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. This is just reaction stoichiometry applied to rates.
\[a\,{\rm A} + b\,{\rm B} \rightarrow c\,{\rm C} + d\,{\rm D}\] \[{\rm overall\;rxn\;rate} = {-\Delta[{\rm A}]\over a\Delta t} = {-\Delta[{\rm B}]\over b\Delta t} = {+\Delta[{\rm C}]\over c\Delta t} = {+\Delta[{\rm D}]\over d\Delta t}\]
Here are three plots showing concentration of reactant A decreasing with time via zero order kinetics, first order kinetics, and second order kinetics. The half-lives are also show as boxes under the curves.
Note how the half-lives vary for each of the different orders. Only first order has a constant half-live regardless of the concentration. Zero order has consecutively shorter and shorter half-lifes, each 1/2 the previous. Second order has consecutively longer and longer half-lives, each 2x (twice) the previous.
| formula | Zero Order | First Order | Second Order |
|---|---|---|---|
| rate law | \[rate = k\] | \[rate = k[{\rm A}]\] | \[rate = k[{\rm A}]^2\] |
| integrated rate law | \[[{\rm A}]_0 - [{\rm A}] = kt\] | \[\ln[{\rm A}]_0 - \ln[{\rm A}] = kt\] \[\ln\left({[{\rm A}]_0\over[{\rm A}]}\right) = kt\] |
\[{1\over[{\rm A}]} - {1\over[{\rm A}]_0} = kt\] |
| straight line plot | \[ [{\rm A}] = -kt + [{\rm A}]_0\] | \[\ln[{\rm A}] = -kt + \ln[{\rm A}]_0\] | \[{1\over[{\rm A}]} = kt + {1\over[{\rm A}]_0}\] |
| half-life | \[t_{1/2} = {[{\rm A}]_0\over 2k}\] | \[t_{1/2} = {\ln 2\over k}\] | \[t_{1/2} = {1\over k[{\rm A}]_0}\] |
Remember that you can always get the OTHER reactants and product concentrations via the stoichiometry of the reaction. You do NOT have to have separate rate equations for the other species. Stoichiometry goes a long way at getting you vital information about a reaction.
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. Know how to use these fast equilibrium steps to “get rid of” your intermediate in your rate equation.
How do you get the kinetic version of an equilibrium constant? You simply 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...
\[{\rm A \rightleftharpoons B}\] \[rate_{\rm f} = rate_{\rm r}\] \[k_{\rm f}[{\rm A}] = k_{\rm r}[{\rm B}]\] \[{k_{\rm f}\over k_{\rm r}} = {[{\rm B}]\over [{\rm A}]}\]
and therefore the equilibrium constant:
\[K = {k_{\rm f}\over k_{\rm r}}\]
These are the plots that show potential energy of the reactants and products and the transition state in between the two. Here is the very simple elementary step of A changing to B in 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 on the diagram? Sometimes the “potential energy” is different thermodynamic state functions, so you could be dealing with ΔH, ΔU, 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. You get as many "humps" as you have steps in the mechanism. The peaks are transition states and the valleys are intermediates.
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 equations?). 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.
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?
How does catalysis effect a distribution plot? The activation energy is smaller when catalyzed. This means that more of the molecules have sufficient energies to react.