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

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Discussion Overview

The discussion revolves around the graph of ln(A/R) versus 1/T as it relates to the Arrhenius equation, focusing on the interpretation of the graph and the constants involved. Participants are exploring the implications of the equation and its graphical representation, with an emphasis on homework-related inquiries.

Discussion Character

  • Homework-related, Technical explanation, Debate/contested

Main Points Raised

  • One participant presents an attempt to derive the graph from the Arrhenius equation, suggesting that ln(k/R) can be treated as a constant in the context of the graph.
  • Another participant questions the treatment of A as a constant, emphasizing that k varies with temperature and challenging the interpretation of ln(k/R) as a constant.
  • A different viewpoint suggests that the graph should be a straight line parallel to the x-axis, indicating a constant y value, but expresses confusion about the appropriateness of the graph.
  • Further clarification is provided that in the equation y = mx + c, 'c' refers to a constant, not just the y-intercept, which adds to the complexity of the discussion.
  • Participants discuss the context of the homework question and the options provided in the textbook, with one participant asserting that the other options are incorrect.
  • There is a consensus that A is a constant in the Arrhenius equation, with one participant arguing that if A varied with temperature, it would complicate the determination of E from the slope.

Areas of Agreement / Disagreement

Participants express disagreement regarding the interpretation of the constants in the Arrhenius equation and the nature of the graph. While some agree that A is a constant, others challenge the implications of this on the graph's representation and the treatment of ln(k/R).

Contextual Notes

There are unresolved assumptions regarding the treatment of constants in the Arrhenius equation and the implications for the graph. The discussion also highlights potential ambiguities in the textbook options provided for the homework question.

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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