Zero- and First-Order Reactions

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. Half of a reactant is used up in a time called the half-life. Now we wait one more half-life. Is the reactant all gone? Half plus half makes the whole, right? Well, it depends on the order! For one order, yes, it is all gone. For the other, a quarter is still left. Today: zero and first order reactions, and their half-lives. These come up often in JEE kinetics numericals. So, let's begin.

0:33Zero order: a straight-line fall

Look at this graph from the course. Along the bottom is time, t. Up the side is the concentration of A over its starting value: the square brackets mean concentration, and the small zero means at the start. All three start at one and reach the dashed line, one half, together, at the mark one. So each mark along the time axis is one starting half-life: the time to fall from one to one half. We follow zero and first order today; second order waits for the next lesson. Okay? So, zero order. The concentration of A equals A naught minus k t, where A naught is the starting concentration and k is the rate constant. So it drops by the same amount in every unit of time, like the line labelled zero order. Right? Now, think. When is half of A naught used up? Put in half of A naught, and t half equals A naught over two k. The t with a small one over two is the half-life, and the sign like a little fish means proportional to. So double the start, and the zero order half-life doubles too. Clear? And when does the concentration become zero, all used up? Very good! Put the concentration equal to zero, and t equals A naught over k: exactly two half-lives! On the graph, the zero order line crosses one half at one, and hits zero at two. So a zero order reaction is complete after two half-lives. Got it? And plot the concentration of A against t: for zero order, a straight line. With me?

2:12First order: the integrated rate equation

Now, first order. Natural log of A naught over the concentration of A equals k t. On screen, l n is the natural log, the log to the base e. Okay? In numericals we often use log to the base ten: k equals two point three zero three over t, times log of a over a minus x. Here a is the starting amount and x the amount reacted, so a minus x is what remains. And two point three zero three is just the natural log of ten. Clear? We can also write it this way: the concentration of A equals A naught times e to the power minus k t. It falls fast at first, then slower and slower, like the curve labelled first order. Right? Now, can e to the power minus k t ever become exactly zero? Correct, it never can; it only gets closer and closer to zero. So first order is never strictly complete; its curve never touches the axis. Got it? And the straight line test: plot the natural log of the concentration against t, and first order gives a straight line with slope minus k. Okay?

3:24First order: half-life and half-lives

Now the first order half-life. Put the concentration equal to half of A naught. What do you get? The ratio becomes two, so natural log of two equals k times t half. The double arrow means so: t half equals zero point six nine three over k, since the natural log of two is zero point six nine three. With me? Now, think. Double the starting concentration. What happens to the first order half-life? Yes, exactly: nothing, because there is no A naught in the formula. Zero order was different: its half-life grows with A naught. Please don't mix these two up. Clear? Every half-life halves whatever is present: after n half-lives, one half to the power n remains. Now, our puzzle! After two half-lives, one quarter is still left. On the graph, at two, the first order curve is at one quarter, while the zero order line is at zero. So our puzzle is solved. Right? The small subscripts with the percent sign show how much has reacted. Seventy five percent complete leaves one quarter: two half-lives. Eighty seven point five percent complete leaves one eighth: three half-lives. Okay? And ninety nine point nine percent takes about ten half-lives, since one half to the power ten is about one in a thousand. The wavy equals sign means approximately. Got it? Last, the average life: tau, the Greek letter like a small t, equals one over k. Clear?

4:58Example 1: zero order half-life and time to fall to 0.30 M

Here is a zero order reaction with k equal to zero point zero two moles per litre per minute; on screen, m o l is mole, capital L is litre, m i n is minute, and the minus one powers mean per. The starting concentration is one point zero molar, where the capital M means moles per litre. Find the half-life, and the time to fall to zero point three zero molar. Pause the video and try it first. Done? Okay, let's check. The half-life is A naught over two k: one point zero over two times zero point zero two, that is one over zero point zero four. Twenty five minutes. Okay? Now part two. Some students say: zero point five after twenty five minutes, then half again, zero point two five. So we find zero point three zero by halving, right? Oh no, careful here! Halving again and again is first order behaviour. Zero order falls in a straight line, so use the concentration of A equals A naught minus k t. Put in the numbers: zero point three zero equals one point zero, minus zero point zero two times t. With me? So t equals one point zero minus zero point three zero, over zero point zero two. That is zero point seven zero over zero point zero two, so the answer is thirty five minutes. Got it?

6:28Example 2: first order 20% done in 10 min, k and 90% time

Now a slightly harder one. We have a first order reaction that is twenty percent complete in ten minutes. Find k, and the time for ninety percent completion. Pause the video and try it first. Done? Okay, let's check. Twenty percent complete, so the concentration of A over A naught is zero point two, right? No, no, careful here! Twenty percent complete means twenty percent has reacted, so eighty percent remains. So A naught over the concentration of A is one over zero point eight. Okay? So, with t equal to ten minutes, k equals one over ten, times natural log of one over zero point eight. That log is zero point two two three one, so k is zero point zero two two three per minute. Right? Now ninety percent completion. The same trap is waiting. What is A naught over A now? Yes, exactly: ten percent remains, so the ratio is ten. Okay? So t ninety, the time for ninety percent completion, is natural log of ten over k. And natural log of ten is two point three zero three, the number from our log formula! Two point three zero three over zero point zero two two three: one hundred and three minutes. Always keep k and t in the same time unit. Got it?

7:57Quick revision: zero and first order laws, half-lives

So, quick revision. Zero order: the concentration of A equals A naught minus k t, the half-life is A naught over two k, and it is complete at two half-lives. First order: natural log of A naught over A equals k t, and the half-life is zero point six nine three over k, independent of A naught. And x percent complete means one hundred minus x percent remains; keep k and t in the same time unit. Now go and try the practice problems on this page. All the best!