 Okay, so let's try to do this. This is the Le Chatelier's principle of problem. It says consider the reaction below or the reaction pictured. Which change would cause the equilibrium to shift to the right? Okay, so is everybody cool with what we're doing? So it gives us four choices here and we want it to shift to the right. Okay, so hopefully three of them shifted to the left and the other one shifts it to the right then it's not ambiguous. Okay, so Le Chatelier's principle, remember what happens if you remove something from the reaction, right, the side of the reaction it shifts towards that way. Okay, so this one says removing some sulfur trioxide. Okay, where is sulfur trioxide? It's here, right? So if we decrease the amount that's removing it, what's going to happen to this equilibrium arrow? It's going to shift to the left. Okay, is that to the right? No, so that's not the answer. Are we cool with that one? Yes. Okay, good. So let's just erase this. So remember we want to shift to the right. Let's go down to this one. Okay, adding some sulfur dioxide since we just went over the other one, right? So we're adding sulfur dioxide. We're increasing the concentration of that and that's going to do what? Shift it to the left again, right? Because if we increase the concentration it's going to try to offset it by shifting to the left, okay? So did that shift it to the right? No. No, so we can cancel that one. Okay, so decreasing the temperature. So this one might not be so obvious, but remember when we look over here it says delta H equals plus 198 kilojoules. What kind of reaction is this? XO or endothermic? XO means it's releasing heat. The heat is exiting the reaction, right? Okay, so endothermic, that means we have to add heat to this reaction to make it work. Does that make sense? Okay, we have to add heat to this reaction to make it work. So another way we can think of this is that heat has to be supplied as a reactant or the reaction won't go. Does that make sense? Yeah. Okay, so let's just think of heat as a reactant here, okay? So if we decrease the temperature that means we're removing heat, right? So that's going to decrease the heat. So which way does it go? To the left, right? So is that to the right? No. No, so that's not the right answer either. Okay, so we know that C has got to be the right answer, right? Because that's process of elimination, but let's just understand why. Okay, so can I erase this? Yes. But that's the way you want to think about those types, okay? So the last thing we're saying, okay, increasing the container volume. So this sulfur trioxide is a gas, sulfur dioxide is a gas, and molecular oxygen is a gas, okay? Remember, molar volume, right, that all gases occupy 22.4 liters at standard temperature pressure. So at any other pressure, we can think of these as all ideal gases. They're all occupying the same amount per mole, okay? So that's per mole we're talking about, 22.4 liters. So if we look here, right, how many moles of gas do we have on the left side of the reaction? Two, right, so two moles. And how many moles of gas do we have on the right side of the reaction? Three moles, yeah. Okay, so which one is going to be a bigger volume? Two moles or three moles? Three moles. Because remember, per mole, each one of these gases is the same volume, okay? So if, let's just say this was at standard temperature pressure, it would be 22.4 times two, which would be 44.8, right? This would be 22.4 times three, okay? So this set, the product side, needs a bigger volume to have the same amount of moles being contained in it. Does that make sense? Yes. As the reactants, okay? So what we're going to do is if we increase the volume of the container, right, we're going to do something that looks like this. We're going to have a small container and we're going to make it into a bigger container like that, okay? Which one of these containers can hold more gas moles? The one on the right, okay? So since this can hold more moles of gas, the reaction itself is going to want to go to the side that produces more moles of gas, okay? So if we did this to the volume of the container, it's going to shift the equilibrium to the right, okay? So that one actually doesn't really have anything to do with the types, like removing a particular reactant or product or something like that, but the actual total number of moles of gas, okay? So let's just box the answer just to make it official. Any questions on that one? No. Okay, good. So a pretty thorough lecture actually on Lushat-Lay's principles.