Graph of ln(A/R) vs 1/T: Arrhenius Equation and Homework Equations

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SUMMARY

The discussion centers on the graph of ln(A/R) versus 1/T as derived from the Arrhenius equation, k = Ae^(-E/RT). Participants clarify that A and R are constants independent of temperature, leading to the conclusion that the graph should represent a straight line parallel to the x-axis, with a constant value of ln(k/R) + E/RT. Confusion arises regarding the interpretation of the y-intercept and the nature of the graph, with participants emphasizing that ln(k/R) cannot be treated as a constant due to the temperature dependence of k.

PREREQUISITES
  • Understanding of the Arrhenius equation and its components (A, E, R, T)
  • Knowledge of natural logarithms and their properties
  • Familiarity with linear equations in the form y = mx + c
  • Basic concepts of graphing functions and interpreting intercepts
NEXT STEPS
  • Study the derivation and implications of the Arrhenius equation in chemical kinetics
  • Learn about the significance of the activation energy (E) in reaction rates
  • Explore the graphical representation of logarithmic functions and their slopes
  • Investigate the relationship between temperature and reaction kinetics in detail
USEFUL FOR

Chemistry students, educators, and anyone involved in chemical kinetics or thermodynamics who seeks to understand the graphical representation of the Arrhenius equation.

subhradeep mahata
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Homework Statement


Draw a probable graph of ln (A/R) vs 1/T.

Homework Equations


Arrhenius equation: k= Ae-E/RT

The Attempt at a Solution


Taking natural log on both sides of arrhenius equation :
ln k = ln A - E/RT
ln k - ln R = ln A - E/RT - ln R (subtracting ln R from both sides)
ln (k/R) = ln (A/R) - E/RT
rearranging terms and comparing it with y=mx + c,
c= ln (k/R) and m = E/R
upload_2018-7-16_20-21-51.png

According to me the graph should be like this. In the answers, the graph is a similar one, except that there is no intercept given. I am confused. Any help is appreciated.
 

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Are you sure you've got your terms right? A in the Arrhenius equation is a constant independent of T. And you can't write ln(k/R) = c because k certainly varies with T.
 
@mjc123
ln(k/R)=c, by 'c' i don't mean constant, i mean the y-intercept.
Regarding the terms, yes i am sure. Though A and R are both independent of T, the graph could also have been straight line parallel to x axis. Any help is appreciated.
 
The graph must be a straight line parallel to the x axis, with a constant y value of ln(k/R) + E/RT. But it seems a strange graph to plot anyway.
BTW, c in y = mx + c is a constant. It's not just the value of the y-intercept, it's telling you that the only x-dependence of y is in the term mx.
 
Ok, got it. Should i send a screenshot of that page of the book? (for confirmation)
 
If it has the text of the question, that would help.
 
The question is : Which of the following graphs is correct ?
the answer : fig 8.8 first from left
upload_2018-7-17_20-52-13.png

Even the very next graph is wrong.
log in which base 10 is not mentioned is natural log according to this book.
There were only 3 options, the other two are absolutely wrong, simple arrhenius plots.
 

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That's definitely wrong. A is a constant in the Arrhenius expression. If A varied with temperature, you couldn't get E from the slope of the ln k vs. 1/T plot.
 
Got it, thanks.
 

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