#Le Châtelier's Principle: Why It Works (Using Q)

Hey there, future AP Chem master! 👋 Ready to dive deeper into Le Châtelier's Principle? Last time, we used logic to predict shifts in equilibrium. Now, let's see how the reaction quotient, Q, justifies these shifts. This is your go-to guide for understanding the why behind the what.

#Review of the Reaction Quotient, Q

Remember Q, the reaction quotient? It's like a snapshot of your reaction at any given moment, not just at equilibrium.

• Q tells us which way a reaction will shift to reach equilibrium.

• If Q < K, the reaction shifts towards products.

• If Q > K, the reaction shifts towards reactants.

• If Q = K, the system is at equilibrium.

#Applying Q to Le Châtelier's Principle

#Concentration

Changing concentrations messes with the balance, and Q helps us see how the system responds.

• Adding products increases the numerator of Q, making Q > K. The system shifts to the reactants.

• Adding reactants increases the denominator of Q, making Q < K. The system shifts to the products.

Image: Reaction quotient (Q) vs. equilibrium constant (K). Notice how the system shifts to make Q = K.

#Pressure

Pressure changes also cause shifts, but here, the stoichiometry of the reaction is crucial. We use Qp (partial pressure quotient) here.

• Increasing pressure favors the side with fewer moles of gas.
• Decreasing pressure favors the side with more moles of gas.

Let's break it down with an example:

Example 1: A + B ⇌ C

$$Q_p = \frac{P(C)}{P(A)P(B)}$$

If pressure doubles, all partial pressures double:

$$Q_p' = \frac{2P(C)}{2P(A)2P(B)} = \frac{1}{2} \frac{P(C)}{P(A)P(B)}$$

Qp decreases, so the reaction shifts towards the products.

Example 2: 3A + B ⇌ 2C

$$Q_p = \frac{P(C)^2}{P(A)^3P(B)}$$

If pressure doubles:

$$Q_p' = \frac{(2P(C))^2}{(2P(A))^3(2P(B))} = \frac{4P(C)^2}{16P(A)^3P(B)} = \frac{1}{4} \frac{P(C)^2}{P(A)^3P(B)}$$

Qp decreases, so the reaction shifts towards the products.

Image: Effect of pressure on equilibrium. Note how changes in pressure shift the reaction to the side with fewer moles of gas.

#Temperature as the Exception

Here's where things get interesting: Q does NOT explain temperature changes! 🤯

• Temperature changes actually alter the value of K itself.

• K is temperature-dependent; it's only constant at a constant temperature.

• For exothermic reactions, increasing temperature decreases K (shifts to reactants).

• For endothermic reactions, increasing temperature increases K (shifts to products).

#Final Exam Focus

• High-Value Topics: Le Châtelier's Principle, Reaction Quotient (Q), Equilibrium Constant (K), and their interconnections.
• Common Question Types: - Predicting shifts in equilibrium based on changes in concentration, pressure, and temperature. - Calculating Q and comparing it to K. - Explaining why temperature changes affect equilibrium differently than concentration or pressure.
• Time Management: Quickly analyze the given reaction and the changes, then apply the principles. Don't get bogged down in complex calculations unless necessary.
• Common Pitfalls: - Forgetting stoichiometric coefficients in pressure-related questions. - Confusing Q and K. - Thinking temperature changes are explained by Q.

#Practice Questions

 ii. 1 point for predicting a shift to the left (towards reactants)
  1 point for explaining that an increase in temperature favors the endothermic direction, which is the reverse reaction since the forward reaction is exothermic.

 iii. 1 point for predicting a shift to the right (towards products)
  1 point for explaining that a decrease in volume increases pressure, favoring the side with fewer moles of gas (2 moles on the product side vs. 3 moles on the reactant side)

You've got this! Keep practicing, and you'll ace that exam! 💪