IB DP Chemistry - R2.2.13 Arrhenius factor - Study Notes - New Syllabus - 2026, 2027 & 2028
IB DP Chemistry – R2.2.13 Arrhenius factor – Study Notes – New Syllabus
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Reactivity 2.2.13 – The Arrhenius Factor, A
Reactivity 2.2.13 – The Arrhenius Factor, A
The Arrhenius factor (also called the frequency factor), denoted by \( A \), is a constant that represents the frequency of collisions between reactant molecules with the correct orientation for a reaction to occur. It reflects both:
- The number of collisions per second (collision frequency)
- The fraction of collisions with the correct geometry to form products
Context in the Arrhenius Equation:
The Arrhenius equation is:
\( k = A \cdot e^{-E_a/RT} \)
Where:
- \( k \) = rate constant
- \( A \) = Arrhenius factor (frequency factor)
- \( E_a \) = activation energy (J/mol)
- \( R \) = universal gas constant = 8.314 J/mol·K
- \( T \) = temperature (K)
Explanation:
- Even if particles have sufficient energy (i.e. \( E \geq E_a \)), a reaction may not occur unless the molecules collide in the correct orientation.
- \( A \) captures how often reactants approach each other in the right way to form products.
- Higher values of \( A \) suggest that the reaction occurs readily once energy criteria are met.
- For complex molecules, \( A \) is typically smaller because proper orientation is less likely during collisions.
Importance in Reaction Kinetics:
The factor \( A \) helps explain why two reactions with similar activation energies can still have different rate constants. One reaction might have a higher frequency of correctly-oriented collisions.
Example
Consider two reactions with similar activation energies, but one involves very bulky molecules and the other involves simple molecules. Even though both have enough energy:
▶️Answer/Explanation
The reaction with simple molecules is more likely to proceed faster because the molecules can easily align in the correct geometry during collision. Therefore, it has a larger Arrhenius factor \( A \), leading to a higher rate constant \( k \).
The bulky molecules have more restricted geometry, leading to fewer effective collisions, a lower \( A \), and a slower reaction even if \( E_a \) is the same.
Determination of the Activation Energy and the Arrhenius Factor
Linear Form of the Arrhenius Equation:
\( \ln k = \ln A – \frac{E_a}{R} \cdot \frac{1}{T} \)
This has the form of a straight-line equation:
\( y = mx + c \)
where:
- \( y = \ln k \)
- \( x = \frac{1}{T} \)
- \( m = -\frac{E_a}{R} \) → slope of the line
- \( c = \ln A \) → y-intercept
Steps to Determine \( E_a \) and \( A \):
- Convert temperatures to Kelvin (if not already)
- Calculate \( \frac{1}{T} \) for each temperature
- Take natural logarithms of the rate constants to get \( \ln k \)
- Plot \( \ln k \) vs. \( \frac{1}{T} \)
- Determine the slope \( m \) of the line: \( m = -\frac{E_a}{R} \)
- Calculate \( E_a = -m \cdot R \)
- Use the y-intercept \( \ln A \) to find \( A \) by taking the inverse logarithm: \( A = e^{\ln A} \)
Data Needed:
- Temperatures \( T \) in Kelvin
- Corresponding rate constants \( k \) in consistent units (e.g. s\(^{-1}\))
Example
The following rate constants were measured for a reaction at different temperatures:
| Temperature (K) | Rate constant, \( k \) (s\(^{-1}\)) |
|---|---|
| 1400 | 0.022 |
| 1460 | 0.070 |
| 1520 | 0.210 |
| 1580 | 0.686 |
| 1620 | 1.578 |
▶️ Answer/Explanation
Step 1: Convert to \( \frac{1}{T} \) and \( \ln k \) for each row:
- For example, at \( T = 1400 \, \text{K} \): \( \frac{1}{T} = 7.14 \times 10^{-4} \), \( \ln k = \ln(0.022) = -3.819 \)
- Repeat for all values.
Step 2: Plot a graph of \( \ln k \) vs. \( \frac{1}{T} \). The plot is approximately linear.
Step 3: From linear regression using all data points, slope \( m = -43761.75 \), and y-intercept = 27.36
Step 4: Calculate activation energy:
\( E_a = -m \cdot R = 43761.75 \times 8.314 = 363835 \, \text{J/mol} = \boxed{363.8 \, \text{kJ/mol}} \)
Step 5: Calculate pre-exponential factor \( A \):
\( A = e^{27.36} \approx \boxed{7.59 \times 10^{11} \, \text{s}^{-1}} \)