Talk:Arrhenius plot

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What are the numerical values of the constants in the equation that is plotted? The text says:

the value of the extrapolated "y-intercept" will correspond to ln(A)

What is the numerical value of ln(A) at the "y-intercept" for the plot shown?

Truncating the graph at x=0.0015 seems to suggest that the y-intercept at 4.1e-4. The value of the y-intercept would be higher at x=0. The graph seems misleading to me. -Ac44ck (talk) 00:28, 18 June 2009 (UTC)

The value of "A" here is A=4.08299x10^9 L/mol-s. The extrapolated value in the graph is 4.1e-4 cm^3/mol-s = 4.1e-7 L/mol-s. One of the exponents seems way off -- if the value at 1/T = 0.0015 can be called the "y-intercept." -Ac44ck (talk) 07:03, 1 September 2010 (UTC)

Added worked example based on the given graph. Agreed, the entire process could be done better and more easily in a spreadsheet without ever producing a graph. As the example gives neither the data points nor the regression equation for the graph, the added example uses an 'old school' procedure. - Ac44ck (talk) 12:50, 16 March 2013 (UTC)

I agree, the graph is misleading since the x-axis stops at 0.0015. It would be wise to add the // in the x-axis to indicate a part of that axis is "missing". -- 192.12.88.13 (talk) 18:26, 7 September 2010 (UTC)

Yes, the graph is misleading, but many textbooks and research papers have similar Arrhenius plots which do not show the true y-axis. The reason is that k varies very rapidly with T so that it is usually only convenient to study a small range of T, here 590-660 K. The true y-axis at x = 1/T = 0 corresponds to infinite T and is very far from the experimental range, or "way off the graph". If we insist on a graph which goes to x = 0, the measured points will cover only a small part of the graph which would make it hard to understand. (It would also have led to inaccurate determination of E_a in the days when slopes were determined manually with graph paper, although this is no longer true since slopes are now determined by linear regression with a spreadsheet such as Excel.)

So instead of replacing the graph, I will explain in the article that the left side of the graph is not in fact the y-intercept, and that ln A corresponds to the true y-intercept which must be calculated. Putting a // on the x-axis is also a good idea which I have seen in some textbooks, if someone has the software to re-draw the graph that way. Dirac66 (talk) 20:03, 15 March 2013 (UTC)

To Ac44ck: Yes, the article is better now with the worked example and the // on a new graph. :Not much can be done about the uses of 'k' and 'K' as most (or all) chemistry textbooks use these same letters for rate constant (from Konstante in German), Boltzmann constant, kilo and Kelvin, not to mention equilibrium constant which is often mentioned in the same discussion. Perhaps writing the value of Ea in both J and kJ will help a little. Dirac66 (talk) 21:07, 17 March 2013 (UTC)

Plots of k(T) are incorrect at high T

The article now contains one Arrhenius plot of ln k vs. 1/T which appears to be correct, and also two non-Arrhenius plots of k vs. T which appear to contain a serious mathematical error. These graphs show that at large T, k(T) increases rapidly like an exponential function. But this is NOT what the Arrhenius equation predicts. If $k=Ae^{-E_{a}/RT}$, then as T → ∞, E_a/RT → 0, e^{-E_a/RT} → 1 and k → A which is constant. This agrees with simple collision theory: at high enough T all the pairs of colliding molecules would have enough energy to react so k is just the total collision frequency.

I propose that we delete the two incorrect (and unsourced) graphs from the article. The correct linear graph is also now unsourced, but this is easy to fix as many general chemistry and physical chemistry texts contain similar graphs which we can cite, using words such as "similar linear graphs are found in ...". Dirac66 (talk) 23:29, 12 September 2017 (UTC) pra

I understand now. These graphs are actually valid in the range T << E_a/R, which is true for most reactions (unless E_a is very small) at practical temperatures. At much higher T, k will level off to a constant value or plateau. So I will not delete the 2 graphs, but instead I will explain in the text that at much higher T the function k(T) levels off. (Assuming that the reaction is unchanged at much higher T, which is rarely true.) Dirac66 (talk) 02:30, 13 September 2017 (UTC)