Heterogeneous Equilibria

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Put some limestone in an empty, sealed box, and heat it to a fixed high temperature. The limestone, calcium carbonate, breaks down and gives carbon dioxide gas. Now do it again, same box, same temperature, but with ten times as much limestone. Which box ends up with more carbon dioxide pressure? Keep that thought; we will settle it today, with one important condition. In JEE, these solid and gas equilibria come up often. So, let's begin.

0:37Heterogeneous: solids and a gas together

So, our limestone reaction. Calcium carbonate, C a C O three, where the small s in brackets means solid, gives solid calcium oxide, C a O, and carbon dioxide gas; the small g means gas. The two half arrows, one pointing each way, mean it is reversible and at equilibrium. Two solids and a gas: more than one phase. That is why we call it a heterogeneous equilibrium. With me? Now we want K p: capital K with a small p below, the equilibrium constant written with partial pressures. Three substances here. Which of them go into K p? Only the gas, yes, exactly! Pure solids and pure liquids do not appear in K, okay? Their activity is taken as one, so they drop out. Clear? So K p equals p C O two: the partial pressure of carbon dioxide, with C O two written small below the p. Just one term. Right? At a given temperature, capital T, K p is fixed, so the pressure of carbon dioxide over the solids is fixed too, however much of each solid is present. So, our puzzle: a little limestone, or ten times as much? The same pressure in both boxes! With me? But one condition, and questions do test it. Some of each solid must remain. If all the calcium carbonate breaks down before the pressure gets there, there is no equilibrium. We will see exactly this in example two. Got it? Quick check. The mixture is at equilibrium, in a closed vessel, at constant temperature. I add more calcium carbonate. More reactant, so more carbon dioxide, right? No, no, careful! The pressure is unchanged. The pure solid does not appear in K p, which is just p C O two, so adding more does not disturb the equilibrium.

2:29Writing Kp: check every state label

Okay? Alright, one more from the course. Three iron, solid, plus four water, gives the iron oxide F e three O four, solid, plus four hydrogen gas. Now, many students throw the water out, because water is a liquid, right? You write K p. Does the water stay? Let's see. K p equals p H two to the power four, over p H two O to the power four. The fours up top come from the coefficients, and only the two solids have dropped out. With me? So yes, the water stays! Look at its label: small g. Here water is steam, a gas, so it goes in. So always read the state labels first, then write K p. Got it?

3:12Starting from the solid alone

Now, a type that comes up often. Solid ammonium hydrogen sulphide, N H four H S, breaks into two gases: ammonia, N H three, and hydrogen sulphide, H two S, okay? We put only the solid into an empty flask. Right? At equilibrium, call the total pressure capital P, okay? Tell me, how is it shared between the two gases? Equally! Yes, exactly. Every time one ammonia forms, one hydrogen sulphide forms with it, so their partial pressures are always equal. Each one is P over two. Clear? So K p equals P over two, times P over two: P squared over four. The small two up top means squared. Okay? Next, solid ammonium carbamate, N H two COONH four. It gives two ammonia and one carbon dioxide, both gases. With me? Again we start from the solid alone. How is the total pressure shared now? Two to one! Two ammonia for every carbon dioxide. So ammonia has two P over three, and carbon dioxide has P over three. Right? So K p equals two P over three, in brackets, squared, times P over three. That gives four P cubed over twenty seven; the small three up top means cubed. But careful: both results hold only when the flask starts with the solid alone. If the flask already has some ammonia, the gases are no longer shared in this ratio. Got it?

4:43Example 1: Kp for NH4HS from total pressure 0.84 atm

Now, example one, from the course. Solid ammonium hydrogen sulphide is placed in an evacuated flask, with all the air pumped out. At equilibrium, the total pressure is zero point eight four atmospheres; a t m on screen means atmospheres. Find K p. Pause the video and try it first. Done? Okay, let's check. Some students put the total pressure straight in, and square zero point eight four. Is that right? No, no, wait! K p needs each gas's own partial pressure. We started from the solid alone, so the pressure splits equally: zero point eight four over two, zero point four two atmospheres each. Right? So K p equals zero point four two squared: approximately zero point one seven six atmospheres squared; the little two on the a t m means squared. Clear?

5:36Example 2: does all the CaCO3 decompose in 5 L?

Now a harder one, and it brings back our puzzle. For calcium carbonate going to calcium oxide and carbon dioxide, K p is zero point two zero atmospheres at one thousand and seventy three kelvin; the capital K after the number means kelvin. Calcium carbonate is heated to that temperature in an evacuated vessel of five point zero litres, capital L for litres. What happens with zero point five zero grams, and with two point zero grams? Pause the video and try it. Done? Let's check. To reach equilibrium, the pressure of carbon dioxide must build up to K p, zero point two zero atmospheres. Okay? So, how many moles of carbon dioxide is that? Which law will you use? The ideal gas law, yes! n, the moles, equals p V over R T, with V the volume and R the gas constant. Zero point two zero times five point zero, over zero point zero eight two one times one thousand and seventy three. That is zero point zero one one four moles of carbon dioxide. Right? Now part a. The molar mass, capital M, is one hundred grams per mole; the g mol with a small minus one on screen means grams per mole. So zero point five zero grams is zero point zero zero five zero moles. Even if all of it breaks down, that gas gives only zero point zero eight eight atmospheres. With me? So, is equilibrium reached? No! The pressure needs zero point two zero, but the solid runs out at zero point zero eight eight. So all of it decomposes, and equilibrium is never reached. Remember our condition: some of each solid must remain. Here, none does. Clear? Part b. Two point zero grams is zero point zero two zero moles, more than enough. Only zero point zero one one four moles decomposes, just enough to build the pressure up to zero point two zero. Okay? So zero point zero zero eight six moles, that is zero point eight six grams, remains, in equilibrium with zero point two zero atmospheres of carbon dioxide. More solid would not change that pressure, just as in our puzzle. Got it?

7:46Quick revision: solids out of K, CaCO3, NH4HS, carbamate

So, quick revision. Pure solids and pure liquids stay out of K, so read the state labels first. For calcium carbonate, K p is just p C O two, fixed at a given temperature, as long as some of each solid remains. And from the solid alone, K p is P squared over four for ammonium hydrogen sulphide, and four P cubed over twenty seven for ammonium carbamate, with P the total pressure. Now go and try the practice problems on this page. You will find them easy. All the best!