Note that the x axis is officially titled "reaction coordinate." This is a somewhat complicated measure of how far the reaction has progressed, or how much of the reaction has happened. It is often more convenient to assume the reaction is "going" at a somewhat constant rate and think of the x axis simply as time.

Thermodynamics has nothing to do with time. The thermodynamic information on the figure above can be represented in the following way:

See how much simpler it looks? What this tells you is that the products are at a lower free energy, or more stable, than the reactants. Because of this, the reactants "want" to be converted into the products. For example, graphite and diamond are both forms of carbon, but graphite has a lower free energy. Therefore, diamond wants to convert into graphite. Another example is that your skin wants to dissolve in the soap when it is washed. In this case, the products of the dissolution reaction (namely, your skin dissolved in the soap) are more stable than the reactants (your undissolved skin and the soap separately).

The above examples were chosen to specifically because even though the reaction should go (the products really are more stable than the reactants, and nature always tries to get things as stable as they can be), it doesn’t seem to. Your hands are safe when you wash them as is the diamond on your finger. Even though the reaction is thermodynamically favorable, it is slow. It’s just plain too difficult to get the diamond to break all of its bonds and re-form them in the different, more stable graphite configuration. To answer the question of why this is so we must go back to the first reaction diagram featuring the hump between reactants and products. That hump is a measure of how difficult it is to get the reaction to go in any reasonable amount of time. For diamond, the hump for the conversion into graphite is high. Even though the reaction should go thermodynamically, it does not because it is kinetically unfavorable.

Why is the hump so high? It’s because the transition state, or activated complex, that is at the top of the hump is so unstable (high energy). In the case of the diamond --> graphite reaction, getting to the transition state probably involves, among other things, breaking all the carbon-carbon bonds in diamond. Breaking bonds always requires the input of energy, and there are a lot of carbon-carbon bonds in diamond. Hence the height of the hump for the diamond --> graphite conversion.

This is the difference between thermodynamics and kinetics. Thermodynamics can tell you only that a reaction should go because the products are more stable (have a lower free energy) than the reactants. Another way of saying this is that the reaction has a negative free energy change: DG is negative and therefor the reaction is spontaneous. Yet another way of saying this is to say that the reaction has a large equilibrium constant, signifying that if nature could ever attain equilibrium, there would be many more products present than there are reactants. Notice that the DG = -RTlnK equation says all this at once. Kinetics, on the other hand, can tell you how fast the reaction will go but doesn't tell you anything about the final state of things once it gets there. Consider the following reaction sequence:

The equilibrium constant, capital K, is a thermodynamic quantity. As such, it depends only on the overall reaction. What the individual reaction steps that changed A and D into B and E were doesn't matter in the slightest. Which combination is more stable: A + D or B + E? Well, that's the one that should thermodynamically predominate at equilibrium. Notice that the reaction intermediate C doesn't even enter into the equation for K. For the purposes of thermodynamic analysis, the individual steps above the overall reaction might as well not even be there.

One side note about K is that it depends on how the overall reaction equation is written. If the stoichiometric coefficients were multiplied by two, then the value of K for the original overall reaction would have to be squared in order to be appropriate for the new overall reaction. Thus, in order to determine K for a given reaction, you must be aware of exactly which stoichiometric coefficients you are using. If a chemist experimentally finds K for the synthesis of water from H₂ and O₂ and gives her number to a colleague, and the colleague mistakenly thinks that the reaction was written H₂ + ½O₂ --> H₂O instead of 2H₂ + O₂ --> 2H₂O, then when the colleague goes to reproduce the experiment, she will get the wrong number. Assuming the experiment is reproducable in the first place, the number that the colleague gets will be the square root of the number that the original chemist got.

A second side note about K is that pure solids and pure liquids don't get included in it. Their concentrations are so big that we can just take them as being constant throughout the reaction. Thirdly, people often analyze perturbations to equilibrium in terms of Le Chatelier's principle, which says that a system will shift to counter any change you try to make. For example, if a reaction is exothermic, that means that it gives off heat. If you raise the temperature, you are effectively adding a product to the reaction, which will cause it to shift back to the reactant side. Be careful about this though because temperature can change equilibrium constants. The change is slight, but it’s definitely there.

Kinetics, on the other hand, does not depend in the slightest on what the situation looks like at equilibrium. The rate of the reaction has no dependence on the overall reaction equation but instead depends on the reaction mechanism, the elementary steps. (This was the part of the reaction sequence that we ignored for thermodynamics.) The molecules on the left of each elementary step must collide in order to react so that the products on the right are formed. Notice that in the first step of the reaction sequence above, the reactant A doesn't have to collide with anything. Instead, it simply breaks apart, producing B + C. This is what is called a "unimolecular," "first order" elementary step because only one atom is involved. In the second step of the reaction sequence, C and D do have to collide in order to produce E. This is what is called a "bimolecular" step because two atoms have to come together for the reaction to occur.

The rate of an elementary step depends on the concentration of species available to react. For example, in the second step, if there are many molecules of C and D around, then the likelihood of a molecule of C colliding with a molecule of D with sufficient energy and the right orientation to make the elementary step go is high. Therefore, the rate of the elementary step is proportional to the concentrations of the reactant molecules. Here are expressions for the rates of the two elementary steps for the reaction sequence above:

Notice that if for instance one of the elementary steps were to involve two molecules of C colliding, then the rate of that step would be proportional to [C]².

Elementary steps of higher molecularity (termolecular and on up) are very rare because in any real scenario, it is unlikely that three molecules would hit each other in exactly the right way and with exactly enough energy for the step to happen.

The rate constant k is a quantity that students love to confuse with the equilibrium constant K. Don’t do this!! K tells you the ratio of products to reactants at equilibrium while k tells you the rate of an elementary step in the reaction mechanism! Although it is most usual to find little k experimentally, it can also be found from the Arrhenius equation,

The Arrhenius equation does not tell you the rate of the reaction; it tells you the rate constant for an elementary step of the reaction. The variable Ea is the activation energy for the step, or the height of the hump on the reaction diagram at the beginning of the section. The constant R is our old friend the gas constant, and T is the temperature at which the elementary step is performed. The large sensitivity of k to T is the reason that it is extremely difficult experimentally to find rate constants. Most elementary steps either give off or take up heat, and the resulting temperature change changes the rate of the elementary step itself. Thus, the practical utility of the Arrhenius equation is limited.

The constant A is the "Arrhenius factor." It takes into account the fact that molecules not only must collide, they must collide with the right orientation. When you take Chem 33, you will learn that for some reactions classified as "SN₂" the collision must involve one molecule putting electron density into an antibonding orbital on another molecule. This is a very precise place to have to put electron density! Therefore, only a small fraction of collisions result in reaction. The Arrhenius factor is also called an "entropic factor" to stress that it accounts for how random collisions can be if they are to result in a reaction.

From experimental data, it is often possible to find the rate law for a reaction. For example, given the following numerical data, you can deduce that the overall reaction is first order in A, first order in B, second order in C, and that the rate constant is 6 x 10² M^-3s^-1. This is because the rate of reaction doubles when you double the concentration of either A or B (leaving all other initial concentrations constant) and quadruples when you double C (leaving all other initial concentrations constant). Once you know the exponents, you can plug in to the equation to obtain k. Keep in mind that this procedure finds the initial rate of the reaction. Often, as the reaction progresses, the rate changes.

If the first elementary step is standing in line at the ATM and the second step is getting a coke, and if it takes a long time to get money but relatively little time to get a coke, then we can write that the rate of the entire reaction (here modeled by the coke-obtaining process) is equal to the rate of the first elementary step:

Note that there is no way we could have predicted which step would be the slowest. The bottleneck step must be determined experimentally. If there were a long line at the coke machine but no line at the ATM, then the rate of the overall reaction would be the rate of the second elementary step:

However, it is considered bad form to write the rate of an overall reaction in terms of a reaction intermediate. Reaction intermediates are constantly being created and being consumed, so [C] varies greatly from time to time during the reaction. We can get around this by using what is called the "steady state approximation" to solve for [C] in terms of the concentrations of the other reactants.

In order to understand the steady-state approximation, we have to realize that thus far we have only considered the rate of an elementary step going forward. Each elementary step has a corresponding back reaction that also has an associated rate. The real situation can be represented as follows:

We are going to assume that the rate of change of the concentration of the intermediate C, d[C]/dt, is approximately zero. (This is why it is called the "steady state approximation.") Now, d[C]/dt is equal to the rate of formation of C, k₁[A] + k_-2[E], minus the rate of destruction of C, k_-1[B][C] + k₂[C][D]. So the math for this scenario is as follows:

 Pretty complicated rate expression, eh?! Depending on what tricks you use (the steady state approximation is just one of them) you can get some very crazy expressions. See for example Professor Zare’s lecture example of H₂(g) + Br₂(g) à 2HBr(g). If you understood the preceding example, you already understand all of the important ideas behind Michaelis-Menton kinetics. Michaelis-Menton does the same steady-state approximation math for a biological enzyme-substrate system.

Enzymes are special proteins that catalyze biological reactions. Many enzymes break down food molecules into material from which your body can get energy. The enzyme (called E) has a little niche into which the substrate food molecule (called S) fits just perfectly. When it gets into the enzyme’s protective cleft, it is broken down into the usable-for-energy product (called P). We can represent the system as follows:

 To solve this system, use the fact that the second step is the slow step to invoke the steady-state approximation. You will also use the all-important fact that [E] = [E_o] – [ES], where E_o is the initial amount of enzyme. Note that there is no k_-2 for this reaction: the enzyme will not catalyze the conversion of the product back to the substrate. Given all of these constraints, the math looks like what follows. To make sure that you understand it, try to reproduce the answer from the given reaction sequence and the constraints detailed in this paragraph.

If [S] is really big (i.e. [S] >> k_-1 + k₂), then d[P]/dt = k₂ [E_o], which is effectively a zero-order reaction because k₂ and [E_o] are both constant. So, as in the kinetics problem set, if you know the rate constant for the second elementary step (k₂) and you’re also given the rate of the zero-order product formation reaction (d[P]/dt), then you can solve for [E_o], which is the enzyme (on the problem set that was LADH).

The last topic to consider before we leave kinetics and go back for a last look at thermo is integrated rate laws. Given the elementary steps, it is possible to integrate the corresponding rate law, using calculus to solve for the concentration of some reaction species as a function of time. Test yourself heavily on both first order and second order rate law integration. From the equation for the elementary step, you should be able to figure out the concentration of the species as a function of time. Remember, this is just math. The chemical parts are only the first line and the last line of each derivation. Here are the answers you should be able to derive:

Actively test yourself on this!! You may be able to follow all the math, but could you reproduce it? It’s also important to have a feel for what is happening chemically for first order and second order reactions.

Common examples of first order reactions are radioactive decay processes. As must be the case for first order reactions, the rate of decay depends on the amount of radioactive stuff that's around at any one time. This results in an exponential decrease in the decay rate with time. When less stuff is around, it does not decay anywhere nearly as fast.

When chemists talk about radioactive decay, they typically like to talk about half-lives. The half life of a substance is how long it takes for half of whatever's there to decay. At time t_1/2, [A]_t1/2= (1/2) [A] _o. Therefore, if you bring the [A]o over to the left side of the first order equation, the left side becomes simply ½. Thus you have ½ = e^-kt. Take the natural log of both sides and solve for t_1/2to obtain t_1/2= .7/k.

Now back to thermo to wrap up this integrated discussion of thermodynamics and kinetics. The last thing to talk about with respect to thermodynamics is free energies and electrochemistry. It is usually not a temptation to mix these topics up in kinetics, which is why they are tacked on the end in this section.

There are three main kinds of free energy:

• U = q – w_PV – w_elec = heat in minus work out (or equivalently, heat in plus work in). Note that w_PV is pressure volume work and w_elec is electrical work.
• H = U + PV = enthalpy. Note that if a reaction has a negative enthalpy change, it is exothermic and if it has a positive enthalpy change it is endothermic. Note also that breaking bonds is always endothermic (positive DH) and forming them is always exothermic (negative DH). In common terms, it always takes energy to break bonds and it always releases energy to form them.
• G = H-TS. If you take the delta of both sides and do the math for this situation you come out with DG = w_elec, or electrical work. Note that q = TDS and we are assuming volume is constant, so that DV = 0. The math is shown below:

Notice that the heat flow TDS and the pressure-volume work terms cancel so that at constant temperature and pressure we are left simply with electrical work. For us, this means work that a chemical system can do by causing electrons to flow. The oxidation states of the chemicals in the system change as the electrons flow between them (more on this when we do redox). If we put the chemical that wants to give up electrons into one beaker and the chemical that wants to take electrons into another beaker we can force the electrons to flow along the wire connecting them (and light up a lightbulb or something along the way . . .). This is the principle behind an electrochemical cell. Be cognizant of the following equations and what they are telling you:

The cell potential E⁰_cell is measured in Volts. Note that DG^o is defined to be the amount of electrical work you can get out of the system when you start with every concentration being 1M. (Note also that this is one of those cases where we can't treat K as though it had units, because you can't take the log of anything that has units.) E⁰_cell is "reduction potential," which is a measure of how much a species wants to be reduced at STP. If a chemical really wants to be reduced (gain electrons) then it's a good oxidizer (loser of electrons). Also in these equations, n is the number of moles of electrons that is flowing, F is the Faraday constant in coulombs per mole, and E⁰_cell has units of Joules/Coulomb, which makes it work out so that DG has units of Joules/mole, as it should. Since negative DG signals a spontaneous reaction, it follows that positive E⁰_cell signals a spontaneous reaction too. E⁰_cell = E⁰_{oxidized substance} - E⁰_{reduced substance} for a total reaction, but it is important to realize that if you're not doing a total reaction but instead using the E⁰'s to calculate the E⁰_cell for a new half-cell, you have to go through DG. Professor Zare gave a handout on the mathematical reason why this is so; it boils down to the fact that DG is what is called a "state function" and E⁰ is not. Like DG, positive E⁰_cell is a thermodynamic quantity.