I am working through the Chapter 14 homework problems and was wondering if you would like us to know how to plot the Arrhenius plot for questions like 65, 67, and 99. I tried working through them algebraically multiple times and cannot come to the right answer, so I am assuming I must plot them in order to find the true slope.

Ella_M.....You're correct. In concept, these problems should be done graphically by plotting ln(k) vs 1/T as in Example 14.7 (Tro, p 645). This is summarized in slides 22-23 of the Chapter 14 PowerPoint presentation. A least squares fit gives you the slope, which is equal to -Ea/R. I wouldn't expect you to do all of that on an exam but it's fair game and worthwhile in the homework. The 2-point form of the Arrhenius equation, using any two data points will provide a quicker estimate (just be careful with the math) of Ea (slide 23).

Would you like us to include all k values in the predicted rate law's or just sum them all up to an all inclusive value of k?

Ella_M.....When predicting a rate law based on a proposed mechanism, the most import thing is understanding how to determine the proper exponent (order) for each concentration term. The algebraic process for doing so will lead you to how the overall k equates to the individual k's for the various steps. Example 14.9 (Tro, pp 651-653) is a good illustration of this process.

Also, in general, keep in mind the overall "rules" for mechanisms:
1. Steps must add up correctly to the actual overall reaction. (Keep things simple! Look at the overall reaction and see which bonds need to brake in the reactants and which are being formed in the products.)
2. No step in a mechanism can have more than two reactant particles (molecules, ions, or atoms).
3. The predicted rate law for any individual step in a mechanism has exponents (orders) equal to the coefficients in the balanced equation for that step.

I have a few questions to clear up some of my confusion surrounding Chapter 14. The first is about instantaneous rate. If I interpreted this correctly, instantaneous rate is calculated by finding the slope of the line tangent the curve at the point of interest. I know how to do this looking at a graph, but struggle to do it without one, like was asked on Quiz 5. So I was eager to see your work as to how you did question 1b on Quiz 5, I realized that you just found the average rate of reaction at t = 30 seconds. There was no tangent involved, so now I am left quite confused. Where is my disconnect happening? My second question is about Reaction Mechanisms. We will not be responsible for writing the mechanisms, correct? Just answer questions asked about the mechanisms and what not? Help would be much appreciated. Thank you!

Hi Dr. Patty and Dr. Bob, I hope you are both doing well as well as you guys' family! For number 33 a. iii, I know we would have to find the instantaneous rate by finding the slope of the tangent, but in order to find the instantaneous rate of formation for the product, how do we plot the values? I know for this problem, HBr has a coefficient of 2 so it would be twice the slope of the reactant (Br2) and would be positive. Thanks for your time and help!

Ruwayd.....The instantaneous rate is slope of the tangent of the line drawn that touches at the given time (25 sec for this one). With graphing software drawing that line is easy, but here just approximate a tangent the best you can. Your answer may not be exactly the same as in the book, but if it is of the same order of magnitude, then go on to the next problem. You can also pick data points on the curve a very short distance on either side of the 25 sec point, draw that line and use that slope. That line should be parallel to the actual tangent point.

Coleton, Ruwayd, and others asking about instantaneous rates.... See Dr. Patty's response just above. On Quiz 5, to which Coleton refers, there was a table of data provided. Just take the two points closest to the 30 sec point and calculate the average rate between them, i.e., (delta M)/(delta time) as shown on the answer key. This provides the best estimate of the slope of the tangent at the 30 sec point. See Figure 14.1 (Tro, p 625) for a good example. Also, that is essentially the same as the HBr reaction in question 14.33a.

For the question on the outline asking for the possible mechanism of rate = k[NO]^2 I was wondering if the following would work?
Slow: 2NO —-> N2O + O
Fast: NO + O —-> NO2
Overall: 3NO ——> N2O + NO2

Joe,

Yes that is a perfectly plausible mechanism. The answer on the Annotated Lecture Outline as I did it on the recorded lecture, uses the option of "fusing" the two reactants to make an intermediate and your example uses the option of making one product and then the "left over" O is the intermediate.

Good job!

14.33
Have you looked at Figure 14.2 on page 625 of on page 3 of the Chapter 14 outline? That is for HI, but the principle is the same. The HI or the or HBr forms at twice the rate that the Br2 (or I2 or H2) disappears so they throw in a ½ factor. On Figure 14.2 at 50 seconds the H2 has decreased by about 0.62 M, and the HI has increased by about 1.24 M. That is how you draw the curve for the appearance of HI. Then the rate of formation of HI is twice that of the rate of disappearance of Br2. I’d personally rather see the rate in iii as 0.026 because that is the rate of formation of HI at 50 seconds while the rate of disappearance of Br2 is 0.013 (the answer they gave). Bottom line is the most important thing is that you understand why the curve for HBr formation grows twice as fast.

14.49, 50, 51

Complete answers are posted on D2L.

Hi Dr. Bob and Dr. Patty, I hope you are both doing well! For problems like 63 where they give us the rate constant values as well as the temperature, should we worry about plotting the points or should we just know how to solve the problem with a given equation? Thanks for your help and time!

For question 77 c, I know that we can't use the rate law for the slow step since it includes the intermediate which isn't ideal for the overall reaction. Would we have to find the overall reaction by getting rid of the rate law of the slow step rate= K3[Cl][CHCl3] and set [cl] equal to the other rate law so we could substitute the intermediate? If not, how do we find it? Thanks again for your help!

14.63

When given two rate constants k and two temperatures, the activation energy can be calculated using the slope of the line formed by the two points. Most student just want to plug and chug into a carefully-memorized, soon-forgotten equation, but should always remember that the equation is really just delta y/delta x, i.e., the slope of the linear,natural logarithmic form of the Arrhenius equation.

14.77c

You are basically on the right track. Use the rate law for the slow (2nd) step. Replace the intermediate by setting the forward and reverse rate laws for the 1st (fast) step equal and solving for the intermediate.

The third fast step does not affect the rate, just as second fast steps that follow a slow first step do not affect the rate.

Frequency Factor

In the lecture where everyone was too excited because it was almost spring break, I said that "frequency factor" was essentially a useless number in terms of showing how a reaction happened. It is simply the intercept of a plot of the natural logarithmic form of the Arrhenius equation, i.e, a plot of ln k versus 1/TEMP (not time!). What IS important about the plot of the linear equation is the slope because the slope of the line allows for determination of the activation energy, Ea. Slope = -Ea/R.

Note that in the Dr. Patty annotated outlines (page 23-24 as numbered in the upper right) (posted on D2L or the drive link you were sent), the slope of that line is derived and gives the often carefully-memorized, but soon-forgotten equation that most do not associate with the Arrhenius equation, a line, or slope. Be sure to understand how and why this equation works.

117

For this problem, can you just use the last fast equilibrium equation and the slow equation to determine the rate law?

14.117

This is a real algebra mind stretcher. You need to use the slow step for the rate law, as you stated, but then you must use the 2nd step as a fast equilbrium to solve for the intermediate [ClCO]. However, the rate expression for the second step has the intermediate [Cl] and you must then use the fast equilibrium first step rate laws to get [Cl] so can substitute into the second step law to solve for [ClCO]. After that you can substitute into the third step rate law. It's the same process we used on two steps, just repeated done more time.

It's not that hard if you, keep track of things. There is a square root which accounts for a 1/2 power.

Chapter 14 mechanisms to rate law problems.

Please listen to the lecture and follow the annotated lecture outlines to lead you through these problems. Some examples are # 7 on page 44 of the outline, # 4 on page 41, # 1c, page 38, bottom of page 29. Again, none of this makes sense without listening to the lecture, but once you get it, it is easier than it originally looks.

Ch.14 Question 35, it asks you to make a rough sketch of the plot of [A] vs time. Since this is a 1st order reaction, I wrote this as a negative sloped straight line - the book says for 1st order rxn, a plot of the natural log of the reaction conc. yields a straight line.... and now as I typed that out I think I may have found my answer - is this because I overlooked the stipulation that the question didn't ask for the natural log of [A], it just wanted [A], not making it a straight line?

Here is a question many are asking and my answer

For #2 at the end of the chapter 14 powerpoint (and I guess for any other similar questions), how do you know which two rates to use to find the order? For example, in finding [A]^x in this specific problem.

Dr. P.'s answer:

I suggest looking at problem # 1 first since there are only 2 reactants. To decide which two experiments to compare, you pick one where one reactant does not change and the other does. Then the only reason the rate changes is because that one reactant concentration changed. Then it is a valid comparison. If both reactants change, it is tough to know which one affected the rate the most and how much.

Now problem # 2 is a bit more challenging because there are 3 reactants, so more strategy is needed. Here choosing experiments 2 and 3 works to find out how C is affecting the rate because A and B concentrations are the same. That is the red comparison. Next I see that in experiment 1 and 3 only [A] changes while [B] and [C] are the same. This is the blue comparison. Fortunately A is not affecting the rate, so we no longer have to worry about its concentration. No matter what happens with A, the rate is not affected. So then to find out how B is affecting the rate, experiments 2 and 4 show that B has changed, but C is the same (and A does not matter), so that is the gray comparison.

Ch.14 Question 35,

Yes you answered your question. All they wanted in the plot was [A] versus time and not ln[A] versus time.

On textbook page 632, it gives you the equation for rate with multiple reactants - Rate=k[A]^m[B]^n, where m and n are the respective rxn order numbers. Alternatively (on the test for instance), could we write something like Rate=k[A]^n(sub 1)[B]^n(sub 2), only using n but adding subscripts to denote the different rxn order numbers? Obviously this is merely syntactic and doesn't make a difference in the answer, but this just makes a lot more sense in my head.

Chem 10123 Students,

IMPORTANT MESSAGE:

"FREAK"QUENCY FACTOR:

I said in lecture, that this is a meaningless number. It tells us nothing useful. Please bury it somewhere. See page 24 of the annotated outlines for the derivation of the useful equation from the slope of lnk vs 1/Temp. That is what is important - the relationship of rate constant (hence rate) and temperature, and how this is related to the activation energy needed for molecules to react, i.e, for a reaction to go.

All radioactive decay follows first order kinetics. See page 14 of the Chapter 14 outlines.

Counts from a Geiger counter are proportional to concentration. Use the counts as concentration.