Topic 5.9: Pre-Equilibrium Approximation

Pre-equilibrium: when the slow step comes second

Everything so far had its slow step first, and you could read the rate law straight off it. Now let’s put our suspect on trial, the reaction that turns colorless nitric oxide brown. Watch what happens when nitric oxide meets the oxygen in air. It’s colorless coming out of the nozzle, and it blooms brown right where it mixes, two N O plus O two making two N O two.

Chemists measured how fast this goes more than a hundred years ago, and the rate law is k times N O squared times O two, which makes it third order overall. Now, the simplest story would be two N Os and an O two all slamming together at the same instant. But three-way collisions are very rare, fewer than one in a thousand compared with ordinary two-molecule collisions, so chemists propose two steps instead.

So let’s go into the edge of that brown cloud, where the two gases mix. In the first step, two nitric oxides stick together into a pair called N two O two, and almost right away the pair falls apart again. Pairs form and fall apart over and over, like a revolving door, and both directions are fast.

Every once in a while, an oxygen molecule catches a pair in the moment it exists, and that makes two nitrogen dioxides. That’s the slow step, and it’s where the brown comes from. And this is what makes this mechanism different from the last one. The pairs form just about as fast as they fall apart, so the first step stays very close to equilibrium, and treating it as an equilibrium that sits in front of the slow step is called the pre-equilibrium approximation.

Here’s the mechanism on paper. First comes a fast step that goes both ways, two N Os pairing up into N two O two, and then a slow step, where an O two catches that pair and makes two nitrogen dioxides. The slow step sets the pace, so we start there, and its rate law is k two times N two O two times O two.

But we’ve got a problem, because N two O two is an intermediate. You can’t buy it in a bottle or set its concentration, so we want a rate law that uses only things you can control, like N O and O two. So we have to replace it, and the fast step tells us how.

With that approximation, we set the forward rate equal to the reverse rate. The forward rate is k one times N O squared, because two N Os have to collide, and the reverse rate is k minus one times N two O two, because one pair just falls apart. Solve that for N two O two. Which expression do you get?

Now swap that into the slow step’s rate law in place of N two O two, and we get k two times k one over k minus one, times N O squared, times O two. All those k’s are just constants, so we roll them into one k, and the rate law is k times N O squared times O two, which is exactly what chemists measured.

Let me say that a second way, because it’s the heart of this whole idea. The fast step decides how many pairs are around at any moment, and that goes as N O squared, and the slow step decides how often an oxygen catches one, which multiplies by O two.

Let’s check it in the world. Double the nitric oxide, and there are about four times as many pairs at any moment, so the brown comes in four times as fast. Put the nitric oxide back and double the oxygen instead, and it’s twice as fast, because the same number of pairs get caught twice as often. Your turn with the controls. Make the brown come in eight times faster.

So now you’ve got both cases. When the first step is the slow one, you read the rate law straight off it. And when a fast, reversible step comes first, you write the slow step’s rate law, then use that fast equilibrium to swap out the intermediate.

Two stories, one rate law

But does matching the measured rate law prove the N two O two route is the real one? Here’s a completely different route for the same reaction, where a nitric oxide first hooks onto an oxygen molecule to make an N O three, and then that N O three meets a second nitric oxide in the slow step. Double the nitric oxide, and it’s four times as fast again, and if you work it out on paper, this route gives the very same rate law, k times N O squared times O two.

Let’s put this route on paper. First comes the fast step, which runs both ways, and then the slow step, where the N O three meets a second N O. Its rate is k two times N O three times N O. But N O three is an intermediate, so we swap it out, the same way we did with N two O two.

In the fast step, the forward rate, k one times N O times O two, equals the reverse rate, k minus one times N O three. Divide both sides by k minus one, and N O three equals k one over k minus one, times N O times O two.

Swap that into the slow step’s rate law, and the two N Os multiply into N O squared. The rate is k times N O squared times O two, the same law the N two O two route gave.

So a rate law can rule a mechanism out, but it can’t prove one. To choose between these two, you’d want to catch the intermediate, N two O two or N O three, which is the same kind of evidence we talked about earlier.

And this is exactly what the AP exam asked in twenty nineteen. Nitrogen dioxide breaks down into nitric oxide and oxygen, and its measured rate law is k times N O two squared.

Then the question gave two mechanisms and asked whether each one is consistent with that rate law. Mechanism one has a slow step first and a fast step after it, and Mechanism two has a fast equilibrium first, making N two O four, and then a slow step. In Mechanism one the first step is the slow one, and it’s two nitrogen dioxides colliding, so we read the rate law straight off it, k times N O two squared, and it’s consistent.

Now Mechanism two on your own. What rate law does it predict, in terms of things you can measure? Both mechanisms predict the rate law chemists measured. What can we conclude?

Back to Los Angeles

Now let’s finish the trial we promised. We’re back in the air over Los Angeles, and we’ve got our suspect, two N O plus O two, and its rate law. Picture a flask of air where nitric oxide is, say, one percent of the gas, and in there this reaction is quick, with half of that N O turning brown in less than a minute.

But out in city air, nitric oxide is more like, say, a tenth of a part per million. One percent is ten thousand parts per million, so that’s a hundred thousand times less N O. The rate law squares that. A hundred thousand squared is ten billion, so each liter of air makes N O two ten billion times more slowly.

But there’s also a hundred thousand times less N O to turn brown. How much longer does the city wait? Ten billion times slower, but a hundred thousand times less to do. Ten billion divided by a hundred thousand is a hundred thousand. It takes a hundred thousand times longer, not ten billion.

That still turns under a minute into weeks. The cloud over Los Angeles rolled in over a single morning. So this reaction is real, and we’ve got its rate law right, but it’s far too slow to make a city’s brown on its own. The air over Los Angeles had faster routes, and they ran on sunlight, which is what Haagen-Smit worked out.

That’s the same test the city ran when it shut down the plant, only this time with a rate law instead of a factory. A story about hidden steps has to survive the measurement, and now you can run that test yourself.

So check that the steps add up, let the slow step set the rate law, and when a fast, reversible step comes before it, use the equilibrium to swap out the intermediate. Next time, we’ll draw these steps as hills and valleys on an energy diagram, and you’ll see what a slow step actually looks like.