Hi there! My name is Jeremy Krug, and this is my ten minute review of AP Chemistry Unit 7 – which covers Equilibrium. This is THE PLACE for AP Chemistry, don’t forget to check out my full-length daily videos for the ENTIRE AP course, as well as free-response walkthroughs, and more! And if you’re looking for a comprehensive practice and review program as you get ready for the AP exam, then check out my practice materials over at UltimateReviewPacket.com – I’ve got full-length unit summary videos, complete study guides, hundreds of practice questions – AP level multiple-choice and free-response, and more. The link is in the description down below! Now let’s get right into Equilibrium! Equilibrium refers to reversible processes. Water can be boiled, and it can be condensed. Two ions can form a precipitate, or that precipitate can dissolve. Reversible processes can attain equilibrium, and we use a double-headed arrow to show that reversibility. For most reactions, the rate starts out fast, and eventually slows down and levels out. Reversible reactions eventually reach a point where the rate of the forward reaction is equal to the rate of the reverse reaction. THAT’S equilibrium. Notice, the reaction has not stopped. You won’t see the concentration of anything changing, but that’s just because the net change is zero. The direction in which equilibrium will lie has to do with relative rates of the forward and reverse reactions. If the forward reaction is faster than the reverse reaction, then you’ll end up with more products than reactants. On the other hand, if the reverse reaction is faster, you’ll end up with more reactants, and relatively little conversion to products. We write the equilibrium constant expression for any reaction by making the equilibrium constant Kc equal to the concentration of the products over the reactants, all raised to the power of the coefficients, just like we see here. For Kp, it’s the partial pressure of the products over the reactants, raised to the power of the coefficients. In all cases, we leave out any solids or pure liquids from equilibrium calculations. Be aware that the reaction quotient, Q, is written the same way as the equilibrium constant, but it includes values that aren’t necessarily at equilibrium. To calculate the equilibrium constant, plug in the equilibrium concentrations into the expression, and compute the answer. Likewise, if you know the equilibrium constant and all but one of the equilibrium concentrations, you can solve for the missing value. This process works the same for Kc as it does for Kp. Remember that all equilibrium constants are temperature dependent. That means that if you change the temperature, you change the equilibrium constant. In fact, changing the temperature is the only way you can change the value of the equilibrium constant. The magnitude of the equilibrium constant tells us to what extent a reversible reaction takes place. A very large equilibrium constant, like we see here, tells us that we’re going to have almost 100% of the reactants converted into products. We’d say equilibrium lies to the right, or favors the products. A very small equilibrium constant, like in this example, tells us that almost 0% of the reactants convert into products. So we say equilibrium lies to the left, or that it favors the reactants. The larger the equilibrium constant, the greater the ratio of products to reactants you’ll generally have. When we manipulate a reaction, the equilibrium constant changes accordingly. When we flip a reaction around, the equilibrium constant becomes the reciprocal of what it was before. When we double all the coefficients in a reaction, the equilibrium constant becomes the square of what it was before. And if we add reactions together to create a new reaction, the two individual equilibrium constants are multiplied by each other to become the constant of the new reaction. Make sure you can use an equilibrium constant and initial concentrations to calculate equilibrium concentrations of a reaction. I recommend using a chart to keep your data organized, like this ICE box, where you’ll write in the initial, change, and equilibrium concentrations of all the species present. We start by writing in the initial concentrations like this. Since we don’t know the equilibrium concentrations, we can say the concentration of NO goes down by 2X. Then the concentration of N2 goes up by X, since it has to move in the opposite direction of the reactants, and there’s a 2 to 1 ratio here. Then the concentration of O2 goes up by X. So here are our product concentrations that we can plug into the equilibrium constant expression. We use algebra to solve for X, and conversely, for the equilibrium concentrations of all our substances. We can use a particle diagram to represent the relative numbers of reactants and products in a reversible process. In this example, we’re given a particle diagram, and we’re asked to describe the equilibrium constant. Well, we see that there are a whole lot more carbon monoxide and hydrogen molecules than there are methanol molecules. And since there are so many more product molecules than reactants, we’d say the equilibrium constant has a very large value. And if we’re asked to estimate the partial pressure of, say, the hydrogen, we could multiply the mole fraction, which is six out of ten, or six-tenths, by the total pressure. If the total pressure is 2.0 atmospheres, we’d say the partial pressure of hydrogen gas is 1.2 atmospheres. If we have a system at equilibrium, and we disturb that system, the system will re-adjust itself to get back to equilibrium. So if we add a substance, that substance will react with something on its side of the equation, and the opposite side of the reaction will increase, decreasing the side to which you added the substance. And if you take something away, the other side will react, causing those concentrations to decrease. And the side on which you removed the substance will increase. If you decrease the volume, the equilibrium shifts to the side that has fewer moles of gas, and if you increase the volume, it shifts to the side that has more moles of gas. In an exothermic reaction, since heat is a product, adding heat shifts the reaction toward the reactants, while lowering the temperature shifts it toward the products. The opposite would be true for an endothermic reaction. If you have a system that’s not at equilibrium, you can plug those pressures or concentrations into the reaction quotient, which we call Q, and determine in which direction the reaction will proceed to attain equilibrium. If Q is greater than K, the reaction proceeds to the left, and it will make more reactants, at the expense of the products. If Q is less than K, the reaction will proceed to the right and produce more products. If you have trouble keeping that straight, just take the symbol and turn it into a little Pac-man character, and you can see which direction it’s going. An important application of equilibrium is the solubility of ionic compounds. If we write the balanced equation for the dissociation of lead(II) bromide, the equilibrium constant expression is what we call the Ksp, or the solubility product constant. If we’re trying to find the molar solubility of this compound, well, we can call it X. The lead ion concentration would be X, and the bromide would be 2X. We can plug these values into the Ksp expression and solve for X. So we see the molar solubility of lead(II) bromide, as 0.012 moles per liter. We can convert that to grams per liter if we wish, with a simple mole to gram conversion. Now let’s take that same compound, and instead of dissolving it into distilled water, let’s try to dissolve it into a 0.10 molar solution of sodium bromide. As you can see, the bromide is a common ion that’s also in the lead(II) bromide, and the bromide from the other compound reduces the capacity of lead(II) bromide to dissolve. We can calculate this change, by setting up an ICE box. We set the initial concentration of lead(II) ions equal to zero, but this time bromide is 0.10 molar. Our equilibrium values are X and 0.10 plus 2X. So when you plug this into the equilibrium expression, you can ignore the 2X, since the equilibrium constant is very small. So the X, or the molar solubility, is much less than it was before, 6.6 x 10-4 moles per liter. So that’s Equilibrium in about ten minutes! If you found this video helpful, I would really appreciate a LIKE, and if you really want to score a FIVE on the AP exam, then check out all of my AP Chemistry materials – both right here on YouTube and my comprehensive review and practice program over at UltimateReviewPacket.com. I’m Jeremy Krug, thanks for watching! And I hope to see you soon over on my 10 minute review of Unit 8, which will cover Acids and Bases. See you soon!