What is the Arrhenius equation and what does it calculate?

The Arrhenius equation is a fundamental mathematical relationship in chemical kinetics that quantifies the exponential dependence of a chemical reaction's rate constant (k) on absolute temperature (T) and activation energy (Ea): k = A · e^(-Ea / RT), where A is the pre-exponential frequency factor and R is the universal gas constant (8.314 J/(mol·K)).

What is the derivation of the Arrhenius equation?

Svante Arrhenius derived his equation in 1889 by combining Jacobus van 't Hoff's thermodynamic equilibrium relation (d ln K / dT = ΔH° / RT²) with Maxwell-Boltzmann collision theory. Arrhenius proposed that reactants must pass through an activated complex (transition state) with energy barrier Ea. The fraction of molecular collisions having kinetic energy >= Ea is given by the Boltzmann distribution factor f = e^(-Ea/RT). Multiplying this by collision frequency and steric orientation factor (A) yields k = A·e^(-Ea/RT).

What is the two-point form of the Arrhenius equation?

The two-point Arrhenius equation relates rate constants k1 and k2 measured at two distinct absolute temperatures T1 and T2: ln(k2 / k1) = (Ea / R) · (1/T1 - 1/T2). This form eliminates the pre-exponential factor A and allows direct experimental calculation of activation energy Ea from two temperature data points.

What is the pre-exponential factor (A) in the Arrhenius equation?

The pre-exponential factor (A, also known as the frequency factor) represents the total frequency of molecular collisions per unit time that possess the correct steric spatial orientation required for a successful chemical transformation.

What value of the gas constant (R) should be used in the Arrhenius equation?

In Arrhenius kinetic calculations, always use the molar energy gas constant R = 8.31446 J/(mol·K) (or 8.31446 × 10^-3 kJ/(mol·K), or 1.9872 cal/(mol·K)). Do not use the volume gas constant 0.082057 L·atm/(mol·K).

How does temperature affect the rate constant in the Arrhenius equation?

Increasing temperature causes an exponential increase in the rate constant (k) because higher thermal energy dramatically increases the fraction of molecular collisions that exceed the activation energy threshold (Ea), as dictated by the Boltzmann factor e^(-Ea/RT).

What is the Q10 temperature coefficient in chemical kinetics?

The Q10 temperature coefficient is a measure of the rate of change of a biological or chemical system as a consequence of increasing the temperature by 10 °C (or 10 K): Q10 = (k2 / k1)^(10 / (T2 - T1)). For many standard organic and biochemical reactions, Q10 typically ranges between 2.0 and 3.0 (meaning the reaction rate doubles or triples per +10 °C rise).

How does an Arrhenius plot determine activation energy experimentally?

Taking the natural logarithm of both sides yields the linear equation: ln(k) = (-Ea / R) · (1/T) + ln(A), matching y = mx + c. Plotting ln(k) on the y-axis against 1/T on the x-axis yields a straight line with slope m = -Ea / R and y-intercept ln(A). Multiplying the slope by -R provides the activation energy Ea.

How does a catalyst affect the Arrhenius equation parameters?

A catalyst provides an alternative reaction mechanism with a lower activation energy barrier (Ea,cat < Ea,uncat). Lowering Ea exponentially increases the term e^(-Ea/RT), multiplying the rate constant k by several orders of magnitude without altering thermodynamic equilibrium or Gibbs free energy.

How do pharmaceutical and battery engineers use the Arrhenius Equation Calculator?

Pharmaceutical scientists use the Arrhenius equation in accelerated stability testing (e.g. heating drug samples to 40 °C to project 2-year room-temperature shelf life), while battery engineers model solid electrolyte interphase (SEI) growth rates and thermal runaway thresholds in lithium-ion cells.

⚡ 100% Free • Single-Point & Two-Point Kinetics, Derivation Steps & Boltzmann Factor

Arrhenius Equation Calculator

Calculate chemical reaction rate constants ($k = A \cdot e^{-E_a/RT}$), activation energy ($E_a$), pre-exponential frequency factor ($A$), and two-point temperature jumps with complete logarithmic derivation steps.

Kinetics Presets:

Calculation Mode

Target Variable to Solve

Activation Energy ($E_a$)

Pre-Exponential Factor ($A$)

s⁻¹

Temperature ($T$)

Calculated Rate Constant (k)

1.74 × 10⁻³ s⁻¹

Standard exponential Arrhenius form: k = A·e^(-Ea/RT)

Boltzmann Factor ($e^{-E_a/RT}$)

1.74 × 10⁻⁹

Fraction $\ge E_a$

$Q_{10}$ Factor

2.01 ×

Rate speedup / +10 °C

1st-Order Half-Life ($t_{1/2}$)

398.4 s

$\ln(2) / k$

Arrhenius Plot Parameters ($\ln k = -\frac{E_a}{R}\frac{1}{T} + \ln A$) Linear Form

Slope ($m = -E_a/R$): -6013.7 K

Y-Intercept ($c = \ln A$): +25.33

Step-by-Step Mathematical & Logarithmic Derivation

What is the Arrhenius Equation and How is it Derived?

Formulated in 1889 by Swedish chemist Svante Arrhenius, the Arrhenius Equation is the cornerstone relationship in physical chemistry and chemical kinetics describing how the rate constant ($k$) of a chemical reaction changes exponentially with absolute temperature ($T$) and activation energy ($E_a$):

$$k = A \cdot e^{-\frac{E_a}{R T}}$$

Where:
• $k$ is the reaction rate constant (units depend on reaction order: $\text{s}^{-1}$, $\text{M}^{-1}\text{s}^{-1}$).
• $A$ is the pre-exponential frequency factor, representing the frequency of collisions with proper steric spatial orientation.
• $E_a$ is the activation energy ($\text{J/mol}$ or $\text{kJ/mol}$), the minimum energy required to form the activated transition state.
• $R$ is the universal molar gas constant ($8.31446\text{ J}/(\text{mol}\cdot\text{K})$).
• $T$ is the absolute thermodynamic temperature in Kelvin ($\text{K} = ^\circ\text{C} + 273.15$).

Derivation of the Arrhenius Equation from Collision & Transition State Theory

Arrhenius connected Jacobus Henricus van 't Hoff's thermodynamic equilibrium relationship with James Clerk Maxwell and Ludwig Boltzmann's statistical energy distributions:

1. Van 't Hoff Isochore Connection: The temperature dependence of chemical equilibrium constant $K = k_f / k_r$ is given by $\frac{d \ln K}{dT} = \frac{\Delta H^\circ}{RT^2}$.

2. Splitting Forward & Reverse Rates: Arrhenius proposed that $\frac{d \ln k_f}{dT} = \frac{E_{a,f}}{RT^2} + C$ and $\frac{d \ln k_r}{dT} = \frac{E_{a,r}}{RT^2} + C$, where $C \approx 0$.

3. Indefinite Integration: Integrating $\int d \ln k = \int \frac{E_a}{RT^2} dT$ yields $\ln k = -\frac{E_a}{RT} + \ln A$.

4. Exponentiation to Standard Form: Exponentiating both sides yields the famous expression: $$k = A \cdot e^{-\frac{E_a}{RT}}$$

Two-Point Arrhenius Equation & Arrhenius Plot Linearization

01. Two-Point Form (Eliminating Frequency Factor A)

When rate constants $k_1$ and $k_2$ are measured at temperatures $T_1$ and $T_2$, subtracting the two logarithmic equations eliminates $\ln A$:

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

02. Linear Arrhenius Plot ($y = mx + c$)

Plotting $\ln k$ against $1/T$ yields a straight line with slope $m = -E_a/R$ and y-intercept $c = \ln A$. Multiplying the slope by $-R$ gives the exact activation energy:

$$E_a = -R \cdot (\text{Slope}) \qquad \text{and} \qquad A = e^{\text{Y-Intercept}}$$

Comprehensive Worked Arrhenius Equation Examples

Example 1: Calculating Rate Constant at 25 °C

Single-Point Form

Problem: An organic substitution reaction has activation energy $E_a = 50.0\text{ kJ/mol}$ ($50{,}000\text{ J/mol}$) and pre-exponential factor $A = 1.0 \times 10^{11}\text{ s}^{-1}$. Calculate $k$ at $25\text{ }^\circ\text{C}$ ($298.15\text{ K}$).

1. Calculate Exponent: $-\frac{E_a}{RT} = -\frac{50{,}000}{8.31446 \times 298.15} = -20.170$

2. Compute Boltzmann Factor: $e^{-20.170} = 1.74 \times 10^{-9}$

3. Multiply by Frequency Factor: $k = (1.0 \times 10^{11}) \times (1.74 \times 10^{-9}) = 1.74 \times 10^{-3}\text{ s}^{-1}$

Example 2: Two-Point Temperature Jump Activation Energy

Two-Point Form

Problem: A reaction has rate constant $k_1 = 0.020\text{ s}^{-1}$ at $T_1 = 300\text{ K}$ and $k_2 = 0.065\text{ s}^{-1}$ at $T_2 = 320\text{ K}$. Calculate activation energy $E_a$.

1. Natural Log of Rate Ratio: $\ln(k_2 / k_1) = \ln(0.065 / 0.020) = \ln(3.25) = 1.17865$

2. Reciprocal Temperature Difference: $\frac{1}{300} - \frac{1}{320} = 0.003333 - 0.003125 = 2.0833 \times 10^{-4}\text{ K}^{-1}$

3. Solve for $E_a$: $E_a = \frac{8.31446 \times 1.17865}{2.0833 \times 10^{-4}} = 47{,}035\text{ J/mol} = 47.04\text{ kJ/mol}$

Frequently Asked Questions

Expert physical chemistry answers regarding the Arrhenius equation, activation energy, and chemical reaction kinetics.