Topic 9.5: Free Energy and Equilibrium

Free energy and equilibrium: the forest that flashes in sync

This is Elkmont, in the Great Smoky Mountains of Tennessee, at about nine thirty on a night in late May. Thousands of fireflies flash together, five or six or eight times in a row, then the whole forest goes dark for about eight seconds, and then they do it all again. Every one of those flashes is a chemical reaction, happening right here, near the tip of his body, in a part called his lantern.

What pays for the light? The answer you’ll usually hear is ATP, the molecule every cell in your body spends for energy. It’s a great guess, and it’s wrong, and the reason why is what this lesson is about.

And so many people want to see this that the park hands out passes by lottery. Around 1991, Lynn Faust, whose family had a cabin right here, read in a science magazine that no fireflies in this half of the world flash in sync, and she knew better, because she’d watched this here her whole life, so she got in touch with the scientists, and the next summer they came and saw it.

Here’s a guess to make before we go on. When his lantern flashes, does it get hot, the way a light bulb does? A female waiting down in the leaves answers him with a quick double flash. Why the males flash together is a question for biologists, and ours is a chemistry question. By the end of this lesson, you’ll be able to prove with one line of arithmetic that ATP can’t pay for this light, and you’ll find the reaction that does.

On the way, you’ll learn to read from a single number how far any reaction goes, why a cold pack keeps dissolving even while it gets colder, and how to make a reaction run uphill. So let’s start with the question every reaction in this forest has to answer: if you leave it alone, where does it end up?

How far does a reaction go? ΔG° and K

You know that a negative delta G naught means a reaction is thermodynamically favored, and from Unit 7, that K tells you which side a reaction leans toward at equilibrium. Those are the same idea, and the equation that connects them is delta G naught equals minus R T times the natural log of K. R is eight point three one four joules per mole per kelvin, and T is the temperature in kelvin.

If K is bigger than one, its log is positive, and the minus sign makes delta G naught negative. If K is less than one, the log is negative, and delta G naught comes out positive.

And if K is exactly one, delta G naught is zero, and neither side is favored. So a negative delta G naught and a K bigger than one say the same thing, that the equilibrium mixture leans toward products, and a positive delta G naught leans it back toward reactants. And favored says nothing about speed: a reaction with a huge K can still take years if its barrier is high, which was the story of the last lesson.

This link is a classic AP question. You get K or delta G naught, and the point goes to whoever says why: delta G naught is negative, so K is greater than one, so products are favored. Solve the same equation for K and you get K equals e to the minus delta G naught over RT. That exponent is why small changes in delta G naught make huge changes in K, so let’s watch it happen.

Here’s a model of the simplest reaction there is, A turning into B and back. Each bead is a molecule, jiggling with thermal energy, which is what RT measures. With the wells equally deep, delta G naught is zero, so let’s watch where the beads end up. They settle about half and half, so K is one.

Now I’ll make B deeper by five point seven kilojoules per mole, a bit more than twice RT. And now there are about ten B’s for every A, so K is ten. Another five point seven, and it’s about a hundred to one. So every five point seven kilojoules per mole, at room temperature, multiplies K by ten.

Now I’ll make the hill between the wells much taller, and start every bead back in A. Watch them crawl over. They take much longer to cross, but they end up at the same hundred to one.

So it’s the height of the hill… that decides how fast the beads get there. and the depth of the wells decides how far. Once the counts stop changing, the mixture is at equilibrium, with no push either way, so its actual delta G is zero. Delta G naught is the fixed depth difference that decides where it settles.

So delta G belongs to the mixture you have right now, and it goes to zero as the mixture settles, while delta G naught belongs to the reaction itself, and at a given temperature it never changes. Now you try it. Drag the floor of well B until, at equilibrium, there are about a thousand B’s for every A.

Now real numbers. Nitrogen and hydrogen making ammonia has a delta G naught of minus thirty-three point oh kilojoules per mole at 298 kelvin. That’s negative and more than ten RTs, so I predict a huge K.

First I’ll turn kilojoules into joules, minus thirty-three thousand joules per mole. The two minus signs make a plus, and the units all cancel, so the exponent is just a number. Thirty-three thousand over eight point three one four times two ninety-eight is about thirteen point three, and e to the thirteen point three is about six times ten to the fifth, far toward ammonia, just as we predicted.

Estimating K, and a real AP question

You can estimate K without a calculator. At room temperature, RT is about two and a half kilojoules per mole, and that’s the ruler for delta G naught. If delta G naught is within a couple of kilojoules of zero, K stays close to one. If it’s tens of kilojoules, that’s many RTs, so K lands many powers of ten away from one, one power of ten for every five point seven.

Try this one without a calculator. A reaction at 298 kelvin has a delta G naught of plus twenty-five kilojoules per mole. Which value of K is most reasonable?

Now one that was actually on the AP exam, in 2015. Ethanol vapor turns into ethene and water vapor, and the question gave delta H and delta S and asked whether they support the claim that K is less than one at 298 kelvin. From the last lesson, delta G naught equals delta H minus T delta S. The trap is the units: delta S is in joules, so one hundred twenty-six joules becomes zero point one two six kilojoules.

With those numbers, two ninety-eight times zero point one two six is about thirty-seven point five, and forty-five point five minus thirty-seven point five is plus eight point oh kilojoules per mole. Delta G naught is positive, so K is less than one, and the data support the claim. That one sentence is the point.

If you want the actual number, K is e to the minus eight thousand over RT, about zero point zero four. And ‘I set up the reaction, and nothing happened’ wouldn’t count as evidence… because a reaction that doesn’t seem to happen might just be slow.

Now it’s your turn, and this one comes from inside the firefly. ATP reacts with water to make ADP and phosphate. Biochemists measure it at the pH inside a cell and add a little prime mark, but it works the same way, and it’s minus thirty point five kilojoules per mole. Find K prime for this reaction at 298 kelvin.

Flip it around, and making ATP has the opposite sign, plus thirty point five, and a K of one over that, about five times ten to the minus six, so left alone, a cell would make almost no ATP. And yet that firefly is spending ATP all night long, so something is making it, and I want you to hold that thought. Here’s a different reaction. A mixture with K equal to one hundred has reached equilibrium. Which statement about it is true?