🎯⭐ INTERACTIVE LESSON

Reaction Quotient and Le Chatelier's Principle

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Reaction Quotient and Le Chatelier's Principle - Complete Interactive Lesson

Part 1: The Reaction Quotient (Q)

🔄 The Reaction Quotient Q

Part 1 of 7 — Same Expression as K, but at Any Time

Topics in This Part

Section
📌 Defining Q
Key Distinction
Q at Special Times
🔢 Calculating Q
The Rules for Solids and Liquids

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 1

Understanding the core concepts covered in Part 1
Applying these ideas to solve practice problems
Building toward AP exam readiness for this topic

📌 Defining Q

For the general reaction:

$aA + bB \rightleftharpoons cC + dD$

$Q_{c} = \frac{[C]^{c} [D]^{d}}{[A]^{a} [B]^{}}$

🔢 Calculating Q

Problem: For $\text{N}_2(g) + 3,\text{H}_2(g) \rightleftharpoons 2,\text{NH}_3(g)$ , at 400°C. Current concentrations: M, M, M. Calculate and predict the direction of shift.

Understanding Q 🎯

Calculating Q 🧮

For the reaction: $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\,\text{HI}(g)$ , at 448°C.

Q Concepts 🔍

Exit Quiz — Reaction Quotient ✅

Part 2: Comparing Q and K

🔄 Comparing Q and K

Part 2 of 7 — Predicting the Direction of Shift

Topics in This Part

Section
📌 The Three Cases
Case 1: $Q < K$ — Shift Right (→)
Case 2: $Q > K$ — Shift Left (←)
Case 3: $Q = K$ — At Equilibrium

Part 3: Le Chatelier's Principle

🔄 Le Chatelier's Principle — Concentration Changes

Part 3 of 7 — How the System Responds to Stress

Topics in This Part

Section
📌 Adding or Removing Species
Adding Reactant → Shift Right
Adding Product → Shift Left
Removing a Species → Opposite Shift
Key Insight

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 3

Understanding the core concepts covered in Part 3
Applying these ideas to solve practice problems
Building toward AP exam readiness for this topic

📌 Adding or Removing Species

Adding Reactant → Shift Right

For: $\text{N}_2(g) + 3\,\text{H}_2(g) \rightleftharpoons 2\,\text{NH}_3(g)$

Part 4: Changes in Concentration

🔄 Le Chatelier — Pressure/Volume and Temperature

Part 4 of 7 — How Gases and Heat Affect Equilibrium

Topics in This Part

Section
💨 Pressure and Volume Changes
The Rule
Why?
Example
Special Case: Equal Moles

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 4

Understanding the core concepts covered in Part 4
Applying these ideas to solve practice problems
Building toward AP exam readiness for this topic

💨 Pressure and Volume Changes

The Rule

💡 Tip: For gaseous equilibria, decreasing volume (increasing pressure) shifts toward the side with fewer moles of gas. Increasing volume shifts toward more moles of gas.

Why?

Decreasing volume concentrates all species equally. The side with more moles of gas is affected more. Shifting toward fewer moles relieves the pressure stress.

Example

Part 5: Changes in Temperature & Pressure

🔄 Catalysts and Inert Gas Addition

Part 5 of 7 — Stresses That Don't Shift Equilibrium

Topics in This Part

Section
⚖️ Catalysts and Equilibrium
What Catalysts Do
Key Facts About Catalysts at Equilibrium
Why No Shift?
What Catalysts ARE Useful For

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 5

Understanding the core concepts covered in Part 5
Applying these ideas to solve practice problems
Building toward AP exam readiness for this topic

⚖️ Catalysts and Equilibrium

What Catalysts Do

A catalyst speeds up a reaction by providing an alternative pathway with a lower activation energy ( $E_a$ ).

Part 6: Problem-Solving Workshop

🧮 Problem-Solving Workshop

Part 6 of 7 — Q, K, and Le Chatelier Calculations

Practice Makes Perfect

This workshop features multi-step problems that mirror the AP Chemistry exam format. Each problem requires you to combine concepts from previous parts and show your work clearly.

🔑 Why this matters: The AP Chemistry exam rewards students who can apply concepts to unfamiliar problems — structured practice is the best preparation.

What You'll Master in Part 6

Working through complete multi-step problems from start to finish
Building problem-solving strategies you can apply on the AP exam
Identifying which concepts to apply and in what order

🛠️ Problem-Solving Strategy

🔑 Key Concept: Always start by calculating $Q$ , then compare to $K$ . For Le Chatelier problems, identify the stress type first — only temperature changes $K$ .

Part 7: Synthesis & AP Review

🎓 Synthesis & AP Review

Part 7 of 7 — Reaction Quotient & Le Chatelier's Principle

Bringing It All Together

This comprehensive review connects every concept from Parts 1–6 with AP-style problems. The questions are designed to mirror what you'll see on the actual exam — multi-step, multi-concept, and requiring clear written explanations.

🔑 Why this matters: AP Chemistry exam questions rarely test one concept in isolation — success requires connecting ideas across topics.

What You'll Master in Part 7

Solving AP-style questions that integrate multiple concepts from this unit
Writing clear, concise explanations using proper chemistry terminology
Identifying and avoiding common AP exam traps and mistakes

📋 Complete Concept Summary

Reaction Quotient Q

Same expression as K, using current (not equilibrium) concentrations

$\boxed{Q < K \to \text{shift right} \quad|\quad Q > K \to \text{shift left} \quad|\quad Q = K \to \text{equilibrium}}$

Q_{c} = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}} (using current concentrations)

$\boxed{Q_p = \frac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b}} \quad \text{(using current pressures)}$

	$K$	$Q$
Uses	Equilibrium concentrations only	Any concentrations at any time
Value	Fixed at a given temperature	Changes as concentrations change
Meaning	Where equilibrium lies	Where the system is right now

At $t = 0$ (only reactants): $Q = 0$ (numerator = 0)
At equilibrium: $Q = K$
If only products present: $Q = \infty$ (denominator = 0)

🔑 Key Concept: $Q = K$ means equilibrium. $Q < K$ → reaction shifts forward (right). $Q > K$ → reaction shifts in reverse (left).

[\text{N}_2] = 1.0

[\text{H}_2] = 2.0

[\text{NH}_3] = 3.0

Solution:

$\boxed{Q_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} = \frac{(3.0)^2}{(1.0)(2.0)^3} = \frac{9.0}{8.0} = 1.125}$

Since $Q_c = 1.125 > K_c = 0.50$ :

The system has too many products relative to equilibrium

The reaction will shift to the left (toward reactants) to reach equilibrium

The Rules for Solids and Liquids

💡 Tip: Just like with $K$ , pure solids and pure liquids are excluded from the $Q$ expression — their activities are defined as 1.

Current concentrations: $[\text{H}_2] = 0.10$ M, $[\text{I}_2] = 0.10$ M, $[\text{HI}] = 0.50$ M

1) Calculate $Q_c$ . (Enter as a whole number)

2) Is $Q > K$ , $Q < K$ , or $Q = K$ ? (Enter "Q > K", "Q < K", or "Q = K")

3) If you start with only reactants and no products, what is the initial value of Q? (Enter as a number)

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 2

Understanding the core concepts covered in Part 2
Applying these ideas to solve practice problems
Building toward AP exam readiness for this topic

📌 The Three Cases

Case 1: $Q < K$ — Shift Right (→)

$\boxed{Q < K \implies \text{shift right (forward)}}$

$\frac{\text{products}}{\text{reactants}} < \text{equilibrium ratio}$

There are too few products (or too many reactants)
The system shifts right (forward) to make more products
Q increases until $Q = K$

Case 2: $Q > K$ — Shift Left (←)

$\boxed{Q > K \implies \text{shift left (reverse)}}$

$\frac{\text{products}}{\text{reactants}} > \text{equilibrium ratio}$

There are too many products (or too few reactants)
The system shifts left (reverse) to make more reactants
Q decreases until $Q = K$

Case 3: $Q = K$ — At Equilibrium

$\boxed{Q = K \implies \text{system is at equilibrium}}$

No net change occurs
Forward and reverse rates are equal

Memory Aid

💡 Tip: Think of Q as "chasing" K:

$Q < K$ : Q needs to increase → more products → shift right

$Q > K$ : Q needs to decrease → more reactants → shift left

🔑 Key Concept: $Q < K$ → forward shift. $Q > K$ → reverse shift. $Q = K$ → equilibrium. These three cases are the foundation of predicting reaction direction.

📋 Visual Summary

$\underbrace{Q = 0}_{\text{pure reactants}} \quad \xleftarrow{\text{shift right}} \quad Q < K \quad \longrightarrow \quad \underbrace{Q = K}_{\text{EQUILIBRIUM}} \quad \longleftarrow \quad Q > K \quad \xrightarrow{\text{shift left}} \quad \underbrace{Q = \infty}_{\text{pure products}}$

Worked Example

Problem: $\text{CO}(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CO}_2(g) + \text{H}_2(g)$ , at 700 K. Given: , , , M. Predict the direction of shift.

Solution:

$\boxed{Q_c = \frac{[\text{CO}_2][\text{H}_2]}{[\text{CO}][\text{H}_2\text{O}]} = \frac{(0.20)(0.20)}{(0.10)(0.10)} = \frac{0.04}{0.01} = 4.0}$

Predicting the Direction of Shift 🎯

Q vs K Calculations 🧮

For: $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\,\text{NO}_2(g)$ , $K_c = 0.36$ at 100°C

Current state: $[\text{N}_2\text{O}_4] = 0.50$ M, $[\text{NO}_2] = 0.20$ M

1) Calculate $Q_c$ . (Enter as a decimal to 2 places)

2) Does the reaction shift right or left? (Enter "right" or "left")

3) At equilibrium, will $[\text{NO}_2]$ be higher or lower than 0.20 M? (Enter "higher" or "lower")

Round all answers to 3 significant figures.

Q vs K — Quick Concepts 🔍

Exit Quiz — Comparing Q and K ✅

If we add more N₂:

$[\text{N}_2]$ increases immediately
Q decreases (denominator gets bigger) → $Q < K$
System shifts right to consume the added N₂
At the new equilibrium: $[\text{NH}_3]$ is higher, $[\text{H}_2]$ is lower

Adding Product → Shift Left

If we add more NH₃:

$[\text{NH}_3]$ increases immediately
Q increases (numerator gets bigger) → $Q > K$
System shifts left to consume the added NH₃
At the new equilibrium: $[\text{N}_2]$ and $[\text{H}_2]$ are higher

Removing a Species → Opposite Shift

Action	Effect on Q	Shift Direction
Add reactant	Q decreases	Right →
Remove reactant	Q increases	Left ←
Add product	Q increases	Left ←
Remove product	Q decreases	Right →

⚠️ Warning: The system shifts to partially counteract the change. It never fully restores the original concentrations — it finds a new equilibrium position.

🧪 Worked Example

Problem: $\text{CO}(g) + 2\,\text{H}_2(g) \rightleftharpoons \text{CH}_3\text{OH}(g)$ , $K_c = 14.5$ . Original equilibrium: $[\text{CO}] = 0.20$ , $[\text{H}_2] = 0.30$ , $[\text{CH}_3\text{OH}] = 0.26$ M. Stress: Add CO to make $[\text{CO}] = 0.40$ M. Predict the shift.

Solution:

Verify original: $Q = \frac{0.26}{(0.20)(0.30)^2} = \frac{0.26}{0.018} = 14.4 \approx K$ ✓

Le Chatelier — Concentration 🎯

Predicting Concentration Changes 🧮

For: $\text{H}_2(g) + \text{Cl}_2(g) \rightleftharpoons 2\,\text{HCl}(g)$ at equilibrium

1) If $\text{Cl}_2$ is added, the reaction shifts which direction? (Enter "right" or "left")

2) If HCl is removed, the reaction shifts which direction? (Enter "right" or "left")

3) If $\text{H}_2$ is removed and $\text{Cl}_2$ is simultaneously added, the reaction shifts which direction? (Enter "right" or "left")

Concentration Stress Effects 🔍

Exit Quiz — Concentration Stresses ✅

N_{2} (g) + 3 H_{2} (g) ⇌ 2 NH_{3} (g)

Reactant side: $1 + 3 = 4$ moles of gas
Product side: $2$ moles of gas

Change	Shift	Why
Decrease volume	Right →	Fewer moles on right (2 vs 4)
Increase volume	Left ←	More moles on left (4 vs 2)

Special Case: Equal Moles

$\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\,\text{HI}(g) \quad (\Delta n = 0)$

Both sides have 2 moles of gas. No shift occurs with pressure/volume changes.

🌡️ Temperature Changes

⚠️ Warning: Temperature is unique — it's the only stress that changes the value of $K$ . All other stresses shift the equilibrium position but leave K unchanged.

The "Heat as a Species" Trick

Treat heat as a reactant or product:

Exothermic ( $\Delta H < 0$ ): Heat is a product $A \rightleftharpoons B + \text{heat}$

Endothermic ( $\Delta H > 0$ ): Heat is a reactant $\text{heat} + A \rightleftharpoons B$

Effect of Temperature Changes

Reaction Type	Increase T	Decrease T
Exothermic	Shift left ←, K decreases	Shift right →, K increases
Endothermic	Shift right →, K increases	Shift left ←, K decreases

Example

$\text{N}_2(g) + 3\,\text{H}_2(g) \rightleftharpoons 2\,\text{NH}_3(g) \quad \Delta H = -92$ kJ (exothermic)

$\text{N}_2 + 3\,\text{H}_2 \rightleftharpoons 2\,\text{NH}_3 + \text{heat}$

Increase T → adds heat → shifts left → K decreases
Decrease T → removes heat → shifts right → K increases

Key Point

🔑 Key Concept: Pressure, volume, and concentration changes shift the equilibrium position but do NOT change K. Temperature changes BOTH the position AND the value of K.

Pressure, Volume, and Temperature 🎯

Predicting Shifts 🧮

For: $\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)$ , $\Delta H = +87$ kJ (endothermic)

1) If the volume is decreased, the reaction shifts which direction? (Enter "right" or "left")

2) If the temperature is increased, the reaction shifts which direction? (Enter "right" or "left")

3) Does increasing temperature increase or decrease K for this reaction? (Enter "increase" or "decrease")

Pressure & Temperature Effects 🔍

Exit Quiz — Pressure and Temperature ✅

Key Facts About Catalysts at Equilibrium

Property	Effect
Forward reaction rate	Increased
Reverse reaction rate	Increased equally
Position of equilibrium	No change
Value of K	No change
Time to reach equilibrium	Decreased

⚠️ Warning: A catalyst does not shift equilibrium. It speeds up both the forward and reverse reactions by the same factor. The equilibrium position and K are unchanged — you just get there faster.

What Catalysts ARE Useful For

Reaching equilibrium faster
Making a slow reaction practical (e.g., the Haber process uses an iron catalyst)
Allowing equilibrium to be reached at lower temperatures (which may favor products for exothermic reactions)

The Haber process: $\text{N}_2 + 3\,\text{H}_2 \rightleftharpoons 2\,\text{NH}_3$

Without a catalyst, this reaction is impractically slow at lower temperatures. The iron catalyst allows the reaction to reach equilibrium quickly at moderate temperatures (~450°C), which is a compromise between speed and yield.

💨 Adding an Inert Gas

An inert (noble) gas does not react with any species in the equilibrium. Its effect depends on the conditions:

At Constant Volume

Adding inert gas at constant volume:

Increases total pressure
Does NOT change the partial pressures of any reacting species
Does NOT change concentrations
Q is unchanged → No shift
K is unchanged

At Constant Pressure

Adding inert gas at constant pressure:

The container must expand to maintain constant pressure
This effectively increases the volume
All partial pressures of reacting species decrease
This is equivalent to a volume increase → shifts toward more moles of gas

Summary

Condition	Effect of Adding Inert Gas
Constant volume	No shift (partial pressures unchanged)
Constant pressure	Shifts toward more moles of gas (volume increases)

AP Exam Note

💡 Tip: On the AP exam, "adding an inert gas" typically means at constant volume unless stated otherwise. The answer is usually no effect on equilibrium.

Catalysts and Inert Gas 🎯

📋 Complete Summary of All Stresses

🔑 Key Concept: Only temperature changes K. Catalysts and inert gas (at constant V) cause no shift. All other stresses shift the position but leave K the same.

Stress	Shift Direction	K Changes?
Add reactant	Right →	No
Remove reactant	Left ←	No
Add product	Left ←	No
Remove product	Right →	No
Decrease volume (↑P)	Toward fewer moles of gas	No
Increase volume (↓P)	Toward more moles of gas	No
Increase temperature	Endothermic: right; Exothermic: left	Yes
Decrease temperature	Endothermic: left; Exothermic: right	Yes
Add catalyst	No shift	No
Add inert gas (const V)	No shift	No
Add inert gas (const P)	Toward more moles of gas	No

Stress Identification 🔍

Quick Checks 🧮

1) Does a catalyst change the value of K? (Enter "no")

2) Does adding an inert gas at constant volume shift the equilibrium? (Enter "no")

3) If a catalyst is removed from a system at equilibrium, does the position shift? (Enter "no")

Exit Quiz — Catalysts and Inert Gas ✅

Write the $Q$ expression (same form as $K$ )
Plug in the current concentrations or pressures
Compare $Q$ $Q$ to $K$ $K$ :
- $Q < K$ → shift right
- $Q > K$ → shift left
- $Q = K$ → at equilibrium

For Le Chatelier Problems

Identify the stress (concentration, pressure/volume, temperature, catalyst, inert gas)
Predict the direction of shift:
- Concentration: shift away from added species
- Volume: shift toward side with more/fewer moles of gas
- Temperature: treat heat as a species
- Catalyst/inert gas at const V: no shift
Determine effect on each species' concentration
Determine effect on K (only temperature changes K)

🧪 Worked Example 1

Problem: $\text{CO}(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CO}_2(g) + \text{H}_2(g)$ , $K_c = 5.10$ at 700 K. A flask contains: $[\text{CO}] = 0.200$ , $[\text{H}_2\text{O}] = 0.300$ , $[\text{CO}_2] = 0.400$ , $[\text{H}_2] = 0.500$ M. Determine the direction of shift.

Solution:

$\boxed{Q = \frac{(0.400)(0.500)}{(0.200)(0.300)} = \frac{0.200}{0.060} = 3.33}$

Practice Problem 1 🧮

$\text{N}_2\text{O}_4(g) \rightleftharpoons 2\,\text{NO}_2(g)$ , $K_c = 4.60 \times 10^{-3}$

Current: $[\text{N}_2\text{O}_4] = 0.500$ M, $[\text{NO}_2] = 0.100$ M

1) Calculate Q. (Enter as a decimal to 2 places)

2) Is $Q > K$ , $Q < K$ , or $Q = K$ ? (Enter exactly, e.g. "Q > K")

3) Which direction does the reaction shift? (Enter "right" or "left")

Round all answers to 3 significant figures.

Practice Problem 2 — Le Chatelier Multi-Stress 🎯

$2\,\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\,\text{SO}_3(g)$ , $\Delta H = -198$ kJ (exothermic)

Practice Problem 3 🧮

For: $\text{A}(g) + 2\,\text{B}(g) \rightleftharpoons 3\,\text{C}(g)$ , $\Delta H = +150$ kJ, $K = 0.25$ at 500 K

1) If the volume is halved, which direction does the reaction shift? (Enter "right" or "left")

2) If temperature is increased to 600 K, does K increase or decrease? (Enter "increase" or "decrease")

3) How many moles of gas are on each side? Enter as "left:N, right:M" (e.g., "left:3, right:2")

Quick Stress Review 🔍

Exit Quiz — Problem-Solving Workshop ✅

Q<K→shift right∣Q>K→shift left∣Q=K→equilibrium

Le Chatelier's Principle

Stress	Direction of Shift	K Changes?
Add reactant	Right →	No
Remove reactant	Left ←	No
Add product	Left ←	No
Remove product	Right →	No
↓ Volume (↑ P)	Toward fewer mol gas	No
↑ Volume (↓ P)	Toward more mol gas	No
↑ Temperature	Endothermic: →; Exothermic: ←	Yes
↓ Temperature	Endothermic: ←; Exothermic: →	Yes
Catalyst	No shift	No
Inert gas (const V)	No shift	No

⚠️ Warning: Only temperature changes K. A catalyst speeds up the approach to equilibrium but doesn't change position or K.

🔑 Key Concept: The system partially counteracts a stress — never fully. This is the defining feature of Le Chatelier's Principle.

AP-Style Multiple Choice — Set 1 🎯

AP-Style Multiple Choice — Set 2 🎯

AP Free-Response Style 🧮

$\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)$ , $K_c = 0.040$ at 250°C, $\Delta H = +87$ kJ

Current concentrations: $[\text{PCl}_5] = 0.50$ , $[\text{PCl}_3] = 0.10$ , M

1) Calculate Q. (Enter as a decimal)

2) Which direction does the reaction shift? (Enter "right" or "left")

3) If the temperature is raised to 300°C, does K increase or decrease? (Enter "increase" or "decrease")

Round all answers to 3 significant figures.

Final Concept Review 🔍

Final Exit Quiz ✅

[\text{CO}] = 0.10

[\text{H}_2\text{O}] = 0.10

[\text{CO}_2] = 0.20

[\text{H}_2] = 0.20

(0.10)(0.10)(0.20)(0.20)

Since $Q = 4.0 < K = 5.0$ :

The system shifts right to produce more CO₂ and H₂
$[\text{CO}]$ and $[\text{H}_2\text{O}]$ will decrease
$[\text{CO}_2]$ and $[\text{H}_2]$ will increase

Immediate Q after stress:

$\boxed{Q = \frac{0.26}{(0.40)(0.30)^2} = \frac{0.26}{0.036} = 7.2}$

Since $Q = 7.2 < K = 14.5$ : the system shifts right.

At the new equilibrium:

$[\text{CO}]$ is higher than 0.20 but lower than 0.40 (some consumed)
$[\text{H}_2]$ is lower than 0.30 (consumed)
$[\text{CH}_3\text{OH}]$ is higher than 0.26 (produced)

$Q = 3.33 < K = 5.10$ → shift right

At the new equilibrium:

$[\text{CO}]$ decreases, $[\text{H}_2\text{O}]$ decreases
$[\text{CO}_2]$ increases, $[\text{H}_2]$ increases

[\text{Cl}_2] = 0.10

Reaction Quotient and Le Chatelier's Principle

Key Distinction

Q at Special Times

The Rules for Solids and Liquids

What You'll Master in Part 2

📌 The Three Cases

Case 1: $Q < K$ — Shift Right (→)

Case 2: $Q > K$ — Shift Left (←)

Case 3: $Q = K$ — At Equilibrium

Memory Aid

📋 Visual Summary

Worked Example

Adding Product → Shift Left

Removing a Species → Opposite Shift

Key Insight

🧪 Worked Example

Special Case: Equal Moles

🌡️ Temperature Changes

The "Heat as a Species" Trick

Effect of Temperature Changes

Example

Key Point

Key Facts About Catalysts at Equilibrium

Why No Shift?

What Catalysts ARE Useful For

Example

💨 Adding an Inert Gas

At Constant Volume

At Constant Pressure

Summary

AP Exam Note

📋 Complete Summary of All Stresses

For Q vs K Problems

For Le Chatelier Problems

🧪 Worked Example 1

Le Chatelier's Principle

Key Reminders

Reaction Quotient and Le Chatelier's Principle

Reaction Quotient and Le Chatelier's Principle - Complete Interactive Lesson

Part 1: The Reaction Quotient (Q)

🔄 The Reaction Quotient Q

Topics in This Part

What You'll Master in Part 1

📌 Defining Q

🔢 Calculating Q

Part 2: Comparing Q and K

🔄 Comparing Q and K

Topics in This Part

Part 3: Le Chatelier's Principle

🔄 Le Chatelier's Principle — Concentration Changes

Topics in This Part

What You'll Master in Part 3

📌 Adding or Removing Species

Adding Reactant → Shift Right

Part 4: Changes in Concentration

🔄 Le Chatelier — Pressure/Volume and Temperature

Topics in This Part

What You'll Master in Part 4

💨 Pressure and Volume Changes

The Rule

Why?

Example

Part 5: Changes in Temperature & Pressure

🔄 Catalysts and Inert Gas Addition

Topics in This Part

What You'll Master in Part 5

⚖️ Catalysts and Equilibrium

What Catalysts Do

Part 6: Problem-Solving Workshop

🧮 Problem-Solving Workshop

Practice Makes Perfect

What You'll Master in Part 6

🛠️ Problem-Solving Strategy

Part 7: Synthesis & AP Review

🎓 Synthesis & AP Review

Bringing It All Together

What You'll Master in Part 7

📋 Complete Concept Summary

Reaction Quotient Q

Key Distinction

Q at Special Times

The Rules for Solids and Liquids

What You'll Master in Part 2

📌 The Three Cases

Case 1: Q<KQ < KQ<K — Shift Right (→)

Case 2: Q>KQ > KQ>K — Shift Left (←)

Case 3: Q=KQ = KQ=K — At Equilibrium

Memory Aid

📋 Visual Summary

Worked Example

Adding Product → Shift Left

Removing a Species → Opposite Shift

Key Insight

🧪 Worked Example

Special Case: Equal Moles

🌡️ Temperature Changes

The "Heat as a Species" Trick

Effect of Temperature Changes

Example

Key Point

Key Facts About Catalysts at Equilibrium

Why No Shift?

What Catalysts ARE Useful For

Example

💨 Adding an Inert Gas

At Constant Volume

At Constant Pressure

Summary

AP Exam Note

📋 Complete Summary of All Stresses

For Q vs K Problems

For Le Chatelier Problems

🧪 Worked Example 1

Le Chatelier's Principle

Key Reminders

Case 1: $Q < K$ — Shift Right (→)

Case 2: $Q > K$ — Shift Left (←)

Case 3: $Q = K$ — At Equilibrium