Eyring Equation Calculator
Calculate rate constants and activation parameters using the Eyring equation
About the Eyring Equation
The Eyring equation is derived from transition state theory and relates rate constants to thermodynamic activation parameters. Unlike the Arrhenius equation, it provides insight into both enthalpic and entropic contributions to the activation barrier.
Negative dS* values indicate a more ordered transition state (common in bimolecular reactions), while positive values suggest a looser transition state.
What Is the Eyring Equation?
The Eyring equation is a fundamental relationship in chemical kinetics derived from transition state theory (TST). It relates the rate constant of a chemical reaction to the thermodynamic activation parameters: the enthalpy of activation (ΔH‡) and the entropy of activation (ΔS‡). Unlike the Arrhenius equation, which is empirical, the Eyring equation has a rigorous theoretical foundation based on statistical mechanics and quantum theory.
The equation takes the form: k = κ × (k_BT/h) × exp(ΔS‡/R) × exp(−ΔH‡/RT), where κ is the transmission coefficient (usually ≈ 1), k_B is Boltzmann's constant, h is Planck's constant, R is the gas constant, and T is the absolute temperature. The prefactor k_BT/h represents the universal frequency of crossing the transition state barrier, approximately 6.25 × 10¹² s⁻¹ at 298 K.
The Eyring equation provides deeper insight than the Arrhenius equation because it separates the activation barrier into enthalpic (ΔH‡) and entropic (ΔS‡) contributions. A large positive ΔH‡ indicates a high energy barrier, while a negative ΔS‡ indicates a more ordered transition state (typical of bimolecular reactions). This separation helps chemists understand the mechanism of a reaction at the molecular level.
Eyring Equation
Where:
- k= Rate constant (s⁻¹ for first-order reactions)
- κ= Transmission coefficient (usually ≈ 1)
- k_B= Boltzmann constant, 1.381 × 10⁻²³ J/K
- T= Absolute temperature (K)
- h= Planck constant, 6.626 × 10⁻³⁴ J·s
- ΔH‡= Enthalpy of activation (J/mol)
- ΔS‡= Entropy of activation (J/mol·K)
- R= Gas constant, 8.314 J/(mol·K)
Understanding Activation Parameters
The two activation parameters provide complementary information about the reaction mechanism:
| Parameter | Positive Value Indicates | Negative Value Indicates |
|---|---|---|
| ΔH‡ (Enthalpy) | High energy barrier, strong bonds broken | Low barrier, bonds barely distorted |
| ΔS‡ (Entropy) | Looser TS (bond breaking dominates) | Tighter TS (bond making dominates) |
Negative ΔS‡ is typical of bimolecular reactions where two molecules must come together in a specific orientation to form the transition state. This creates a more ordered (lower entropy) transition state. Values of ΔS‡ around −10 to −30 J/mol·K are common for SN2 reactions and associative mechanisms.
Positive ΔS‡ is characteristic of unimolecular reactions where a molecule fragments or rearranges, increasing the number of independent particles or their freedom of motion. Dissociative mechanisms and elimination reactions often show positive ΔS‡ values.
The Gibbs energy of activation combines both parameters: ΔG‡ = ΔH‡ − TΔS‡. This value determines the overall rate: a larger ΔG‡ means a slower reaction. At the temperature where ΔG‡ = RT × ln(k_BT/hk), the reaction rate is at a characteristic value.
How to Use This Calculator
This calculator provides two modes for working with the Eyring equation:
- Calculate Rate Constant from ΔH‡ and ΔS‡: Enter the temperature (K), enthalpy of activation (kJ/mol), entropy of activation (J/mol·K), and the transmission coefficient κ (default = 1). The calculator determines the rate constant, Gibbs energy of activation, and the first-order half-life.
- Calculate ΔH‡ and ΔS‡ from Rate Data: Enter rate constants at two different temperatures. The calculator uses the linearized Eyring equation (ln(k/T) vs. 1/T) to extract the activation parameters from the slope and intercept.
Transmission coefficient (κ): This factor accounts for the probability that a trajectory crossing the transition state barrier will actually form products rather than return to reactants. For most reactions, κ ≈ 1. Values less than 1 indicate significant recrossing, which is rare in solution-phase reactions.
Temperature range: The Eyring equation is most accurate near room temperature (280–350 K). At very high or very low temperatures, quantum tunneling effects and non-classical behavior can cause deviations from the predicted rate.
The Linearized Eyring Equation
The Eyring equation can be rearranged into a linear form that is useful for extracting activation parameters from experimental data:
ln(k/T) = −(ΔH‡/R) × (1/T) + [ΔS‡/R + ln(k_B/h)]
This has the form y = mx + b, where:
- y = ln(k/T)
- x = 1/T
- Slope (m) = −ΔH‡/R → ΔH‡ = −slope × R
- Intercept (b) = ΔS‡/R + ln(k_B/h) → ΔS‡ = (intercept − ln(k_B/h)) × R
By measuring rate constants at two or more temperatures and plotting ln(k/T) vs. 1/T, the slope gives ΔH‡ and the intercept gives ΔS‡. The calculator implements this analysis automatically when you enter rate constants at two temperatures. The method requires that ΔH‡ and ΔS‡ are approximately constant over the temperature range studied.
The linearized form also reveals the relationship between the Eyring and Arrhenius equations. Comparing k = A × exp(−Ea/RT) with k = (k_BT/h) × exp(ΔS‡/R) × exp(−ΔH‡/RT), we see that Ea = ΔH‡ + RT and A = (k_BT/h) × exp(ΔS‡/R). The Arrhenius activation energy includes a small temperature-dependent term RT.
Real-World Applications
The Eyring equation is widely used in physical organic chemistry to elucidate reaction mechanisms. By determining ΔH‡ and ΔS‡ experimentally, chemists can distinguish between associative (SN2, ΔS‡ < 0) and dissociative (SN1, ΔS‡ > 0) substitution mechanisms. This mechanistic insight guides the design of synthetic routes and catalysts.
In biochemistry, the Eyring equation analyzes enzyme-catalyzed reactions. The activation parameters reveal whether the rate-limiting step involves bond making (negative ΔS‡), bond breaking (positive ΔS‡), or conformational changes. Temperature-dependent enzyme kinetics using the Eyring plot (ln(k/T) vs. 1/T) provide mechanistic information that Arrhenius analysis alone cannot.
Pharmaceutical development uses the Eyring equation to predict drug shelf life and degradation rates. By measuring degradation rates at multiple elevated temperatures and extrapolating to room temperature using the Eyring equation, researchers estimate the long-term stability of drug formulations without waiting years for real-time data.
In materials science, the Eyring equation models diffusion rates, crystallization kinetics, and polymer relaxation processes. The activation parameters provide insight into the molecular-level processes controlling mass transport and structural rearrangements in materials.
Worked Examples
Rate Constant from Activation Parameters
Problem:
Calculate the rate constant for a reaction with ΔH‡ = 75 kJ/mol, ΔS‡ = −10 J/mol·K at T = 298 K, κ = 1.
Solution Steps:
- 1Convert ΔH‡ to J/mol: 75,000 J/mol
- 2Calculate prefactor: k_BT/h = (1.381×10⁻²³ × 298) / (6.626×10⁻³⁴) = 6.22×10¹² s⁻¹
- 3Calculate entropy term: exp(ΔS‡/R) = exp(−10/8.314) = exp(−1.203) = 0.300
- 4Calculate enthalpy term: exp(−ΔH‡/RT) = exp(−75000/(8.314×298)) = exp(−30.27) = 7.2×10⁻¹⁴
- 5k = 1 × 6.22×10¹² × 0.300 × 7.2×10⁻¹⁴ = 0.134 s⁻¹
Result:
k = 0.134 s⁻¹ (half-life ≈ 5.2 s). The negative ΔS‡ reduces the rate by a factor of ~3 compared to ΔS‡ = 0.
Extracting Activation Parameters from Two Temperatures
Problem:
k₁ = 0.001 s⁻¹ at T₁ = 298 K, k₂ = 0.01 s⁻¹ at T₂ = 323 K. Find ΔH‡ and ΔS‡.
Solution Steps:
- 1ln(k₁/T₁) = ln(0.001/298) = ln(3.36×10⁻⁶) = −12.60
- 2ln(k₂/T₂) = ln(0.01/323) = ln(3.10×10⁻⁵) = −10.38
- 31/T₁ = 1/298 = 3.356×10⁻³ K⁻¹, 1/T₂ = 1/323 = 3.096×10⁻³ K⁻¹
- 4Slope = (−10.38 − (−12.60)) / (3.096×10⁻³ − 3.356×10⁻³) = 2.22 / (−2.60×10⁻⁴) = −8538
- 5ΔH‡ = −slope × R = 8538 × 8.314 = 70,994 J/mol ≈ 71.0 kJ/mol
Result:
ΔH‡ ≈ 71.0 kJ/mol. The slope of the Eyring plot directly yields the enthalpy of activation.
Comparing Unimolecular vs. Bimolecular
Problem:
Reaction A has ΔS‡ = +15 J/mol·K, Reaction B has ΔS‡ = −25 J/mol·K. What do these values suggest?
Solution Steps:
- 1Positive ΔS‡ (+15) indicates a looser transition state → likely unimolecular (bond breaking)
- 2Negative ΔS‡ (−25) indicates a tighter transition state → likely bimolecular (two molecules coming together)
- 3The magnitude |ΔS‡| = 25 J/mol·K is typical for associative mechanisms like SN2
- 4The magnitude |ΔS‡| = 15 J/mol·K is consistent with dissociative mechanisms like SN1 or elimination
Result:
Reaction A (ΔS‡ = +15) likely proceeds by a dissociative mechanism. Reaction B (ΔS‡ = −25) likely proceeds by an associative mechanism. The sign and magnitude of ΔS‡ provide mechanistic information.
Tips & Best Practices
- ✓Use at least three temperatures for a reliable Eyring plot; two temperatures give ΔH‡ and ΔS‡ but no linearity check.
- ✓Negative ΔS‡ indicates a more ordered transition state (bimolecular, associative mechanisms).
- ✓Positive ΔS‡ indicates a looser transition state (unimolecular, dissociative mechanisms).
- ✓ΔG‡ = ΔH‡ − TΔS‡ combines both parameters into a single measure of the activation barrier.
- ✓The prefactor k_BT/h ≈ 6.25 × 10¹² s⁻¹ at 298 K sets the frequency of barrier crossing.
- ✓Convert ΔH‡ from kJ/mol to J/mol before using the Eyring equation.
Frequently Asked Questions
Sources & References
Last updated: 2026-06-06
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Editorial Note
MyCalcBuddy Editorial Team
This page is maintained as an educational calculator reference.
Formula Source: Chemistry: The Central Science
by Brown, LeMay, Bursten