Understanding Arrhenius Plots: ln(rate) vs 1/T for Determining Activation Energy

  • Thread starter Thread starter pwnzorz
  • Start date Start date
  • Tags Tags
    Arrhenius Plot
Click For Summary
SUMMARY

Arrhenius plots can be constructed using ln(rate) versus 1/T to determine activation energy, particularly in heterogeneous catalysis. This approach is valid because the rate constant (k) is related to the reaction rate, allowing ln(rate) to serve as a proxy for ln(k). The Arrhenius equation, k=A*exp(-Ea/(R*T)), underpins this relationship, where Ea represents activation energy, R is the gas constant, and T is temperature in Kelvin. Clarification is necessary as some practitioners may interchangeably use "rate" and "rate constant," which can lead to confusion.

PREREQUISITES
  • Understanding of the Arrhenius equation and its components (k, A, Ea, R, T).
  • Familiarity with the concept of activation energy in chemical kinetics.
  • Knowledge of heterogeneous catalysis and its significance in reaction rates.
  • Basic skills in interpreting logarithmic plots and their implications in scientific data.
NEXT STEPS
  • Study the derivation and applications of the Arrhenius equation in various chemical reactions.
  • Explore the differences between homogeneous and heterogeneous catalysis and their impact on reaction rates.
  • Learn how to construct and interpret Arrhenius plots using experimental data.
  • Investigate the implications of using ln(rate) versus ln(k) in kinetic studies and the potential for misinterpretation.
USEFUL FOR

Chemists, chemical engineers, and researchers in the field of kinetics and catalysis will benefit from this discussion, particularly those involved in analyzing reaction mechanisms and activation energy determination.

pwnzorz
Messages
1
Reaction score
0
I have been noticing that in some cases that Arrhenius plots are sometimes plotted ln(rate) vs 1/T as opposed to ln(k) vs 1/T to find activation energy especially in hetereogeneous catalysis.

Isn't the Arrhenius law k=A*exp(-Ea/(R*T))? why can ln(rate) also be plotted vs T^-1 to find activation energy?
 
Engineering news on Phys.org
pwnzorz said:
I have been noticing that in some cases that Arrhenius plots are sometimes plotted ln(rate) vs 1/T as opposed to ln(k) vs 1/T to find activation energy especially in hetereogeneous catalysis.

Isn't the Arrhenius law k=A*exp(-Ea/(R*T))? why can ln(rate) also be plotted vs T^-1 to find activation energy?
Is it possible that some people refer to the rate constant simply as "rate?" Otherwise, the rate would have to be constant for the equation to apply.

Chet
 

Similar threads

Replies
8
Views
4K
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
8K
Replies
3
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
9
Views
8K
  • · Replies 2 ·
Replies
2
Views
4K
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 131 ·
5
Replies
131
Views
10K