⚖️ Dynamic Equilibrium

Part 1 of 7 — Forward and Reverse Rates

Chemical reactions don't always go to completion. Many reactions are reversible — the products can react to re-form the reactants. When the forward and reverse reactions occur at the same rate, the system reaches dynamic equilibrium.

Reversible Reactions

Consider the reaction:

$\text{N}_2\text{O}_4(g) \rightleftharpoons 2\,\text{NO}_2(g)$

• The forward reaction: N₂O₄ decomposes into NO₂
• The reverse reaction: NO₂ molecules recombine to form N₂O₄

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 — it is constantly reacting in both directions

This is why we call it dynamic equilibrium.

Rates Over Time

Before Equilibrium

At Equilibrium

$\text{Rate}_{\text{forward}} = \text{Rate}_{\text{reverse}}$

• Concentrations of reactants and products remain constant (not necessarily equal!)
• The ratio $[\text{products}]/[\text{reactants}]$ stays fixed at a given temperature

Key Misconception

Equilibrium does NOT mean:

• The reaction has stopped
• Concentrations of reactants and products are equal
• Nothing is happening

It means the rates are balanced so there is no net change.

Concept Check — Dynamic Equilibrium 🎯

Conditions for Equilibrium

For a system to reach equilibrium, several conditions must be met:

1. Closed System

The system must be closed — no matter can enter or leave. (Energy transfer is allowed.)

2. Reversible Reaction

The reaction must be able to proceed in both directions.

3. Constant Temperature

Temperature must remain constant. (Changing temperature shifts the equilibrium position.)

4. Sufficient Time

The system needs time to reach equilibrium. Some reactions reach it in milliseconds; others take days.

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 🔍

Equilibrium Practice 🧮

Consider the reaction: $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\,\text{HI}(g)$

At a certain temperature, the following data are collected at equilibrium:

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 $2.0 \times 10^{-3}$ 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.

Exit Quiz — Dynamic Equilibrium ✅

⚖️ Equilibrium Expressions: $K_c$ and $K_p$

Part 2 of 7 — Writing and Using Equilibrium Constants

The equilibrium constant quantifies the ratio of product concentrations to reactant concentrations at equilibrium. There are two forms: $K_c$ (using molar concentrations) and $K_p$ (using partial pressures for gaseous systems).

The Equilibrium Constant $K_c$

For the general reaction:

$aA + bB \rightleftharpoons cC + dD$

The equilibrium constant expression is:

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

Rules for Writing $K_c$

1. Products go in the numerator, reactants in the denominator
2. Each concentration is raised to the power of its stoichiometric coefficient
3. $K_c$ uses molar concentrations (mol/L)
4. $K_c$ is dimensionless by convention on the AP exam

Example

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

$K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$

The Equilibrium Constant $K_p$

For gaseous reactions, we can use partial pressures instead of concentrations:

$K_p = \frac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b}$

Relationship Between $K_c$ and $K_p$

$K_p = K_c(RT)^{\Delta n}$

Where:

• $R = 0.08206$ L·atm/(mol·K)
• $T$ = temperature in Kelvin
• $\Delta n = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)}$

Example

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

$\Delta n = 2 - (1 + 3) = -2$

$K_p = K_c(RT)^{-2}$

Special Case: $\Delta n = 0$

When $\Delta n = 0$, then $K_p = K_c$ because $(RT)^0 = 1$.

Writing Equilibrium Expressions 🎯

Calculating $K_c$ 🧮

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

At equilibrium: $[\text{H}_2] = 0.10$ M, $[\text{I}_2] = 0.20$ M, $[\text{HI}] = 0.40$ M

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

2. What is $\Delta n$ for this reaction?

3. If $K_c = 8.0$ at this temperature, what is $K_p$? (Enter as a number)

Round all answers to 3 significant figures.

Worked Example: Converting $K_c$ to $K_p$

Problem: For $\text{N}_2(g) + 3\,\text{H}_2(g) \rightleftharpoons 2\,\text{NH}_3(g)$, $K_c = 0.500$ at $T = 400$ K. Find $K_p$.

Solution:

$\Delta n = 2 - (1 + 3) = -2$

$K_p = K_c(RT)^{\Delta n} = 0.500 \times (0.08206 \times 400)^{-2}$

$K_p = 0.500 \times (32.82)^{-2} = 0.500 \times \frac{1}{1077.4}$

$K_p = 0.500 \times 9.28 \times 10^{-4} = 4.64 \times 10^{-4}$

Notice that $K_p < K_c$ when $\Delta n < 0$ (fewer moles of gas on the product side).

$K_c$ vs $K_p$ Concepts 🔍

Exit Quiz — Equilibrium Expressions ✅

⚖️ Heterogeneous Equilibrium

Part 3 of 7 — Solids and Liquids in Equilibrium Expressions

When the reactants and products are in different phases (solid, liquid, gas, aqueous), the equilibrium is called heterogeneous. The key rule: pure solids and pure liquids are excluded from the equilibrium expression.

Why Exclude Solids and Liquids?

The equilibrium constant is defined in terms of activities, not concentrations:

• For gases: activity ≈ partial pressure (in atm)
• For dissolved species: activity ≈ molar concentration (in M)
• For pure solids and pure liquids: activity = 1 (by definition)

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 — 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

$\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)$

$K_p = P_{\text{CO}_2}$

Both CaCO₃ and CaO are solids — they are excluded. Only the gaseous CO₂ appears.

Example 2: Water Equilibrium

$\text{H}_2\text{O}(l) \rightleftharpoons \text{H}^+(aq) + \text{OH}^-(aq)$

$K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14} \text{ at 25°C}$

Liquid water is excluded from the expression.

Writing Heterogeneous Equilibrium Expressions 🎯

Important Clarifications

Solids Must Still Be Present!

Even though solids and liquids don't appear in the $K$ expression, they must still be present for the equilibrium to exist.

For $\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)$:

• If all the CaCO₃ decomposes (none left), the system is NOT at equilibrium
• Some solid CaCO₃ must remain for the reverse reaction to be possible

Amount of Solid Doesn't Matter

As long as some solid is present:

• Adding more solid does NOT shift the equilibrium
• Removing some solid (as long as some remains) does NOT shift the equilibrium
• The equilibrium partial pressure of CO₂ is the same whether you have 1 g or 1 kg of CaCO₃

Aqueous Species ARE Included

Don't confuse dissolved species with liquids:

• $\text{H}_2\text{O}(l)$ → pure liquid → excluded
• $\text{Na}^+(aq)$ → dissolved species → included
• $\text{NaCl}(s)$ → solid → excluded
• $\text{NaCl}(aq)$ → dissolved → included

Include or Exclude? 🔍

For each species, determine whether it appears in the equilibrium expression.

Heterogeneous Equilibrium Calculations 🧮

1. For $\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)$, if $K_p = 0.040$ atm at a certain temperature, what is $P_{\text{CO}_2}$ at equilibrium? (in atm)

2. How many species appear in the $K_c$ expression for $\text{C}(s) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CO}(g) + \text{H}_2(g)$? (Enter a number)

3. For $\text{NH}_4\text{Cl}(s) \rightleftharpoons \text{NH}_3(g) + \text{HCl}(g)$, if $P_{\text{NH}_3} = P_{\text{HCl}} = 0.30$ atm at equilibrium, what is $K_p$? (Enter to 3 significant figures)

Exit Quiz — Heterogeneous Equilibrium ✅

⚖️ Manipulating Equilibrium Constants

Part 4 of 7 — Reversing, Multiplying, and Adding Reactions

When you modify a chemical equation, the equilibrium constant changes in a predictable way. These rules are essential for combining known $K$ values to find unknown ones.

Rule 1: Reversing a Reaction

If you reverse a reaction, the new $K$ is the reciprocal of the original:

$\text{Forward: } A \rightleftharpoons B \quad K_{\text{fwd}}$

$\text{Reverse: } B \rightleftharpoons A \quad K_{\text{rev}} = \frac{1}{K_{\text{fwd}}}$

Example

$\text{N}_2(g) + 3\,\text{H}_2(g) \rightleftharpoons 2\,\text{NH}_3(g) \quad K_c = 4.0 \times 10^8$

$2\,\text{NH}_3(g) \rightleftharpoons \text{N}_2(g) + 3\,\text{H}_2(g) \quad K_c = \frac{1}{4.0 \times 10^8} = 2.5 \times 10^{-9}$

The products and reactants switch — the fraction flips.

Rule 2: Multiplying a Reaction by a Factor

If you multiply all coefficients by a factor $n$, the new $K$ is raised to the $n$th power:

$\text{Original: } A \rightleftharpoons B \quad K$

$\text{Multiplied by } n: \quad nA \rightleftharpoons nB \quad K' = K^n$

Example

$\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\,\text{HI}(g) \quad K_c = 50$

Multiply by $\frac{1}{2}$:

$\frac{1}{2}\text{H}_2(g) + \frac{1}{2}\text{I}_2(g) \rightleftharpoons \text{HI}(g) \quad K_c' = 50^{1/2} = \sqrt{50} \approx 7.07$

Multiply by 2:

$2\,\text{H}_2(g) + 2\,\text{I}_2(g) \rightleftharpoons 4\,\text{HI}(g) \quad K_c' = 50^2 = 2500$

Rule 3: Adding Reactions (Hess's Law for K)

If you add two reactions together, the overall $K$ is the product of the individual $K$ values:

$\text{Reaction 1: } A \rightleftharpoons B \quad K_1$ $\text{Reaction 2: } B \rightleftharpoons C \quad K_2$ $\text{Overall: } A \rightleftharpoons C \quad K_{\text{overall}} = K_1 \times K_2$

Why Multiply?

When you add reactions, the equilibrium expressions multiply (it's algebra — you're multiplying fractions). Intermediates cancel out.

Example

$\text{Reaction 1: } \text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2\,\text{NO}(g) \quad K_1 = 4.7 \times 10^{-31}$

$\text{Reaction 2: } 2\,\text{NO}(g) + \text{O}_2(g) \rightleftharpoons 2\,\text{NO}_2(g) \quad K_2 = 1.8 \times 10^{6}$

$\text{Overall: } \text{N}_2(g) + 2\,\text{O}_2(g) \rightleftharpoons 2\,\text{NO}_2(g)$

$K_{\text{overall}} = K_1 \times K_2 = (4.7 \times 10^{-31})(1.8 \times 10^{6}) = 8.5 \times 10^{-25}$

Summary Table

Manipulating K — Concept Quiz 🎯

Manipulating K — Calculations 🧮

Given: $\text{A}(g) \rightleftharpoons 2\,\text{B}(g)$, $K_c = 25$

1. What is $K_c$ for $2\,\text{B}(g) \rightleftharpoons \text{A}(g)$? (Enter as a decimal)

2. What is $K_c$ for $\frac{1}{2}\text{A}(g) \rightleftharpoons \text{B}(g)$? (Enter as a whole number)

3. Given also: $\text{B}(g) \rightleftharpoons \text{C}(g)$, $K_c = 2.0$. What is $K_c$ for $\text{A}(g) \rightleftharpoons 2\,\text{C}(g)$? (Enter as a whole number)

Round all answers to 3 significant figures.

Operation Identification 🔍

Exit Quiz — Manipulating K ✅

⚖️ Magnitude of K and Extent of Reaction

Part 5 of 7 — What K Tells Us About the Reaction

The numerical value of the equilibrium constant tells you how far a reaction proceeds toward products before reaching equilibrium. Understanding the magnitude of K is crucial for predicting whether products or reactants dominate at equilibrium.

Large K: Products Favored

When $K \gg 1$ (say, $K > 10^3$):

$K = \frac{[\text{products}]}{[\text{reactants}]} \gg 1$

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"

Examples

Small K: Reactants Favored

When $K \ll 1$ (say, $K < 10^{-3}$):

$K = \frac{[\text{products}]}{[\text{reactants}]} \ll 1$

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

Examples

Intermediate K

When $K \approx 1$ (roughly $10^{-3} < K < 10^3$):

• Significant amounts of both reactants and products present
• Neither side is strongly favored
• The equilibrium position is roughly in the middle

Interpreting K Values 🎯

K Depends on Temperature

The equilibrium constant is a function of temperature only.

What Changes K?

• Temperature — the ONLY factor that changes K

What Does NOT Change K?

• Changing concentrations
• Changing pressure/volume
• Adding a catalyst
• Adding an inert gas

These factors may shift the equilibrium position (where Q moves relative to K), but K itself remains constant at a given temperature.

Temperature and K Direction

Think of heat as a "reactant" (endothermic) or "product" (exothermic).

K Value Interpretation 🔍

K Magnitude Practice 🧮

1. A reaction has $K = 2.0 \times 10^{-20}$. Is the reaction product-favored or reactant-favored? (Enter "reactant-favored")

2. For the reaction $\text{A} \rightleftharpoons \text{B}$, $K = 100$ 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 — Magnitude of K ✅

🧮 Problem-Solving Workshop

Part 6 of 7 — Equilibrium Expression and K Calculations

This workshop brings together everything from Parts 1–5: writing equilibrium expressions, calculating K, manipulating K values, and interpreting results. These multi-step problems mirror AP-level questions.

Problem-Solving Strategy

Steps for Equilibrium Expression Problems

1. Write the balanced equation
2. Identify phases — exclude solids (s) and liquids (l)
3. Write the $K$ expression: products over reactants with coefficient exponents
4. Plug in equilibrium values
5. Check — does the magnitude of K make sense?

Key Formulas

Worked Example 1: Calculating $K_c$

Problem: At 450°C, the equilibrium concentrations for the reaction

$\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\,\text{HI}(g)$

are: $[\text{H}_2] = 0.0050$ M, $[\text{I}_2] = 0.0050$ M, $[\text{HI}] = 0.040$ M.

Solution:

$K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \frac{(0.040)^2}{(0.0050)(0.0050)}$

$K_c = \frac{1.6 \times 10^{-3}}{2.5 \times 10^{-5}} = 64$

Since $K > 1$, products (HI) are favored at this temperature.

Practice Problem 1 🧮

For the reaction: $\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)$

At equilibrium: $[\text{PCl}_5] = 0.20$ M, $[\text{PCl}_3] = 0.30$ M, $[\text{Cl}_2] = 0.30$ M

1. Calculate $K_c$ (Enter to 3 significant figures)

2. Is the reaction product-favored or reactant-favored? (Enter "product-favored" or "reactant-favored")

3. What is $\Delta n$ for this reaction? (Enter as an integer with sign, e.g. +1)

Worked Example 2: Combining K Values

Problem: Given:

$\text{Reaction 1: } \text{NO}(g) + \frac{1}{2}\text{O}_2(g) \rightleftharpoons \text{NO}_2(g) \quad K_1 = 1.3 \times 10^{6}$

Find $K$ for:

$2\,\text{NO}_2(g) \rightleftharpoons 2\,\text{NO}(g) + \text{O}_2(g)$

Solution:

Step 1: The target is the reverse of Reaction 1, multiplied by 2.

Step 2: Reverse Reaction 1: $K_{\text{rev}} = 1/K_1 = 1/(1.3 \times 10^6) = 7.7 \times 10^{-7}$

Step 3: Multiply by 2: $K = (K_{\text{rev}})^2 = (7.7 \times 10^{-7})^2 = 5.9 \times 10^{-13}$

Practice Problem 2 — Combining K Values 🎯

Practice Problem 3 — $K_c$ to $K_p$ Conversion 🧮

For $2\,\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\,\text{SO}_3(g)$ at $T = 1000$ K, $K_c = 280$.

1. What is $\Delta n$? (Enter as an integer with sign)

2. Calculate $RT$ using $R = 0.08206$ L·atm/(mol·K). (Round to 3 significant figures)

3. Calculate $K_p$. (Round to 3 significant figures)

Exit Quiz — Problem-Solving Workshop ✅

🎓 Synthesis & AP Review

Part 7 of 7 — Introduction to Equilibrium

This final part reviews all key concepts: dynamic equilibrium, $K_c$/$K_p$ expressions, heterogeneous equilibrium, manipulating K, and the meaning of K's magnitude. These questions mirror AP Chemistry free-response and multiple-choice formats.

Concept Summary

Dynamic Equilibrium

• Forward rate = reverse rate
• Concentrations are constant but not necessarily equal
• System must be closed

Equilibrium Expressions

• $K_c = \frac{[\text{products}]^{\text{coefficients}}}{[\text{reactants}]^{\text{coefficients}}}$
• $K_p = \frac{(P_{\text{products}})^{\text{coeff}}}{(P_{\text{reactants}})^{\text{coeff}}}$
• $K_p = K_c(RT)^{\Delta n}$

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

Magnitude of K

• $K \gg 1$: product-favored
• $K \ll 1$: reactant-favored
• Only temperature changes K

AP-Style Multiple Choice — Set 1 🎯

AP-Style Multiple Choice — Set 2 🎯

AP Free-Response Style 🧮

The reaction $2\,\text{SO}_3(g) \rightleftharpoons 2\,\text{SO}_2(g) + \text{O}_2(g)$ has $K_c = 1.6 \times 10^{-10}$ at 900 K.

1. Is this reaction product-favored or reactant-favored at 900 K? (Enter "reactant-favored")

2. What is $K_c$ for $\text{SO}_2(g) + \frac{1}{2}\text{O}_2(g) \rightleftharpoons \text{SO}_3(g)$? (Enter in scientific notation, e.g. 7.9e4)

3. The decomposition of SO₃ is endothermic. If temperature increases, does $K_c$ for the decomposition increase or decrease? (Enter "increase")

Round all answers to 3 significant figures.

Final Concept Review 🔍

Final Exit Quiz ✅