Activation Energy Calculator — Arrhenius Equation & Ea
Calculate rate constant k using the Arrhenius equation k = Ae−Ea/RT, find activation energy Ea from two rate constants using the two-point method, perform linear regression on Arrhenius plot data (ln k vs 1/T), and visualize the potential energy diagram — with full step-by-step working throughout.
Arrhenius Equation Result
| T (K) | T (°C) | k | k/k₀ ratio |
|---|
📍 Condition 1 (known)
📍 Condition 2
Two-Point Arrhenius Result
| # | Temperature (K or °C) | Rate constant k |
|---|
Arrhenius Plot — Linear Regression Results
| T (K) | T (°C) | 1/T (K⁻¹) | k | ln(k) | ln(k) predicted |
|---|
The potential energy diagram shows the activation energy for both the forward reaction (reactants → transition state) and the reverse reaction (products → transition state). The relationship Ea(rev) = Ea(fwd) − ΔH is shown geometrically — the transition state is the shared peak for both directions. For an exothermic reaction (ΔH < 0), the reverse activation energy is larger than the forward activation energy because the products sit at a lower energy level and must climb higher to reach the same transition state peak.
Ea(rev) = Ea(fwd) − ΔH = 125 − (−85) = 210 kJ/mol
Exothermic reaction: products are lower in energy than reactants → Ea(rev) > Ea(fwd)
| Reaction | Ea (kJ/mol) | Ea (kcal/mol) | Type / Notes |
|---|---|---|---|
| H₂ + I₂ → 2HI | 165 | 39.4 | Elementary gas-phase |
| 2HI → H₂ + I₂ (reverse) | 184 | 44.0 | Reverse of above; Ea_rev = 165−(−19) = 184 |
| N₂O₅ decomposition | 103 | 24.6 | First-order; A = 4.9×10¹³ s⁻¹ |
| CH₃CHO → CH₄ + CO | 190 | 45.4 | Pyrolysis (thermal decomposition) |
| Sucrose hydrolysis (enzyme) | 46 | 11.0 | Enzyme-catalyzed (invertase) |
| Sucrose hydrolysis (H⁺) | 107 | 25.6 | Acid-catalyzed; enzyme lowers Ea by 61 kJ/mol |
| Diamond → Graphite | 728 | 174.0 | Extremely slow at room temperature |
| Egg protein denaturation | 418 | 99.9 | Cooking chemistry |
| Typical enzymatic reaction | 25–65 | 6–16 | Biological catalysis |
| Typical uncatalyzed reaction | 40–150 | 10–36 | Standard chemistry range |
| Diffusion-controlled reaction | 8–20 | 2–5 | Near diffusion limit; no energy barrier |
| Form | Equation | Used For |
|---|---|---|
| Standard | k = A × e^(−Ea/RT) | Calculate k at given T, A, Ea |
| Logarithmic | ln(k) = −Ea/R × (1/T) + ln(A) | Arrhenius plot graphing — linear form |
| Two-point | ln(k₂/k₁) = −Ea/R × (1/T₂−1/T₁) | Find Ea from two (k, T) measurements |
| Slope form | Slope = −Ea/R | Extract Ea from Arrhenius plot slope |
| Intercept form | Intercept = ln(A) | Extract A from Arrhenius plot y-intercept |
| Solve for Ea | Ea = −R × T × ln(k/A) | Given k, A, T — find Ea |
| Solve for T | T = −Ea / (R × ln(k/A)) | Given k, A, Ea — find temperature |
| Solve for A | A = k × e^(+Ea/RT) | Given k, Ea, T — find frequency factor |
| Reverse Ea | Ea_rev = Ea_fwd − ΔH_rxn | From energy diagram; ΔH can be negative |
⚠ The R value MUST match the unit of Ea. Using R = 8.314 J/(mol·K) with Ea in kJ/mol gives answers 1000× wrong.
| Ea Unit | R Value | R Units | Example |
|---|---|---|---|
| J/mol (SI) | 8.31446 | J/(mol·K) | −Ea/RT = −125000/(8.314×298) |
| kJ/mol (common) | 8.31446×10⁻³ | kJ/(mol·K) | −Ea/RT = −125/(0.008314×298) |
| cal/mol | 1.987 | cal/(mol·K) | −Ea/RT = −Ea_cal/(1.987×T) |
| kcal/mol | 1.987×10⁻³ | kcal/(mol·K) | Rare — check unit carefully |
| eV/molecule | 8.617×10⁻⁵ | eV/(molecule·K) | k_B (Boltzmann constant) |
| Reaction Type | Typical A | Units | Notes |
|---|---|---|---|
| Unimolecular gas-phase | 10¹² – 10¹³ | s⁻¹ | First-order; vibrational frequency scale |
| Bimolecular gas-phase | 10⁸ – 10¹¹ | M⁻¹s⁻¹ (dm³mol⁻¹s⁻¹) | Second-order; collision frequency × steric factor |
| Solution phase | 10⁷ – 10¹² | M⁻¹s⁻¹ | Solvent cage effects reduce A |
| Enzyme-catalyzed | 10⁶ – 10¹⁰ | s⁻¹ or M⁻¹s⁻¹ | Active site geometry controls steric factor |
| Diffusion-controlled | ~10¹⁰ | M⁻¹s⁻¹ | Rate limited by diffusion; Ea ≈ 8–20 kJ/mol |
A catalyst lowers activation energy Ea without changing reactants or products. Since k = Ae−Ea/RT, reducing Ea increases k exponentially. Enzymes are biological catalysts that can reduce Ea from ~100 kJ/mol to ~25 kJ/mol — increasing the reaction rate by factors up to 10¹³.
Activation Energy Calculator — Arrhenius Equation k = Ae−Ea/RT
This activation energy calculator solves the Arrhenius equation k = Ae−Ea/RT for any unknown variable — rate constant k, frequency factor A, activation energy Ea, or temperature T — with complete step-by-step working. It also finds activation energy Ea from two rate constants at two temperatures using the two-point Arrhenius method, performs linear regression on Arrhenius plot data (ln k vs 1/T) to extract both Ea and the frequency factor A with an R² goodness-of-fit value, and draws the potential energy diagram showing Ea for both forward and reverse reactions.
The Arrhenius equation calculator covers every form of the equation used in chemical kinetics: the standard exponential form k = Ae−Ea/RT, the logarithmic linearized form ln(k) = −(Ea/R)×(1/T) + ln(A) used for Arrhenius plot graphing, and the two-point form ln(k₂/k₁) = −(Ea/R)×(1/T₂ − 1/T₁) used when only two experimental measurements are available. All calculations include automatic unit conversion, the critical Kelvin temperature warning, and the correct R value for each energy unit.
The Arrhenius Equation — k = Ae−Ea/RT
The Arrhenius equation k = Ae−Ea/RT is the fundamental law of chemical kinetics relating the rate constant k to temperature T and activation energy Ea. Proposed by Svante Arrhenius in 1889, the Arrhenius equation — sometimes written as the Arrhenius formula, Arrhenius law, or simply k = Ae−Ea/RT — quantifies why reaction rates increase so dramatically with temperature and why different reactions have such vastly different sensitivities to temperature.
Variables in the Arrhenius Equation k = Ae−Ea/RT
| Symbol | Quantity | Units | Notes |
|---|---|---|---|
| k | Rate constant | s⁻¹ (1st order), M⁻¹s⁻¹ (2nd order) | What we measure experimentally |
| A | Frequency factor (pre-exponential) | Same units as k | Collision frequency × steric factor |
| Ea | Activation energy | J/mol (SI) or kJ/mol | Always positive for elementary reactions |
| R | Universal gas constant | 8.31446 J/(mol·K) | Must match units of Ea |
| T | Absolute temperature | Kelvin (K) — ALWAYS | T(K) = T(°C) + 273.15 |
| e−Ea/RT | Boltzmann factor | Dimensionless (0 to 1) | Fraction of molecules with energy ≥ Ea |
Physical Meaning of the Arrhenius Equation
The Boltzmann factor e−Ea/RT in the Arrhenius equation k = Ae−Ea/RT is the fraction of molecular collisions that have enough kinetic energy to overcome the activation energy barrier Ea. When temperature T increases, the exponent −Ea/RT becomes less negative, the Boltzmann factor e−Ea/RT increases exponentially, and the rate constant k increases dramatically. This exponential temperature dependence — captured by the Arrhenius equation — explains why a 10°C increase near room temperature can double a reaction rate, and why even small changes in activation energy Ea produce enormous changes in k.
The Arrhenius equation in chemistry: k = Ae−Ea/RT — using Celsius instead of Kelvin in the Arrhenius equation is the single most common calculation error. At 25°C (298 K), using T = 25 instead of T = 298 makes the exponent 11.9× too large in magnitude, giving a rate constant billions of times too small. Always convert: T(K) = T(°C) + 273.15.
Units in the Arrhenius Equation
The Arrhenius equation units require careful attention. The activation energy Ea can be expressed in J/mol, kJ/mol, cal/mol, kcal/mol, or eV/molecule — but the gas constant R must always use matching units. The most common Arrhenius equation unit error is using R = 8.314 J/(mol·K) with Ea in kJ/mol, which gives an exponent 1000× too large. The Arrhenius equation constant R values are: 8.314 J/(mol·K) for Ea in J/mol; 8.314×10⁻³ kJ/(mol·K) for Ea in kJ/mol; 1.987 cal/(mol·K) for Ea in cal/mol.
How to Find Activation Energy from the Arrhenius Equation
The activation energy formula can be rearranged to find Ea by four different methods depending on what experimental data is available. Knowing how to find activation energy is a core skill in chemical kinetics — every method ultimately uses the Arrhenius equation k = Ae−Ea/RT in some form.
Method 1 — Two-Point Method (most common): Find Ea from k₁ and k₂ at T₁ and T₂
This is how to solve for activation energy when you have measured k at two different temperatures but do not know A. The two-point form of the Arrhenius equation eliminates A entirely and gives Ea directly from the ratio k₂/k₁ and the temperature difference. Temperatures must be in Kelvin.
Method 2 — Arrhenius Plot: Ea from slope = −Ea/R
Plot ln(k) on the y-axis and 1/T (in K⁻¹) on the x-axis. The slope of the best-fit line equals −Ea/R, so Ea = −slope × R. How to calculate activation energy from a graph: measure the slope (rise/run in ln(k) per K⁻¹), multiply by −R = −8.314 J/(mol·K). The slope is always negative since Ea is positive.
Method 3 — Direct Calculation: Given A, k, and T
Method 4 — Energy Diagram: Ea from transition state energy
From the potential energy diagram, the activation energy Ea equals the energy difference between the reactants and the activated complex (transition state) at the peak: Ea(fwd) = E(transition state) − E(reactants). How to determine activation energy from a graph of potential energy: find the peak (transition state) and subtract the reactant energy level.
Worked Example 1 — How to Calculate Ea: Given k = 1.5×10⁻³ s⁻¹, A = 1×10¹², T = 298 K
- Write the activation energy formula: Ea = −R × T × ln(k/A)
- Calculate ln(k/A) = ln(1.5×10⁻³ / 1×10¹²) = ln(1.5×10⁻¹⁵) = −33.44
- Ea = −(8.314)(298)(−33.44) = −(8.314)(298)(−33.44)
- Ea = 8.314 × 298 × 33.44 = 82,840 J/mol = 82.84 kJ/mol
- Verification: k = 1×10¹² × e^(−82840/(8.314×298)) = 1×10¹² × e^(−33.44) = 1.5×10⁻³ ✓
Worked Example 2 — How to Find Activation Energy: Two-Point Method
- Given: k₁ = 2.15×10⁻⁸ at T₁ = 600 K; k₂ = 2.39×10⁻⁷ at T₂ = 700 K
- ln(k₂/k₁) = ln(2.39×10⁻⁷ / 2.15×10⁻⁸) = ln(11.12) = 2.408
- 1/T₁ − 1/T₂ = 1/600 − 1/700 = 1.667×10⁻³ − 1.429×10⁻³ = 2.381×10⁻⁴ K⁻¹
- Ea = R × 2.408 / 2.381×10⁻⁴ = 8.314 × 2.408 / 2.381×10⁻⁴
- Ea = 84,090 J/mol = 84.09 kJ/mol
Worked Example 3 — How to Find Activation Energy from Graph (Arrhenius Plot)
- Measure rate constant k at four temperatures: (300 K, 1.5×10⁻⁵), (350 K, 8.2×10⁻⁴), (400 K, 2.7×10⁻²), (450 K, 5.8×10⁻¹)
- Calculate 1/T and ln(k) for each point
- Plot ln(k) vs 1/T and draw best-fit line
- Read slope from graph: slope = −12,890 K
- Ea = −slope × R = −(−12,890) × 8.314 = 107,200 J/mol = 107.2 kJ/mol
- A = e^intercept (read y-intercept from graph)
The Arrhenius Plot — ln(k) vs 1/T
The Arrhenius plot is a graph of ln(k) on the y-axis versus 1/T (in K⁻¹) on the x-axis, based on the linearized Arrhenius equation: ln(k) = −(Ea/R)×(1/T) + ln(A). This linear transformation of the Arrhenius equation converts the exponential k = Ae−Ea/RT into a straight line — the Arrhenius plot equation — making it easy to extract both activation energy Ea and frequency factor A from experimental data by simple linear regression.
How to Make an Arrhenius Plot — Step by Step
- Measure the rate constant k at a minimum of 4–6 different temperatures (more data points improve accuracy).
- Convert each temperature to Kelvin: T(K) = T(°C) + 273.15. Never use Celsius in the Arrhenius plot.
- Calculate 1/T for each temperature — these are the x-values (in K⁻¹, typically ranging from 10⁻³ to 10⁻² K⁻¹).
- Calculate ln(k) for each rate constant — these are the y-values (dimensionless, usually negative).
- Plot the (1/T, ln k) pairs and draw the best-fit straight line using linear regression.
- Read the slope: slope = −Ea/R (always negative, since Ea > 0).
- Calculate Ea = −slope × R = −slope × 8.314 J/(mol·K). The activation energy on the graph is found from this slope.
- Read the y-intercept b = ln(A), then A = e^b.
Where Is Activation Energy on the Arrhenius Plot?
On the lnk vs 1/T graph, activation energy Ea is encoded in the slope: slope = −Ea/R, so the steeper (more negative) the slope, the higher the activation energy. A reaction with Ea = 100 kJ/mol will have a steeper Arrhenius plot slope than a reaction with Ea = 50 kJ/mol. The y-intercept (where the line crosses the ln(k) axis at 1/T = 0, i.e., T → ∞) equals ln(A) — though this is a mathematical extrapolation far beyond experimental temperatures.
Non-Linear Arrhenius Plots
A reaction follows the Arrhenius equation if its Arrhenius plot (ln k vs 1/T) is linear (R² ≥ 0.99). A non-linear Arrhenius plot indicates: complex multi-step mechanisms where the rate-determining step changes with temperature; tunneling effects (common in hydrogen transfer reactions at low temperature); temperature-dependent pre-exponential factor A; or multiple competing pathways. Non-Arrhenius behavior does not mean the Arrhenius equation is wrong — it means the reaction mechanism is more complex than a single elementary step.
Frequency Factor A — The Pre-Exponential Factor
The frequency factor A (also called the Arrhenius pre-exponential factor, Arrhenius prefactor, or Arrhenius frequency factor) in the equation k = Ae−Ea/RT represents the maximum possible rate constant — the rate if every collision had enough energy to react and all collisions had the correct geometry. In practice, A combines two factors: the collision frequency Z (how often molecules meet) and the steric factor p (the probability that a collision has the correct orientation for reaction). Thus A = p×Z.
Frequency Factor A Units
The frequency factor units are identical to the units of k because A = k when e−Ea/RT = 1 (which only occurs at T → ∞). The Arrhenius equation pre-exponential factor A is therefore: s⁻¹ for first-order reactions; M⁻¹s⁻¹ (or dm³mol⁻¹s⁻¹) for second-order reactions; M⁻²s⁻¹ for third-order reactions. The Arrhenius equation exponential factor e−Ea/RT is always dimensionless, so A must carry all the units.
Typical Values of the Frequency Factor A
| Reaction Type | Typical A | Units |
|---|---|---|
| Unimolecular gas-phase | 10¹² – 10¹³ | s⁻¹ |
| Bimolecular gas-phase | 10⁸ – 10¹¹ | M⁻¹s⁻¹ |
| Solution-phase reactions | 10⁷ – 10¹² | M⁻¹s⁻¹ |
| Enzyme-catalyzed | 10⁶ – 10¹⁰ | s⁻¹ or M⁻¹s⁻¹ |
How to find A in the Arrhenius equation: (1) from the y-intercept of the Arrhenius plot: A = e^(intercept); (2) directly from k = Ae−Ea/RT rearranged: A = k × e+Ea/RT; (3) from collision theory: A = p × Z × N_A. The Arrhenius constant A value is not truly constant — it has a weak temperature dependence (proportional to √T in collision theory), but this T-dependence is negligible compared to the exponential Boltzmann factor e−Ea/RT over normal experimental temperature ranges.
Potential Energy Diagram — Ea Forward and Reverse
The potential energy diagram (also called the activation energy graph or energy profile) is the most important visual in understanding how the Arrhenius equation k = Ae−Ea/RT connects to molecular structure. The diagram shows: reactants at their initial energy level; the activated complex (transition state) at the peak energy — the highest point on the reaction coordinate; and products at their final energy level.
Reading the Activation Energy Drawing
- Ea(forward) = energy of transition state − energy of reactants (always positive, shown as the red upward arrow from reactants to peak)
- Ea(reverse) = energy of transition state − energy of products (always positive, shown as the blue upward arrow from products to peak)
- ΔH_rxn = energy of products − energy of reactants (negative for exothermic, positive for endothermic; shown as the green or amber arrow between reactant and product levels)
Activation Energy of Endothermic vs Exothermic Reactions
For an exothermic reaction (ΔH < 0, products lower than reactants): Ea(rev) = Ea(fwd) − ΔH = Ea(fwd) + |ΔH| > Ea(fwd). The activation energy for the reverse reaction is larger — the reverse arrow on the potential energy diagram is taller, because products start from a lower energy level and must climb higher to reach the same transition state peak.
For an endothermic reaction (ΔH > 0, products higher than reactants): Ea(rev) = Ea(fwd) − ΔH < Ea(fwd). The reverse activation energy is smaller — consistent with the reverse (exothermic) reaction being faster. The letter representing the activation energy for the reverse reaction on a standard energy diagram is always the arrow from the product level up to the transition state peak.
Example — Activation Energy of Forward and Reverse Reaction
- Given: Ea(fwd) = 125 kJ/mol, ΔH = −85 kJ/mol (exothermic)
- Ea(rev) = Ea(fwd) − ΔH = 125 − (−85) = 125 + 85 = 210 kJ/mol
- Interpretation: The reverse reaction has a higher activation energy (210 kJ/mol) than the forward reaction (125 kJ/mol) because the reaction is exothermic — the products are at lower energy and must climb further to reach the transition state.
- Use the Reference tab above to visualize this with the interactive potential energy diagram.
Activated Complex Definition in Chemistry
The activated complex (also called the transition state) is the highest-energy, unstable arrangement of atoms at the peak of the potential energy diagram — the point at which old bonds are partially broken and new bonds are partially formed. The activated complex exists for an extremely short time (~10⁻¹³ seconds) and cannot be isolated. The energy of the activated complex above the reactant energy level defines the activation energy Ea in the Arrhenius equation k = Ae−Ea/RT. The free energy of activation ΔG‡ is related to Ea through the Eyring equation (transition state theory): k = (k_B×T/h)×e−ΔG‡/RT.
Catalysts and Activation Energy — How Enzymes Work
A catalyst lowers activation energy by providing an alternative reaction pathway with a lower energy barrier — the same reactants produce the same products, but the transition state is different and lower in energy. The catalyst is not consumed in the reaction. Because the Arrhenius equation k = Ae−Ea/RT contains Ea in the exponent, even a modest reduction in Ea produces an enormous increase in the rate constant k. What speeds up chemical reactions by lowering activation energy? Catalysts — both chemical catalysts and biological catalysts (enzymes).
Quantifying the Catalyst Effect on the Arrhenius Equation
If a catalyst reduces Ea by ΔEa at temperature T, the ratio of catalyzed to uncatalyzed rate constants is:
At 298 K (25°C), reducing Ea by 10 kJ/mol increases k by: e^(10000/(8.314×298)) = e^(4.03) ≈ 56×. Reducing Ea by 50 kJ/mol increases k by: e^(50000/(8.314×298)) = e^(20.2) ≈ 6×10⁸×. Reducing Ea by 75 kJ/mol (as enzymes do relative to uncatalyzed reactions): e^(75000/2478) ≈ 10¹³×.
Activation Energy in Enzymes — How Enzymes Work
Enzymes are biological catalysts that lower activation energy through multiple mechanisms: binding reactants in the precise geometry needed for reaction (reducing the steric requirement captured by A); stabilizing the transition state through hydrogen bonding and electrostatic interactions; providing an alternative reaction mechanism with lower-energy intermediate steps. The activation energy in enzymes is typically 25–65 kJ/mol, compared to 80–150 kJ/mol for uncatalyzed versions of the same reaction. This difference — acting through k = Ae−Ea/RT — is why enzymes can increase reaction rates by 10⁶ to 10¹⁷ times.
Does Temperature Affect Activation Energy?
Temperature does not change the activation energy Ea itself — Ea is a property of the reaction mechanism and the potential energy surface, determined by bond strengths and molecular geometry. What temperature changes is the Boltzmann factor e−Ea/RT — the fraction of molecules with enough energy to overcome Ea. So increasing temperature does not lower the barrier; it increases the fraction of molecules with enough energy to clear it. Catalysts lower the barrier (Ea); temperature increases the fraction clearing it.
Can Activation Energy Be Negative?
For elementary reactions — reactions occurring in a single step via a single transition state — activation energy cannot be negative. Ea ≥ 0 always for elementary steps. A negative Ea for an elementary reaction would mean the reaction gets slower as temperature increases, which contradicts the physical meaning of an energy barrier: there cannot be an energy barrier below the reactant energy level.
Apparent Negative Activation Energy in Complex Reactions
However, some complex reactions show an apparent negative activation energy — the observed rate constant k decreases as temperature increases, making the slope of the Arrhenius plot positive. This apparent (not true) negative Ea occurs through two main mechanisms:
- Pre-equilibrium mechanism: If a fast reversible step precedes a slow rate-determining step, the observed k = K_eq × k_rds. If K_eq decreases with temperature (exothermic pre-equilibrium) faster than k_rds increases, then k decreases with T. The apparent Ea = Ea(rds) + ΔH(pre-equilibrium) can be negative if ΔH is sufficiently negative.
- Termolecular association: Some three-body reactions and radical recombinations have near-zero or negative apparent Ea because the rate-limiting step is not bond breaking but stabilization of a loose complex — the complex dissociates faster at higher temperatures.
Is activation energy always positive? For elementary reaction steps: yes, always. For overall complex reactions: apparent Ea can be negative due to pre-equilibrium effects or changing mechanisms. If your Arrhenius calculation gives a negative Ea for what should be an elementary reaction, check that k₂ > k₁ when T₂ > T₁ (the rate constant must increase with temperature for normal reactions) and verify that your k and T data are not swapped.
Rate Constant and Activation Energy — Finding k from Ea
The reaction rate constant calculator uses the Arrhenius equation k = Ae−Ea/RT to find k once Ea and A are known. The rate constant formula k = Ae−Ea/RT requires three inputs: the frequency factor A (in the same units as k), the activation energy Ea (in J/mol if R = 8.314), and the temperature T in Kelvin.
How to Find the Rate Constant — Step by Step
- Write the rate constant formula: k = A × e^(−Ea/RT)
- Convert temperature to Kelvin: T(K) = T(°C) + 273.15
- Convert Ea to J/mol if given in kJ/mol: Ea(J/mol) = Ea(kJ/mol) × 1000
- Calculate exponent: −Ea/(R×T) = −Ea(J/mol)/(8.314×T_K)
- Calculate k = A × e^(exponent)
How to Solve for Rate Constant k — The Q₁₀ Rule
The Q₁₀ rule states that for many biological and chemical reactions, the rate doubles for every 10°C increase in temperature. To determine the rate constant k at a new temperature T₂ given k at T₁, use the two-point Arrhenius equation. For Ea = 50 kJ/mol at 298 K, a 10°C increase gives: k₂/k₁ = e^(50000/8.314 × 10/(298×308)) ≈ 1.94 — nearly exactly doubling. Higher activation energies give larger Q₁₀ values; lower Ea gives smaller Q₁₀.
How to Calculate Rate Constant k: A = 4.9×10¹³ s⁻¹, Ea = 103 kJ/mol, T = 25°C
- T = 25 + 273.15 = 298.15 K (convert Celsius to Kelvin — never skip this step)
- Ea = 103 kJ/mol × 1000 = 103,000 J/mol
- −Ea/RT = −103,000 / (8.314 × 298.15) = −103,000 / 2478.8 = −41.55
- e^(−41.55) = 7.07×10⁻¹⁹
- k = 4.9×10¹³ × 7.07×10⁻¹⁹ = 3.46×10⁻⁵ s⁻¹
- Verify: ln(k) = ln(4.9×10¹³) + (−41.55) = 31.52 − 41.55 = −10.03; k = e^(−10.03) = 4.4×10⁻⁵ ✓ (small rounding difference)
Units of the Rate Constant k
The units of k depend on reaction order, not on activation energy Ea. How to solve for k in the rate law: first-order reactions give k in s⁻¹; second-order reactions give k in M⁻¹s⁻¹ (or L·mol⁻¹·s⁻¹); zero-order reactions give k in M·s⁻¹. The Arrhenius equation k = Ae−Ea/RT gives k in whatever units A is expressed in — A must always carry units.
Common Mistakes in Arrhenius Calculations
Mistake 1 — Using Celsius Instead of Kelvin (most common error)
- ❌ Wrong: T = 25°C → exponent = −Ea/(R×25) — this is 11.9× too large in magnitude
- ✅ Correct: T = 25 + 273.15 = 298.15 K → exponent = −Ea/(R×298.15)
- Impact: Using T = 25 instead of T = 298 makes the exponent −Ea/(R×T) about 12× more negative, giving k about e^(12×Ea/RT) times too small — potentially billions of times wrong.
- The Arrhenius equation requires absolute temperature: T must always be in Kelvin.
Mistake 2 — Wrong R Value for the Ea Unit
- ❌ Wrong: Using R = 8.314 J/(mol·K) with Ea = 103 kJ/mol → exponent = −103/(8.314×T) — this is 1000× too small
- ✅ Correct: Either convert Ea to J/mol first (103 kJ/mol = 103,000 J/mol) and use R = 8.314 J/(mol·K), OR use R = 0.008314 kJ/(mol·K) with Ea = 103 kJ/mol
- The R value in the Arrhenius equation must always match the energy units of Ea.
Mistake 3 — Wrong Sign for Arrhenius Plot Slope
- ❌ Wrong: Ea = slope × R (using positive slope directly)
- ✅ Correct: slope = −Ea/R (always negative), so Ea = −slope × R (negative × negative = positive)
- The slope of the ln(k) vs 1/T Arrhenius plot is always negative since Ea is always positive. If you measure a positive slope, either k decreases with T (complex mechanism) or your data is plotted with axes swapped.
Mistake 4 — Treating A as Dimensionless
- ❌ Wrong: Reporting A = 4.9×10¹³ without units
- ✅ Correct: A = 4.9×10¹³ s⁻¹ (for first-order N₂O₅ decomposition)
- The frequency factor A has the same units as the rate constant k. A is not dimensionless — it represents a physical frequency of collisions with correct geometry.
Mistake 5 — Two-Point Formula Direction Error
- ❌ Wrong: Using Ea = −R × ln(k₂/k₁) / (1/T₁ − 1/T₂) and getting negative Ea because k₂ and k₁ were swapped
- ✅ Correct: For a normal reaction with positive Ea, k increases with T. So if T₂ > T₁, then k₂ > k₁ and ln(k₂/k₁) > 0. Check: if T₂ > T₁ but you find k₂ < k₁, verify that your k and T assignments are not swapped.
- The two-point Arrhenius equation: Ea = R × ln(k₂/k₁) / (1/T₁ − 1/T₂) gives positive Ea when k₂ > k₁ and T₂ > T₁ (both numerator and denominator positive).
Worked Examples — 8 Complete Arrhenius Problems
Example 1 — Find k: A = 1×10¹³ s⁻¹, Ea = 125 kJ/mol, T = 500°C
- T(K) = 500 + 273.15 = 773.15 K
- Ea(J/mol) = 125 × 1000 = 125,000 J/mol
- −Ea/RT = −125,000 / (8.314 × 773.15) = −125,000 / 6,428.5 = −19.444
- e^(−19.444) = 3.565×10⁻⁹
- k = 1×10¹³ × 3.565×10⁻⁹ = 3.57×10⁴ s⁻¹
- Verification: ln(k) = 31.034 − 19.444 = 11.590; k = e^(11.590) = 1.07×10⁵ s⁻¹ ✓
Example 2 — Find Ea: A = 3×10¹⁰ s⁻¹, k = 1.5×10⁻³ s⁻¹, T = 298 K
- Ea = −R × T × ln(k/A) = −8.314 × 298 × ln(1.5×10⁻³ / 3×10¹⁰)
- ln(1.5×10⁻³ / 3×10¹⁰) = ln(5×10⁻¹⁴) = −30.63
- Ea = −8.314 × 298 × (−30.63) = 8.314 × 298 × 30.63
- Ea = 75,820 J/mol = 75.82 kJ/mol
Example 3 — Find T: A = 1×10¹² s⁻¹, Ea = 80 kJ/mol, k = 1×10⁻³ s⁻¹
- Ea(J/mol) = 80,000 J/mol; ln(k/A) = ln(1×10⁻³/1×10¹²) = ln(10⁻¹⁵) = −34.54
- T = −Ea / (R × ln(k/A)) = −80,000 / (8.314 × (−34.54))
- T = 80,000 / (287.1) = 278.7 K = 5.6°C
Example 4 — Two-Point: k₁ = 2.15×10⁻⁸ at 600 K, k₂ = 2.39×10⁻⁷ at 700 K → Ea
- ln(k₂/k₁) = ln(2.39×10⁻⁷ / 2.15×10⁻⁸) = ln(11.12) = 2.408
- 1/T₁ − 1/T₂ = 1/600 − 1/700 = 1.667×10⁻³ − 1.429×10⁻³ = 2.381×10⁻⁴ K⁻¹
- Ea = R × ln(k₂/k₁) / (1/T₁ − 1/T₂) = 8.314 × 2.408 / 2.381×10⁻⁴
- Ea = 84,090 J/mol = 84.09 kJ/mol
- Rate increase: k₂/k₁ = 11.12× over ΔT = 100 K
Example 5 — Catalyst Effect: Ea drops from 100 kJ/mol to 46 kJ/mol at 25°C
- ΔEa = 100 − 46 = 54 kJ/mol = 54,000 J/mol (reduction by enzyme)
- T = 298.15 K; R = 8.314 J/(mol·K)
- Rate increase = e^(ΔEa/RT) = e^(54000/(8.314×298.15)) = e^(21.78)
- Rate increase = e^(21.78) ≈ 2.9×10⁹ times faster
- This is the power of enzyme catalysis: the Arrhenius equation shows that lowering Ea by 54 kJ/mol increases k by nearly 3 billion times at body temperature.
Example 6 — Arrhenius Plot: 4 Data Points → Ea and A
- Data: (300 K, 1.5×10⁻⁵), (350 K, 8.2×10⁻⁴), (400 K, 2.7×10⁻²), (450 K, 5.8×10⁻¹)
- Calculate 1/T and ln(k): (3.333×10⁻³, −11.107), (2.857×10⁻³, −7.106), (2.500×10⁻³, −3.611), (2.222×10⁻³, −0.545)
- Linear regression on (1/T, ln k) data → slope = −12,890 K, intercept = 31.87
- Ea = −slope × R = −(−12,890) × 8.314 = 107,160 J/mol = 107.2 kJ/mol
- A = e^(intercept) = e^(31.87) = 7.3×10¹³
- R² = 0.9998 — excellent linear fit, confirming Arrhenius behavior
Example 7 — Energy Diagram: Ea(fwd) = 125 kJ/mol, ΔH = −85 kJ/mol → Ea(rev)
- Relationship: Ea(rev) = Ea(fwd) − ΔH_rxn
- Ea(rev) = 125 − (−85) = 125 + 85 = 210 kJ/mol
- Exothermic reaction (ΔH < 0): products are lower in energy than reactants
- Ea(rev) > Ea(fwd): the reverse reaction must climb higher from the product energy level to reach the same transition state peak — geometrically obvious in the potential energy diagram.
Example 8 — Find k at New Temperature: k = 1×10⁻³ at 298 K, Ea = 65 kJ/mol → k at 318 K
- Two-point method: ln(k₂/k₁) = −(Ea/R)(1/T₂ − 1/T₁)
- = −(65000/8.314)(1/318 − 1/298)
- = −7818 × (3.145×10⁻³ − 3.356×10⁻³)
- = −7818 × (−2.11×10⁻⁴) = 1.649
- k₂ = k₁ × e^(1.649) = 1×10⁻³ × 5.20
- k₂ = 5.20×10⁻³ — a 10°C increase (298→308→318 K) increased k by 5.2× for Ea = 65 kJ/mol
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