Okay, something else we can do with this, the graph, especially that comparison of molar masses, is think about differences in properties due to differences in molar masses. And something we're going to do is the law of effusion and Graham's law. So Graham's law is going to give us the rate of relative a fusion.
That's what we're going to do in the next few slides, but let me explain what a fusion is. So effusion is this process where if we have a sample of gas in a container, right, remember these particles are bouncing all around, bouncing off the walls, and we have a very small hole here, a pinhole, just really big enough for particles to get through. What's going to happen is we don't get airflow in the way we would expect, like out of a nozzle or out of a... you know, a hair dryer, right?
But because it's not a big enough hole for the particles to flow through, what's going to happen is we'll get particles passing through this pinhole only when, if they try to bounce off this wall, when they go to run into the wall, they don't run into the wall, they run into where the pinhole is and then they just kind of sneak out without even knowing it, right? That's what effusion is. is. So, and then they'll effuse over to the other side. So we want to look at what the comparative rates of effusion are, and the comparative rates of effusion are going to be directly proportional to the comparative rates of speed of different molecules.
So let's think about what comparative rates of speeds are. Graham's law tells us that the rate of effusion of a gas is directly proportional to its RMS speed. And remember, the RMS speed is dependent on the temperature, is dependent on the average temperature.
So the faster the molecules are moving, the more collisions they'll have with the walls of the container, the higher chance they'll have of hitting the hole in the wall and effusing. So the rate of effusion is proportional to the RMS speed. And like we said earlier, the RMS speed is the square root of 3 RT over the molar mass of an individual gas.
So let's look at comparative rates of effusion. We're not going to calculate absolute rates of effusion. So let's imagine there's more than one gas in our container and they're both bouncing off of the walls, bouncing off of each other, and we want to know what's their relative rates of effusion.
Because they're in the same container, they have the same pressure. The only difference in the variables that matter, since temperature is the same, is going to be the molar mass. So the relative rate of effusion of gas number one to gas number two is the RMS speed of gas number one divided by the RMS speed of gas number two, right? So the rate of effusion of gas one to gas two, it's one over two for the RMS speeds.
Okay, well that is equal to three halves RT divided by the molar mass of gas one square root and three halves RT divided by the molar mass of the second molecule square root of right one divided by the other. Well I can cancel out the numerators the three RTs because those are the same for both molecules. So now I have the square root of one over the molar mass of molecule one divided by the square root of one over the molar mass of molecule number two clean this up because they're both square roots i can put them under the same square root symbol and because it's one over one over i can flip-flop them that is oh i'm in the way all right so that means that when we look at the relative rates of effusion of molar mass of the gases where we're looking at hydrogen and nitrogen.
The molar mass of hydrogen is 2.016 grams per mole, and the molar mass of nitrogen is 28.02 grams per mole. The rate of effusion... of the hydrogen divided by the rate of effusion of the nitrogen is equal to, and this is what I meant on the last slide that got covered up, we flip-flop the square root, the molar masses, because we had one over the red one on the numerator and one over the blue one in the denominator.
Well, I can flip those to get rid of the one over part. So mathematically speaking, the relative rates of effusion of hydrogen 2 nitrogen is equal to the square root of the molar mass of nitrogen divided by the molar mass of hydrogen. This is not a typo because the rate of effusion is inversely proportional to molar mass. That's why the molar mass of the nitrogen is in the numerator and the hydrogen is in the denominator. Now we just plug in those numbers, molar mass of nitrogen.
28.02 molar mass of hydrogen 2.0168 divide those out and we get the square root of 13.8988 and i'm covering up the answer but the square root of this number is sorry i can say it out loud but that's the answer in and it's a relative rate so it's the rate of one to the other so it's a unitless value 13.8988 square root 3.728 3.728 um hydrogen there it is hydrogen effuses 3.728 times faster than the nitrogen right that's what this tells us if we wanted to flip it around we could invert the 3.72 and that would give us i don't know if you can read that that would give us that number and then multiply that by a hundred to get 26.82 percent that is nitrogen is only 22 Nitrogen is only 26.82% of the rate of effusion of hydrogen. So the lighter molecule always effuses faster than the heavier molecule. Because the lighter molecule is moving faster, it has more collisions with everything, it has a higher chance of leaking out of the box.
Okay, so that is Graham's Law of Effusion. Do the calculation on the... Laws of effusion because it's harder to explain diffusion, but let me try to explain diffusion.
It's very similar to a fusion. It has to do with how fast a molecule is moving in space and it's having all of these collisions as it moves. Sometimes it travels a long distance and has a collision. Sometimes it's a short distance and it's bouncing.
It can go backwards. It can be go forwards up down back and forth and it does this random walk and overtime. That random walk will add up to an average distance for a collection of molecules. Some of them won't move, some of them will move far, right?
So there is a sort of rate at which a cloud of gas will spread out without the air being stirred. It'll just diffuse away from the center. Also works with liquids. If you put a drop of food coloring in a glass of water and then don't touch it, don't stir it, it'll diffuse out and it takes a... long time for liquids to effuse, so it'll just, it'll take a long time for that dot of food coloring for those molecules to disperse throughout the whole glass of water.
So that's what's going on with our model of diffusion, but the thing is each one of these steps is a speed, right, not the root mean squared speed, but every time it has a collision and it's a different speed in different direction. And on average, that speed is going to be the characteristic average speed of the particles. So how far it will diffuse depends on what its average speed is, just like in Grams'law.
So that's what diffusion is. It's harder to do calculations with diffusion, so that's why I introduced the effusion problem. I'm not going to ask any diffusion questions because it's too hard to calculate.
Okay, so this is the figure in the textbook that explains diffusion. It's a really good figure, but It's hard to understand just by looking at it what's happening here. But I want to explain it because it's a really nice demonstration of effusion.
And it may not look like a demonstration of anything because it's a picture of two bottles sitting on a counter. What's important is what the bottles are. And you can read on them. One of them is ammonia and one of them is hydrochloric acid. So this is concentrated ammonium hydroxide.
So that's got our NH4 plus and OH minus. And over here it's our concentrated hydrochloric acid, HCl dissolved. Okay, so what happened is they took the caps off of each of these bottles, set up a camera, and then walked away and allowed these to diffuse.
So what's happening is the NH4OH is dissociating, well is becoming ammonia, not ammonium, but ammonia gas, and it's that is diffusing out into the air. So here's a little cloud of ammonia gas hovering over the bottle, and it has a molar mass of 17.03 grams per mole. On this side, the HCl is also becoming a vapor, right?
So instead of being hydrochloric acid, it's hydrogen chloride gas. And there's a cloud of that hovering over the bottle of hydrochloric acid. It has molar mass of 36.46. So what's happening here is these clouds are diffusing outward from the openings of the bottle.
And when they meet, they're going to react with each other. Now, if we do our diffusion calculations or our effusion calculations, we'll see that the ammonia, because it is so much lighter, is diffusing at 1.46 times the rate of the hydrogen chloride gas. So that means the cloud of ammonia. Gas is spreading out faster, one and a half times, or 50% faster than the HCl cloud is spreading out.
And at a certain point, the cloud of ammonia is like this big, but the cloud of HCl is only like this big. And so what's happening is, in the air, we have essentially a precipitation reaction. We have our HCl. are NH3 reacting with HCl to form NH4Cl. This is ammonium chloride solid, and it's forming in the air.
And when we form these little particles of solid floating in the air, they're just like a precipitation reaction. They turn into a cloud. So this is ammonium chloride solid floating in the air right where it's being formed. And it's... being formed over the hydrogen chloride side not over the ammonium side because when this picture was taken the hydrogen chloride gas which is heavier didn't have time to diffuse to be over the mouth of the ammonium hydroxide bottle right so what's happening is this picture is showing that The lighter molecule, the NH3, is diffusing faster than the heavier molecule so that this cloud of ammonium chloride that we can see is happening on the side where the HCl is.
So that's, I think it's a wonderful demonstration and it's a great picture to put in a textbook but does need quite a bit of explanation. in order to understand why this is happening. Okay, so that's kinetic molecular theory. I've talked a little bit longer than I should have, but I just wanted to make sure that I covered all this material as in-depth as I could. Okay, I will see you in class.