Mechanisms and the Rate-Determining Step

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Picture a wide highway with eight lanes, and in the middle, one narrow bridge with a single lane. What decides how fast the cars get across? Keep your answer in mind; we will come back to it. Many reactions also happen in small steps, one after another. Today we will see how these steps decide the rate law. In JEE, questions on this come up often. So, let's begin.

0:31Elementary steps: order equals molecularity

So, first, two words. Many reactions happen in a series of simple steps; each one is an elementary step, and the whole series is the mechanism. In an elementary step, the species written really do react together, all in one go. So how is the order of such a step fixed? By the number of reacting species taking part, exactly! That number is called the molecularity. So for an elementary step, order equals molecularity. Right? For example, A and B collide in one elementary step and give products; this arrow means gives. Then rate equals k times A times B, where the square brackets mean concentration, and k is the rate constant. With me? But careful, this works only for an elementary step. For an overall reaction, the order is found by experiment, not read from its balanced equation. This is a classic trap. Clear?

1:31Slow step first: the rate-determining step

Now, the simplest case. The first step of a mechanism is slow, and the steps after it are fast. Remember our highway with the narrow bridge? So which step sets the overall rate? The slow step, yes, exactly! It is our narrow bridge. So the rate law is just the rate law of the slow step, which we call the rate-determining step. Got it? Now, inside a mechanism, some species are formed in one step and used up in a later one. These are intermediates. Okay? And a catalyst is the other way round: used up in one step, and regenerated, given back, in a later one. So when you add up all the steps, do these two appear in the overall equation? No, they cancel out! Neither appears in the overall equation. To tell them apart, see which comes first: formed first means intermediate; used up first means catalyst. Right?

2:30Fast pre-equilibrium, then the slow step

Now the second case: a fast step comes first. Here, A plus B quickly form an intermediate, called I, and these two half arrows mean the step runs both ways and sits at equilibrium, with equilibrium constant capital K, okay? Then I gives the product P slowly, with rate constant k two written on the arrow, a small two below the k. Right? The slow step decides, so its rate law is k two times the concentration of I, okay? But I is an intermediate, so it must be replaced. So what do we replace it with? The fast equilibrium, exactly! It gives the concentration of I as capital K times A times B, in concentrations. Got it? Put it in: rate equals k two, times capital K, times A and B in concentrations. So the observed rate constant, k obs, with obs written small below, is k two times capital K, the two constants multiplied. Clear? Now, E a two, capital E with a small a and two below, is the slow step's activation energy, and delta H one, with the little triangle delta for change, is the fast step's enthalpy change. Since k obs is k two times capital K, the observed activation energy, E a obs, equals E a two plus delta H one. If the fast step gives out heat, delta H one is negative. With me?

3:53Steady state: the intermediate stays nearly constant

Now the same scheme with all three rate constants: k one, written above the half arrows, for the forward step; k minus one, written below, for the backward step; and k two, from I to the product. Okay? Here we use the steady-state approximation. The intermediate is used up about as fast as it is formed, so its concentration stays nearly constant. So we set d of the concentration of I, over d t, its rate of change with time t, equal to zero. Right? Solve that, and the rate equals k one times k two, over k minus one plus k two, times A times B, in concentrations. Still first order in A and in B, just with a combined rate constant. Got it?

4:35Energy profiles, and the big trap

Now picture the energy profile of a two-step reaction. Each peak on it is an activated complex, also called a transition state. Okay? And each valley between two peaks is an intermediate. So which step is rate-determining? The step with the highest peak, correct! But measure that height from the starting reactants, not from the valley just before it. That step is usually the rate-determining one. Clear? Now, the big trap. Suppose your mechanism gives exactly the rate law found by experiment. Does that prove the mechanism? No, no, careful here! Different mechanisms can give the same rate law. So a matching rate law means the mechanism is possible, never that it is proved. With me?

5:26Example 1: rate law of 2NO2 + F2, F atom as intermediate

Now, example one. Two N O two plus F two gives two N O two F, and it goes in two steps. First, N O two plus F two gives N O two F plus an F atom: slow. Then N O two plus that F atom gives more N O two F: fast. Find the rate law, and name the intermediate. Pause the video and try it first. Done? Okay, let's check. The balanced equation has two N O two and one F two. So the rate is k times N O two squared times F two, right? Oh no, careful here, that is the trap! The overall equation is not an elementary step. The rate-determining step is the slow one: N O two plus F two. Okay? This slow step is elementary, so order equals molecularity. Rate equals k times N O two times F two, in concentrations: first order in each. Right? Now the intermediate: the F atoms, formed in step one and used up in step two. Add the two steps, and they cancel out of the overall equation. Perfect!

6:38Example 2: ozone pre-equilibrium, order minus 1 in O2

Now a harder one. Ozone, O three, decomposes: two O three gives three O two. Step one is fast and reversible: O three gives O two plus an oxygen atom, with equilibrium constant capital K, okay? Step two is slow: the oxygen atom plus O three gives two O two, rate constant k. Find the rate law, and the order with respect to O two. Pause the video and try it first. Done? Okay, let's check. The slow step decides: rate equals k times the oxygen atom times O three, in concentrations. But the oxygen atom is an intermediate, so we replace it. With me? Use the fast equilibrium: capital K equals O two times the oxygen atom, over O three, all in concentrations. So the oxygen atom's concentration is capital K times O three, over O two. Clear? Put it in. Rate equals k times capital K times O three squared, over O two; the small two up top means squared. Now, O two is a product, so its order must be zero, right? No, no, careful here! O two sits in the denominator, so its order is minus one. Yes, orders can be negative. Got it?

7:58Quick revision: slow step, pre-equilibrium, energy profile

So, quick revision. For an elementary step, order equals molecularity. Slow step first: its rate law is the rate law. Fast pre-equilibrium: replace the intermediate using capital K, and k obs is k two times capital K, okay? The highest peak, measured from the reactants, is usually rate-determining. And a matching rate law never proves a mechanism. Now go and try the practice problems on this page. All the best!