Topic 5.8: Reaction Mechanism and Rate Law

Mechanism and rate law: the slowest step sets the pace

Now that we can read a mechanism, let’s ask how fast the reaction goes. Think about a full airplane emptying through one narrow door. You can hurry down the aisle as fast as you like, but people still leave at the speed of the door. A reaction works the same way, because the whole thing can’t go any faster than its slowest step, and that’s why we call the slow one the rate-limiting step.

Back in our container, let’s give the two steps some made-up times, so the first step takes about a second and the second one happens in a blink. Now let’s run a lot of molecules at once. See all those carbon monoxides in front? They’re just waiting around, because N O three is rare, and the instant one shows up, a carbon monoxide grabs it. This meter counts the carbon dioxide made every second, at the pace of that slow first step.

We’re about to double the carbon monoxide. What do you think happens to the rate? Let’s see. Twice the carbon monoxide, and the rate doesn’t budge, because the extra C O just joins the crowd, waiting for an N O three that isn’t showing up any faster.

Now put the C O back and double the nitrogen dioxide instead. Look at that, four times as fast, because the slow step needs two N O twos to find each other, and with twice as many around, those meetings happen two times two, four times as often.

So let’s write the rate law. Remember from our lesson on elementary reactions that each step’s rate law comes straight from the molecules that take part in it. Step one, the slow one, is two nitrogen dioxides colliding with each other, so its rate is k times N O two times N O two, which is k times N O two squared.

The slow step sets the pace for the whole reaction, so this is the rate law for the whole reaction. Carbon monoxide isn’t in it at all, which means the reaction is zero order in C O, and that’s exactly what chemists measure for this reaction below two hundred twenty-five degrees Celsius.

And here’s the trap I never want you to fall into. The overall equation has one N O two and one C O, so it’s tempting to write rate equals k times N O two times C O. But the overall equation only tells you what reacts and how much, and it’s the slow step that tells you how fast.

Let’s put some numbers in. Say the rate constant is zero point five oh oh per molar per second, the nitrogen dioxide is zero point oh one oh oh molar, and the carbon monoxide is zero point oh two oh oh molar. The rate is zero point five oh oh times zero point oh one oh oh squared, and the carbon monoxide never enters, because it isn’t in the rate law.

The units work out too, because one over molar per second, times molar squared, leaves molar per second, so the rate is five point oh oh times ten to the minus fifth molar per second. Now let’s do one together. This time the nitrogen dioxide is zero point oh three oh oh molar and the carbon monoxide is zero point oh one oh oh molar, with the same rate constant.

With k at zero point five zero zero and the nitrogen dioxide at zero point zero three zero zero molar, what’s the rate? And now one on your own, with letters instead of molecules, where A plus B makes I in a slow first step, and then I plus B makes C in a fast second step. What’s the overall equation for this mechanism, and what’s the rate law?