Chemistry II Final Exam 20 June, 2014 (Total Score: 100 points

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Chemistry II
Final Exam
20 June, 2014 (Total Score: 100 points)
Periodic Table of Elements
R = 8.314 J/mol∙K = 0.08314 L·bar/ mol∙K = 0.0821 L∙atm/ mol∙K = 8.314 L·kPa/ mol∙K
 Pay attention to the significant figures!! 
1.
The rate constant of the reaction O(g) + N2(g) → NO(g) + N(g), which takes place in the
stratosphere, is 2.50×109 L mol-1∙s-1 at 700°C and 9.70×1010 L∙mol-1∙s-1 at 800°C. Calculate the
activation energy and pre-exponential factor of the reaction.
2.
(5%)
The contribution to the destruction of the ozone layer caused by high-flying aircraft has been
attributed to the following mechanism:
Step 1 O3 + NO → NO2 + O2
Step 2 NO2 + O → NO + O2
(a) Write the overall reaction. (b) Write the rate law for each step and indicate its molecularity. (c)
What is the reaction intermediate? (d) A catalyst is a substance that accelerates the rate of a
reaction and is regenerated in the process. Is there a catalyst in this reaction?
(total 8%)
1
3.
Dinitrogen pentoxide, N2O5, decomposes with a rate constant of 5.20×10-3 s-1 at 65°C. (a) What
is the initial rate of decomposition of N2O5 when 3.45 g of N2O5 is confined in a 0.750-L container
and heated to 65°C? (b) What is the half-life for the decomposition of N2O5 at this temperature?
(c) How much time will elapse before the N2O5 concentration decreases to 0.0375 mol∙L-1?
(2%
each, total 6%)
4.
How does a catalyst affect:
(2% each, total 6%)
(a) the rate of the reverse reaction;
(b) H of the reaction;
(c) the equilibrium constant?
5.
A six-coordinated complex ML6 is shown in Figure 1 with the definition of x, y, and z-axes. The
relative position of ligands in spectrochemical series is shown in Figure 2. And the MO diagram
of ML6 complex is shown in Figure 3. The following questions are related to this model complex.
Figure 1
Figure 2
2
Metal
Complex
Ligand
Figure 3
(a) Draw an orbital energy-level diagram showing the configuration of d-electrons on the metal
ion in each of the following complexes: (i) [Fe(CN)6]3– ; (ii) a tetrahedral complex [ZnCl4]2– .
(3%)
(b) Which of the complexes, [Fe(CN)6]3– or [ZnCl4]2– has the smaller energy gap between
orbitals for its d-electrons?
(3%)
(c) The solution with [ZnCl4]2– ion is colorless. Suggest a reason why this solution is colorless?
(3%)
(d) In a ML6 complex, which d-orbitals on the metal ion are used to form -bonds between
octahedral metal ions and ligands?
(3%)
(e) Is the best description of the eg-orbitals in [Fe(CN)6]3– (a) bonding, (b) antibonding, or (c)
nonbonding? Explain how you reached that conclusion.
(3%)
(f) What kind of orbital interaction(s) to metal can be observed for the ligand CN–?
(3%)
(g) Base on the ligand field theory, what electronic configurations for [Fe(CN)6]3– and
[Fe(H2O)6]3+ ions in terms of their d-orbital electrons? Is their any difference in terms of
their magnetic property?
(3%)
(h) Draw all possible isomers of [CoCl4(NH3)2] and check if there is any stereoisomer.
(4%)
3
6.
Determine whether each of the following systematic name is correct or not. If a name is correct,
draw the line structure. If not, please correct it without changing its composition of elements.
(1+3% for each)
(a) 1,3,5-trinitrophenol
(b) 2,3,5-trimethyl-4-propyloctane
(c) tetramethyl amine
7.
Fill each blank with the line structure of proper compound. (3% for each blank)
(a) 1 C2H2 + 1 HCl → _(i)_
(b) methanoic acid + methylamine and heat to 200°C → _(ii)_
(c) benzene + _(iii)_ + _(iv)_ → nitrobenzene, then add one equivalent Br2 and FeBr3 → the
major product is _(v)_
(write chemical formulas for (iii) & (iv))
(d) 2-bromopropane + OH– → _(vi)_ , then add an oxidizing agent such as KMnO4 → a water
soluble organic compound _(vii)_ , then add Tollens reagent (containing Ag+) → _(viii)_
(e) polymerization reaction of HO2CC6H4CO2H + HOCH2CH2OH → _(ix)_
(f) polymerization reaction of styrene C6H5CH=CH2 → _(x)_
 8.-11. are multiple choice, one answer questions. 
8.
Regardless of cyclic compounds, how many isomers does C5H10 have?
(2%)
(A) 3 , (B) 4 , (C) 5 , (D) 6 , (E) 7
9.
How many of the following compounds have optical isomers?
(2%)
1,2-dichloropropane , 2-bromo-2-methylbutane , cis-1,2-dimethylcyclohexane ,
(CH3)2CHCOOC2H5
(A) 0 , (B) 1 , (C) 2 , (D) 3 , (E) 4
10. What is the product and the type of the reaction CH3CH2CHBrCH3 + CH3CH2O–
taking
place in ethanol? (2%)
(A) no reaction
(B) CH3CH=CHCH3 , bromination reaction
(C) CH3CH=CHCH3 , elimination reaction
(D) CH3CH2CH(OC2H5)CH3 , ether reaction
(E) CH3CH2CH(OC2H5)CH3 , ethanol adding reaction
4
11. If one says “-helix” in a protein, he is talking about
(2%)
(A) the primary structure of the protein
(B) the tertiary structure of the protein
(C) the product of the denaturation of a “-sheet”
(D) the formation of disulfide link, –S–S– , between amino acids
(E) the polypeptide chain held by hydrogen bonds between amino acids
5
102B Chemistry (II) final exam
Answers
計算題答案有效位數多/少一位不扣分,多/少兩位以上扣 1 分
1.
We use ln(k’/k) = (Ea/R)(1/T - 1/T’) =(Ea/R)((T’-T)/T’∙T)
k’(973K) = 2.50×109
;
k(1073K) = 9.70×1010
ln(2.50×109/9.70×1010) = (Ea/8.314×10-3 kJ∙mol-1∙K-1)((973K-1073K)/973K∙1073K)
ln(0.0258) = Ea∙ (-0.0115)
;
Ea = 318 kJ∙mol-1 (2%)
k= A exp(-Ea/RT) (1%)
A = k / exp(-Ea/RT)
= 2.50×109 L mol-1 s-1 /exp(-318 kJ∙mol-1/8.314×10-3 kJ mol-1∙K-1∙973K)
= 2.95×1026 L mol-1∙s-1 (2%)
2.
(a) O3 + O → 2 O2
(b)
(2%)
Rate Law
Molecularity
Step 1
rate = k[O3][NO]
bimolecular (2%)
Step 2
rate = k[NO2][O]
bimolecular (2%)
(c) The reaction intermediate is NO2
3.
(1%)
(d) The Catalyst is NO molecule.
(1%)
From the units of the rate constant,
it follows that the reaction is first order, thus rate = k[N2O5]
 3.45 g N 2 O5   1 mol N 2 O5 
1
(a) [N 2 O5 ]  
  0.0426 mol  L

 0.750 L   108.02 g N 2 O5 
rate = 5.20 × 10−3 s −1 × 0.0426 mol ∙ L−1 = 2.22 × 10−4 (mol N2 O5 ) ∙ L−1 ∙ s−1
(b) half-life = ln2/k = 0.693/5.20×10-3 s-1 = 1.33×102 s
(c) ln([N2O5]0/[ N2O5]t) = kt
t = 24.5 s
4.
;
(2%)
(2%)
ln(0.0426/0.0375) = 5.20×10-3 s-1 ∙ t
(2%)
(a) increase the rate of the reverse reacion
(b) does NOT change H of the recation
(c) does NOT change the equilibrium constant
5.
(a)-(g), 3% each ; 4% for (h)
(a) [Fe(CN)6]3–
[ZnCl4]2–
6
(b) [ZnCl4]2– . There are two reasons: (1) The energy gap of a tetrahedral complex is typically
smaller than of an octahedral complex; (2) Cl– is a weak-field ligand and CN– is a strongfield ligand. The weak-field ligand makes a smaller energy gap.
(c) In Zn 2+ , the 3d-orbitals are filled (d 10 ). Therefore, there can be no electronic transitions
between the t and e levels; hence, no visible light is absorbed and the aqueous ion is colorless.
The d 10 configuration has no unpaired electrons, so Zn compounds would be diamagnetic,
not paramagnetic.
(d) eg set, which comprises the d x
2
and d z orbitals.
- y2
2
(e) Nonbonding or slightly antibonding. In a complex that forms only -bonds, the t2g set of
orbitals is nonbonding. If the ligands can function as weak -acceptors (those close to the
middle of the spectrochemical series, such as the t2g set becomes slightly bonding by
interacting with the empty *-orbitals on the ligands.
(f) -type (to dz2, dx2-y2) and -type (to dxy, dxz, dyz) interactions.
(g) [Fe(CN)6]3– : t2g5eg0 , paramagnetic ;
[Fe(H2O)6]3+ : t2g4eg2 , paramagnetic
Cl
Cl
(h)
6.
(each correct/incorrect 1% , each corrected name/structure 3%)
(a) incorrect ,
2,4,6-trinitrophenol
(b) correct ,
(c) incorrect ,
7.
tetramethylammonium ion
(3% for each blank)
NH
Cl
(i)
(ii)
O
(iii) & (iv) HNO3 & H2SO4 順序可顛倒
O 2N
Br
OH
(v)
(vi)
7
O
O
(vii)
(viii)
O
O
O
O
n
(ix)
8. D
9. B
10. C
11. E
or
(x)
no reaction
n
(8-11, 2% for each)
8
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