Chem 104
Exam 3
Spring '03 Name: 88 Points (3 pts multiple choice and 10 pts problems)
MULTIPLE CHOICE QUESTIONS: only one choice is correct
1. When a strip of zinc is placed in a solution of Cu(2+) ion, the voltage is initially +1.1 V. If the voltage is measured every hour, its value will be observed to:
a) be the same as the initial value of +1.1 V
b) decrease from the initial value to -1.1 V
d) none of the above
2. Enzymes are catalysts for biochemical reactions, such as the conversion of sucrose to glucose and fructose. This means that enzymes:
a) have no effect on the reaction mechanism.
b) decrease Ea for the reaction.
c) are consumed after each reaction.
d) can do none of the above.
3. For the reaction: 2 NO2 + O3 --> N2O5 + O2
which of the following statements is true?
a) The rate law is: rate = k[NO2]2[O3]
b) The reaction is third order.
c) The instantaneous reaction rate = - d[O3]/dt
d) none of the above
4. If the potential of a redox reaction is +1.8 V,
a) DG < 0
b) DStotal > 0
c) the reaction cell is galvanic
d) all of the above
5. Which is the anode in the cell diagram below?
Pt(s) | H+(aq), H2(g) || Fe3+(aq), Fe2+(aq) | Pt(s)
a) Pt(s) (on left side) (In a cell diagram the anode is always on the left. Anode refers to the actual electrode not necessarily the material being oxidized.)
b) (H+(aq), H2(g) ) (Solutions and gases do not make good electrodes!)
c) (Fe3+(aq), Fe2+(aq) )
d) Pt(s) (on right side)
QUESTIONS & PROBLEMS:
Choose and complete any seven of the remaining s9 questions and problemMs.
MARK WHICH QUESTIONS YOU WANT GRADED.
Q1. Which metal is expected to react most violently with protons (H+) in when the metal is dropped into water? Use the reductive half reactions (given below) to explain your answer.
E° Li+/Li0 = -3.05 V
E° Na+/Na0 = -2.71 V
E° Cs+/Cs° = -2.92 V
E° Pt2+/Pt° = +1.20 V
The two half reactions that combine and react spontaneously with the largest Eocell are:
2H+ + 2e- => H2(g) 0 V Reduction
2Lio => 2Li+ + 2e- 3.05 V Oxidation
Eocell: 3.05 V
The free energy (DG = -nFE = -585 kJ/mol!) released from this reaction is enough to cause a violent reaction in the presence of the product hydrogen gas.
Q2. Label the seven parts, [ (A), (B), (C), (D), (E), (F), (G)] on the diagram below.
What is the name of the diagram below?__Reaction Profile________________
(A) - reactants
(B) - products
(C) - activation energy
(D) - energy( delta G, potential energy)
(E) - reaction progress
(F) - delta G
(G) - intermediate
Q3. Keq = 1025 for the ligand substitution reaction on cobalt below:
[Co(NH3)6]3+ + 6 H3O+ --> [Co(H2O)6]3+ + 6 H4N+
a) What is the free energy change for this reaction?
DG = -RTln Keq
DG = -(8.341 Jmol-1K-1)(298 K)ln 1025
DG = -143 kJ/mol
b) Is this reaction predicted to be spontaneous?
Yes
c) This reaction fails to react, even after three days.
Does this fact agree with your prediction of spontaneity in (b) above?
Explain what this observation means.
No, a spontaneous reaction should go! But failure to react tells you that the activation energy must be large for this reaction, making the reaction very slow. A catalyst could speed this reaction up by lowering the activation energy.
Q4. Dr. B. has talked about and used the concept of "the great cycle of energy".
Describe what this is and using a specific example, explain why it is so important (and useful).
The great cycle of energy refers to the way deltaG, E and Keq can be interconverted. For example, in group work you showed how a concentration gradient ( an equilibirum difference) could be expressed in terms of potential ( membrane potential) . It is a useful concept to see how energy is incterconverted in processes, not created nor destroyed. Another example is the use of measured potentials for Ni(O)/Ni(2+) in water and in thre presence of ammonia to get the Kf for [Ni(NH3)6]2+.
Q5. The table below shows the change in initial rates of reaction as the initial concentrations (in M units) of the reactants were varied.
a) What is the initial rate law for the reaction?
BrO3- + 5Br- +6H+ --> 3 Br2 + 3 H2O
[BrO3-]o [Br-]o [H+]o rateo [molL-1sec-1] Experiment #
0.10 0.10 0.10 8.0 x 10-4 1
0.20 0.10 0.10 1.6 x 10-3 2
0.20 0.20 0.10 3.2 x 10-3 3
0.10 0.10 0.20 3.2 x 10-3 4
rate = k[BrO3-]x[Br-]y[H+]z
Find x, y, and z:
Experiments 1 and 2 show that when [BrO3-] was doubled the rate doubled so x = 1.
Experiments 2 and 3 show that when [Br-] was doubled the rate doubled so y = 1.
Experiments 1 and 4 show that when [H+] was doubled the rate quadrupled so z = 2.
rate = k[BrO3-][Br-][H+]2
b) What will the rate be if the initial concentrations are:
[BrO3-]o = 0.40 M, [Br-]o = 0.20 M, [H+]o = 0.05 M ?
Calculate k by choosing one data set and substituting the values into the rate law from part a:
rate = k[BrO3-][Br-][H+]2
k = 8.0x10-4/0.104 = 8.0 m-3s-1
Calculate the rate with the given date and the rate law:
Rate = 8.0 m-3s-1 * 0.40 M * 0.20 M * 0.052 M
Rate = 1.6x10-3 M/s
Q6. a) Draw a picture to show a galvanic electrochemical cell made from combining the redox couples
Cu2+/Cu+ (E°red = +.153 V) and Ni2+/Ni° (E°red = -.250 V).
The copper redox couple has higher reduction potential so that reaction will occur at the cathode using an inert electrode (e.g. graphite or platinum) while the Ni(s) will be oxidized to Ni2+ at the anode.
Don’t forget the salt bridge to balance the charge.
b) Write the cell diagram that summarizes the galvanic cell you have completed above.
Ni(s) | Ni2+ (aq) || Cu+ (aq), Cu2+ (aq) | Pt(s)
The anode (oxidation reaction) is written on the left. An inert electrode must be used for the cathode since the redox species are in solution and cannot be used as an electrode like in the case of solid nickel.
c) Cl- is added to the beaker containing the Cu2+/Cu+ couple causing the precipitation of
insoluble CuCl.
What is the predicted change in Ecell° after chloride is added to the Cu2+/Cu+ half-cell?
Ecell will increase as Cl is added.
Cu2+ + e- => Cu+
The addition of chloride removes Cu+ by forming insoluble CuCl:
Cu+ (aq) + Cl- (aq) => CuCl (s)
The cell potential will increase as predicted by the Nernst equation (Ecell = Eo - (0.0591 V/n)log Q) where Q will decrease as Cl- reacts with with Cu+ and lowers the concentration of products.
Q7. The transition metal technetium has two radioactive isotopes, 99Tc and 99mTc. Like other nuclear decay processes, the decay reactions of these Tc isotopes are first order reactions:
99mTc ----> b + 99Tc k = 3.2 x 10-5 sec-1
99Tc ----> b + 99Ru k = 1.10 x 10-13 sec-1.
a) Define what is meant by the term "half-life".
The time it takes for half of the initial amount of material to decay into products.
b) Calculate the half-life for each of the two Tc isotopes.
99mTc:
t1/2 = ln 2/k
t1/2 = 0.653 / 3.2 x 10-5 sec-1
t1/2 = 2.2 x 104 s = 6 hours
99Tc:
t1/2 = 0.653 / 1.10 x 10-13 sec-1
t1/2 = 6.3 x 1012 s = 200,000 years
c) One isotope is used as an imaging agent to make internal organs ( heart, liver, brain) visible. The b particles emitted expose photographic film making a picture of the organ.
Which isotope is the useful one for medical imaging? Explain your choice.
The half life of 99mTc is short enough that a sufficient number of b particles will be produced to expose the photographic film to make the picture of the organ in a reasonable amount of time. With a half life of 200,000 years a patient could wait for a long while for the results of their test if 99Tc was used as the imaging agent.
Q8. (a) CoCl3 + 6NH3 ---> [Co(NH3)6]Cl3
(b) Co is +3 and has d6 electron configuration. NH3 is a strong field ligand and teh complex is low spin.
so then dxy, dxz and dyz are all filled with paired electrons and dz2 and d x2-y2 are empty.
(c) see book!
(d) CoCl3 + 6 H2O ---> [Co(H2O)6]Cl3
Q9. 4. If you can imagine it, you might be able to make it.
Can you imagine how to invent a device that changes color when it become magnetic?
In other words, can you imagine under what conditions coordination complexes could change color as well as also change their magnetic properties from diamagnetic to paramagnetic? Use any metal, any geometry, and any ligands you wish in your invention. To convince me–the fund-raiser for your project–also give specific examples of colors which may pair with the diamagnetic and paramagnetic behaviors.
It is the ligand field strength that determines both the color and the magnetic properties. Therefore it should be possible to change both simultaneously.
For example, Fe(2+) has 6 d electrons.
If the ligand is the very strong field strength CN, making Fe(CN)6 4-would be a diamagnetic complex. Since the delta o is expected to be large, it would absorp in the short wavelength violet-blue and reflect yellow light.
If the ligand were H2O which has a weak ligand field, the 6 d electrons would not be all paired and a paramagnetic complex with 2 paired and 4 unparied electrons. The small delta o menas that long wavelength, or red light, woudl be absorbed and greenish blue light reflected.