Le Chatelier's Principle: Concentration, Pressure and Inert Gases

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take a closed bottle of soda water: no bubbles at all. Now open the cap. Fizz! Bubbles everywhere. Same soda, same temperature. So what changed? Keep your guess; we will answer it soon. The rule behind it is Le Chatelier's principle, and in JEE, questions on it come up often, especially the tricky ones with inert gases. So, let's begin.

0:31Le Chatelier's principle

So, here is the rule. Take a system at equilibrium, and disturb it: change a concentration, or the pressure, or the temperature. We have given it a push, okay? Now think. How will the system respond to that push? It pushes back, yes, exactly! The equilibrium shifts in the direction that partly counteracts the change. Look at that word, partly: it never undoes the change completely. We will see this on a graph in a moment. Right? One more thing. In this lesson the temperature is constant, so capital K, the equilibrium constant, stays the same. Only the mixture moves. Clear?

1:12Concentration: compare Q with K

Let's start with concentration. This graph from the course shows concentration against time; capital M means molar, moles per litre. The reaction: H two plus I two, the twos written small below, giving two H I, hydrogen iodide. The two half arrows mean reversible, and K c, capital K with a small c, is sixty four. Before t one, the little one below the t, the lines are flat: H I at zero point eight, hydrogen and iodine at zero point one each. That is equilibrium. With me? Now, at t one, we add hydrogen. See the dark blue line jump up? The note on the graph already tells us the system shifts forward. But why forward? Use capital Q, the reaction quotient: the same expression as K, but with the concentrations at this moment. Just after we add hydrogen, a reactant, is Q less than K, or greater than K? Less than K, yes, exactly! More reactant makes Q smaller, so Q drops below K, and to bring it back up, the reaction goes forward: some of the extra hydrogen reacts with iodine to make more hydrogen iodide. So the rule on screen: add a reactant, or remove a product, and Q becomes less than K; that sign is less than. Forward. Okay? And the other way round: add a product, and Q becomes greater than K, so the shift is backward. Right? Now the course's numbers. We add zero point two zero moles per litre of hydrogen, so hydrogen becomes zero point three zero. Q is zero point eight zero squared, the small two up top, over zero point three zero times zero point one zero. About twenty one; the wavy equals sign means approximately. Less than sixty four, so forward. Got it? Now the graph after t one. Square brackets mean concentration, and the arrows mean goes to. Hydrogen jumped from zero point one zero to zero point three zero, then fell back, but only to zero point two five, not to zero point one zero! Partly undone, just as the principle says. Meanwhile iodine falls from zero point one zero to zero point zero five zero, and hydrogen iodide climbs to zero point nine zero, until Q again equals K, okay?

3:39Pressure: count the moles of gas

Now, pressure. Suppose we compress the mixture into a smaller volume, so the pressure goes up. How can the system push back against that? By making fewer gas molecules! Very good! Fewer moles of gas give a lower pressure. So on compression, the equilibrium shifts towards the side with fewer moles of gas. Okay? But what if both sides have the same moles of gas? We write delta n g: the Greek capital delta, the little triangle, means change in, and n g, with the small g below, means moles of gas, counting gases only. When delta n g equals zero, pressure has no effect at all. With me? And the opposite: expand the mixture to lower the pressure, and the shift is towards more moles of gas. Right? Now, our soda bottle! Inside, carbon dioxide gas, C O two with g in brackets, is in equilibrium with carbon dioxide dissolved in water, a q in brackets. So, when you open the cap, which way does this equilibrium shift? Opening the cap lets the gas above the liquid escape, so p C O two, small p for its partial pressure, falls. The equilibrium shifts towards the gas, and dissolved carbon dioxide comes out as bubbles. Fizz! Mystery solved. Got it?

5:01Inert gases, solids and catalysts

Now the favourite trap: inert gases. An inert gas, like argon, takes no part in the reaction. We pump argon into a rigid flask, so the volume is constant, and the total pressure goes up. So the shift is towards fewer moles of gas, right? No, no, careful! The volume is the same, so the partial pressures of the reacting gases do not change. Only those partial pressures go into Q, so Q still equals K, and there is no shift at all. Okay? Now add argon at constant total pressure. To keep the pressure constant, what must happen to the volume? It must increase, correct! The volume grows, so every partial pressure falls, just like a pressure decrease. So the shift is towards more moles of gas; and again, if delta n g is zero, there is no shift. Clear? And two quick ones. Adding a pure solid or a pure liquid: no shift, because they do not appear in K, right? Adding a catalyst: no shift either; it only helps the system reach equilibrium sooner. With me?

6:09Example 1: H2 + I2 to 2HI, add H2, remove HI, compress

Alright, example one, from the course. Hydrogen gas plus iodine gas gives two hydrogen iodide, all gases. State the effect of: part a, adding hydrogen; part b, removing hydrogen iodide; and part c, compressing the mixture. Pause the video and try it first. Done? Okay, let's check. Part a: hydrogen is a reactant. Adding it makes Q less than K, so the shift is forward. Easy, right? Part b: hydrogen iodide is a product. Removing it also makes Q less than K, so forward again; the system replaces what we took away. Okay? Part c: compressing. The pressure goes up, so the shift is towards fewer moles of gas. Which side is that? Count them. Left side: one hydrogen plus one iodine, two moles of gas. Right side: two moles of hydrogen iodide. Two and two! Neither side has fewer. Clear? So delta n g is zero, and compressing gives no shift at all. Just saying forward here loses the mark. Always count the gas moles first. Got it?

7:21Example 2: ammonia equilibrium, argon at constant V or P

Now a harder one; in JEE, questions like this come up often. Nitrogen plus three hydrogen gives two ammonia, N H three, at equilibrium. Predict the effect of: part a, compressing; part b, adding argon at constant volume; and part c, adding argon at constant total pressure. Pause and try. Watch out for part c! Done? Okay, let's check. Part a: on the left, one nitrogen plus three hydrogen, four moles of gas. On the right, two moles of ammonia. Four goes to two, so compressing shifts it forward, towards ammonia. Clear? Part b: argon at constant volume. The total pressure went up, so forward again, right? No, no, wait! Remember the trap. Capital V, the volume, is constant, so the partial pressures of nitrogen, hydrogen and ammonia do not change. No effect! Okay? Part c: argon at constant total pressure; that is the capital P on screen. To keep the pressure constant, the volume must grow. And when the volume grows, every partial pressure falls. With me? So it behaves like a pressure decrease. The double arrow means so: the system favours more moles of gas, the four on the left. Backward! Forward, none, backward: three parts, three different answers. Got it?

8:52Quick revision: concentration, pressure and inert gas

So, quick revision. Add a reactant or remove a product: forward; add a product: backward. Raise the pressure: towards fewer moles of gas, but nothing happens when delta n g is zero. And an inert gas: no shift at constant volume, but towards more moles of gas at constant pressure. Now go and try the practice problems on this page. You will find them easy. All the best!