Rate Law, Order and Molecularity

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle to start. Ammonia gas breaks down on a hot platinum surface. Now we double the pressure of the ammonia, so there is twice as much gas over the metal. How much faster does it break down? Twice as fast? Hold that thought. Today we learn the rate law, order and molecularity, and you will see why the balanced equation alone cannot tell you how fast a reaction goes. These ideas come up often in JEE kinetics questions. So, let's begin.

0:34The rate law: found by experiment

So, here is the rate law. Small r is the rate, and small k is the rate constant. The square brackets around A mean the concentration of A, in moles per litre. So r equals k, times the concentration of A to the power m, times the concentration of B to the power n. And look at the words on the right: found by experiment. Okay? The powers m and n are called the orders. m is the order with respect to A, and n is the order with respect to B, okay? Add them, and you get the overall order, m plus n. Clear? Now tell me something. Can we simply read m and n from the coefficients of the balanced equation? No, we cannot! The orders come only from experiment. And they can be almost anything: zero, a fraction, or even negative. Got it? So they need not match the coefficients in the balanced equation at all. The one exception is a single elementary step, which we will meet in a moment. This is a classic JEE trap: students copy the coefficients as the powers. Please don't do that! With me? And now, our puzzle. For N H three, ammonia, where the small three counts hydrogen atoms, on hot platinum at high pressure, the rate equals just k, with no concentration term at all. That is zero order! The metal surface is already saturated with adsorbed ammonia, so extra gas cannot speed it up. That is why doubling the pressure changes nothing. Right?

2:10The rate constant k and its units

Next, the rate constant, k. Put every concentration equal to one mole per litre in the rate law, and only k is left. So k is the rate at unit concentrations. Okay? Now, think. If we double the concentration of A, the rate changes. But does k change? No, it does not! k depends on the temperature and on a catalyst, but not on concentration. The rate changes; the rate constant stays the same. Got it? Now its units. For a reaction of order n, k equals the rate over the concentration to the power n; the slash means divided by. So the units of k are moles per litre, to the power one minus n, times per second. On screen, m o l is mole, capital L is litre, and the minus one powers mean per litre and per second. Clear? Zero order first. n equals zero, so the power is one minus zero, which is one. Moles per litre per second, the same units as the rate itself. Okay? Now you try one. First order, n equals one. What are the units of k? One minus one is zero, so the whole bracket disappears! Just per second. Very good! And second order: one minus two is minus one, so the bracket flips over. Litre per mole per second. Careful here: many students give every k the units of a rate. Only a zero order k has those units, right?

3:41Molecularity: count the colliding species

Now, a new word: molecularity. An elementary step is a single step, in which the reacting species collide together. Molecularity is the number of species colliding in that one step. So it is defined only for elementary steps. With me? Quick question. Can a molecularity ever be zero? Or three over two? No, it can never be zero, and it can never be a fraction. Zero species cannot collide, and half a molecule cannot take part in a collision. So it is one, two, or rarely three, since three species rarely collide at the same moment. Okay? And here is the link between the two ideas. For an elementary step, and only for an elementary step, the order equals the molecularity. For an overall reaction, the order still comes from experiment. Clear?

4:36Pseudo-first-order reactions

One last idea. In a hydrolysis, water is one of the reactants, but it is present in a huge excess. As the reaction runs, does the concentration of water change much? Hardly at all, exactly! So its concentration stays effectively constant. Right? So the rate, k times the concentration of A times the concentration of H two O, is approximately k prime times the concentration of A, okay? k prime, with the little dash, is the new constant, and the wavy equals sign means approximately. Two molecules react, but the rate behaves as first order. We call it pseudo first order, and pseudo means false. Got it?

5:19Example 1: order 5/2, rate when both are doubled

Now let's solve a problem. The rate equals k, times the concentration of A to the power one half, times the concentration of B squared. Find the order in each, the overall order, and how the rate changes when both concentrations are doubled. Pause the video and try it first. Done? Okay, let's check. The orders are just the powers: one half with respect to A, and two with respect to B, okay? So the overall order is one half plus two: five over two. See, a fractional order, just as we said. Got it? Now we double both concentrations. Everything is doubled, so the rate simply doubles too, right? Oh no, careful here! Each concentration carries its own power. Doubling A multiplies the rate by two to the power one half, and doubling B multiplies it by two squared. Together: two to the power five over two. With me? Two to the power five over two is four root two; the tick-shaped sign is the square root. That is about five point six six. So the rate becomes about five point six six times, not two times. Clear?

6:31Example 2: units of k for third order, order from units

Now a slightly harder one. Part one: give the units of k for a third order reaction. Part two: a reaction has k in litre to the power one half, mole to the power minus one half, per second. What is its order? Pause the video and try it first. Here is a hint: use the units formula both ways. Done? Okay, let's check. Part one: put n equals three in the formula. One minus three is minus two, so we get moles per litre, to the power minus two, per second. Okay? The minus two flips the bracket and squares it: litre squared, per mole squared, per second. That is the third order answer. Right? Part two goes backwards. Can you write litre to the power one half, times mole to the power minus one half, as moles per litre raised to some power? Yes, it is moles per litre, to the power minus one half! Check it: mole gets the power minus one half, and litre gets minus one times minus one half, which is plus one half. Got it? So compare the powers: one minus n equals minus one half. The double arrow means so, and n equals three over two. An order of three over two, a fraction again; that is fine for an order, but never for a molecularity. Clear?

7:51Quick revision: rate law, units of k, molecularity

So, quick revision. The rate law: r equals k times the concentrations raised to the orders, and the orders come from experiment, not from the balanced equation. The units of k: moles per litre to the power one minus n, per second; and k changes with temperature, not with concentration. Molecularity is one, two or rarely three, and it equals the order only for an elementary step. Now go and try the practice problems on this page. All the best!