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function calculateActivationEnergy() {
// Get inputs
var t1Input = document.getElementById('temp1').value;
var k1Input = document.getElementById('k1').value;
var t2Input = document.getElementById('temp2').value;
var k2Input = document.getElementById('k2').value;
var unit = document.getElementById('tempUnit').value;
// Parse floats
var t1 = parseFloat(t1Input);
var k1 = parseFloat(k1Input);
var t2 = parseFloat(t2Input);
var k2 = parseFloat(k2Input);
// Validation
if (isNaN(t1) || isNaN(k1) || isNaN(t2) || isNaN(k2)) {
alert("Please enter valid numerical values for all fields.");
return;
}
if (k1 <= 0 || k2 <= 0) {
alert("Rate constants (k) must be positive values.");
return;
}
// Convert to Kelvin if necessary
if (unit === "celsius") {
t1 = t1 + 273.15;
t2 = t2 + 273.15;
}
if (t1 <= 0 || t2 A = k / e^(-Ea/RT)
// Using T1 and k1
var A = k1 / Math.exp(-Ea_J / (R * t1));
// Format output
// Handle scientific notation for A if it's very large
var aDisplay = A > 10000 || A < 0.0001 ? A.toExponential(4) : A.toFixed(4);
document.getElementById('resJoules').innerHTML = Ea_J.toLocaleString('en-US', {maximumFractionDigits: 2}) + " J/mol";
document.getElementById('resKiloJoules').innerHTML = Ea_kJ.toFixed(3) + " kJ/mol";
document.getElementById('resArrheniusA').innerHTML = aDisplay + " s⁻¹"; // Assuming 1st order for unit simplicity
// Show results
document.getElementById('results').style.display = 'block';
}
How to Calculate Activation Energy Given Rate Constant and Temperature
Calculating activation energy ($E_a$) is a fundamental task in chemical kinetics, allowing chemists to understand the energy barrier that must be overcome for a reaction to occur. This calculator utilizes the Arrhenius Equation to determine activation energy based on rate constants observed at two different temperatures.
Understanding the Arrhenius Equation
The relationship between the rate constant ($k$), absolute temperature ($T$), and activation energy ($E_a$) is given by the Arrhenius equation:
k = A × e-Ea / (R × T)
Where:
- k: The rate constant of the reaction.
- A: The frequency factor (or pre-exponential factor), representing the frequency of collisions.
- Ea: The activation energy (usually in Joules per mole, J/mol).
- R: The universal gas constant ($8.314\ \text{J}/(\text{mol}\cdot\text{K})$).
- T: The absolute temperature in Kelvin.
The Two-Point Form Calculation
While the standard equation is useful, experimentally determining activation energy usually involves measuring the rate constant at two different temperatures. By comparing these two points ($T_1, k_1$) and ($T_2, k_2$), we can eliminate the frequency factor $A$ and solve directly for $E_a$ using the logarithmic form:
ln(k2 / k1) = (-Ea / R) × (1/T2 – 1/T1)
Rearranging this formula to solve for Activation Energy yields:
Ea = -R × ln(k2 / k1) / (1/T2 – 1/T1)
Step-by-Step Calculation Example
To demonstrate how this calculation works practically, let's assume a chemical reaction has the following data:
- State 1: At 300 K ($27^\circ\text{C}$), the rate constant $k_1$ is $2.5 \times 10^{-3}\ \text{s}^{-1}$.
- State 2: At 310 K ($37^\circ\text{C}$), the rate constant $k_2$ increases to $5.0 \times 10^{-3}\ \text{s}^{-1}$.
1. Calculate the natural log of the rate ratio:
$\ln(k_2/k_1) = \ln(0.005 / 0.0025) = \ln(2) \approx 0.693$
2. Calculate the difference in inverse temperatures:
$(1/310) – (1/300) \approx 0.0032258 – 0.0033333 = -0.0001075\ \text{K}^{-1}$
3. Apply the Gas Constant ($R$):
$E_a = -8.314 \times 0.693 / (-0.0001075)$
$E_a \approx 53,600\ \text{J/mol}$ or $53.6\ \text{kJ/mol}$
Why is Activation Energy Important?
Understanding $E_a$ helps predict how reaction rates change with temperature. A reaction with a high activation energy is very sensitive to temperature changes, meaning a small increase in temperature will lead to a significant increase in the reaction rate. Conversely, reactions with low activation energy are less affected by temperature fluctuations.
Common Units
Always ensure your units are consistent before calculating. The gas constant $R$ is typically $8.314\ \text{J}/(\text{mol}\cdot\text{K})$, which results in energy in Joules. If you are working in kilojoules ($kJ$), remember to convert $R$ or divide your final result by 1000.