Kinetics of the Iodide–Hydrogen Peroxide Reaction

Transcript

0:00Introduction

Hello everyone! Here is a little lab magic. We mix the solutions and start a stopwatch. For a while, there is no blue colour at all. Then, in an instant, the blue colour appears, like an alarm clock going off. This is the iodine clock. Why does the blue appear suddenly, and not gradually? And how can a stopwatch give the rate of a reaction? This experiment is part of JEE practical chemistry, so let's begin.

0:29Two reactions: one slow, one fast

Two reactions run together here. The first is slow: hydrogen peroxide, H two O two, reacts with two iodide ions and two hydrogen ions. The small twos written low are subscripts; the raised minus and plus are charges. The arrow means gives: iodine, I two, and two water molecules. Okay? The second reaction is fast. Iodine reacts with two thiosulphate ions, S two O three, with a raised two minus charge. Out come two iodide ions again, and the ion S four O six, also two minus. Right? In the experiment, we add a small, fixed amount of sodium thiosulphate, and some starch, to the acidified iodide solution. Remember those two words, small and fixed. With me? Now let me ask you this. Iodine forms slowly, but the thiosulphate is already waiting. So what happens to each bit of iodine the moment it forms? Yes, exactly, the thiosulphate removes it at once. So no free iodine stays, and there is no blue colour yet. Clear? But the thiosulphate is only a small, fixed amount. Once it is used up, free iodine appears and turns the starch blue, all of a sudden. So the blue comes only at the instant the thiosulphate runs out, and that solves our first puzzle.

1:50The stopwatch gives the rate

Got it? Now the stopwatch. Let t, the small italic letter, be the time for the blue colour to appear. In that time, a fixed amount of iodine has formed, the same in every run. Okay? So tell me. In one run the blue comes quickly, and in another it takes much longer. Which run had the faster reaction? That is correct, the quick run was faster: the same iodine, in less time. So the initial rate is proportional to one over t; the curly sign on screen means is proportional to. Right? Now, rate equals delta of the iodine concentration, over t. Delta, the Greek letter like a triangle, means change in, and square brackets mean concentration. This equals the starting thiosulphate concentration over two times t; the small zero written low means at the start. With me? Now, where did that two come from? Look back at the fast reaction. Very good, one iodine molecule uses up two thiosulphate ions. So the iodine formed is half the thiosulphate, and we divide by two times t. Keep this two in mind for example one, okay?

3:06The rate law: first order in each reactant

Now, how does the rate depend on the concentrations? Here is the slow reaction again, and it has two iodide ions. So is the rate second order in iodide? No, and this is a classic trap: orders come from experiment, not from the equation. Experiments show that rate equals k, the rate constant, times the hydrogen peroxide concentration, times the iodide concentration. Clear? So each concentration appears to the power one, first order in each reactant. Add the powers, one plus one, and the reaction is second order overall. Okay? To show first order in iodide, keep the hydrogen peroxide fixed, change the iodide, and plot one over t against the iodide concentration; the slash means over. Got it? If the rate is first order in iodide, what shape should this graph have?

4:00The graph that gives the order

Yes, a straight line through the origin. Look at the graph in the figure: one over t rises in a straight line with the iodide concentration, with the hydrogen peroxide fixed, so double the iodide, and one over t doubles too. The box beside it sums up the experiment: the slow and fast reactions, and the blue colour that gives us the time t. Right? In dilute acid, the rate is almost independent of the hydrogen ion concentration, even though H plus is in the slow reaction in the figure. So never read orders from an equation. With me?

4:35Keeping every run fair

Two more points keep every run fair. First, the fast reaction turns the iodine back into iodide, so the iodide concentration stays constant during a run. Okay? Second, keep the temperature constant, because the rate roughly doubles for a ten degree Celsius rise; the small raised circle and capital C mean degrees Celsius. Example two is built on this fact. Clear?

4:58Example 1: initial rate of iodine formation in a 40 s run

Now, example one. The mixture initially contains two point zero times ten to the power minus four molar thiosulphate; capital M means molar, moles per litre. The blue colour appears after forty seconds, small s for seconds. Find the initial rate of formation of iodine. Pause the video and try it first. Done? Okay, let's check. Many students take two point zero times ten to the power minus four, over forty, and write five point zero times ten to the power minus six. Is that right? That is exactly the trap, because this number is the thiosulphate, not the iodine. Each iodine needs two thiosulphate ions, so the change in iodine concentration is half, which is one point zero times ten to the power minus four molar. Okay? This iodine forms in forty seconds, so the rate is one point zero times ten to the power minus four, over forty, which gives two point five times ten to the power minus six moles per litre per second; the raised minus ones mean per litre and per second. See, that little two halved the answer!

6:07Example 2: activation energy, 80 s at 20 C, 40 s at 30 C

Now example two, a harder one. In this question, a run takes eighty seconds at twenty degrees Celsius, and forty seconds at thirty degrees Celsius. Estimate the activation energy, with R, the gas constant, as eight point three one four joules per kelvin per mole. Pause the video and try it. Hint: the rate is proportional to one over t. Done? Okay, let's check. We use the Arrhenius equation for two temperatures. Natural log of k two over k one equals E sub a over R, times one over T one, minus one over T two. Here l n is the natural log, k is the rate constant at each temperature, E sub a is the activation energy, and capital T one and T two are the two temperatures. With me? Only the temperature changes, and the rate is proportional to one over t. So k two over k one is eighty over forty, which gives two. The rate doubled for a ten degree rise, just as we said earlier, right? T one is twenty and T two is thirty, so can we simply put them in? That would be wrong, because this equation needs kelvin, not degrees Celsius. Add two hundred and seventy three: T one is two hundred and ninety three kelvin, and T two is three hundred and three kelvin; capital K means kelvin. Got it? Now put everything in. Natural log of two equals E sub a over R, times one over two hundred and ninety three, minus one over three hundred and three. Okay? Natural log of two is zero point six nine three, and the bracket is one point one two six times ten to the power minus four. So E sub a is eight point three one four times zero point six nine three, over that number, in joules per mole. That is about fifty one kilojoules per mole; the wavy equals sign means approximately. Well done if you got it!

8:01Quick revision: sudden blue, rate as 1/t, rate law

Quick revision now. First, the blue appears only when the small, fixed amount of thiosulphate is used up. Second, the initial rate is proportional to one over t, and equals the starting thiosulphate concentration over two times t; don't forget the two. Third, rate equals k times the two concentrations, first order in each, at constant temperature. Now try the practice problems on this page. All the best!