🎯⭐ INTERACTIVE LESSON

Activation Energy and Temperature Effects

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Activation Energy and Temperature Effects - Complete Interactive Lesson

Part 1: Collision Theory

💥 Collision Theory

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

Two Requirements for an Effective Collision

Requirement	What It Means	If Missing
Sufficient energy	KE ≥ $E_a$	Molecules bounce off — no reaction
Proper orientation	Reactive sites align	Collision is wasted

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 ( $E_a$ ):

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

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:

The peak shifts to higher energy and becomes lower and broader
The fraction of molecules with $KE \geq E_a$ increases

Maxwell-Boltzmann Distribution 🔍

⏱️ Collision Theory Rate Equation

Putting it all together, collision theory predicts:

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

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 . ( J/(mol·K); round to 3 significant figures)

Exit Quiz — Collision Theory ✅

Part 2: Activation Energy

⛰️ Activation Energy & Energy Diagrams

Part 2 of 7 — The Energy Landscape of Reactions

Reading an Energy Diagram

Feature	What It Represents	How to Find It
Y-axis height of reactants	Potential energy of reactants	Starting level
Y-axis height of products	Potential energy of products	Ending level
Peak height	Transition state energy	Highest point on curve
$E_a$ (forward)	Activation energy	Peak − Reactants

Part 3: Energy Diagrams

📐 The Arrhenius Equation

Part 3 of 7 — Connecting Rate Constants to Temperature

The Arrhenius Equation

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

Symbol	Meaning	Units

Part 4: Arrhenius Equation

📉 Linearized Arrhenius Equation

Part 4 of 7 — Finding Ea from Graphical Data

From Exponential to Linear

Form	Equation	Graph
Exponential	$k = Ae^{-E_a/RT}$

Part 5: Catalysts & Catalysis

🔄 Two-Point Arrhenius Equation

Part 5 of 7 — Finding Ea from Two Temperatures

When You Have Just Two Data Points

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

Part 6: Problem-Solving Workshop

🧬 Catalysts

Part 6 of 7 — Lowering the Energy Barrier

Catalyst Effects on the Energy Diagram

Feature	Without Catalyst	With Catalyst
$E_a$	Higher	Lower (new pathway)
$\Delta H$	Unchanged	Unchanged
(equilibrium constant)

Part 7: Synthesis & AP Review

🎓 Synthesis & AP Review

Part 7 of 7 — Comprehensive Arrhenius and Catalyst Problems

Equations You Must Know

Equation	When to Use
$k = Ae^{-E_a/RT}$

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 + NO₃ → 2NO₂

✅ O of NO hits O of NO₃ → bonds can rearrange → reaction!
❌ N of NO hits N of NO₃ → wrong atoms in contact → no reaction

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.

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.

$Z$ = 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.

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)

🔑 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

Feature	Description
Reactants	Starting energy level (left side)
Products	Final energy level (right side)
Transition state (activated complex)	The peak — highest energy point
$E_a$ (forward)	Energy from reactants to transition state
$E_a$ (reverse)	Energy from products to transition state
$\Delta H$	Energy difference between products and reactants

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 NH₄NO₃

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

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

Feature	Transition State	Intermediate
Energy	Maximum (peak)	Minimum (valley between peaks)
Stability	Unstable	Somewhat stable
Lifetime	~10⁻¹³ s	Can sometimes be detected
On diagram	Top of a hill	Bottom of a valley

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 ✅

🔑 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

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

Symbol	Name	Units
$k$	Rate constant	Depends on order
$A$	Frequency factor (pre-exponential factor)	Same as $k$
$E_a$

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

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 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

Reaction Type	Typical $A$	Why
Gas-phase, simple molecules	$10^{10}$ – $10^{14}$ s⁻¹	High collision frequency
Solution-phase	$10^{8}$ – s⁻¹

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 ✅

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 $y = mx + b$ !

Variable	Corresponds To
$y$	$\ln k$
$x$	$1/T$
$m$ (slope)

Key Result

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

Slope $= -E_a/R$ → $E_a = -R \times \text{slope}$

Linearized Arrhenius Quiz 🎯

🧪 Worked Example: Determining Ea from Data

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

$T$ (K)	$k$ (s⁻¹)	$1/T$ (K⁻¹)	$\ln k$
300	1.0 × 10⁻⁷	3.33 × 10⁻³	−16.12
350	3.0 × 10⁻⁵	2.86 × 10⁻³	−10.41
400	1.5 × 10⁻³	2.50 × 10⁻³	−6.50
450	2.0 × 10⁻²	2.22 × 10⁻³	−3.91

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

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

Arrhenius Plot Analysis 🔍

Exit Quiz — Linearized Arrhenius ✅

Given	Can Solve For
$k_1$ , $k_2$ , $T_1$ , $T_2$	$E_a$
$k_1$ , $E_a$ , $T_1$ ,
$k_1$ , $k_2$ , $E_a$ ,

🔑 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 at :

🧪 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 ✅

🔑 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 → larger

What Catalysts Do NOT Do

❌ Catalysts do NOT...	✅ Catalysts DO...
Change $\Delta H$ or $\Delta G$	Lower $E_a$
Shift equilibrium	Speed up both forward and reverse equally
Get consumed (overall)	Participate in mechanism, then regenerate

📂 Types of Catalysts

1. Homogeneous Catalysts

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

Feature	Detail
Phase	Same as reactants
Example	H⁺ catalyzing ester hydrolysis
Advantage	Better mixing, uniform activity
Disadvantage	Hard to separate from products

2. Heterogeneous Catalysts

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

Feature	Detail
Phase	Different from reactants
Example	Pt surface in catalytic converters
Mechanism	Adsorption → reaction → desorption
Advantage	Easy to separate, reusable
Disadvantage	Can be poisoned (blocked)

3. Biological Catalysts (Enzymes)

Proteins that catalyze specific biochemical reactions.

Feature	Detail
Specificity	Very high — lock-and-key or induced fit
Conditions	Mild (body temperature, neutral pH)
Rate increase	$10^6$ to $10^{12}$ times faster
Sensitivity	Can be denatured by heat, pH extremes

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 . 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 ✅

🔑 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

Form	Equation	Use Case
Standard	$k = Ae^{-E_a/(RT)}$	Relates rate constant to temperature
Linearized	$\ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A$
Two-Point	$\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$

📈 Energy Diagram Relationship

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

📌 Constants & Units

Constant	Value	Watch Out
$R$	8.314 J/(mol·K)	Use J, not kJ
$E_a$	J/mol	Convert from kJ/mol if needed (× 1000)
$T$	Kelvin	Convert from °C:

⚠️ 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 N₂O₅ 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 10¹³ 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:

T (K)	k (M⁻¹s⁻¹)
300	0.050
350	0.85

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 ✅

Products (lower energy)

Products (higher energy)

E_a \uparrow

E_a/(RT) \uparrow

, so

e^{-E_a/(RT)} \downarrow