Review Sheet - Exam 4 - CH302 - McCord


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Textbook Chapters for Exam 4

Chapter 15 sections 1-6, 8-9 (skip 7). Chapter 20, sections 1-3 (See Christina's Help Sheet and refer to H15). You might also want to refer to the equation sheet that is available on our web site for all the formulas you need to know for kinetics and electrochemistry.

Chapter 15 - Kinetics

The Various Rates within an Overall Reaction

Below is a typical generic reaction. The rates in each reactant and product are then given and normalized to all be equal to the overall reaction rate.

\[a\,{\rm A} + b\,{\rm B} \rightarrow c\,{\rm C} + d\,{\rm D}\] \[{\rm overall\;rxn\;rate} = {-\Delta[{\rm A}]\over a\Delta t} = {-\Delta[{\rm B}]\over b\Delta t} = {+\Delta[{\rm C}]\over c\Delta t} = {+\Delta[{\rm D}]\over d\Delta t}\]

Collision Theory

Know the two criteria for an effective collision. (section 15.8 and class notes). Part of this theory involves the graphical depiction of the energy criteria (aka, the activation energy). This is shown as Ea on a potential energy diagram (Figure 15.11) or on a distribution diagram (Figure 15.12).

Reaction Mechanisms

Know how to interpret and write a plausible Reaction Mechanism for a chemical reaction. What are elementary steps? What is the molecularity of a reaction (uni- or bimolecular?). What is the rate-determining step and how does that affect your version of the rate law. What is a pre-equilibrium condition when looking at a multistep mechanism? This is a fast reaction preceding a slow reaction and is in Zumdahl's book in section 15.6. Know how to use these fast equilibrium steps to “get rid of” your intermediate in your rate equation. DON’T worry about section 15.7 - the Steady State Approximation is the section we are skipping for now although you may see it in future courses.

How do you get the kinetic version of an equilibrium constant? Zumdahl comes close to showing this but he stops a little short. The bottom line is that you set the forward and reverse rate equations equal to each other and then collect concentration terms on the right and the rate constants on the left. You get the following...

A ⇌ B

\[k_{\rm f}[{\rm A}] = k_{\rm r}[{\rm B}]\] \[{k_{\rm f}\over k_{\rm r}} = {[{\rm B}]\over [{\rm A}]}\]

and therefore the equilibrium constant:

\[K_{\rm eq} = {k_{\rm f}\over k_{\rm r}}\]

Potential Energy Diagrams

These are the plots that show potential energy of the reactants and products and the transition state in between the two. See Figure 15.11 in Zumdahl for a good example, plus I did go over this in class. Here’s a A goes to B example as a single step:

Know HOW to interpret the diagram. How to get ΔE and Ea,f , and Ea,r. Know transition state theory (activated complex) and how it relates to a potential energy diagram - where does it occur?

Sometimes the “potential energy” is different thermodynamic state functions, so you could be dealing with ΔH, ΔE, or ΔG on this diagram (the energy difference between A and B in the diagram above).

Also know how a multistep reaction would look on a diagram. I showed this in class. Zumdahl only shows an example of a 2 step reaction scheme in Exercise 75(d) at the end of the chapter - there is a picture there. Notice the 2 humps in the middle. Those “humps” are the 2 transition states for the 2 steps in the mechanism.

Temperature Dependence of k

The Arrhenius Equation:

\(k=A{\rm e}^{-E_{\rm a}/RT}\)

 This shows the temperature dependence and the activation energy dependence of k. And, there’s that special pre-exponential factor A in it. This will be different for every reaction - of course Ea is different too. We know that for relatively small temperature changes that A and Ea do not change much witch allows us to algebraically get rid of the A and write:

\[\ln \left({k_2\over k_1}\right) = {E_{\rm a}\over R}\left({1\over T_1} - {1\over T_2}\right)\]

This is another form of the Arrhenius equation - one in which you should now be very familiar with (remember the Claussius-Clapeyron and the van’t Hoft eqations?). And, let’s remember that half-life is inversely proportional to k so that we can also use half-life ratios in place of the k2/k1 ratio...

\[\ln \left({t_1\over t_2}\right) = {E_{\rm a}\over R}\left({1\over T_1} - {1\over T_2}\right)\]

Note the “flipping” of the positions for t1 and t2 which are the half-lives at T1 and T2. One more thing... those times don’t have to be half-life times. Any kind of time will work: quarter-life, third-life, 3/7th’s-life... any convenient stopping point. Think of it like running a race. You can put the finish line anywhere you want and time the runners. As long as the reference points are the same, you can use ratios of times (t1/t2) in place of the rate constant ratios (k2/k1). Just make sure you remember to put them in the correct positions. Remember, the FASTEST reaction will have the SMALLEST time and vice-versa. No matter what your “finishing line” is, rate is always inversely proportional to time which means k is proportional to 1/t.

Catalysis

How do catalysts affect reaction rates? How is this represented on a potential energy diagram? What is the difference in a homogeneous and a heterogeneous catalyst? How does a catalyst affect the thermodynamics of a reaction ? What are catalysts in living systems called? There is a whole 2 page spread on this in your book called “Chemical Insights” (in section 15.9).

How does catalysis effect the distribution plot? Take a look at Figure 15.16 in this section. Note how the activation energy is smaller when catalyzed. This means that more of the molecules have sufficient energies to react.

Another reminder to take a look a the Kinetics Formula sheet that I have on our web page under help sheets.

Chapter 20 - Nuclear Chemistry

Please refer to Christina's Help/Review Sheet for the Nuclear stuff.