Arrhenius Equation Calculator
Calculate a reaction rate constant with the Arrhenius equation. Enter the frequency factor, activation energy in kJ/mol, and temperature in kelvin.
Calculate Reaction Rate Constant Using Arrhenius Equation
Arrhenius Equation Formula
Where:
- \(k\) — reaction rate constant
- \(A\) — frequency factor, or pre-exponential factor
- \(E_a\) — activation energy entered in kJ/mol
- \(R\) — gas constant, 8.314 J/(mol·K)
- \(T\) — absolute temperature in kelvin (K)
The factor of 1000 converts the entered activation energy from kJ/mol to J/mol so its units match \(R\). The equation describes how the rate constant depends on temperature when \(A\) and \(E_a\) are treated as constant.
Arrhenius Equation – Calculation Example
Example: A reaction has a frequency factor of \(1\times10^7\) s⁻¹ and an activation energy of 50 kJ/mol. Find its rate constant at 298.15 K.
\[ E_a=50\times1000=50{,}000\ \text{J/mol} \]
\[ k=(1\times10^7)e^{-\frac{50{,}000}{8.314\times298.15}}\approx1.7374\times10^{-2}\ \text{s}^{-1} \]
Result: \(k\approx0.017374\) s⁻¹, displayed by the calculator as approximately 1.7374e-2 sec⁻¹.
How to use the calculator
- Enter the Frequency Factor (A). For a first-order reaction, use a value in s⁻¹.
- Enter the Activation Energy (Ea) in kJ/mol. The calculator converts it to J/mol internally; do not convert it before entering it.
- Enter the Temperature (T) in kelvin. To convert from degrees Celsius, add 273.15.
- Read the Reaction Rate Constant (k). The result is shown in scientific notation and labeled sec⁻¹.
What the Arrhenius equation tells you
The Arrhenius equation connects temperature with a chemical reaction’s rate constant. Activation energy represents an energy barrier associated with the reaction. At a higher temperature, the exponential factor \(e^{-E_a/(RT)}\) is larger, so the calculated rate constant increases when the other inputs stay the same. The frequency factor \(A\) sets the scale of the result; it is a fitted parameter, not simply a count of collisions.
A rate constant is not the same as a reaction rate. The rate constant is a parameter in a rate law, while the rate at a particular moment can also depend on reactant concentrations and the reaction orders. A larger \(k\) generally indicates a faster reaction under comparable conditions, but comparing constants from different rate laws requires care. Balancing a reaction with the Chemical Equation Balancer establishes its stoichiometry; it does not determine its rate law or activation energy.
Activation energy controls how strongly the rate constant responds to a temperature change. When two reactions have the same starting temperature and comparable activation energies, the one with the higher activation energy generally has the larger proportional change in \(k\) for a given temperature increase. The equation is therefore useful for estimating a rate constant at a specified temperature when appropriate values of \(A\) and \(E_a\) are known.
Assumptions and practical limits
- Use absolute temperature. Enter kelvin, not degrees Celsius or Fahrenheit. Temperature must be greater than 0 K.
- Check the units of \(A\). The rate constant has the same units as \(A\). This calculator always labels its output sec⁻¹, which is appropriate for a first-order rate constant. For other reaction orders, the numerical calculation still follows the equation, but the displayed unit may be incorrect.
- Use parameters for the same reaction and conditions. Values of \(A\) and \(E_a\) are usually obtained from experimental data. A reaction mechanism, catalyst, or medium change can make previously fitted values unsuitable.
- Avoid extrapolating too far. Treating \(A\) and \(E_a\) as constant may give a poor estimate outside the temperature range used to determine them.
- Expect scientific notation. A result such as 1.7374e-2 means \(1.7374\times10^{-2}\), or 0.017374. The calculator displays four digits after the decimal point in its scientific-notation value; this formatting does not indicate the precision of the inputs.
The gas constant also appears in calculations such as the Ideal Gas Law, but here it must be expressed in J/(mol·K) to match the activation energy after conversion. If you are working from measured temperatures, check the temperature scale before using either equation.
Frequently Asked Questions
How can I calculate an Arrhenius rate constant by hand?
First convert the activation energy from kJ/mol to J/mol. Then divide it by \(8.314T\), change the sign, and evaluate the exponential. Multiply that result by \(A\). A scientific calculator’s \(e^x\) key evaluates the exponential; its \(10^x\) key does not give the same result.
How do I find the rate constant at a second temperature?
If you know \(k_1\) at \(T_1\) and the activation energy, you do not need the frequency factor. Use \(k_2=k_1\exp[(1000E_a/R)(1/T_1-1/T_2)]\), with \(E_a\) in kJ/mol and both temperatures in kelvin. This comparison assumes the same Arrhenius parameters apply at both temperatures.
How can I determine activation energy from measured rate constants?
Measure rate constants at several temperatures and plot \(\ln k\) against \(1/T\), using kelvin. If the Arrhenius model fits, the points are approximately linear and the slope equals \(-E_a/R\) when \(E_a\) is in J/mol. Divide the resulting value by 1000 for kJ/mol. Curvature can signal that a single fitted activation energy is inadequate.
What is a normal or good value for a reaction rate constant?
There is no universal good value: it depends on the reaction, temperature, and rate law. For a first-order process, \(k\) has units of s⁻¹ and its half-life is \(t_{1/2}=\ln(2)/k\). For example, \(k=0.01\) s⁻¹ corresponds to a half-life of about 69 seconds under first-order conditions. Other reaction orders need a different interpretation.
Why might the displayed sec⁻¹ unit be wrong for my reaction?
The Arrhenius calculation gives \(k\) the same units as \(A\), but this calculator always displays sec⁻¹. That unit suits a first-order rate law. For a second-order rate law expressed using mol/L and seconds, \(k\) and \(A\) typically have units of L/(mol·s). Check the experimentally determined rate law before assigning units to the numerical result.
What if the activation energy is zero or negative?
At zero activation energy, the exponential factor is one, so \(k=A\) at every positive temperature in this model. A negative fitted activation energy makes the calculated \(k\) decrease as temperature rises. Such an apparent value can occur in more complex kinetics; it should not automatically be interpreted as a simple negative energy barrier.
Can I recover the frequency factor from a known rate constant?
Yes. Rearranging the equation gives \(A=k\exp[1000E_a/(RT)]\) when \(E_a\) is in kJ/mol. Use the temperature at which \(k\) was measured, and give \(A\) the same units as \(k\). A single measurement can only determine \(A\) this way if you already know an appropriate activation energy.