Activation Energy and Temperature Effects

💥 Collision Theory

Part 1 of 7 — Why Do Molecules Need to Collide?

Two Requirements for an Effective Collision

Most collisions fail — only a tiny fraction have both enough energy AND the right orientation.

🔑 Why this matters: Collision theory explains why reactions have activation energies and why temperature dramatically affects reaction rates.

What You'll Master in Part 1

• Explaining why molecules must collide with sufficient energy and proper orientation
• Interpreting Maxwell-Boltzmann distribution curves
• Predicting how temperature changes shift the fraction of effective collisions

📊 Requirements for an Effective Collision

For a collision to result in a reaction, two conditions must be met simultaneously:

Condition 1: Sufficient Energy

The colliding molecules must have kinetic energy at least equal to the activation energy ($E_a$):

$\boxed{KE_{\text{collision}} \geq E_a}$

If the collision energy is below $E_a$, the molecules simply bounce off each other without reacting.

Condition 2: Proper Orientation

Even with enough energy, the molecules must collide with the correct geometric orientation. The reactive parts of the molecules must be facing each other.

Example: NO + $NO_{3}$ → $2NO_{2}$

• ✅ O of NO hits O of $NO_{3}$ → bonds can rearrange → reaction!
• ❌ N of NO hits N of $NO_{3}$ → wrong atoms in contact → no reaction

The Steric Factor

The fraction of collisions with correct orientation is called the steric factor ($p$), typically $0 < p < 1$. For simple atoms, $p \approx 1$; for complex molecules, $p$ can be very small.

Collision Theory Concepts 🎯

📌 Maxwell-Boltzmann Distribution

At any temperature, molecules have a distribution of kinetic energies. The Maxwell-Boltzmann distribution shows:

• Most molecules have moderate energies
• A few have very low or very high energy
• The area under the curve beyond $E_a$ represents the fraction of molecules that can react

Effect of Temperature

When temperature increases:

1. The peak shifts to higher energy and becomes lower and broader
2. The fraction of molecules with $KE \geq E_a$ increases dramatically
3. This is why higher temperature → faster rate

The Boltzmann Factor

The fraction of molecules with energy $\geq E_a$ is approximately:

$\boxed{f = e^{-E_a/(RT)}}$

💡 This exponential dependence explains why even small temperature changes can produce large rate changes.

Maxwell-Boltzmann Distribution 🔍

⏱️ Collision Theory Rate Equation

Putting it all together, collision theory predicts:

$\boxed{\text{Rate} = Z \cdot p \cdot e^{-E_a/(RT)}}$

where:

• $Z$ = collision frequency (depends on concentration and temperature)
• $p$ = steric factor (orientation)
• $e^{-E_a/(RT)}$ = fraction with sufficient energy

Connection to the Arrhenius Equation

This leads directly to the Arrhenius equation:

$\boxed{k = A \cdot e^{-E_a/(RT)}}$

where $A = Z \cdot p$ is the frequency factor (also called the pre-exponential factor). We will derive this in detail in Part 3.

🔑 Key Connection: Collision theory → Arrhenius equation. The frequency factor $A$ captures collision frequency and orientation; the exponential captures the energy requirement.

Collision Theory Calculations 🧮

1) At 300 K, the Boltzmann factor for a reaction with $E_a = 50$ kJ/mol is $e^{-E_a/(RT)}$. Calculate $E_a/(RT)$. ($R = 8.314$ J/(mol·K); round to 3 significant figures)

2) Calculate $e^{-20.0}$ to 2 significant figures. (Use scientific notation: e.g., 2.1e-9)

3) If temperature increases from 300 K to 310 K ($E_a = 50$ kJ/mol), calculate $E_a/(RT)$ at 310 K. (to 3 significant figures)

Exit Quiz — Collision Theory ✅

⛰️ Activation Energy & Energy Diagrams

Part 2 of 7 — The Energy Landscape of Reactions

Reading an Energy Diagram

🔑 Why this matters: Energy diagrams appear frequently on the AP exam — you must be able to read, label, and calculate $E_a$ and $\Delta H$ from them.

What You'll Master in Part 2

• Labeling all parts of an energy diagram (reactants, products, $E_a$, transition state)
• Distinguishing exothermic ($\Delta H < 0$) from endothermic ($\Delta H > 0$) diagrams
• Calculating forward and reverse activation energies from diagram data

🏗️ Anatomy of an Energy Diagram

An energy diagram plots potential energy (y-axis) vs. reaction progress (x-axis):

Key Features

Mathematical Relationship

$\boxed{E_a(\text{reverse}) = E_a(\text{forward}) - \Delta H}$

Or equivalently:

$\boxed{\Delta H = E_a(\text{forward}) - E_a(\text{reverse})}$

📌 Exothermic vs. Endothermic Diagrams

Exothermic ($\Delta H < 0$): Products LOWER than Reactants

• Products are more stable (lower energy)
• $E_a(\text{forward}) < E_a(\text{reverse})$
• Energy is released to surroundings
• Example: Combustion reactions

$\text{Reactants} \xrightarrow{E_a} \text{Transition State} \rightarrow \text{Products (lower energy)}$

Endothermic ($\Delta H > 0$): Products HIGHER than Reactants

• Products are less stable (higher energy)
• $E_a(\text{forward}) > E_a(\text{reverse})$
• Energy is absorbed from surroundings
• Example: Dissolving $NH_{4}NO_{3}$

$\text{Reactants} \xrightarrow{E_a} \text{Transition State} \rightarrow \text{Products (higher energy)}$

Important

⚠️ $E_a$ is always positive — it is always an energy barrier that must be overcome, regardless of whether the reaction is exo- or endothermic.

Energy Diagram Quiz 🎯

🧊 The Transition State

The transition state (or activated complex) is the configuration of atoms at the energy maximum. It is:

• Not a real molecule — it cannot be isolated or observed directly
• Fleeting — exists for approximately $10^{-13}$ seconds
• Characterized by partial bonds — old bonds are partially broken, new bonds are partially formed
• Denoted with a double dagger: $\ddagger$ (e.g., $[ABC]^\ddagger$)

Example: SN2 Reaction

$\text{HO}^- + \text{CH}_3\text{Br} \rightarrow [\text{HO---CH}_3\text{---Br}]^\ddagger \rightarrow \text{CH}_3\text{OH} + \text{Br}^-$

In the transition state, both O−C and C−Br bonds are partial.

Transition State vs. Intermediate

Reading Energy Diagrams 🧮

An energy diagram shows:

• Reactants at 100 kJ
• Transition state at 250 kJ
• Products at 60 kJ

1) What is $E_a$ (forward)? (in kJ)

2) What is $\Delta H$? (in kJ, include sign)

3) What is $E_a$ (reverse)? (in kJ)

Energy Diagram Concepts 🔍

Exit Quiz — Energy Diagrams ✅

📐 The Arrhenius Equation

Part 3 of 7 — Connecting Rate Constants to Temperature

The Arrhenius Equation

$k = Ae^{-E_a/RT}$

🔑 Why this matters: The Arrhenius equation is one of the most important in kinetics — it quantitatively connects rate constants to temperature and activation energy.

What You'll Master in Part 3

• Understanding each variable in $k = Ae^{-E_a/RT}$
• Explaining how $k$ changes with temperature and $E_a$
• Calculating rate constants at different temperatures

📌 The Arrhenius Equation

$\boxed{k = Ae^{-E_a/(RT)}}$

What Each Part Means

$A$ (frequency factor): Related to how often molecules collide with correct orientation

• $A = Z \cdot p$ (collision frequency × steric factor)
• Large $A$ → favorable collision geometry
• $A$ is approximately temperature-independent

$e^{-E_a/(RT)}$ (Boltzmann factor): Fraction of collisions with sufficient energy

• As $T \uparrow$: $E_a/(RT) \downarrow$, so $e^{-E_a/(RT)} \uparrow$, so $k \uparrow$
• As $E_a \uparrow$: $E_a/(RT) \uparrow$, so $e^{-E_a/(RT)} \downarrow$, so $k \downarrow$

Arrhenius Equation Concepts 🎯

🌡️ Temperature Sensitivity and Ea

How Much Does k Change with Temperature?

The sensitivity of $k$ to temperature depends on $E_a$:

• Large $E_a$: $k$ is very sensitive to temperature changes (reaction speeds up dramatically)
• Small $E_a$: $k$ is less sensitive to temperature changes

Example Calculation

Problem: For $E_a = 100$ kJ/mol, comparing $k$ at 300 K and 310 K:

Solution:

$\frac{k_{310}}{k_{300}} = \frac{e^{-E_a/(R \cdot 310)}}{e^{-E_a/(R \cdot 300)}} = e^{(E_a/R)(1/300 - 1/310)}$

$= e^{(100{,}000/8.314)(1/300 - 1/310)}$

$= e^{(12{,}027)(1.075 \times 10^{-4})} = e^{1.293} = 3.64$

A 10°C increase nearly quadruples the rate for this high-$E_a$ reaction!

For $E_a = 20$ kJ/mol: $\frac{k_{310}}{k_{300}} = e^{(20{,}000/8.314)(1.075 \times 10^{-4})} = e^{0.259} = 1.30$

Only a 30% increase — much less sensitive.

Arrhenius Calculations 🧮

1) Calculate $E_a/(RT)$ for $E_a = 75.0$ kJ/mol at $T = 500$ K. ($R = 8.314$ J/(mol·K); to 3 significant figures)

2) A reaction has $A = 1.0 \times 10^{13}$ $s^{-1}$ and $E_a = 100$ kJ/mol. Calculate $k$ at 300 K. (in $s^{-1}$, to 1 significant figure in scientific notation: e.g., 3e-5)

3) If $k = 0.010$ $s^{-1}$ at 300 K and $k = 0.040$ $s^{-1}$ at 310 K, by what factor does k increase? (to 3 significant figures)

📊 The Frequency Factor A

The frequency factor $A$ represents the maximum possible rate constant — the value $k$ would have if every collision were effective ($E_a = 0$).

Typical Values

Key Point for AP

🔑 $A$ is approximately independent of temperature — all the temperature dependence of $k$ comes from the $e^{-E_a/(RT)}$ term.

Arrhenius Equation Review 🔍

Exit Quiz — Arrhenius Equation ✅

📉 Linearized Arrhenius Equation

Part 4 of 7 — Finding Ea from Graphical Data

From Exponential to Linear

Plot $\ln k$ vs $1/T$ → slope = $-E_a/R$, y-intercept = $\ln A$

🔑 Why this matters: The AP exam often provides data as a table of temperatures and rate constants — you need to know how to plot and analyze it graphically.

What You'll Master in Part 4

• Deriving the linearized Arrhenius equation from the exponential form
• Determining $E_a$ from the slope of a $\ln k$ vs $1/T$ plot
• Interpreting Arrhenius plots for AP exam data analysis questions

📌 Deriving the Linear Form

Starting from: $k = Ae^{-E_a/(RT)}$

Take the natural log of both sides:

$\ln k = \ln A + \ln(e^{-E_a/(RT)})$

$\boxed{\ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A}$

This is $y = mx + b$!

Key Result

A plot of $\ln k$ vs $1/T$ gives a straight line with:

• Slope $= -E_a/R$ → $E_a = -R \times \text{slope}$
• y-intercept $= \ln A$ → $A = e^{\text{intercept}}$

Linearized Arrhenius Quiz 🎯

🧪 Worked Example: Determining Ea from Data

The rate constant for a reaction was measured at several temperatures:

Finding the Slope

Using the first and last points: $\text{slope} = \frac{-3.91 - (-16.12)}{2.22 \times 10^{-3} - 3.33 \times 10^{-3}} = \frac{12.21}{-1.11 \times 10^{-3}} = -11{,}000 \; \text{K}$

Finding Ea

$E_a = -R \times \text{slope} = -(8.314)(-11{,}000) = 91{,}500 \; \text{J/mol} = 91.5 \; \text{kJ/mol}$

Arrhenius Plot Calculations 🧮

An Arrhenius plot of ln k vs 1/T has two data points:

• Point 1: $1/T = 3.00 \times 10^{-3}$ $K^{-1}$, $\ln k = -8.00$
• Point 2: $1/T = 2.50 \times 10^{-3}$ $K^{-1}$, $\ln k = -4.00$

1) What is the slope of the line? (in K, include sign)

2) What is $E_a$ in kJ/mol? (to 3 significant figures)

3) What is $\ln A$ (the y-intercept)? Use: $\ln A = \ln k + (E_a/R)(1/T)$, evaluated at point 1. (to 3 significant figures)

📌 Practical Tips for AP

Converting Temperature

Always convert °C to K before using the Arrhenius equation:

$T(K) = T(°C) + 273.15$

Units of Ea

• In the Arrhenius equation, use $E_a$ in J/mol (not kJ/mol) when $R = 8.314$ J/(mol·K)
• Convert kJ to J by multiplying by 1000

Common Mistakes

1. ❌ Using temperature in °C instead of K
2. ❌ Mixing units: $E_a$ in kJ/mol with $R$ in J/(mol·K)
3. ❌ Forgetting the negative sign in the slope
4. ❌ Plotting $k$ vs $T$ instead of $\ln k$ vs $1/T$

Arrhenius Plot Analysis 🔍

Exit Quiz — Linearized Arrhenius ✅

🔄 Two-Point Arrhenius Equation

Part 5 of 7 — Finding Ea from Two Temperatures

When You Have Just Two Data Points

$\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$

🔑 Why this matters: This is the most commonly tested Arrhenius equation form on the AP exam — many free-response problems give exactly two (T, k) data points.

What You'll Master in Part 5

• Using the two-point Arrhenius equation to find $E_a$, $k$, or $T$
• Correctly converting temperatures to Kelvin before substituting
• Recognizing which form to use based on given data

📌 Deriving the Two-Point Form

Write the Arrhenius equation at two temperatures:

$\ln k_1 = -\frac{E_a}{R} \cdot \frac{1}{T_1} + \ln A$

$\ln k_2 = -\frac{E_a}{R} \cdot \frac{1}{T_2} + \ln A$

Subtract equation 1 from equation 2:

$\ln k_2 - \ln k_1 = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$

$\boxed{\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)}$

Using This Equation

To find $E_a$: $E_a = \frac{R \cdot \ln(k_2/k_1)}{1/T_1 - 1/T_2}$

To find $k$ at a new temperature: If you know $E_a$, $k_1$, and $T_1$, find $k_2$ at $T_2$: $\ln k_2 = \ln k_1 + \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$

🧪 Worked Example

A reaction has $k_1 = 0.0120$ $s^{-1}$ at $T_1 = 400$ K and $k_2 = 0.150$ $s^{-1}$ at $T_2 = 500$ K.

Find $E_a$:

$\ln\frac{k_2}{k_1} = \ln\frac{0.150}{0.0120} = \ln(12.5) = 2.526$

$\frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{400} - \frac{1}{500} = 0.002500 - 0.002000 = 0.000500 \; \text{K}^{-1}$

$E_a = \frac{R \cdot \ln(k_2/k_1)}{1/T_1 - 1/T_2} = \frac{8.314 \times 2.526}{0.000500}$

$E_a = \frac{21.0}{0.000500} = 42{,}000 \; \text{J/mol} = 42.0 \; \text{kJ/mol}$

Practice: Finding Ea 🧮

A reaction has $k = 2.0 \times 10^{-3}$ $s^{-1}$ at 300 K and $k = 6.0 \times 10^{-2}$ $s^{-1}$ at 400 K.

1) Calculate $\ln(k_2/k_1)$. (to 3 significant figures)

2) Calculate $1/T_1 - 1/T_2$. (in $K^{-1}$, give as decimal: e.g., 0.000833)

3) Calculate $E_a$ in kJ/mol. (to 3 significant figures)

Practice: Predicting k at a New Temperature 🧮

A reaction has $E_a = 50.0$ kJ/mol and $k = 0.010$ $s^{-1}$ at 350 K.

1) Calculate $k$ at 400 K. First find $\ln(k_2/k_1)$. (to 3 significant figures)

2) Now find $k_2$. (in $s^{-1}$, to 3 significant figures)

3) By what factor did k increase from 350 K to 400 K? (to 3 significant figures)

Two-Point Arrhenius Concepts 🎯

Two-Point Arrhenius Review 🔍

Exit Quiz — Two-Point Arrhenius ✅

🧬 Catalysts

Part 6 of 7 — Lowering the Energy Barrier

Catalyst Effects on the Energy Diagram

🔑 Why this matters: The AP exam frequently tests what catalysts do and do NOT change — especially the distinction between kinetics (rate) and thermodynamics (equilibrium).

What You'll Master in Part 6

• Explaining how catalysts lower $E_a$ by providing an alternative reaction pathway
• Distinguishing homogeneous, heterogeneous, and biological (enzyme) catalysts
• Understanding that catalysts speed up both forward and reverse reactions equally

🔧 How Catalysts Work

A catalyst provides an alternative reaction pathway with a lower activation energy:

$\boxed{E_a(\text{catalyzed}) < E_a(\text{uncatalyzed})}$

🔑 Key Principle: A catalyst lowers $E_a$ but does NOT change $\Delta H$ or $\Delta G$.

On an Energy Diagram

The catalyzed pathway shows a lower peak (transition state) while the reactants and products remain at the same energy levels:

• $\Delta H$ is unchanged — the catalyst does not affect thermodynamics
• $E_a$ is reduced — more molecules have sufficient energy to react
• $k$ increases — from the Arrhenius equation: lower $E_a$ → larger $e^{-E_a/(RT)}$ → larger $k$

What Catalysts Do NOT Do

📂 Types of Catalysts

1. Homogeneous Catalysts

Same phase as the reactants (typically all in solution).

2. Heterogeneous Catalysts

Different phase from reactants (typically a solid catalyst with gas or liquid reactants).

3. Biological Catalysts (Enzymes)

Proteins that catalyze specific biochemical reactions.

Catalyst Concepts Quiz 🎯

📋 Heterogeneous Catalysis: The Four Steps

When a gaseous reactant reacts on a solid catalyst surface:

Step 1: Adsorption

Reactant molecules bind to the catalyst surface at active sites. Bonds in the reactant may be weakened.

Step 2: Migration / Diffusion

Adsorbed molecules move along the surface to find each other.

Step 3: Reaction

The weakened bonds allow the reaction to proceed with lower $E_a$. New bonds form.

Step 4: Desorption

Product molecules detach from the surface, freeing active sites for new reactant molecules.

⚠️ Catalyst Poisoning

If a substance binds strongly to active sites and cannot be removed, the catalyst is poisoned:

• Lead poisons Pt catalytic converters (why leaded gas is banned)
• CO poisons iron catalysts in the Haber process
• Heavy metals poison enzymes

Catalyst Types and Properties 🔍

Catalyst Effect on Rate 🧮

An uncatalyzed reaction has $E_a = 120$ kJ/mol and $k = 1.0 \times 10^{-10}$ $s^{-1}$ at 300 K.

1) A catalyst lowers $E_a$ to 80 kJ/mol. What is the ratio $k_{\text{cat}}/k_{\text{uncat}}$ at 300 K? Use $\ln(k_{\text{cat}}/k_{\text{uncat}}) = (E_{a,\text{uncat}} - E_{a,\text{cat}})/(RT)$. Calculate this exponent first. (to 3 significant figures)

2) The catalyzed $k$ is approximately how many times larger? Express as a power of 10. (integer)

3) If the catalyzed half-life is $t_{1/2} = 0.693/k_{\text{cat}}$, and $k_{\text{cat}} \approx 1.0 \times 10^{-3}$ $s^{-1}$, what is the half-life? (in seconds, whole number)

Exit Quiz — Catalysts ✅

🎓 Synthesis & AP Review

Part 7 of 7 — Comprehensive Arrhenius and Catalyst Problems

Equations You Must Know

🔑 Why this matters: AP Chemistry free-response questions often combine energy diagrams, Arrhenius calculations, and catalyst effects in a single multi-part problem.

What You'll Master in Part 7

• Solving comprehensive problems that combine collision theory, energy diagrams, and Arrhenius
• Interpreting how catalysts affect energy diagrams and rate constants
• Working through AP-style free-response questions under timed conditions

📋 Key Equations Summary

🧪 Arrhenius Equations

📈 Energy Diagram Relationship

$\Delta H = E_a(\text{forward}) - E_a(\text{reverse})$

📌 Constants & Units

⚠️ AP Trap: Mismatched units between $E_a$ (often given in kJ/mol) and $R$ (in J) is the #1 calculation error.

AP Problem 1: Energy Diagram Analysis 🎯

A reaction energy diagram shows:

• Reactants: 50 kJ
• Transition state (uncatalyzed): 150 kJ
• Transition state (catalyzed): 100 kJ
• Products: 30 kJ

AP Problem 2: Arrhenius Calculation 🧮

The rate constant for the decomposition of $N_{2}O_{5}$ is $k = 3.46 \times 10^{-5}$ $s^{-1}$ at 298 K and $k = 4.87 \times 10^{-3}$ $s^{-1}$ at 338 K.

1) Calculate $E_a$ in kJ/mol. (to 3 significant figures)

2) Calculate the frequency factor $A$. (order of magnitude: enter the exponent, e.g., for $10^{13}$ enter 13)

3) What would $k$ be at 310 K? (in $s^{-1}$, to 1 significant figure in scientific notation, e.g., 2e-4)

AP Problem 3: Catalyst and Arrhenius 🎯

Comprehensive Review 🔍

Challenge: Complete Analysis 🧮

A catalyzed reaction has the following data:

1) Calculate Ea for the catalyzed reaction. (in kJ/mol, to 3 significant figures)

2) The uncatalyzed reaction has Ea = 100 kJ/mol. By how many kJ/mol does the catalyst lower Ea? (to 3 significant figures)

3) At 300 K, what is the ratio k(cat)/k(uncat)? (to 1 significant figure, scientific notation: e.g., 3e5)

Final Exit Quiz — Activation Energy & Arrhenius ✅