Activation energy from Arrhenius plot; effect of catalyst

Overview

The Arrhenius equation connects reaction rate to temperature through activation energy, and its graphical form—the Arrhenius plot—lets us extract Ea experimentally. Catalysts work by lowering this activation energy barrier without changing thermodynamics.

Core Concepts

[!intuition] Why Temperature Affects Reaction Rate

Molecules need a minimum energy (the activation energy) to react. As temperature increases:

• More molecules have kinetic energy ≥ Ea
• Collision frequency increases slightly
• But the fraction of successful collisions increases exponentially (this dominates)

Think of it like a high-jump bar: if you lower the bar (Ea), way more athletes (molecules) can clear it, even if their jumping ability (temperature) only increases a little.

[!definition] Activation Energy (Ea)

The minimum energy required for reactant molecules to overcome the energy barrier and form products. It's the "height" of the transition state above the reactant energy level.

Units: kJ/mol or J/mol

[!definition] The Arrhenius Equation

$k = A e^{-E_a/RT}$

Where:

• $k$ = rate constant (units depend on reaction order)
• $A$ = pre-exponential factor or frequency factor (same units as $k$)
• $E_a$ = activation energy (J/mol)
• $R$ = gas constant = 8.314 J/(mol·K)
• $T$ = absolute temperature (K)

WHY this form? The exponential $e^{-E_a/RT}$ represents the Boltzmann distribution fraction of molecules with energy ≥ Ea. As T increases or Ea decreases, this fraction grows exponentially.

Deriving the Arrhenius Plot

WHAT we want: A straight-line graph to find Ea experimentally.

HOW: Take the natural logarithm of both sides of the Arrhenius equation.

$k = A e^{-E_a/RT}$

Apply $\ln$ to both sides: $\ln k = \ln\left(A e^{-E_a/RT}\right)$

WHY this step? Logarithms convert exponentials to linear forms, which are easier to analyze graphically.

Using logarithm properties: $\ln(xy) = \ln x + \ln y$ and $\ln(e^x) = x$

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

$\ln k = \ln A - \frac{E_a}{RT}$

Rearrange into the form $y = mx + c$:

[!formula] Arrhenius Plot Equation

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

This is the equation of a straight line:

• x-axis: $\frac{1}{T}$ (reciprocal temperature, K⁻¹)
• y-axis: $\ln k$ (natural log of rate constant)
• Slope: $m = -\frac{E_a}{R}$
• y-intercept: $c = \ln A$

Therefore: $E_a = -m \times R = -\text{(slope)} \times 8.314 \text{ J/(mol·K)}$

WHY negative slope? Higher Ea means the rate constant decreases more steeply with decreasing temperature (increasing 1/T), so the line slopes downward.

[!example] Example 1: Calculating Ea from Arrhenius Plot Data

Given: A reaction's rate constant was measured at different temperatures:

Find: Activation energy Ea.

Solution:Step 1**: Create the Arrhenius plot data:

WHY convert? We need 1/T (not T) on the x-axis to get a linear relationship.

Step 2: Plot ln k vs 1/T and find the slope (or calculate using two points).

Using the first and last points: $m = \frac{\ln k_2 - \ln k_1}{(1/T_2) - (1/T_1)} = \frac{-2.083 - (-5.404)}{0.002778 - 0.003333}$

$m = \frac{3.321}{-0.000555} = -5984 \text{ K}$

WHY this formula? It's the slope formula: rise over run.

Step 3: Calculate Ea: $E_a = -m \times R = -(-5984) \times 8.314 = 49,748 \text{ J/mol} = 49.7 \text{ kJ/mol}$

Answer: The activation energy is approximately 49.7 kJ/mol.

[!example] Example 2: Comparing Two Reactions

Given: Two reactions A and B have the same pre-exponential factor. At 298 K:

• Reaction A: $E_a = 50$ kJ/mol, $k_A = 2.0 \times 10^{-3}$ s⁻¹
• Reaction B: $E_a = 75$ kJ/mol

Find: Rate constant for reaction B at 298 K.

Solution:

Step 1: Since A is the same, we can use the ratio: $\frac{k_B}{k_A} = \frac{A e^{-E_{a,B}/RT}}{A e^{-E_{a,A}/RT}} = e^{-(E_{a,B} - E_{a,A})/RT}$

WHY? The A terms cancel, leaving only the exponential difference.

Step 2: Convert Ea to J/mol and calculate: $E_{a,B} - E_{a,A} = 75,000 - 50,000 = 25,000 \text{ J/mol}$

$\frac{k_B}{k_A} = e^{-25,000/(8.314 \times 298)} = e^{-10.09} = 4.15 \times 10^{-5}$

WHY exponential? Small differences in Ea lead to huge differences in rate because of the exponential relationship.

Step 3: Calculate kB: $k_B = k_A \times 4.15 \times 10^{-5} = 2.0 \times 10^{-3} \times 4.15 \times 10^{-5}$

$k_B = 8.3 \times 10^{-8} \text{ s}^{-1}$

Answer: Reaction B is ~24,000 times slower due to its higher activation energy.

Effect of Catalyst

[!definition] Catalyst

A catalyst is a substance that increases the rate of a chemical reaction without being consumed in the process. It achieves this by providing an alternative reaction pathway with a lower activation energy.

Key principle: Catalysts do NOT change:

• ΔH (enthalpy change)
• ΔG (Gibbs free energy)
• Equilibrium position (Keq)

They only change how fast equilibrium is reached.

[!intuition] How Catalysts Work

A catalyst provides "stepping stones" across the energy barrier:

• Uncatalyzed: Reactants → [high energy transition state] → Products
• Catalyzed: Reactants → [intermediate1] → [intermediate 2] → Products

Each intermediate step has a lower peak than the original transition state, so more molecules can cross at a given temperature.

[!formula] Arrhenius Equation with Catalyst

Uncatalyzed: $k_{uncat} = A e^{-E_{a,uncat}/RT}$

Catalyzed: $k_{cat} = A' e^{-E_{a,cat}/RT}$

Where $E_{a,cat} < E_{a,uncat}$ (this is the key!)

Rate enhancement factor: $\frac{k_{cat}}{k_{uncat}} = \frac{A'}{A} \cdot e^{-(E_{a,cat} - E_{a,uncat})/RT}$

If we assume $A' \approx A$ (reasonable for many reactions): $\frac{k_{cat}}{k_{uncat}} \approx e^{-\Delta E_a/RT}$

where $\Delta E_a = E_{a,cat} - E_{a,uncat} < 0$

WHY exponential? Even a modest decrease in Ea (say, 20 kJ/mol) can increase the rate by factors of 1000+ at room temperature.

[!example] Example 3: Catalyst Effect on Rate

Given: A reaction has Ea = 85 kJ/mol at 300 K. A catalyst reduces Ea to 60 kJ/mol.

Find: By what factor does the rate increase?

Solution:

$\Delta E_a = 60 - 85 = -25 \text{ kJ/mol} = -25,000 \text{ J/mol}$

WHY negative? The catalyzed Ea is lower.

$\frac{k_{cat}}{k_{uncat}} = e^{-\Delta E_a/RT} = e^{-(-25,000)/(8.314 \times 300)}$

$= e^{25,000/2494} = e^{10.02} = 22,400$

Answer: The catalyst increases the rate by a factor of ~22,400.

Interpretation: At 300 K, a reaction that took 1 hour uncatalyzed would take only 0.16 seconds with the catalyst!

On an Arrhenius Plot: Catalyst Effect

When we plot ln k vs 1/T for both catalyzed and uncatalyzed reactions:

1. Both lines are straight (both follow Arrhenius behavior)
2. The catalyzed line has a smaller (less negative) slope because $E_{a,cat} < E_{a,uncat}$
3. The catalyzed line is shifted upward (higher ln k at every temperature)
4. The lines are roughly parallel IF the reduction in Ea is constant across temperatures

Visual: The diagram shows two lines diverging as 1/T increases (T decreases), with the catalyzed reaction maintaining higher rates even at lower temperatures.

[!mistake] Common Mistake 1: "Catalysts shift equilibrium"

Wrong thinking: "Adding a catalyst will produce more products because the rate increases."

Why it feels right: We see more product forming faster, so it seems like there's "more" product.

The fix: Catalysts speed up both forward and reverse reactions equally. They help reach equilibrium faster, but don't change where that equilibrium sits. Keq depends only on ΔG, which catalysts don't alter.

Steel-man the mistake: The confusion arises because in practice, we often do get more product when we add a catalyst to an industrial reaction—but that's because we can remove products continuously before reverse reaction kicks in. In a closed system, same equilibrium amount, just faster.

[!mistake] Common Mistake 2: Confusing slope sign

Wrong thinking: "Higher Ea means steper upward slope on the Arrhenius plot."

Why it feels right: "High" and "steep" feel like they should go together.

The fix: The slope is $-E_a/R$ (negative!). Higher Ea means more negative slope, so the line tilts downward more steeply. Think of it this way: high Ea reactions are more "temperature sensitive"—their rate drops off faster as you cool down (move right on the 1/T axis).

[!mistake] Common Mistake 3: Using T instead of 1/T

Wrong thinking: Plotting ln k vs T gives a straight line.

Why it feels right: Temperature is the independent variable, so it should be on the x-axis.

The fix: The Arrhenius equation is exponential in T, but linear in 1/T. You must plot against reciprocal temperature to get a straight line. Check your axis label!

[!recall]- Feynman Technique: Explain to a 12-year-old

Imagine you're trying to roll a ball over a hill to get it into a valley on the other side. The height of the hill is the "activation energy"—how much push you need to get the ball over the top.

If the hill is really tall (high Ea), only a few super-fast balls (hot molecules) have enough speed to make it over. Most balls roll back down. But if you heat things up (increase temperature), more balls move faster, so more make it over the hill.

Now, an Arrhenius plot is like a special graph where we plot how many balls make it over at different temperatures, but we use a trick: instead of plotting temperature directly, we plot "1/temperature." When we do that, we get a straight line! The steper the line tilts down, the taller the hill (bigger Ea).

A catalyst is like building a tunnel through the hill instead of going over the top. Now the path is way shorter (lower Ea), so even slow balls (cold molecules) can get through. The ball still ends up in the same valley (same products, same equilibrium), but way more balls get through per second (faster rate). The catalyst doesn't get used up—it's like the tunnel stays there for every ball.

[!mnemonic] Remembering Arrhenius Plot Features

"Elephants Are Really Reliable"

• Ea from slope: $E_a = -\text{slope} \times R$
• Arrhenius plot: ln k vs 1/T
• Reciprocal temperature on x-axis (not T!)
• Reduced slope (less steep) means catalyst present

Connections

• Arrhenius Equation - the fundamental equation this plot visualizes
• Collision Theory - explains why activation energy exists
• Transition State Theory - deeper model of activation complex
• Rate Constant Temperature Dependence - quantitative relationship
• Catalysis Mechanisms - homogeneous vs heterogeneous catalysts
• Enzyme Kinetics - biological catalysts follow similar principles
• Maxwell-Boltzmann Distribution - fraction of molecules with E ≥ Ea
• Reaction Energy Diagrams - visual representation of Ea
• Equilibrium vs Kinetics - catalysts affect only kinetics
• Industrial Catalysis - practical applications (Haber, Contact process)

Flashcards

What is an Arrhenius plot? :: A graph of ln k (y-axis) vs 1/T (x-axis) that gives a straight line with slope = -Ea/R, allowing experimental determination of activation energy.

What are the units of the slope of an Arrhenius plot? :: Kelvin (K), since slope = -Ea/R and Ea is in J/mol, R is in J/(mol·K), so their ratio has units of K.

Concept Map

Hinglish (regional understanding)