exam 3
4/17
exam 3
4/17
The exam will be on Wednesday, 4/17, from 7:30 pm to 9:00 pm in both BUR 106 and UTC 2.112A
Students will split between two rooms on campus according to the first letter of their last name.
A - L go to UTC 2.112A
M - Z go to BUR 106
Make sure you go to the right room according to your letter or face an exam penalty.
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 main Periodic Table Handout for CH302 exams. There will also be a table of Ksp values that are needed for certain problems. Actual handout will be adjusted for the topics on the exam.
Coverage: Exam 3 covers all the material that was covered on LE's 21-29 and HW's 07-09. The exam covers Chapter 10 on solubility equilibria, Chapter 11 on chemical kinetics, and Chapter 13 on nuclear chemistry.
Questions: The exam has 20 multiple choice questions. The questions will have an equal weight of 4 points each. The breakdown is: 6 solubility questions, 9 kinetics, and 5 nuclear questions. 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.
For a saturated solution of the salt MxAy
MxAy(s) = xM+y(aq) + yA-x(aq)
Ksp = [M+y]x · [A-x]y
None of this will be printed on the exam. You need to KNOW this on your own.
| salt ratio | salt formula | aqueous cation | aqueous anion | equilibrium formula | Ksp(x) | molar solubility | ||
|---|---|---|---|---|---|---|---|---|
| 1:1 | MX(s) | ⇌ | M+(aq) | + | X–(aq) | Ksp = [M+][X–] | \(K_{\rm sp} = x^2\) | \(x = \sqrt{K_{\rm sp}}\) |
| 1:2 | MX2(s) | ⇌ | M2+(aq) | + | 2 X–(aq) | Ksp = [M2+][X–]2 | \(K_{\rm sp} = 4x^3\) | \(x = \root 3 \of {K_{\rm sp}\over 4}\) |
| 1:3 | MX3(s) | ⇌ | M3+(aq) | + | 3 X–(aq) | Ksp = [M3+][X–]3 | \(K_{\rm sp} = 27x^4\) | \(x = \root 4 \of {K_{\rm sp}\over 27}\) |
| 2:3 | M2X3(s) | ⇌ | 2 M3+(aq) | + | 3 X2–(aq) | Ksp = [M3+]2[X2–]3 | \(K_{\rm sp} = 108x^5\) | \(x = \root 5 \of {K_{\rm sp}\over 108}\) |
forward\(\rightarrow\) \(Q < K\) which means more solid will dissolve if there is any or nothing happens because the solution is undersaturated.
equil\(\rightleftharpoons\)ibrium \(Q = K\) which means the solution concentrations will hold constant and there is no net change.
\(\leftarrow\)reverse \(Q > K\) means the solution is oversaturatured and a precipitation will occur until the concentrations drop to saturation level.
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).
Binding Energy holds the nucleus together - all those protons are held in place with the help of just as many or even more neutrons. The binding energy will be the energy equivalent of the mass defect in the nucleus. A free protons and neutrons have a slightly greater mass than they do within the nucleus. The difference in mass is called the mass defect... put that change in mass (\(\Delta m\)) into the Einstein equation (\(E=mc^2\)) and you can calculate the binding energy (\(\Delta E_{\rm BE}\)) of that nucleus. \(\Delta E_{\rm BE} = \Delta m c^2\)
Fusion is when two smaller atoms combine to make a larger atom. Fusion is most common with those elements with atomic numbers less than that of iron (26).
Fission is when an unstable large atom (typically Z > 82) splits into to give two new medium sized atoms. The two "new" atoms are rarely the same atom - meaning the split is not perfectly symmetrical. An example is uranium-235 gaining one neutron to undergo fission and produce barium and krypton. This is more of a 60/40 split than a 50/50 one.
Alpha decay (α) is when the nucleus ejects an α-particle which is the nucleus (no electrons) of a helium atom. The parent atom decreases in mass number by 4 and decreases in atomic number by 2. It has the lowest penetration depth of all the nuclear radiation types - paper / outer layer of skin can stop it.
Beta decay (β) is the ejection of a β-particle (electron) from the nucleus. The mass number stays the same, but the atomic number increases by one. Beta particles penetrate slightly better than alpha rays, but still does not make it far. Wood and normal walls will stop this radiation.
positron emission (β+) is the ejection of a positron (β+) from the nucleus. The mass number stays the same, but the atomic number decreases by one.
Gamma rays/emission (γ) is the only emission that is pure energy and no mass - it is electromagnetic radiation of very high frequency and very short wavelength. It occurs when certain nuclei are in excited nuclear states and then return to nuclear ground state ejecting a gamma photon. This is analogous to excited electrons relaxing to ground state and emitting visible or UV radiation. Gamma emission often accompanies other types of decay as part of the energy release. Gamma rays are the most penetrating radiation there is. The only way to stop it is to use a lot of dense material - like thick lead sheets or other heavy metals.
Students will be able to...
Students will be able to...
Students will be able to...