Topic 5.2: Introduction to Rate Law

Rate laws: let the experiment decide

There’s only one way to find out, and that’s to measure. Chemists track something that changes as a reaction goes, like the gas it gives off, the mass it loses, its color, or the time it takes to reach a finish line. That last one is my favorite. So let’s go back to the lab, after dark.

This is the peroxide clock, hydrogen peroxide reacting with iodide ions in a little acid, a reaction Augustus Vernon Harcourt and William Esson timed at Oxford in the 1860s, back when chemists were just starting to measure rates. Every beaker also gets the same small amount of thiosulfate, plus starch. Let’s look inside.

As the reaction makes iodine, thiosulfate grabs each one the instant it forms and turns it back into iodide, so iodine keeps getting made while the thiosulfate runs down. The moment the thiosulfate runs out, the next iodine survives, finds the starch, and the whole beaker turns blue-black. So the blue is a finish line, and with the same thiosulfate in every beaker, it marks the same amount of reaction. A shorter time to blue means a faster rate.

Only about one percent of the peroxide has reacted by then, so this measures the rate right at the start, the initial rate. Here’s trial one, sped up four times. The stopwatch is running, and the liquid still looks like plain water. In our simulation, trial one turns blue at forty seconds. In the next beaker I’ll double the peroxide and keep everything else the same.

Trial one took forty seconds. With twice the peroxide, when do you think this one will turn blue? Here’s a fresh trial one on the left, and double peroxide on the right, both starting at the same moment.

Twenty seconds, half the time, so the rate doubled when the peroxide doubled. Double the peroxide, double the rate, so the rate is directly proportional to the peroxide, and chemists call that first order in peroxide. Now let’s triple the iodide instead. Everything else stays the same as trial one, and here they go.

Thirteen point three seconds, a third of the time, so tripling the iodide tripled the rate, and it’s first order in iodide too. Together, the rate is k, times peroxide, times iodide, each to the first power. That’s a rate law, and each power is a reactant’s order. The k is just a number for this reaction at this temperature. We’ll find it soon.

Now look back at the equation, with its two iodide ions. The experiment says first order in iodide, not second. The coefficients don’t tell you the orders; only an experiment can. Add the orders: one plus one makes this reaction second order overall. We kept the acid and the temperature the same in every beaker, so this law holds for those conditions.

Twice, four times, or no change

Doubling the peroxide doubled the rate. That isn’t always what happens. Double one concentration, and the rate might not change, might double, or might go up four times. The rate factor is the concentration factor, raised to the order.

Take a reactant that’s second order, and triple it. Three squared is nine, so the reaction goes nine times faster. Say a reactant is first order, and you triple its concentration. How much faster does the reaction go now?

Here’s a classic AP-style problem. Nitrogen monoxide and hydrogen react to make nitrogen gas and water. There are four trials, and for each one it gives the starting concentrations of nitrogen monoxide and hydrogen, and then how fast nitrogen gas forms right at the very start of the trial, in molar per minute. For hydrogen, I want two trials where only the hydrogen changes, and that’s trials one and two.

Hydrogen doubles, from point oh oh one to point oh oh two, and the rate doubles too, from one point eight to three point six. When doubling in gives doubling out, the reaction is first order in hydrogen.

Look at hydrogen’s coefficient in the equation. It’s two, but its order is one. Now nitrogen monoxide. First we need two trials that hold the hydrogen steady. That’s trials three and four, where nitrogen monoxide doubles, and the rate goes from point three to one point two, four times as big.

Two to what power gives four? Two squared, so it’s second order in nitrogen monoxide. The rate law is k times nitrogen monoxide squared, times hydrogen. Two plus one makes it third order overall.

Sometimes a concentration doesn’t matter at all. For nitrogen dioxide and carbon monoxide at a hundred degrees Celsius, the rate depends on nitrogen dioxide squared and not on carbon monoxide. Carbon monoxide still gets used up. The rate just doesn’t depend on how much of it is there, and that’s zero order. At a hundred degrees, if you triple the nitrogen dioxide and double the carbon monoxide, what happens to the rate?

Here’s some practice data for a reaction between A and B, three trials, each with its starting concentrations of A and B and the initial rate it gave, in molar per second. Which rate law fits all three trials?

The rate constant

One number is still missing to make the law match the real rate, and that’s k, the rate constant. Let’s solve for it with trial one. Rearranging the law, k is the rate divided by the nitrogen monoxide concentration squared, and by the hydrogen concentration. Point oh oh six squared, times point oh oh one, is three point six times ten to the minus eight. That makes k five point oh times ten to the third.

Units get their own line, because on the AP exam they’re often their own point. Molar per minute on top, molar squared times molar on the bottom, one molar cancels, leaving molar to the minus two per minute.

Let’s check it with trial four, where both concentrations are different. Point oh oh two squared, times point oh oh six, is two point four times ten to the minus eight. One point two times ten to the minus four, over that, is five point oh times ten to the third again. The rate changed, but k didn’t. That’s why it’s called the rate constant.

Look at the units of k here: molar to the minus two, per minute. That’s what any third-order law gives, with time in minutes. Every order works the same way. Cancel the concentrations, and see what’s left.

One more thing changes k, and that’s the temperature. Let’s go back to the clock and watch it happen. Here’s trial one’s exact mixture again. Make this beaker turn blue sooner, without changing any of its concentrations.

Room temperature on the left, thirty-five degrees on the right. Same mixture, same concentrations, and the only thing we changed is the temperature. The warm one turned blue in about half the time. The concentrations and the orders didn’t change, so what got bigger is k. So concentration changes the rate through the brackets, and temperature changes k itself.

AP style: vitamin C meets triiodide

Let’s finish with a real question from the 2025 AP exam. Vitamin C, ascorbic acid, reacts with triiodide, and the exam gives you the rate law. Here are three trials. The first part asks you to explain how the data show it’s first order in ascorbic acid, so I’ll pick trials one and three.

In trials one and three the triiodide stays the same, the ascorbic acid doubles, and the rate doubles too. So the sentence that earns the point goes like this. In trials one and three, the triiodide is held constant, and doubling the ascorbic acid doubles the rate, so the reaction is first order in ascorbic acid. The sentence that loses the point just says ‘first order’ without saying what stayed the same.

The next part asks for the value of k. Use trial one’s rate and concentrations, shown here. Now use the law. What would the rate be with these new concentrations?

Back to the beetle

So here we are, back at the creek. Now we can finally answer the question from the start. In the storage sac, the peroxide reacts so slowly that the beetle can carry it around. In the chamber, the enzymes are a catalyst, giving it a much faster path, and the reaction’s heat is what boils the mixture and blasts it out.

And in the reaction that makes its gas, oxygen appears exactly half as fast as peroxide disappears. But what’s the rate law in that chamber? You can’t read it off the equation, and counting pulses won’t tell you either.

So which experiment would tell you the order in peroxide inside this chamber? So now you can do what I promised. You can put a number on how fast a reaction goes, and explain how this beetle makes a slow reaction split-second fast, and you can look at a few experiments and tell whether doubling a concentration makes a reaction twice as fast, four times as fast, or no faster at all. Next time, we’ll use rate laws to work out how a concentration changes over time, but for now, our beetle has places to be.