Quick answer The Arrhenius equation gives the rate constant k = A·e^(−E_a/(R·T)) (A = pre-exponential factor, E_a = activation energy, R = 8.314 J/(mol·K)). Plotting ln k against 1/T yields a straight line of slope −E_a/R, and the half-life is t₁/₂ = ln2/k.
Adjust temperature, activation energy and frequency factor to compute the rate constant
Parameters
Results
Rate constant k
1.24e-3
s⁻¹
Half-life t₁/₂
559
s
Q₁₀ (T → T+10)
2.84
—
Thermal energy RT
2.48
kJ/mol
Reaction coordinate
ln(k) vs 1000/T
[A] vs time
Theory & Key Formulas
Arrhenius equation: $k = A\,e^{-E_a/RT}$, with $R = 8.314$ J/(mol·K). A 10 °C temperature rise typically doubles the reaction rate (varies with Ea).
FAQ
What is activation energy?
Activation energy is the minimum energy barrier a reaction must overcome. Higher Ea means slower reaction; raising temperature has a larger effect on high-Ea reactions.
How does a catalyst affect the Arrhenius equation?
A catalyst lowers activation energy Ea, dramatically increasing the rate constant k at any temperature without being consumed.
Why does refrigeration preserve food?
The Arrhenius equation shows that lowering temperature exponentially slows reaction rates. Both spoilage and oxidation reactions are dramatically slowed in the cold.
What is an Arrhenius plot?
Plotting ln(k) vs 1/T gives a straight line with slope = -Ea/R, allowing experimental determination of activation energy.
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I can see the simulation updating, but what exactly is being calculated here?
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Great question! The simulator solves the governing equations in real time as you move the sliders. Each parameter you control directly affects the physical outcome you see in the graph. The key is to build an intuitive feel for how each variable influences the result — that's how engineers develop physical judgment.
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So when I increase this parameter, the curve shifts significantly. Is that a linear relationship?
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It depends on the model. Some relationships are linear, but many engineering phenomena are nonlinear. Try moving the sliders to extreme values and see if the output changes proportionally — if the graph shape changes, that's a sign of nonlinearity. This hands-on exploration is exactly what simulations are best for.
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Where is this kind of analysis actually used in practice?
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Constantly! Engineers run these calculations during the design phase to quickly screen parameters before investing in expensive physical tests or detailed finite element simulations. Getting comfortable with these simplified models is a real engineering skill.
Enter activation energy (Ea) in kJ/mol—typical values range 50–300 kJ/mol for organic reactions
Input pre-exponential factor (A) in appropriate units; for gas-phase reactions use s⁻¹
Set temperature in Kelvin; simulator recalculates rate constant k using k = A·exp(−Ea/RT) where R = 8.314 J/mol·K
Observe real-time curves showing exponential sensitivity to temperature changes
Worked Example
For ester hydrolysis: Ea = 65 kJ/mol, A = 2.5×10¹¹ s⁻¹. At T = 298 K (25°C), k = 2.5×10¹¹ × exp(−65000/8.314×298) = 8.3×10⁻⁴ s⁻¹. Increasing temperature to 323 K (50°C) yields k = 4.2×10⁻³ s⁻¹—a five-fold increase. This demonstrates why pharmaceutical stability testing uses accelerated conditions at 40°C rather than 25°C.
Practical Notes
Catalysts lower Ea without changing A; a zeolite catalyst might reduce Ea from 180 to 95 kJ/mol, dramatically accelerating ammonia synthesis
Use van 't Hoff plots (ln k vs 1/T) to extract Ea from experimental kinetic data across multiple temperatures
For food science shelf-life prediction, a 10°C increase typically doubles reaction rate; confirm with your system's actual Ea value
Pre-exponential factor A encompasses collision frequency and steric factors; literature values vary widely—validate against real experimental k measurements