Reaction Quotient and Le Chatelier's Principle

🔄 The Reaction Quotient Q

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

Topics in This Part

🔑 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:

$aA + bB \rightleftharpoons cC + dD$

$\boxed{Q_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}} \quad \text{(using current concentrations)}$

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

Key Distinction

Q at Special Times

• 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).

🔢 Calculating Q

Problem: For $\text{N}_2(g) + 3,\text{H}_2(g) \rightleftharpoons 2,\text{NH}_3(g)$, $K_c = 0.50$ at 400°C. Current concentrations: $[\text{N}_2] = 1.0$ M, $[\text{H}_2] = 2.0$ M, $[\text{NH}_3] = 3.0$ M. Calculate $Q$ and predict the direction of shift.

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.

Understanding Q 🎯

Calculating Q 🧮

For the reaction: $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\,\text{HI}(g)$, $K_c = 50$ at 448°C.

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)

Q Concepts 🔍

Exit Quiz — Reaction Quotient ✅

🔄 Comparing Q and K

Part 2 of 7 — Predicting the Direction of Shift

Topics in This Part

🔑 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: $\text{CO}(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CO}_2(g) + \text{H}_2(g)$, $K_c = 5.0$ at 700 K. Given: $[\text{CO}] = 0.10$, $[\text{H}_2\text{O}] = 0.10$, $[\text{CO}_2] = 0.20$, $[\text{H}_2] = 0.20$ M. Predict the direction of shift.

Solution:

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

Since $Q = 4.0 < K = 5.0$:

• The system shifts right to produce more $CO_{2}$ and $H_{2}$
• $[\text{CO}]$ and $[\text{H}_2\text{O}]$ will decrease
• $[\text{CO}_2]$ and $[\text{H}_2]$ will increase

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 ✅

🔄 Le Chatelier's Principle — Concentration Changes

Part 3 of 7 — How the System Responds to Stress

Topics in This Part

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

If we add more $N_{2}$:

• $[\text{N}_2]$ increases immediately
• Q decreases (denominator gets bigger) → $Q < K$
• System shifts right to consume the added $N_{2}$
• At the new equilibrium: $[\text{NH}_3]$ is higher, $[\text{H}_2]$ is lower

Adding Product → Shift Left

If we add more $NH_{3}$:

• $[\text{NH}_3]$ increases immediately
• Q increases (numerator gets bigger) → $Q > K$
• System shifts left to consume the added $NH_{3}$
• At the new equilibrium: $[\text{N}_2]$ and $[\text{H}_2]$ are higher

Removing a Species → Opposite Shift

Key Insight

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

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)

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 ✅

🔄 Le Chatelier — Pressure/Volume and Temperature

Part 4 of 7 — How Gases and Heat Affect Equilibrium

Topics in This Part

🔑 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

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

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

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

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 ✅

🔄 Catalysts and Inert Gas Addition

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

Topics in This Part

🔑 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 ($E_a$).

Key Facts About Catalysts at Equilibrium

Why No Shift?

⚠️ 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)

Example

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

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 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 ✅

🧮 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 $K$.

For Q vs K Problems

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

For Le Chatelier Problems

1. Identify the stress (concentration, pressure/volume, temperature, catalyst, inert gas)
2. 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
3. Determine effect on each species' concentration
4. 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:

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

$Q = 3.33 < K = 5.10$ → shift right

At the new equilibrium:

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

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 ✅

🎓 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

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

Le Chatelier's Principle

Key Reminders

⚠️ 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: $[\text{PCl}_5] = 0.50$, $[\text{PCl}_3] = 0.10$, $[\text{Cl}_2] = 0.10$ 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 ✅