Topic 5.3: Concentration Changes Over Time

Concentration changes over time: the man in the ice

On the nineteenth of September, 1991, two hikers from Germany, Erika and Helmut Simon, were crossing this pass high in the Alps, right on the border between Austria and Italy, when they spotted something brown in a little gully full of ice and meltwater. It was a man, and everyone assumed he was a climber who’d died in an accident, maybe a few years earlier.

The next day, the rescuers couldn’t free him from the ice, so they carried down the one thing they could, his axe, and that axe was strange, because its blade was copper. When an archaeologist got to see the body a few days later, he guessed it was at least four thousand years old, and today we know this man lived about five thousand three hundred years ago.

Today everyone calls him Ötzi, after the mountains where he was found, and he lived and died centuries before anyone built the Great Pyramid in Egypt. But nobody wrote down when he died, and there’s no date on anything he carried, so how could anyone possibly know?

The answer is that he carried a clock, and it’s still running. Living things are built from carbon, and about one carbon atom in every trillion is carbon-14, a radioactive kind that slowly turns into nitrogen. As long as you’re alive, every bite you eat tops your carbon-14 back up, but the day Ötzi died, the topping up stopped and his carbon-14 started to run down, at a pace that cold and ice and five thousand years can’t change at all.

By the end of this lesson, you’ll be able to take a few measurements of how much of something is left, figure out what kind of clock is running inside it, and work out how long it’s been ticking. And at the very end, you’ll take the radiocarbon labs’ real result for Ötzi, work out how long ago he died, yourself, and then see the one correction the scientists still had to make.

The labs really did date tiny samples of his tissue and bone, and even blades of grass found with his tissue, so let’s go into one blade of that grass, all the way down to its atoms. Almost all of these are ordinary carbon atoms, and about once in every trillion you’d find one like this, a carbon-14.

A trillion is far too many to show, so from now on I’ll only show you the carbon-14, sixty-four of them, as a model of the same clock that was running in his body. While he was alive, every now and then one of his carbon-14 nuclei would decay, turning into nitrogen-14, and fresh carbon-14 from his food would drift in to take its place, so the count stayed about the same.

Then he died, and from that moment on nothing new came in, so the clock started running. Let’s let it run, a few hundred years every second, and every flash you see is one nucleus decaying, at a completely random moment, with no warning at all. After five thousand seven hundred and thirty years, about half of them have decayed, and thirty-three carbon-14s are left.

You might expect the next five thousand seven hundred and thirty years to finish off the rest, but the clock only takes half of what’s left, so about sixteen remain, a quarter of what we started with. That time, five thousand seven hundred and thirty years, is carbon-14’s half-life, and it stays the same every time, no matter how many atoms are left.

And this is a classic AP question, what fraction is left after three half-lives? It’s a half of a half of a half, one eighth, and even after two half-lives a quarter is still there, so two half-lives never means everything is gone. If we run the clock again from the start, completely different atoms decay at completely different moments, so no single atom knows when its turn will come, but the crowd still halves every five thousand seven hundred and thirty years.

And notice how the flashes thin out as the clock runs, because with fewer carbon-14 atoms left, fewer of them decay each century. How fast it goes is proportional to how much is left, so half as many atoms means half as many flashes, and that’s exactly what chemists mean by first order.

Now I want you to drag the clock to the moment when seventy-five percent of the carbon-14 has decayed. Carbon-14 isn’t the only thing that disappears like this, though, because a chemical in a flask can halve the same way, at a pace that does depend on its temperature, and the AP exam once gave us one.

Equal fractions in equal times

On the 2022 AP exam, students got this reaction, dinitrogen pentoxide breaking down, along with four measurements of how much was left, taken every hour and forty minutes. It goes from point one six molar, to point oh eight, to point oh four, to point oh two, so in every interval half of what’s left disappears, the same staircase as Ötzi’s clock.

Now compare a reaction that starts in the same place but loses the same amount every time, point oh four molar, and on the graph it falls in a straight line. Losing the same amount every time means the speed never changes, so the rate is just k, and that’s zero order. Losing the same fraction means the speed depends on how much is left, so the rate is k times the concentration, and that’s first order, like the clock.

Here are two new runs, measured every ten minutes, so which one is first order, and what’s the evidence? Now let’s watch a real reaction fade, and find an even sharper test for first order, one you can see at a glance on a graph.

A reaction you can watch fade

This purple liquid is crystal violet, a dye, and fading it is one of the labs in the AP Chemistry lab manual. When you add sodium hydroxide, the purple slowly disappears. Each crystal violet ion is shaped like a little propeller, with its electrons spread out over the whole thing, and that’s what makes it purple. When a hydroxide ion bonds to the carbon in the middle, the electrons can’t spread out like that anymore, and a colorless molecule forms.

We add so much hydroxide that it hardly runs low, so the only concentration that really changes is the crystal violet’s. Out here, a beam of light shines through the liquid, and the paler it gets, the more light gets through to the detector, so the instrument measures how much light the dye absorbs, its absorbance, and that tells us the concentration at every moment.

I sped this up, so five minutes go by in about twenty seconds, and every minute the instrument drops a measurement onto our graph. If we divide each measurement by the one before it, we get about point five five every single time, so the same fraction disappears every minute, which looks like first order.

But a falling curve on its own can fool you, because reactions that aren’t first order can make curves that look a lot like this one, so what we want is a plot that comes out straight only when the reaction is first order. So let’s take the natural log of every measurement and plot that instead, and look at that, the curve snaps into a straight line.

It works because every minute the concentration gets multiplied by the same number, and the log turns multiplying into subtracting, so each point sits exactly point six oh oh below the one before. So whenever a plot of the natural log of concentration against time comes out straight, the reaction is first order. And every straight line has a slope, which turns out to be the most useful number in this whole lesson, so let’s take this graph to the board.

The slope gives k

The AP exam gives you this equation on its formula sheet, the first-order integrated rate law, and it says the natural log of the concentration at time t, minus the natural log of the starting concentration, equals negative k times t. Move the starting term to the other side, and it looks just like the equation of a line, y equals m x plus b, with the natural log of concentration as y and time as x, and negative k as the slope.

So let’s get k from our crystal violet line, using the first and last points so the triangle is as big as possible. The natural log falls from negative ten point eight two oh to negative thirteen point eight two oh, a change of negative three point oh oh oh, over three hundred seconds.

Negative three point oh oh oh divided by three hundred seconds is negative point oh one oh oh per second, and since the slope is negative k, k is positive point oh one oh oh per second. The line slopes down, so its slope is negative, but k itself is never negative, and its unit is just per second, because a log has no units.

Now let’s find the half-life, the time it takes to lose half. Start from the sheet’s equation, with its two logs. One log rule joins them. Take the log of a minus the log of b. You get the log of a over b. Use it on the left side, and the two logs become one log of a ratio.

That ratio is what’s left, over what we started with. After one half-life, it’s one half. The natural log of one half is negative point six nine three. Divide both sides by negative k. The half-life is point six nine three over k.

There’s no concentration in that equation, which is why a first-order half-life is the same however much you start with. So now you work out the half-life of our crystal violet run, in seconds.

The dinitrogen pentoxide from the AP exam halved every one point six seven hours, so its k is point six nine three over one point six seven hours, which is point four one five per hour, because k always carries the time unit you used. The natural log straightened crystal violet, but not every reaction is first order, so what happens when the log doesn’t work?

Three kinds of clocks

Here are three different reactions, each starting at point one oh oh molar, with tidy, idealized data so the patterns are easy to see. Run A is amber, run B is purple, and run C is teal.

On a plain concentration graph, run A is already a straight line, losing the same amount every twenty seconds, so run A is zero order. Switch to the natural log, and run B straightens out, so B is first order, halving every twenty seconds. Run C is sneaky, because its first two points are exactly the same as run B’s, and since two points always make a straight line, two points can never prove the order.

The third point gives it away, point oh three three three instead of point oh two five oh. And look at its halvings, because the first takes twenty seconds but the next takes forty, so run C’s halvings keep getting longer, which never happens in first order. Run C won’t straighten on the concentration graph or the log graph, so we need a third kind of plot.

See if you can find the plot that turns run C into a straight line. So a straight line in plain concentration means zero order, a straight line in the natural log means first order, and a straight line in one over the concentration means second order.

Reading the slopes

Let’s put all three side by side, and you don’t need to memorize the laws, because they’re all on the AP formula sheet. For zero order, concentration against time is straight, the slope is negative k, and k is in molar per second. For first order the natural log is straight, the slope is negative k again, and k is just per second.

For second order one over the concentration is straight, and this line climbs, so the slope is positive k, and k is in per molar per second. So a straight line on its own doesn’t tell you it’s zero order. The first thing to read on any of these graphs is what’s on the vertical axis.

Zero order is the easiest to use. Run A starts at point one oh oh molar and loses point oh oh one oh oh molar every second, so after seventy seconds, point oh three oh molar is left. This plot of the natural log of concentration against time, in minutes, has a slope of negative point oh two five oh per minute, so which value is k?

Now let’s do a second-order one together, with run C. One over the concentration starts at ten and reaches forty at sixty seconds, so the slope is thirty divided by sixty seconds, point five oh oh per molar per second, and that’s k. To predict the concentration at eighty seconds, we start at ten and add k times t, point five times eighty, which is forty, so one over the concentration is fifty.

Fifty isn’t the answer yet, though, because it’s one over the concentration, so we flip it and get point oh two oh oh molar. In 2024 the AP exam showed two graphs for nitrogen dioxide, the natural log of its concentration against time and one over its concentration against time, and asked why the reaction is second order, and the sentence that earns the point says that the plot of one over the concentration of nitrogen dioxide against time is the most linear, so the reaction is second order in nitrogen dioxide.

Now I want you to try a first-order one on your own. Run B has a k of point oh three four seven per second and starts at point one oh oh molar, so when does it reach point oh three oh oh molar? How many seconds does that take?

If we start run B at twice the concentration, at the same temperature, what happens to k and to the half-life? And now we have everything we need to answer the question from the very start.

Back to the ice

When the hikers found him, Ötzi’s clock had been running for more than five thousand years, in the ice and out of it. Radiocarbon labs in Zurich and Oxford measured tiny samples of him, and the Zurich lab’s results for his tissue, bone and grass work out to about fifty-seven percent of the carbon-14 in the modern reference that every lab compares against.

So we know what fraction is left compared with that modern level, and for now we’ll assume Ötzi started at exactly that level, with five thousand seven hundred and thirty years for the half-life. First we need the rate constant, and point six nine three divided by five thousand seven hundred and thirty years is one point two one times ten to the minus four per year.

Then we use the first-order law, with the fraction that’s left in place of the two concentrations, because only their ratio matters. So if he started at that modern level, how many years has his clock been running?

That’s close to the five thousand three hundred years we started with, but not the same, because the air back then held a little more carbon-14 than the modern reference, so his clock started higher than we assumed. A higher start means his clock has been running down for longer, so he’s older than our number, and scientists work out how much older with tree rings, wood whose year they can count ring by ring.

With that correction, he died sometime between about 3350 and 3100 BCE, roughly five thousand three hundred years ago, and notice it was the starting amount that needed fixing, the N-zero in our equation, while the clock itself was right all along. At the start I promised you’d be able to look at how much of something is left and work out how long it’s been ticking, and you just did it for a man who died up here more than five thousand years ago. And the clock inside him is still running, losing half of what’s left every five thousand seven hundred and thirty years, whether anyone is watching or not.