MathSci Problems
Intro to Kinetics and Thermodynamics
Students often assume that if a reaction is “favourable,” it happens instantly. Or if a process is “fast,” they assume it leads to a stable result. This intuition is wrong. Let’s find out why.
Journey vs. Destination
To understand chemical behaviour, you have to separate the questions you are asking.
Thermodynamics asks: “Where does the universe want this system to go?”
Thermodynamics compares the energy of where you started to where you end up. It doesn’t care if the path is a straight line or a maze (it is path-independent).
Kinetics asks: “How fast can we get there?”
It is the science of the barrier and the mechanism. It focuses entirely on the obstacles between the start and the finish (it is path-dependent).
Think of a mountain climb. Thermodynamics tells you that the view from the valley on the other side is beautiful (lower energy state). Kinetics tells you that there is a vertical cliff you have to scale to get there (activation energy). You can have a destination that is thermodynamically amazing, but if the kinetic barrier is insurmountable, you will not arrive on any reasonable timescale.
Are Diamonds Really Forever?
Let’s look at a case where kinetics and thermodynamics oppose each other.
“Diamonds are forever” is a thermodynamic lie. Relative to graphite, diamond is thermodynamically unstable. The standard Gibbs free energy change () for the conversion of diamond to graphite is negative (). This means the universe wants your diamond ring to turn into pencil lead.
So, why hasn’t it happened yet? For the carbon atoms in a rigid diamond lattice to rearrange into the sheets of graphite, they must break incredibly strong covalent bonds. The activation energy () for this rearrangement is astronomically high. At room temperature, there isn’t enough thermal energy to overcome this barrier. Kinetics got in the way.
Thermodynamic vs. Kinetic Equations
Thermodynamics: Gibbs free energy
Changes in Gibbs free energy tell us about spontaneity. If is negative, the product is more stable than the reactant. This determines the equilibrium constant (), the ratio of products to reactants at equilibrium.
Kinetics: the Arrhenius equation
This dictates rate. The rate constant () depends exponentially on the barrier height (Ea):
If the barrier is high, is tiny, and the reaction effectively stops.
Here’s a random analogy I thought of before: Being a millionaire is a thermodynamically stable state (low stress, high resources). However, the activation energy (capital, risk, effort) required to transition from “broke” to “rich” is massive. Most people lack the sufficient “thermal energy” to surmount that barrier. On the other hand, “Get Rich Quick” schemes are low- pathways. They are “kinetically accessible”, but they usually lead to products that are “thermodynamically unstable” (i.e. losing all your money).
Thermodynamic Sinks
Sometimes a product is so stable that doing the reverse reaction is like trying to climb out of a massive pit. This is a thermodynamic sink.
Consider the base-mediated hydrolysis of an ester (saponification). Hydroxide attacks the ester and, after the tetrahedral intermediate collapses, you form a carboxylic acid and an alkoxide in the mechanistic sense, but under basic conditions the acid is immediately deprotonated to give a carboxylate anion (and the alkoxide becomes an alcohol). The carboxylate is stabilized by resonance, and that acid–base step is so strongly product-favoured that it effectively locks the system “downhill” and pulls the overall equilibrium forward. The molecule falls into a deep energy well (the sink) and does not return to the starting ester under the same conditions. Many ‘driving forces’ are important in metabolic pathways that keep us alive, such as glycolysis and the Krebs cycle! For example, a certain step is paired to something with a big negative ΔG, such as ATP hydrolysis or the release of CO, so the overall sequence has a clear preferred direction rather than hovering near equilibrium.
Kinetic vs. Thermodynamic Control
Kinetic control is when the conditions of a reaction favour the product that forms the fastest. It’s a bit like when you’re very hungry, you choose the food that’s quickest to get, even if it isn’t the best option overall.
Thermodynamic control is when the conditions of a reaction favour the most stable product. This is like when you aren’t super hungry, so you’re willing to wait and choose the highest-quality option rather than the fastest one.
In organic chemistry, enolate formation is a classic example of kinetic versus thermodynamic control. Under kinetic control, typically achieved using a strong, bulky base at low temperature, the reaction forms the kinetic enolate by removing the proton that is most accessible to the base. This pathway dominates because it has the lowest activation energy (). Under thermodynamic control, the reaction is reversible, usually at higher temperature and over a longer time, allowing the system to reach equilibrium. As a result, the more stable thermodynamic enolate predominates, even though it forms more slowly than the kinetic enolate.
Lesson in Life
A well-lived life (and good chemistry) requires distinguishing between the speed of the journey and the value of the destination. Don’t let a high activation barrier scare you away from a thermodynamically stable valley. And more importantly, don’t let a fast, low-barrier path lead you into a kinetic trap.
Reflect & Explore
Here are some open-ended questions to help you think more deeply about this material and connect it to related ideas.
- If you are given only an equilibrium constant , what can you conclude about kinetics? If you are given only a rate constant , what can you conclude about thermodynamic stability?
Imagine a reaction network (A B C) where B is high energy but forms quickly. Under what conditions might B accumulate, and under what conditions might it never build up to detectable levels?
- Many students treat “rate-determining step” as a synonym for “driving force.” What is the conceptual mistake in that, and how would you correct it with one example?
In the saponification reaction mentioned in this post, why does deprotonation of the carboxylic acid pull the overall process forward?