Introduction to Chemical Equilibrium - Complete Interactive Lesson
Part 1: Dynamic Equilibrium
โ๏ธ Dynamic Equilibrium
Part 1 of 7 โ Forward and Reverse Rates
Topics in This Part
Section
โ๏ธ Reversible Reactions
What "Dynamic" Means
โฑ๏ธ Rates Over Time
Before Equilibrium
At Equilibrium
๐ 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
โ๏ธ Reversible Reactions
Consider the reaction:
N2โO4โ(g)โ2NO
โฑ๏ธ Rates Over Time
Before Equilibrium
Time Period
Forward Rate
Reverse Rate
Net Change
t=0
Maximum
Zero
Products forming rapidly
Early
Decreasing
Increasing
Products still forming
Approaching eq.
Converging
Converging
Slowing net change
At Equilibrium
Concept Check โ Dynamic Equilibrium ๐ฏ
โ๏ธ Conditions for Equilibrium
For a system to reach equilibrium, several conditions must be met:
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Quick frame: equilibrium is about stable macroscopic behavior, not a stopped reaction.
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<h3 class="mt-0 mb-2 text-xl font-extrabold text-emerald-900 dark:text-emerald-100">1) Closed System</h3>
<p class="mb-2 text-emerald-900/90 dark:text-emerald-100/90">No matter enters or leaves (energy transfer can still occur).</p>
<p class="mb-0 text-emerald-900 dark:text-emerald-100"><strong>Key idea:</strong> if matter escapes, concentrations cannot stabilize.</p>
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<h3 class="mt-0 mb-2 text-xl font-extrabold text-violet-900 dark:text-violet-100">2) Reversible Reaction</h3>
<p class="mb-0 text-violet-900 dark:text-violet-100">The process must proceed in both directions so forward and reverse rates can eventually match.</p>
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<h3 class="mt-0 mb-2 text-xl font-extrabold text-amber-900 dark:text-amber-100">3) Constant Temperature</h3>
<p class="mb-0 text-amber-900 dark:text-amber-100">Temperature must stay fixed. Changing temperature shifts the equilibrium position.</p>
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<h3 class="mt-0 mb-2 text-xl font-extrabold text-rose-900 dark:text-rose-100">4) Sufficient Time</h3>
<p class="mb-0 text-rose-900 dark:text-rose-100">Some systems equilibrate in milliseconds; others require hours or days.</p>
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Recognizing Equilibrium
You know a system is at equilibrium when:
All macroscopic properties (concentration, pressure, color, pH) remain constant
The system is closed
The reaction is reversible
Equilibrium Conditions ๐
Equilibrium Practice ๐งฎ
Consider the reaction: H2โ(g)+I2โ(g)โ
Exit Quiz โ Dynamic Equilibrium โ
Part 2: Equilibrium Constant (Keq)
โ๏ธ Equilibrium Expressions: Kcโ and Kpโ
Part 2 of 7 โ Writing and Using Equilibrium Constants
Topics in This Part
Section
โ๏ธ The Equilibrium Constant
Part 3: Writing Equilibrium Expressions
โ๏ธ Heterogeneous Equilibrium
Part 3 of 7 โ Solids and Liquids in Equilibrium Expressions
Topics in This Part
Section
๐ค Why Exclude Solids and Liquids?
Physical Reasoning
Example 1: Decomposition of Calcium Carbonate
Example 2: Water Equilibrium
๐ Important Clarifications
๐ 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
๐ค Why Exclude Solids and Liquids?
The equilibrium constant is defined in terms of activities, not concentrations:
For gases: activity โ partial pressure (in atm)
For dissolved species: activity โ molar concentration (in M)
For pure solids and pure liquids: activity = 1 (by definition)
๐ Key Concept: Pure solids and liquids have an activity of 1, so they don't affect the value of and are left out of the expression.
Part 4: Kp vs Kc
โ๏ธ Manipulating Equilibrium Constants
Part 4 of 7 โ Reversing, Multiplying, and Adding Reactions
Topics in This Part
Section
๐ Rule 1: Reversing a Reaction
Example
๐ Rule 2: Multiplying a Reaction by a Factor
Example
๐ Rule 3: Adding Reactions (Hess's Law for K)
๐ 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
๐ Rule 1: Reversing a Reaction
If you reverse a reaction, the new K is the reciprocal of the original:
Part 5: Heterogeneous Equilibria
โ๏ธ Magnitude of K and Extent of Reaction
Part 5 of 7 โ What K Tells Us About the Reaction
Topics in This Part
Section
๐ Large K: Products Favored
Interpretation
Examples
๐ Small K: Reactants Favored
Interpretation
๐ 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
๐ Large K: Products Favored
When Kโซ1 (say, ):
Part 6: Problem-Solving Workshop
๐งฎ Problem-Solving Workshop
Part 6 of 7 โ Equilibrium Expression and K 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.
๐ Why this matters: The AP Chemistry exam rewards students who can apply concepts to unfamiliar problems โ 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
๐ ๏ธ Problem-Solving Strategy
Steps for Equilibrium Expression Problems
Write the balanced equation
Identify phases โ exclude solids (s) and liquids (l)
Write the K expression: products over reactants with coefficient exponents
Plug in equilibrium values
Check โ does the magnitude of K make sense?
๐ก Tip: On the AP exam, always verify your answer: if , products should dominate; if , reactants should dominate.
Part 7: Synthesis & AP Review
๐ Synthesis & AP Review
Part 7 of 7 โ Introduction to Equilibrium
Bringing It All Together
This comprehensive review connects every concept from Parts 1โ6 with AP-style problems. The questions are designed to mirror what you'll see on the actual exam โ multi-step, multi-concept, and requiring clear written explanations.
๐ Why this matters: AP Chemistry exam questions rarely test one concept in isolation โ 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
๐ Concept Summary
Dynamic Equilibrium
Forward rate = reverse rate
Concentrations are constant but not necessarily equal
System must be closed
Equilibrium Expressions
2
โ
(
g
)
The forward reaction: NโOโ decomposes into NOโ
The reverse reaction: NOโ molecules recombine to form NโOโ
Initially, only the forward reaction occurs. As products build up, the reverse reaction begins and accelerates. Eventually, both reactions proceed at the same rate.
What "Dynamic" Means
At equilibrium:
Both forward and reverse reactions continue to occur
There is no net change in concentrations
The system is NOT static โ it is constantly reacting in both directions
This is why we call it dynamic equilibrium.
๐ Key Concept: Dynamic equilibrium means both forward and reverse reactions continue at equal rates โ the system is NOT static, it is constantly reacting.
Rateforwardโ=Ratereverseโโ
Concentrations of reactants and products remain constant (not necessarily equal!)
The ratio [products]/[reactants] stays fixed at a given temperature
โ ๏ธ Common Misconception: Equilibrium does NOT mean the reaction has stopped, that concentrations of reactants and products are equal, or that nothing is happening. It means the rates are balanced so there is no net change.
๐ก Tip: On the AP exam, look for phrases like "constant concentration," "no further change," or "sealed container" as clues that a system is at equilibrium.
2
HI
(
g
)
At a certain temperature, the following data are collected at equilibrium:
Species
Concentration (M)
Hโ
0.10
Iโ
0.20
HI
0.40
1) What is the rate of the forward reaction compared to the reverse reaction at equilibrium? (Enter "equal")
2) If the forward reaction rate is 2.0ร10โ3 M/s, what is the reverse reaction rate in M/s? (Enter as a decimal, e.g. 0.002)
3) Is the concentration of HI changing at equilibrium? (Enter "no")
Round all answers to 3 significant figures.
Kcโ
Rules for Writing Kcโ
Example
โ๏ธ The Equilibrium Constant Kpโ
Relationship Between Kcโ and Kpโ
๐ 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
โ๏ธ The Equilibrium Constant Kcโ
For the general reaction:
aA+bBโcC+dD
The equilibrium constant expression is:
Kcโ=[A]
Rules for Writing Kcโ
Products go in the numerator, reactants in the denominator
Each concentration is raised to the power of its stoichiometric coefficient
Kcโ uses molar concentrations (mol/L)
Kcโ is dimensionless by convention on the AP exam
๐ Key Concept: Stoichiometric coefficients become exponents, not multipliers, in the equilibrium expression.
Example
N2โ(g)+3H2โ(g)โ
Kcโ=[N2โ
โ๏ธ The Equilibrium Constant Kpโ
For gaseous reactions, we can use partial pressures instead of concentrations:
๐ก Tip: When ฮn=0, then Kpโ=Kcโ because . This is a useful shortcut on the AP exam!
Writing Equilibrium Expressions ๐ฏ
Calculating Kcโ ๐งฎ
For the reaction: H2โ(g)+I2โ(g)โ2HI(g)
At equilibrium: [H2โ]=0.10 M, [I2โ]= M, M
1) Calculate Kcโ. (Enter as a whole number)
2) What is ฮn for this reaction?
3) If Kcโ=8.0 at this temperature, what is Kpโ? (Enter as a number)
Round all answers to 3 significant figures.
๐งช Worked Example: Converting Kcโ to Kpโ
Problem: For N2โ(g)+3,H2โ(g)โ2,NH3โ(g, Kcโ=0.500 at T=400 K. Find Kpโ.
Solution:
ฮn=2โ(1+3)=โ2
Kpโ=Kcโ(RT)
Kpโ=0.500ร(32.82)โ2=0.500ร
Kpโ=0.500ร9.28ร10โ4=
โ ๏ธ Warning: Notice that Kpโ<Kcโ when ฮn< (fewer moles of gas on the product side). Always check the sign of before converting!
Kcโ vs Kpโ Concepts ๐
Exit Quiz โ Equilibrium Expressions โ
K
Since pure solids and liquids have an activity of 1, they don't affect the value of K and are left out.
Physical Reasoning
The "concentration" of a pure solid or liquid is its density divided by its molar mass โ this is a constant that doesn't change as the reaction proceeds. Since it doesn't vary, it's absorbed into the equilibrium constant.
Example 1: Decomposition of Calcium Carbonate
CaCO3โ(s)โCaO(s)+CO2โ(g)
Kpโ=PCO2โโ
Both CaCOโ and CaO are solids โ they are excluded. Only the gaseous COโ appears.
๐ก Tip: Remember: add reactions โ multiplyK values. This is analogous to Hess's Law for enthalpy, but with multiplication instead of addition!
Why Multiply?
When you add reactions, the equilibrium expressions multiply (it's algebra โ you're multiplying fractions). Intermediates cancel out.
Example
Reactionย 1:ย N2โ(g)+O2โ(
Reactionย 2:ย 2NO(g)+O2โ(g)โ2
Overall:ย N2โ(g)+2O2โ(g)
Koverallโ=K1โร
Summary Table
Operation
Effect on K
Reverse reaction
Kโฒ=1/K
Multiply by n
Manipulating K โ Concept Quiz ๐ฏ
Manipulating K โ Calculations ๐งฎ
Given: A(g)โ2B(g), Kcโ=25
1) What is Kcโ for 2B(g)โA(g)? (Enter as a decimal)
2) What is Kcโ for 21โA(? (Enter as a whole number)
3) Given also: B(g)โC(g), Kcโ=2.0. What is for ? (Enter as a whole number)
Round all answers to 3 significant figures.
Operation Identification ๐
Exit Quiz โ Manipulating K โ
K>103
K=[reactants][products]โโซ1
This means the numerator (products) is much larger than the denominator (reactants).
Interpretation
The reaction lies far to the right
At equilibrium, mostly products are present
The forward reaction is strongly favored
The reaction goes "nearly to completion"
๐ Key Concept:Kโซ1 means products dominate โ the larger the K, the more the equilibrium lies to the right.
Examples
Reaction
K
Interpretation
2H2โ(g)+O2โ(g)โ2H2โO(g)
โผ1080
Essentially complete
Ag+(aq)+2NH3โ(aq
๐ Small K: Reactants Favored
When Kโช1 (say, K<10โ3):
K=[reactants][products]โโช1
The denominator (reactants) is much larger than the numerator (products).
Interpretation
The reaction lies far to the left
At equilibrium, mostly reactants remain
The forward reaction barely proceeds
Very little product forms
๐ Key Concept:Kโช1 means reactants dominate โ the smaller the K, the less the reaction proceeds toward products.
Examples
Reaction
K
Interpretation
N2โ(g)+O
๐ Intermediate K
When Kโ1 (roughly 10โ3<K<103):
Significant amounts of both reactants and products present
Neither side is strongly favored
The equilibrium position is roughly in the middle
Interpreting K Values ๐ฏ
๐ก๏ธ K Depends on Temperature
The equilibrium constant is a function of temperature only.
What Changes K?
Temperature โ the ONLY factor that changes K
What Does NOT Change K?
Changing concentrations
Changing pressure/volume
Adding a catalyst
Adding an inert gas
โ ๏ธ Warning: These factors may shift the equilibrium position (where Q moves relative to K), but K itself remains constant at a given temperature. Only temperature changes K!
Temperature and K Direction
Reaction Type
Increase T
K Changes
Exothermic (ฮH<0)
Shifts left
K decreases
Endothermic (ฮH>0)
Shifts right
K increases
Think of heat as a "reactant" (endothermic) or "product" (exothermic).
K Value Interpretation ๐
K Magnitude Practice ๐งฎ
1) A reaction has K=2.0ร10โ20. Is the reaction product-favored or reactant-favored? (Enter "reactant-favored")
2) For the reaction AโB, K=100 at 300 K. If the reaction is exothermic and temperature increases to 400 K, does K increase or decrease? (Enter "decrease")
3) A catalyst is added to a reaction at equilibrium. Does the value of K change? (Enter "no")
Exit Quiz โ Magnitude of K โ
Kโซ1
Kโช1
Key Formulas
Formula
When to Use
Kcโ=[reactants][products]โ
All K calculations
Kpโ=Kcโ(RT)ฮn
Converting between and
Reverse: Kโฒ=1/K
Flipping the reaction
Multiply by n: Kโฒ=Kn
Scaling coefficients
Add reactions: K=K1โรK2โ
Combining reactions
๐ข Worked Example 1: Calculating Kcโ
Problem: At 450ยฐC, the equilibrium concentrations for H2โ(g)+I2โ(g)โ2HI(g) are: [H2โ]=0.0050 M, [I2โ]=0.0050 M, [HI]=0.040 M. Find Kcโ.
Solution:
Kcโ=[H
Kcโ=2.5ร10โ5
Since K>1, products (HI) are favored at this temperature.
Practice Problem 1 ๐งฎ
For the reaction: PCl5โ(g)โPCl3โ(g)+Cl2โ(g)
At equilibrium: [PCl5โ]=0.20 M, [PCl3โ]= M, M
1) Calculate Kcโ (Enter to 3 significant figures)
2) Is the reaction product-favored or reactant-favored? (Enter "product-favored" or "reactant-favored")
3) What is ฮn for this reaction? (Enter as an integer with sign, e.g. +1)
๐งช Worked Example 2: Combining K Values
Problem: Given NO(g)+21โO2โ(g)โNO2โ(g), K1โ=1.3ร106. Find K for 2NO2โ(g)โ2NO(g)+O2.
Solution:
Step 1: The target is the reverse of Reaction 1, multiplied by 2.
Step 2: Reverse Reaction 1: Krevโ=1/K1โ=1/(
Step 3: Multiply by 2: K=(Krevโ)2=(7.7ร
Practice Problem 2 โ Combining K Values ๐ฏ
Practice Problem 3 โ Kcโ to Kpโ Conversion ๐งฎ
For 2SO2โ(g)+O2โ(g)โ2SO3โ(g) at T=1000 K, Kcโ=280.
1) What is ฮn? (Enter as an integer with sign)
2) Calculate RT using R=0.08206 Lยทatm/(molยทK). (Round to 3 significant figures)
3) Calculate Kpโ. (Round to 3 significant figures)
Exit Quiz โ Problem-Solving Workshop โ
K
cโ
=
[reactants]coefficients[products]coefficientsโ
Kpโ=(Preactantsโ)coeff(Pproductsโ)coeffโ
Kpโ=Kcโ(RT)ฮn
Heterogeneous Equilibrium
Exclude pure solids (s) and pure liquids (l)
Include gases (g) and aqueous species (aq)
Solids must still be present for equilibrium to exist
Manipulating K
Operation
Effect
Reverse
Kโฒ=1/K
Multiply by n
Kโฒ=Kn
Add reactions
Ktotalโ=K1โรK
Magnitude of K
Kโซ1: product-favored
Kโช1: reactant-favored
Only temperature changes K
๐ก Tip: For the AP exam, remember the three operations on K: reverse โ reciprocal, multiply coefficients โ raise to power, add reactions โ multiply K values.
โ ๏ธ Warning: A common AP mistake is confusing what changes K versus what shifts equilibrium position. Only temperature changes K!
AP-Style Multiple Choice โ Set 1 ๐ฏ
AP-Style Multiple Choice โ Set 2 ๐ฏ
AP Free-Response Style ๐งฎ
The reaction 2SO3โ(g)โ2SO2โ(g)+O2โ(g) has Kcโ=1.6ร10โ10 at 900 K.
1) Is this reaction product-favored or reactant-favored at 900 K? (Enter "reactant-favored")
2) What is Kcโ for SO2โ(g)+? (Enter in scientific notation, e.g. 7.9e4)
3) The decomposition of SOโ is endothermic. If temperature increases, does Kcโ for the decomposition increase or decrease? (Enter "increase")