Welcome to Unit 7: Equilibrium!
Welcome to one of the most important chapters in AP Chemistry! If you’ve ever felt like a reaction just "stops" before it’s finished, you’ve actually witnessed Equilibrium. In this unit, we’ll explore how reactions can go forward and backward at the same time. Think of it like a busy store: people are walking in and people are walking out. If they move at the same speed, the number of people inside stays the same, even though everyone is moving. Don’t worry if this feels a bit abstract at first—we’ll break it down step-by-step!
7.1 - 7.3: The Nature of Equilibrium and the Equilibrium Constant
In many reactions, the products can turn back into reactants. This is called a reversible reaction, shown with a double arrow (\(\rightleftharpoons\)).
Dynamic Equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of reactants and products stay constant over time. They aren't necessarily equal to each other; they just stop changing!
The Equilibrium Constant (\(K\)): This is a number that tells us "who is winning" at equilibrium. To find it, we use the Law of Mass Action.
For the reaction: \(aA + bB \rightleftharpoons cC + dD\)
The expression is: \(K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}\)
Important Rules for Writing K Expressions:
1. Products over Reactants: Always put the stuff on the right side of the arrow on top.
2. Exponents: Use the coefficients from the balanced equation as powers.
3. No Solids or Liquids: Only gases (\(g\)) and aqueous solutions (\(aq\)) are included. Pure solids (\(s\)) and liquids (\(l\)) have a constant density and don't change concentration, so we leave them out!
4. \(K_p\) vs \(K_c\): Use brackets \([ ]\) for molarity (\(K_c\)) and parentheses with \(P\) for partial pressures (\(K_p\)).
Quick Review: At equilibrium, Rates are equal, but Concentrations are constant. Only gases and aqueous species make it into the \(K\) expression.
7.4 - 7.6: Calculating and Interpreting K
What does the value of \(K\) actually tell us? It's like a scoreboard for the reaction:
- If \(K > 1\): The numerator (products) is bigger. We say the reaction favors the products.
- If \(K < 1\): The denominator (reactants) is bigger. We say the reaction favors the reactants.
- If \(K \approx 1\): There are roughly equal amounts of both at equilibrium.
Manipulating K: Sometimes you need to change the equation. Here is how \(K\) changes:
- Reverse the reaction: Take the reciprocal (\(1/K\)).
- Multiply coefficients by a factor (n): Raise \(K\) to the power of that factor (\(K^n\)).
- Add two reactions together: Multiply their \(K\) values (\(K_1 \times K_2\)).
Memory Trick: Adding equations = Multiplying Ks. It’s not addition; it’s like compound interest!
Key Takeaway: \(K\) tells you the extent of the reaction. Large \(K\) means lots of product; small \(K\) means the reaction barely happened.
7.7 - 7.8: The ICE Table and Q
What if the reaction isn't at equilibrium yet? We use the Reaction Quotient (\(Q\)). \(Q\) is calculated the exact same way as \(K\), but using concentrations from any moment in time.
Comparing Q and K:
- \(Q < K\): Not enough products yet. The reaction shifts Right (toward products).
- \(Q > K\): Too many products. The reaction shifts Left (toward reactants).
- \(Q = K\): Congrats! You are at equilibrium.
The ICE Table Method: For calculating concentrations when you only know the starting amounts.
I - Initial concentration
C - Change (use \(-x\) for reactants and \(+x\) for products, based on coefficients)
E - Equilibrium concentration (Initial + Change)
Example: If you have \(A \rightleftharpoons 2B\) and start with 1.0M of \(A\), your equilibrium line would be \(1.0 - x\) for \(A\) and \(0 + 2x\) for \(B\).
Common Mistake: Forgetting the coefficients in the "Change" row! If the equation says \(2x\), the change is \(2x\), and when you plug it into the \(K\) expression, you must square it: \((2x)^2\).
7.9 - 7.10: Le Châtelier’s Principle
Le Châtelier’s Principle says: If you stress a system at equilibrium, it will shift to reduce that stress. Think of it like a stubborn toddler—if you push them one way, they push back the other way!
1. Concentration: Add more reactant? The system shifts Right to use it up. Remove product? The system shifts Right to replace it.
2. Pressure/Volume: Only affects gases. If you increase pressure (decrease volume), the system shifts to the side with fewer moles of gas to take up less space.
3. Temperature: Treat heat as a reactant or product!
- Exothermic (\(-\Delta H\)): Heat is a product. Adding heat shifts the reaction Left.
- Endothermic (\(+\Delta H\)): Heat is a reactant. Adding heat shifts the reaction Right.
4. Catalysts: They make the reaction reach equilibrium faster, but they do not change the position of equilibrium or the value of \(K\).
Key Takeaway: Only changing Temperature changes the actual value of \(K\). Other changes just cause a temporary shift until a new equilibrium is reached.
7.11 - 7.13: Solubility Equilibria (\(K_{sp}\))
Even "insoluble" salts like \(AgCl\) dissolve a tiny, tiny bit. We use \(K_{sp}\) (Solubility Product) to measure this.
For \(MgF_2(s) \rightleftharpoons Mg^{2+}(aq) + 2F^-(aq)\)
\(K_{sp} = [Mg^{2+}][F^-]^2\)
Molar Solubility (\(s\)): This is the amount of moles that dissolve per liter. If you set up an ICE table, \(s\) is usually your \(x\).
- For a 1:1 salt (like \(NaCl\)): \(K_{sp} = s^2\)
- For a 1:2 salt (like \(MgF_2\)): \(K_{sp} = (s)(2s)^2 = 4s^3\)
The Common Ion Effect: If you try to dissolve a salt in water that already contains one of the ions (e.g., dissolving \(AgCl\) in \(NaCl\) solution), the salt will be less soluble. This is just Le Châtelier’s Principle in action—the extra product (\(Cl^-\)) shifts the reaction left, back into solid form.
pH and Solubility: If a salt contains a basic ion (like \(F^-\) or \(CO_3^{2-}\)), adding acid (\(H^+\)) will react with that ion, removing it from the equilibrium. This shifts the reaction to the right, increasing the solubility.
7.14: Free Energy of Dissolution
Why do some things dissolve and others don't? It's a balance between Enthalpy (\(\Delta H\)) and Entropy (\(\Delta S\)).
- Breaking solute bonds: Costs energy (Endothermic).
- Breaking solvent-solvent attractions: Costs energy (Endothermic).
- Forming new solute-solvent attractions: Releases energy (Exothermic).
If the new attractions are strong and the disorder (\(\Delta S\)) increases enough, the process will be thermodynamically favored (\(-\Delta G\)).
Did you know? Even if a dissolution is endothermic (the beaker feels cold), it can still be favored because the system becomes much more messy (entropy increases) when a solid turns into moving ions!
Final Summary of Unit 7
1. Equilibrium is dynamic: forward rate = reverse rate.
2. \(K\) is products over reactants, coefficients become exponents, ignore solids and liquids.
3. \(Q\) vs \(K\) tells you which direction the reaction will move to reach equilibrium.
4. Le Châtelier’s Principle predicts shifts: the system hates change and will try to undo whatever you do to it.
5. \(K_{sp}\) is just equilibrium for dissolving solids. The Common Ion Effect always lowers solubility.
Don't worry if the math feels heavy! Practice your ICE tables and always remember to check if your equation is balanced before you start writing your \(K\) expression. You've got this!