Variation of equilibrium constant with temperature

The relationship can be shown by putting together two previously-mentioned equations:

This has the form of y = mx + c. So plotting lnK against (1/T) will give a slope with a gradient of (-ΔH° / R) and an intercept of (ΔS° / R).

Notice that whether the gradient is positive or negative depends on the sign of -ΔH°. So an exothermic reaction has K decrease with temperature, and an endothermic reaction has K increase with temperature.

Le Chatelier didn't know about equilibrium constants or these equations, but now we do, we are able to explain why he observed his principle consistently.

Entropy and acids

We should already expect that adding electronegative atoms to a carboxylic acid will increase acidity, example:

The obvious reason is that electronegative atoms will pull electrons away from the oxygen atoms, delocalizing the negative charge in the carboxylate, hence stabilizing the molecule.

It turns out that this obvious reason is a minor one - the main reason is entropy.

You might guess that entropy increases when the acid disassociates, but this is only true in the gaseous phase. In a solvent such as water, charged species have a "shell" of solvent molecules, which is more ordered, hence total entropy change is negative.

When the charge is spread out over the carboxylate, there are more ways of arranging this shell of solvent molecules, so the entropy change is not as negative.

Diagram:

Entropy and hemiacetal formation

Entropy explains why intramolecular hemiacetals are more stable.

The bonds formed at the same in both cases, but the former reaction has a more negative entropy change. Hence we can expect a more positive ΔG from:

Driving reactions with equilibrium constants

In the reaction above, K₁ is about equal to K₂, and they are both in favor of the carbonyls. So we would expect a 50:50 mixture.

The reason that doesn't happen is due to an additional equilibrium - the deprotonation of the acid:

This drives it over to the right.

The process can also be represented as an energy diagram:

Below is another example of a reaction which can be driven left or right, depending on pH:

How the equilibrium constant varies with the energy change of a reaction

The equilibrium constant of a reaction is related to the difference in energy between reactants and products by this equation:

ΔG° = The standard Gibbs energy of the reaction, the difference between two states in kJ / mol. The ° represents standard conditions.

T = Temperature in kelvin
R = Molar gas constant
K = Equilibrium constant of the reaction

For example, this equation can be used to work out the energy change in the addition of water to isobutyraldehyde:

The equilibrium concentrations of hydrate and water can be measure by comparing UV adsorptions of the compound dissolved in water and the compound dissolved in hexane. As usual, approximately constant concentrations such as H₂O are set to 1.

These experiments reveal K_eq at 25° to be around 0.5.

So ∆G° = –8.314 × 298 × ln(0.5) = +1.7 kJmol^–1

The sign of G tells us about the direction the equilibrium favors. At K = 1, there is a 50:50 mixture of products:reactants, and ln(1) = 0, so ∆G° = 0.

At K > 1, ln(K) gives a positive number. At K < 1, ln(K) is negative.

In other words, equilibrium concentration is shifted in favor of the side with the lowest energy - in favor of the reaction with a negative ∆G. This should be pretty intuitive even without working it out from the equation. 

A small change in ∆G makes a big difference in K, which you might notice by considering the log term. Also have a look at this table:

Every reaction is theoretically at equilibrium. But a typical C-C bond is 350 kJ mol^-1. You might want to input that and see what percentage of products you get... it gives you an appreciation for why we consider these reactions to only operate in one direction.

The point of using Gibbs energy instead of enthalpy is that the Gibbs energy takes entropy into account.

Rotation and energy profile diagrams

An amide bond has double-bond character from conjugation. This tends to prevent rotation, but rotation does happen slowly, and can be measure by NMR.

Depending on the relative sizes of attached R groups, we can expect one form to be more stable than another. This can be shown on an energy profile diagram:

The maximum at 90 degrees is from the lone pair on nitrogen not having the right symmetry to overlap with the pi* of the C=O.

When both substituents on N are the same, we can expect equal energies:

Energy profile diagrams can also be used with alkenes:

The energy at 90 degrees going off the scale makes sense, since cis and trans alkenes are not observed to intercovert.

One way to measure these energies is to use the heat of hydrogenation:

We expect movement to a lower energy state to be spontaneous, ignoring entropy effects. Hence we can often consider the higher energy state to be a reactant, and the lower energy state to be a product.

Making an aldehyde from a carbanion by attacking a carbonyl

This is a problem. Attacking a carbonyl which doesn't have a good leaving group will create an alcohol:

While attacking a carbonyl with a good leaving group will result in it being attacked twice:

What we need is something which forms a stable tetrahedral intermediate, yet which has a group which leaves during the acid workup. By leaving at the same time as the acid workup, the organometallic is destroyed at the same time, preventing another attack.

A solution is DMF, or dimethylformamide:

In otherwords, adding DMF to an organometallic will replace the metal with an aldehyde group. Another example:

You might wonder what makes the tetrahedral intermediate stable, since amine anions can occasionally act as leaving groups. I don't know the answer.