exam 4
12/5
exam 4
12/5
The exam will be on Wednesday, 12/5, from 7:30 pm to 9:00 pm in our regular class room, BUR 106.
What we provide on Exams We will provide all students with:
Note that the periodic table handout is available on the gchem site in the appendix under "Exam Preparation". Here is a direct link to the Periodic Table Handout used for exams.
Coverage: Exam 4 covers all the material that was covered on LE's 24-37 and HW07-HW11. The exam covers Chapter 11 on Kinetics and Chapter 12 on Electrochemistry from the gchem site.
Questions: The exam has 20 multiple choice questions. The questions will have an equal weight of 5 points each. There are 10 kinetics questions and 10 electrochemistry questions. Remember to bring your calculator and come prepared. We will only grade you by what is bubbled in on the answer sheet. We will not look at your exam copy for answers, nor consider them in any way. Bubble carefully and correctly.
We can measure the rate of a reaction via any of the species in the over reaction:
\(a {\rm A} + b {\rm B} \rightarrow c {\rm C} + d {\rm D}\)
The rate of the "overall reaction as balanced" means you are measuring the rate of entire reactions as written. So each reactant and product is running at a scaled up version of the overall reaction. The "scaled up" amount is the coefficient for that species. For example, reactant A is running "a" times faster than the overall reaction. So to get ALL species to match up, we normalize the rates to the rate of one mole of reaction (the whole thing).
rate of reaction = \(-{\Delta[{\rm A}]\over a\Delta t} = -{\Delta[{\rm B}]\over b\Delta t} = {\Delta[{\rm C}]\over c\Delta t} = {\Delta[{\rm D}]\over d\Delta t}\)
Make sure you can convert the rate of any one species into the rate of another species. This is really just saying to know how stoichiometry works when dealing with rates.
Rates of reactions are dependent on the concentrations of the reactant (and sometimes product) species. This is generalized by showing the rate law for a reaction of A, B, and C going to products.
\({\rm rate\;of\;rxn} = k [{\rm A}]^x[{\rm B}]^y[{\rm C}]^z\cdots \)
\(x\) is the order in A, \(y\) is the order in B, and \(z\) is the order in C. Sum those orders all together and you'll have what is called the overall order. The overall order is important to know and the units for the rate constant, \(k\), depends on the overall order.
| name | zero order | first order | second order |
|---|---|---|---|
| rate law | \({\rm rate} = k\) | \({\rm rate} = k[{\rm A}]\) | \({\rm rate} = k[{\rm A}]^2\) |
| 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 form (\(y = mx + b\)) |
\([{\rm A}] = -kt + [{\rm A}]_0 \) | \(\ln[{\rm A}] = -kt + \ln[{\rm A}]_0 \) | \({1\over[{\rm A}]} = kt + {1\over[{\rm A}]_0}\) |
| half life | \(t_{1/2} = {[{\rm A}]_0\over 2k}\) | \(t_{1/2} = {\ln 2\over k}\) | \(t_{1/2} = {1\over [{\rm A}]_0 k}\) |
| to actually plot on x,y graph \(y\) = conc \(x\) = time |
\(y = -kx + {\rm A}_0 \) | \(y = {\rm A}_0 e^{-kx} \) | \(y = \left(kt + {1\over{\rm A}_0}\right)^{-1}\) |
Arrhenius Equation: \(k = A\,e^{-E_{\rm a}/RT} \)
\(E_{\rm a}\) is the activation energy for the reaction and it is always positive. The \(A\) is known as the Arrhenius "pre-exponential factor" and is unique for every reaction - it has the same units as the rate constant, \(k\). The factor can be eliminated by setting a ratio for two values of k at two different temperatures. Dr. McCord calls this the "user friendly" version of the Arrhenius equation.
\[\ln\left({k_2\over k_1}\right) = {E_{\rm a}\over R}\left({1\over T_1} - {1\over T_2}\right)\]
Notice that this is the third time you've seen this basic formula in CH302. The first time, it was the Clausius-Clapeyron equation for vapor pressures at two temperatures. The second time, it was the van't Hoff equation for the equilibrium constant, K, at two temperatures. Now the same set up works again but for rate constants. You should now be a "pro" at using this type of formula.
One more thing... the Arrhenius equation can be rewritten to be in a "straight line plot" form, \(y=mx+b\). That form is the following
\(\ln k = {-E_{\rm a}\over R}\left({1\over T}\right) + \ln A \)
So your plot is ln(k) vs 1/T and the slope will be equal to –Ea/R, and the y-intercept is equal to ln(A).
reduction is the gain of electrons and occurs at the cathode.
this means that electrons always flow into the cathode from the external circuitry
oxidation is the loss of electrons and occurs at the anode.
this means that electrons always flow out of the anode into the external circuitry
If you really "think like a chemist" then you should be thinking about the push/pull nature of the electrochemical cell. The anode is pushing electrons out while the cathode is pulling electrons in. The standard potential is really the addition of two standard potentials, one is the driving force of the reduction (\(E^\circ_{\rm red}\), the pull) and the other is the driving force of the oxidation (\(E^\circ_{\rm ox}\), the push). Put these together and you get:
\[{\cal E}^\circ_{\rm cell} = {\cal E}^\circ_{\rm red} + {\cal E}^\circ_{\rm ox} \]
You must "flip" a reaction from the table of standard potentials in order to get an oxidation potential. Also remember, when you flip a reaction, you change the sign on the state function, which in this case is \(E^\circ\). Realize that you have to flip the oxidation reaction anyway when you write out the balanced equation. So this is the best way to do it.
Shortcut method Unfortunately, we tend to teach you the shortcut first and it leads to misconceptions. None-the-less, the fact is that the standard potential for any electrochemical cell is simply the difference in the standard potentials of the two half cells.
\[{\cal E}^\circ_{\rm cell} = {\cal E}^\circ_{\rm cathode} - {\cal E}^\circ_{\rm anode} \]
Realize that the standard cell potential (or reaction potential) can be positive (+) or negative (–). Also realize that the \(-E^\circ_{\rm anode}\) term IS the oxidation potential which is the push of \(E^\circ_{\rm ox}\) shown in the first equation.
We "tell" you which 1/2 cell is the anode by writing it on the left side in the shorthand notation:
anode \(\bigl|\) anodic solution \(\bigl| \bigr|\) cathodic solution \(\bigr|\) cathode
Each of the vertical lines, |, marks a distinct phase-change in the cell as you traverse from the anode to the cathode through the cell. The double line, ||, is the salt bridge. If the 1/2 reaction has no conductor directly in the reaction (like say the H+/H2 reaction for S.H.E.), you use an inert electrode such as Pt, Au, or C(s,graphite). If a solid or gas is an active component, you have to show it directly next to the electrode. Solution species will be those listed adjacent to the salt bridge and the order of listed species does NOT matter.
To illustrate cases with solids and gases, consider a cell made using the Ag/AgCl (E°= +0.22V) and the S.H.E. (E° = 0.00 V) as the two half reactions. The correct way to show this with the Ag/AgCl as the anode is:
Ag(s) \(\bigl|\) AgCl(s) \(\bigl|\) KCl(aq) \(\bigl| \bigr|\) HCl(aq) \(\bigr|\) H2(g) \(\bigr|\) Pt(s)
Note how in this case the needed Cl- ions in the silver cell are shown as a full ionic formula of KCl(aq) - which would certainly provide the needed chloride ions. In a similar manner, the needed H+ ions are provided via hydrochloric acid, HCl(aq). Sometimes the ions are shown by themselves, and other times they are shown as a complete formula with counter ions. So you see, there is some flexibility in the details in writing out a shorthand cell notation.
standard conditions: \(\Delta G^\circ = -n F {\cal E}^\circ \)
non-standard conditions: \(\Delta G = -n F {\cal E} \)
\(-n F {\cal E}^\circ = \Delta G^\circ = -RT\ln K\)
Non-standard cell potentials are calculated via the Nernst Equation:
\[{\cal E}_{\rm cell} = {\cal E}^\circ_{\rm cell} - {RT\over nF}\ln(Q) \]
The temperature (T) is usually 298.15 K for most of our problems. The faraday is always 96485 C/mol of e-, and R is always 8.314 J/mol K. Therefore you can use either of the following versions of the Nernst equation to save yourself from punching in all those long numbers into your calculator.
\[{\cal E}_{\rm cell} = {\cal E}^\circ_{\rm cell} - {0.02570\over n}\ln(Q) \]
\[{\cal E}_{\rm cell} = {\cal E}^\circ_{\rm cell} - {0.05916\over n}\log(Q) \]
FYI... calculate \(RT/F\) and you'll use (8.314)(298.15)/96485 to get 0.02569 which is where that 0.02570 came from up above. You have to use natural log for that number though. If you want to use log base 10 (log), then you'll need to multiply the log 10 version of Q by ln(e) which is 2.303 and that is why the 0.05916 number is there.
Know your basic logarithm algebra laws
\(x = 10^{\log(x)}\hskip36pt x = e^{\ln(x)}\)
\(\log(xy) = \log(x) + \log(y)\)
\(\log(x/y) = \log(x) - \log(y)\)
\(\log(x)^y = y\log(x)\)
Now lets combine two of the above relations to get our "formula" for natural log using base10 log.
\(x = 10^{\log(x)}\) Now take the natural log (ln) of both sides.
\(\ln(x) = \ln(10)^{\log(x)}\) which will solve to
\(\ln(x) = \log(x)\ln(10) = \log(x)\cdot 2.302585...\)
So, natural log of any number is the same as the log base10 number times 2.303 (4 sig figs). So that is why the Nernst equation can be rewritten as
\({\cal E}_{\rm cell} = {\cal E}^\circ_{\rm cell} - {2.303RT\over nF}\log(Q) \)
\({\cal E}_{\rm cell} = {\cal E}^\circ_{\rm cell} - {0.05916\over n}\log(Q) \)
Potential is given the symbol \({\cal E}\) in equations and is measured in volts.
Volts (or voltage) is a unit of measure and is abbreviated with just V.
Electric current is given the symbol \(I\) in equations and is measured in amps.
Electric current flow is always in the opposite direction of electron flow.
Amps is a unit of measure and is abbreviated with an A.
An amp (A) is also just a rate of charge transfer which is coulombs/second or C/s.
Time is given the symbol t (lowercase!) and is measured on many different scales.
Those scales of time are typically seconds, minutes, hours, days, or years. The SI unit of time is the second (s).
Reminder, temperature is given the symbol T (uppercase!) and has 3 different scales that can be used (°F, °C, or K).
Power is given the symbol P in equations and is measured in watts.
Watts (or wattage) is a unit of measure of power and is abbreviated with just W.
Power is defined as a rate of energy use or energy/time. A watt is really just a joule/second or J/s. One common equation for calculating power is: \(P = I \cdot {\cal E}\). Yes, it spells "pie" but is really power = current × potential.
Charge is given the symbol q in equations and is measured in coulombs.
Charge is calculated via current × time or: \(q = I\cdot t\)
Energy (joules, J) is calculated via charge × potential or: Energy = \(q\cdot {\cal E}\)
Don't confuse energy and potential, they are typically both symbolized with an E. Sometimes (not always) a script looking \({\cal E}\) is used for potential to be different from the E used for energy. The unicode script E looks like this, ℰ. Just note the context in which it is used - it is usually obvious which one we mean (blatantly obvious once you LEARN the concepts).
The main conversion factor to know in order to convert from "electricity world" to "chemistry world" is the faraday constant, F. It tells us the number of coulombs in a mole of ionic unary charge - to be even more specific, for our needs, the faraday is 96485 coulombs which is the total absolute charge on 1 mole of electrons. The faraday doesn't care about being positive or negative, it only cares about the magnitude of the charge. YOU need to remember the signs for charge... but no matter where the charge is coming from... 1 F = 96485 C/mol of electron charge.
So the flow of conversions is the following:
current → charge → moles of e– → moles of reaction
So taking all the conversions needed and putting them all into one big equation we get
\[{I\cdot t\over n\cdot F} = {\rm moles\;of \;reaction}\]
(that is moles of reaction that uses n moles of electrons for the way you balanced it)
Students will be able to...
Students will be able to...
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