reaction cross section $Q(E)$, the magnitude of the relative velocity $\sqrt{E}$ and the thermal distribution of relative speed $f(E)$. The simplest form of $Q(E)$ is to suppose that it is a hard sphere of diameter d (like a billiard ball) in which case it has the form $$Q(E)= \pi d^2(1-E_a/E)$$ when $E \ge E_a$ and zero otherwise. Performing the integration and converting to temperature produces and Arrhenious type equation $$ k(T) \approx \exp(-E_a/k_BT)$$ where $k_B$ is the Boltzmann constant.]