Isolation, Pseudo Orders and Units of k

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Take a reaction whose rate equals k times A times B, okay? So it is second order overall, right? Now, in the lab, we add a huge excess of B to the flask. And suddenly the half-life of A stays the same, whatever amount of A we start with, exactly like a first-order reaction. So which is it, first order or second order? Today we untangle this with the isolation method, and also learn to read the order from the units of k. Questions on both come up often in JEE papers. So, let's begin.

0:36Isolation: flood the reaction with one reactant

So, here is a rate law with two reactants. Rate equals k times A to the power x, times B to the power y. The square brackets mean concentration in moles per litre, and k is the rate constant. The powers x and y are the orders with respect to A and B, and they come from experiment, not from the balanced equation. So how do we separate them? Let's see. Here is the trick. We take B in large excess, so there is far more B than A in the flask. Now tell me, as the reaction runs, does the concentration of B change much? Hardly at all, correct! Only a tiny part of all that B gets used, so its concentration stays practically constant. Okay? Now, B to the power y is a fixed number, so it joins k. We call the new constant k prime: that is the k with a small dash after it. So, rate equals k prime times A to the power x, where k prime equals k times B to the power y. With me? The reaction now behaves as if its order is x alone. We call this a pseudo order; pseudo means it only looks that way. When x is one, it is a pseudo-first-order reaction. And that is our puzzle! With B flooded, k times A times B becomes k prime times A, okay? So the half-life of A no longer depends on where A starts. Right? Now, careful. Our flooded reaction behaves like first order. So it is a first-order reaction, right? No, no, careful here! The overall order is still x plus y. For our puzzle, one plus one, which is two. The order in B has not disappeared; it is hiding inside k prime. Clear? So how do we get both orders? With B flooded, find x by any method from this chapter. Then repeat with A in large excess to find y. Simple, isn't it?

2:34Units of k from the order

Now, the second big idea: the units of k. Rate is in moles per litre per second. The minus one up top just means per: L to the power minus one is per litre, and s to the power minus one is per second. Every concentration is in moles per litre. Okay? For a reaction of order n, rate equals k times A to the power n. So what must the units of k be? Yes, exactly, just divide! k equals rate over A to the power n. Moles per litre per second, divided by moles per litre to the power n, leaves moles per litre to the power one minus n, times per second. Got it? Put n equals zero: the power is one, so k has the same units as the rate, moles per litre per second. Put n equals one: the power is zero, so k is just per second. Right? Second order: the power is minus one, so the moles go down and the litres come up: litre per mole per second. Third order: litre squared per mole squared per second. With me? And here is how toppers read it backwards. The power of mol is always one minus n. Read that power, and you have the order. Clear?

3:53Fractional lives: one more fingerprint

One more quick tool. The t with a small fifty percent below it is t half, the half-life, and t seventy five percent is the time for seventy five percent of A to react. At first order, how many half-lives does that take? Two half-lives, correct! The first one removes half. The second removes half of what is left, one more quarter. Half plus a quarter is seventy five percent, so the ratio is two. Okay? Now the full set: the ratio is one point five for zero order, two for first order, and three for second order. So if t seventy five is three times t fifty, it is second order straight away. Got it?

4:33Example 1: order 3/2 from the units of k

Now, example one. Here, a rate constant has units mol to the power minus one half, litre to the power one half, and per second. Find the order of the reaction. Pause the video and try it first. Done? Okay, let's check. Open up the general units: moles per litre, to the power one minus n, times per second. That is mol to the power one minus n, times litre to the power n minus one, times per second. Right? Now match the power of mol. In the question it is minus one half. So one minus n equals minus one half. With me? So n equals three over two. Order three halves! And a check with the litres: n minus one is one half, exactly the power of litre in the question. Perfect!

5:24Example 2: find y and k from pseudo first order constants

Now a harder one. Rate equals k times A, times B to the power y, and B is kept in large excess. The pseudo-first-order constant is two point zero times ten to the power minus three per second when B is zero point five molar, and eight point zero times ten to the power minus three per second at one point zero molar. Capital M means molar. Find y, and then k, okay? Pause the video and try it first. Done? Okay, let's check. B is flooded, so the pseudo-first-order constant, k prime, equals k times B to the power y. Right? B went from zero point five to one point zero molar, so it doubled. And k prime went from two point zero to eight point zero, four times bigger. So what is y? Yes, exactly! Eight point zero over two point zero is four. And one point zero over zero point five, to the power y, is two to the power y. So four equals two to the power y, and y equals two. The double arrow on screen just means therefore. Clear? Now for k. k equals k prime over B squared. From the second run, that is eight point zero times ten to the power minus three, over one point zero squared, so k has that same value. With me? Now check with the other run. Two point zero times ten to the power minus three, over zero point five squared, gives the same value. Very good! So k equals eight point zero times ten to the power minus three per second, right? Same number as k prime, so same units too? Oh no, careful, this is the trap! The number matches only because B was one molar. The full reaction has order one plus two, which is three. So k is eight point zero times ten to the power minus three, litre squared per mole squared per second. The units of k come from the overall order, not the pseudo order. Got it?

7:29Quick revision: isolation, units of k, t75/t50 ratios

So, quick revision. Flood the reaction with B, and the rate becomes k prime times A to the power x, with k prime equal to k times B to the power y; the overall order is still x plus y. The units of k are moles per litre to the power one minus n, times per second. And t seventy five over t fifty is one point five, two and three, for zero, first and second order. Now go and try the practice problems on this page. All the best!