Parallel, Consecutive and Reversible Reactions

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Suppose substance A turns into B, and then B itself turns into C, okay? You want B, the one in the middle. So the longer you wait, the more B you get, right? We will settle this today. We will study three kinds of complex reactions: parallel, consecutive and reversible, with every step first order. In JEE, questions on these come up often. So, let's begin.

0:30Parallel reactions: one reactant, two paths

First, parallel reactions. Here one reactant, A, can go two ways at once. It turns into B with rate constant k one, written on the arrow as k with a small one below it, and into C with rate constant k two. Both paths are first order. Okay? Molecules of A are leaving by both doors at once. So how fast does A disappear? Faster than by either path alone, yes! The two rate constants simply add. So A decays as a first-order reaction with k equals k one plus k two. Right? So the half-life of A, t half, written t with a small one over two below it, equals zero point six nine three over k one plus k two. Use the total k here, never k one alone. Clear? Now tell me, how do B and C share the product between them? In the ratio of their rate constants, exactly! Square brackets mean concentration. So the concentration of B over the concentration of C equals k one over k two, at all times. Okay? So the fraction of the product that is B equals k one over k one plus k two. Times one hundred gives the percentage. With me? Each path also has its own activation energy, E a one and E a two: capital E, with a small a and the path number below it. The overall E a equals k one times E a one, plus k two times E a two, all over k one plus k two. Got it?

2:03Consecutive reactions: B rises, then falls

Now, consecutive reactions: one after the other. Here A turns into B with k one, and then B turns into C with k two. So B is made in the first step and used up in the second. Okay? At the start, there is only A, so there is no B at all. What about at the very end, after a long, long time? Zero again, correct! Everything has gone on to C by then. So in between, the concentration of B must rise to a maximum, and then fall. Remember our puzzle? Waiting forever gives you no B at all! Right? So when does B peak? At t max, written t with a small max below it. t max equals the natural log of k one over k two, all over k one minus k two. The l n means natural log. Clear? And the peak value, B max, written with the same small max below, equals A naught times k one over k two, raised to the power k two over k two minus k one. Here, A naught, with a small zero below, is the starting concentration of A, okay? Watch the order in that power: k two minus k one. With me? Now, C can only form as fast as both steps allow. So which step decides how fast C forms? The slower one, yes! The slower step controls the rate of forming C, like the rate-determining step of the last section. Got it?

3:32Reversible reactions: two-way traffic

Finally, reversible reactions. See the two half arrows, one pointing each way? Here A turns into B with the forward rate constant, k f, and B turns back with the backward rate constant, k b. Small f for forward, small b for backward. Okay? At equilibrium, the two rates balance. So capital K, the equilibrium constant, equals k f over k b. Right? Now, how fast does the reaction approach equilibrium? Only the forward rate constant, k f, matters, doesn't it? No, no, careful here! Both directions count. k f plus k b, times t, equals the natural log of, A naught minus A e q, over, A at time t minus A e q. Here A e q, with a small e q below, is the concentration of A at equilibrium. Again the rate constants add, just like parallel reactions. Clear?

4:31Example 1: parallel paths, half-life of A and percent B

Now, example one. Here, A decays by two parallel first-order paths. k one is one point five times ten to the power minus three, and k two is five times ten to the power minus four, both per second: that is the s to the power minus one. Find the half-life of A, and the percentage of B in the product. Pause the video and try it first. Done? Okay, let's check. First, which rate constant goes into the half-life of A? The total, of course, because A leaves by both paths. Write five times ten to the power minus four as zero point five times ten to the power minus three, and add: k equals two times ten to the power minus three per second. Right? So t half equals zero point six nine three over two times ten to the power minus three. That is three hundred and forty six point five, about three hundred and forty seven seconds. The wavy equals sign means approximately. With me? Now, let us find the percentage of B in the product. The fraction of B equals k one over the total k: one point five times ten to the power minus three, over two times ten to the power minus three. That is zero point seven five. Clear? So B is seventy five percent of the product, at every moment of the reaction. Perfect!

5:55Example 2: A to B to C, t max and B max, the power trap

Now, example two, and this is our puzzle from the start. For A to B to C, k one is zero point two and k two is zero point one, both per minute: that is the min to the power minus one. When is the concentration of B largest, and what is its maximum value? Pause the video and try it first. Done? Okay, let's check. t max equals the natural log of k one over k two, over k one minus k two. So, natural log of zero point two over zero point one, over zero point two minus zero point one. Right? The top is natural log two, and the bottom is zero point one. So t max is natural log two over zero point one, that is zero point six nine three over zero point one: six point nine three minutes. Clear? Now the peak value. k one over k two is two, and the power is zero point one over zero point two minus zero point one, which is plus one. So B max is twice A naught, right? No, no, careful here, that is the trap! That is more B than the A we started with, which is impossible. The power is k two over k two minus k one: zero point one over zero point one minus zero point two, which is minus one. The dot means times. So B max equals A naught times two to the power minus one. Got it? So B max is zero point five times A naught. And our puzzle is solved: wait about six point nine three minutes, and at best half of A has become B, okay? Wait longer, and B only falls. Perfect!

7:28Quick revision: parallel, consecutive, reversible rules

So, quick revision. Parallel: the rate constants add, and B over C equals k one over k two at all times. Consecutive: B rises, peaks at t max, then falls. Reversible: capital K equals k f over k b, and equilibrium is approached with k f plus k b. Now go and try the practice problems on this page. All the best!