1983 Test
1) Sulfuryl chloride, SO
2
Cl
2
, is a highly reactive gaseous compound. When heated, it decomposes as follows.
SO
2
Cl
2
(g) <===> SO
2
(g) + Cl
2
(g)
This decomposition is endothermic. A sample of 3.509 grams of SO
2
Cl
2
is placed in an evacuated 1.00-liter bulb and the temperature is raised to 375 K.
(a) What would be the pressure in atmospheres in the bulb if no dissociation of the SO
2
Cl
2
(g) occurred?
(b) When the system has come to equilibrium at 375 K, the total pressure in the bulb is found to be 1.43 atmospheres. Calculate the partial pressures of SO
2
, Cl
2
, and SO
2
Cl
2
at equilibrium at
375 K.
(c) Give the expression for the equilibrium constant (either K p
or K c
) for the decomposition of
SO
2
Cl
2
(g) at 375 K. Calculate the value of the equilibrium constant you have given, and specify its units.
(d) If the temperature were raised to 500 K, what effect would this have on the equilibrium constant? Explain briefly.
2)
CO(g)
CO(g) + 2 H
2
(g) <===> CH
3
OH(l)
For this reaction,

H° = -128.1 kilojoules

H f
° (kJ mol¯ 1 )

G f
° (kJ mol¯ 1 ) S° (J mol¯
1
K¯
1
)
-110.5 -173.3 +197.9
CH
3
OH(l) -238.6 -166.2 +126.8
The data in the table above were determined at 25 °C.
(a) Calculate
 G° for the reaction above at 25 °C.
(b) Calculate K eq
for the reaction above at 25 °C.
(c) Calculate
 S° for the reaction above at 25 °C.
(d) In the table above, there are no data for H
2
. What are the values of

H f
°, 
G f
°, and of the absolute entropy, S°, for H
2
at 25 °C?

3) The molecular weight of a monoprotic acid HX was to be determined. A sample of 15.126 grams of HX was dissolved in distilled water and the volume brought to exactly 250.00 milliliters in a volumetric flask. Several 50.00-milliliter portions of this solution were titrated against NaOH solution, requiring an average of 38.21 milliliters of NaOH.
The NaOH solution was standardized against oxalic acid dihydrate, H
2
C
2
O
4
.
2 H
2
O (molecular weight: 126.066 gram mol¯ 1
). The volume of NaOH solution required to neutralize 1.2596 grams of oxalic acid dihydrate was 41.24 milliliters.
(a) Calculate the molarity of the NaOH solution.
(b) Calculate the number of moles of HX in a 50.00-milliliter portion used for titration.
(c) Calculate the molecular weight of HX.
(d) Discuss the effect on the calculated molecular weight of HX if the sample of oxalic acid dihydrate contained a nonacidic impurity.
4) Graphical methods are frequently used to analyze data and obtain desired quantities.
(a) 2 HI(g) -----> H
2
(g) + I
2
(g)
The following data give the value of the rate constant at various temperatures for the gas phase reaction above.
T(k) 647 666 683 700 716 k (liter/mole sec) 8.58 x 10¯
5
2.19 x10¯
4
5.11 x 10¯
4
1.17 x 10¯
3
2.50 x 10¯
3
Describe, without doing calculations, how a graphical method can be used to obtain the activation energy for this reaction.
(b) A(g) -----> B(g) + C(g)
The following data give the partial pressure of A as a function of time and were obtained at 100
°C for the reaction above.
P
A
(mm
Hg) t(sec)
348 247 185 105 58
0 600 1,200 2,400 3,600

Describe, without doing calculations, how graphs can be used to determine whether this reaction is first or second order in A and how these graphs are used to determine the rate constant.
6)
(a) Specify the properties of a buffer solution. Describe the components and the composition of effective buffer solutions.
(b) An employer is interviewing four applicants for a job as a laboratory technician and asks each how to prepare a buffer solution with a pH close to 9.
Archie A. says he would mix acetic acid and sodium acetate solutions.
Beula B. says she would mix NH
4
Cl and HCl solutions.
Carla C. says she would mix NH
4
Cl and NH
3
solutions.
Dexter D. says he would mix NH
3
and NaOH solutions.
Which of these applicants has given an appropriate procedure? Explain your answer, referring to your discussion in part (a). Explain what is wrong with the erroneous procedures. (No calculations are necessary, but the following acidity constants may be helpful: acetic acid, K a
=
1.8 x 10¯ 5
, NH
4
+
, K a
= 5.6 x 10¯
10
)
7) The transition metal complex ion, [Co(NH
2
CH
2
CH
2
NH
2
)
2
Cl
2
]
+
, has an octahedral shape and exists as two geometric isomers.
(a) Draw the structural formulas for the two geometrical isomers.
(b) One of the geometrical isomers exists as two optical isomers. Draw the structural formulas for the two optical isomers.
(c) How many moles of AgCl would be immediately precipitated when excess AgNO
3
solution is added to an aqueous solution containing 1 mole of [Co(NH
2
CH
2
CH
2
NH
2
)
2
Cl
2
]Cl, the chloride salt of the transition metal complex? Explain your reasoning.
8)
Ti
3+
+ HOBr <===> TiO
2+
+ Br¯ (in acid solution)
(a) Write the correctly balanced half-reaction and net ionic equation for the skeletal equation shown above.
(b) Identify the oxidizing agent and the reducing agent in this reaction.

(c) A galvanic cell is constructed that utilizes the reaction above. The concentration of each species is 0.10-molar. Compare the cell voltage that will be observed with the standard cell potential. Explain your reasoning.
(d) Give one example of a property of this reaction, other than the cell voltage, that can be calculated from the standard cell potential, E°. State the relationship between E° and the property you have specified.

1983 Answers
1) a) two points n = 3.509 g ÷ 135.0 g/mol
P = (nRT) / V
= [(0.02600 mol) (0.08205 L atm mol¯ 1 K¯ 1 ) (375 K)] / 1.00 L
P = 0.800 atm b) three points
P
SO2Cl2
= 0.800 atm - y atm
P
SO2
= P
Cl2
= y atm
P tot
= P
SO2Cl2
+ P
SO2
+ P
Cl2
1.43 atm = 0.800 atm - y + y + y (by substitution)
P
SO2
= P
Cl2
= y = 0.63 atm
P
SO2Cl2
= 0.800 - 0.63 = 0.17 atm c) three points
K p
= (P
SO2
x P
Cl2
) / P
SO2Cl2
K p
= (0.63 atm)
2
/ 0.17 atm
K p
= 2.3 atm
Alternative approach in (b) and (c) is to determine the number of moles of each species, then molarity, and finally K c
. n
SO2Cl2
= 0.0260 - z n
SO2 = nCl2 = z mol

ntot = nSO2Cl2 + nSO2 + nCl2 ntot = PV / RT = (1.43 atm x 1.00 L) / (0.08205 L atm mol¯
1
K¯
1
x 375 K) ntot = 0.0465 mol = 0.0260 - z + z + z nCl2 = nSO2 = z = 0.0205 mol nSO2Cl2 = 0.0260 - 0.0205 = 0.00550 mol
Kc = ([SO2] [Cl2] ) / [SO2Cl2]
Kc = [0.0205]
2
/ [0.0055]
Kc = 0.076 M d) one point
An endothermic process absorbs heat during dissociation so K500 > K375 or a stress is placed upon the system and K increases in order to remove the stress.
A maximum of 1 point was deducted for math for unit errors or for reporting an unreasonable number of significant figures. This same procedure was also used for Problems 2 and 3.
2) a) two points
[delta]G° = [sigma] [delta]Gf° (prod) - [sigma] [delta]Gf° (react)
[delta]G° = - 166.2 - (- 137.3 + 2 (0))
[delta]G° = -28.9 KJ/mol b) two points
[delta]G° = - RT ln K (or - 2.3 RT log K)
- 28.9 = - ( 8.31 x 10¯
3
) (298) ln K ln K = 11.67

K = 1.17 x 10
5 c) two points
[delta]G° = [delta]H° - T [delta]S°
- 28,900 = -128,100 - 298 [delta]S°
[delta]S° = - 99,200 / 298 = - 333 J / mol-K d) three points
[delta]Hf° (H2) = 0 and [delta]Gf° (H2) = 0 deltaS° = [sigma]S° (prod) - [sigma]S° (react)
- 333 J / mol-K = 126.8 J / mol-K - 197.9 J / mol-K - 2 S° (H2)
S° = 131 J / mol-K
3) a) three points mol H2C2O4
.
2H2O = 1.2596 g H2C2O4
.
2H2O / 126.066 g/mol
= 9.9916 x 10¯
3
mol
H2C2O4 + 2NaOH ---> Na2C2O4 + 2H2O mol NaOH = 9.9916 x 10¯
3
mol H2C2O4 x (2 mol NaOH / 1 mol H2C2O4 ) = 1.9983 x 10¯
2
MNaOH = 0.019983 mol / 0.04124 L = 0.4846 b) two points mol HX = mol NaOH
0.03821 L x 0.4846 mol/L = 0.01852 mol HX

c) two points as follows: a logical argument plus a quantitative comparison = 2; a logical argument but no quantitative comparison = 1; one error in the quantitative comparison = 1
(0.01852 mol HX / 50.00 mL) x 250.00 mL = 0.09260 mol
MW = (15.126 g x 0.09260 mol) = 163.3 g/mol d) two points
The calculated molecular weight is smaller than true value, because of the following:
Measured g H2C2O4 is larger than true value.
Calculated mol H2C2O4 is larger than true value.
Calculated mol NaOH is larger than true value.
Calculated M NaOH is larger than true value.
Calculated mol HX is larger than true value.
Therefore,
MW = gHX (true value) / mol HX (calculated, and too large)
One point was subtracted for an arithmetic error(s) or for a serious significant figure error.
4) a) three points
Plot ln k or log k vs 1/T
Eact = - R (slope) or - 2,303 R (slope)
For partial credit, if the 2-point equation is given for the activation evergy, the student may receive a point. A student may also receive a point if it is stated that k is plotted vs 1/T or if ln K or log k is plotted vs T. b) five points
Plot ln PA or log PA vs time.
Plot 1/PA vs time.

If ln PA vs time is linear, the reaction is first order. If 1/PA vs time is linear, the reaction is second order.
If first order, slope = - k1 or - k1 / 2.303.
If second order, slope = k2.
5) a) three points: correct filling of a molecular orbital energy diagram for NO
+
, correct filling of a molecular orbital energy diagram for NO¯, correct labeling of both molecular orbital diagrams.
These MO diagrams will be posted when drawn (note added Feb 1997)
(b) four points
Bond order for NO
+
= 3.
Bond order for NO¯ = 2.
Or bond order NO
+
> bond order NO¯.
Bond length of NO
+
is shorter than the bond length of NO¯.
Bond strength of NO
+
is greater than the bond strength of NO¯.
(c) one point
NO¯ is paramagnetic since the anti-bonding pi orbitals are degenerate so the electrons will be unpaired in these orbitals.
6)
(a) three points
A buffer solution resists changes in pH upon the addition of an acid or base.
Preparation:
 Mix a weak acid + a salt of a weak acid.

Or mix a weak base + a salt of a weak base.

 Or mix a weak acid with about half as many moles of strong base.
 Or mix a weak base with about half as many moles of strong acid.
 Or mix a weak acid and a weak base.
(b) five points
Carla has the correct procedure. She has mixed a weak base, NH3, with the salt of a weak base, NH4Cl.
Archie has buffer solution but it has a pH around 5.
Beula does not have a buffer solution, since her solution consists of a strong acid and a salt of a weak base.
Dexter does not have a buffer solution, since his solution consists of a weak base plus a strong base.
7)
(a) four points
Representation of the cobalt complex as an octahedral complex.
Ethylenediamine shown as bidentate ligand.
Representation of the cis isomer.
Representation of the trans isomer.
If ethylenediamine is shown as a monodentate ligand, it is still possible to receive credit for the representation of both the cis and trans isomers.
Diagrams will be added when drawn (note added Feb 1997)
(b) two points
Selection of the cis isomer as the proper geometrical isomer.
Representation of the optical pair.
(c) two points
One mole of AgCl would be precipitated.
One chloride ion is outside the coordination sphere and thus available for precipitation, whereas the other two chloride ions are in the inner coordination sphere and are bonded directly to the Co
3+
ion. These ions are not available for precipitation.

8)
(a) three points
Ti
3+
+ H2O ---> TiO
2+
+ 2 H
+
+ e¯
H
+
+ HOBr + 2e¯ ---> Br¯ + H2O
2 Ti
3+
+ HOBr + H2O ---> 2 TiO
2+
+ 3 H
+
+ Br¯
The third point was available for the correct addition of two half-reactions even if the half-reaction(s) were incorrect.
(b) one point
HOBr is the oxidizing agent and Ti
3+
is the reducing agent.
(c) two points
The observed voltage will be greater than E° since E = E°- 0.059/2 (log [0.1]
3
).
Or 2 points for correctly discussing the relationships between cell potential and concentration for the balanced net equation that the student gave in part (a). One point could be achieved for discussing the cell potential and concentration relationship but omitting H
+
or the exponets.
(d) two points
Identification of the property, [delta]G, K or pH.
Statement of the relationshop [delta]G° = - n F E° or E° = 0.059/n (log K).
Or 1 point was given if the student stated that reaction is spontaneous if E > 0.