Chemistry I
Final Exam
18:00-21:00, 16 January, 2015 (Total Score: 100 points)
1
2
nMvrms
 nRT
3
3RT
vrms 
M
1 2
1  3RT  3RT
E K  mvrms
 m

2
2  M  2N
PV 
3
2
 M  2
f  v   4 
 ve
2

RT



Mv 2
2 RT
Constants
R = 8.314 J/mol∙K = 0.08205 L∙atm/mol∙K = 8.314 L∙kPa/mol∙K
me = 9.10939×10-31 kg
c = 2.99×108 m/s
h = 6.626×10-31 J∙s
1 atm = 760 Torr = 1.01×105 Pa
_________________________________________________________________________________
1.
Name the kinds of intermolecular interaction that must be overcome in order to
(a) Melt solid PE (polyethylene)
(3%)
(b) Boil liquid CH3COOH (acetic acid)
(3%)
(c) Dissolve NaCl in liquid H2O (5%)
1
2.
The unit cell of a cubic crystal containing O (black ball) and Nb (hollow) atoms are shown. The
distance between centers of Nb and O atoms is 1.867 Å. Please write the mathematic expressions
to prove your answers of the following questions.
(a) What is the empirical formula of the ionic crystal?
(3%)
(b) What are the coordination numbers of Nb and O?
(4%)
3.
Calculate the pressure (atm) exerted by 8 kg N2(g) at 300°C in 100 L flask under the following
two conditions:
(a) An ideal gas.
(4%)
(b) A van der Waals gas, where a=1.39 (L2∙atm∙mol-2) and b =0.03913 (L∙mol-1).
(6%)
4.
The crystal structure of brass (an alloy of Cu and Zn) adopts the cesium chloride (CsCl) structure.
The radius of copper atom is 128 pm, and the radius of zinc atom is 138 pm. What is the density
(in g/cm3) of the brass?
(6%)
5.
A bottle contains 1.0 mol H2(g) and a second bottle contains 1.0 mol Ar(g) at the same temperature.
At that temperature, the root mean square speed of H2 is 2522.4 m/sec and that of Ar is 564 m/sec.
Assume that both gases behave ideally.
(a) Please calculate the temperatures (K) of the bottles.
(4%)
(b) If the volumes of the two bottles are the same, the pressure of these two bottles should be
the same because their temperatures are the same. However, the root mean square speed of
H2 is different from that of Ar. How can they have the same pressure? Please use the value
of the root mean square speeds of H2 and Ar to prove that the pressure of these two bottles
should be the same if the volumes of the two bottles are the same.
(4%)
(c) These two bottles are heated separately to double the root mean square speeds of H2 and Ar.
Please calculate the ratio of the final temperatures (K) to double the root mean square speeds
of H2 and Ar.
(4%)
2
(d) What is the ratio of the number of H2 in the first bottle to the number of Ar atoms in the
second bottle having these speeds?
6.
(a) The heat of fusion of ice at the standard melting point (0°C) is 6.01 kJ∙mol -1. Determine the
heat of freezing at –8°C. Given the 𝐶P̅ of water = 75.3 and ice = 38.1 J∙mol-1∙K-1. Assuming that
the 𝐶P̅ s are independent of temperature. (b) Is the value of w >, <, or = 0 when water freezes at
0°C, 1 bar?
7.
(4%)
(3% each, 6% total)
Use the bond enthalpies listed below to determine the heat of combustion （燃燒熱） of (a)
propane C3H8 and (b) H2 at 110°C. (c) Which of these two gases is the more efficient fuel (more
energy release when per gram of fuel is burned)?
(3% each, 9% total)
bond enthalpies in kJ∙mol-1 : H-H = 436, O-O = 157, O=O = 496, C-H = 412, C-C = 348, C=C =
612, C-O = 360, C=O = 743, O-H = 463
Only one answer for each question
8.
(2% each for question 8.-12.)
Excluding vibrational contributional, 𝐶V̅ (CV,m) of O3(g) =
(A) 2.5R
(D) 4R
9.
(B) 3R
(E) 6R
(C) 3.5R
After an adiabatic operation, the temperature of a 2-mol He gas sample rised from 300 K to 350
K. Which of the following statements is false?
(A) q = +624 J.
(B) q = +1247 J.
(C) U = 0
(D) The information is not enough to determine the value of w.
(E) This operation may be a compression without heat inflow and outflow.
10. The lattice energy Elatt of calcium oxide is the energy change for the reaction
(A) CaO(s)  Ca(g) + O(g)
(B) CaO(s)  Ca+(g) + O–(g)
(C) CaO(s)  Ca2+(g) + O2–(g)
(D) Ca(g) + 1/2 O2(g)  CaO(s)
(E) Ca2+(g) + O2–(g)  CaO(s)
11. The standard enthalpy of formation Hf°of diamond at 25°C
(A) = 0 by definition.
(B) = 0 , because diamond is the most stable form of carbon.
(C) < 0 , because graphite is the most stable form of carbon.
(D) < 0 , because diamon is more difficult to react than graphite.
(E) > 0 , because graphite is the most stable form of carbon.
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12. Which of the following statements about the hydration of Cl– ion is true?
(A) exothermic, because ion-dipole interaction is stronger than dipole-dipole interaction.
(B) endothermic, because ion-dipole interaction is stronger than dipole-dipole interaction.
(C) meaningless, because there’s no chemical reaction between Cl– ions and water molecules.
(D) (enthalpy of hydration of Cl– + enthalpy of hydration of Na+) = enthalpy of solution NaCl.
(E) None of above.
13. Briefly describe (a) the second law of thermodynamics
(2%)
(b) the third law of thermodynamics. (2%)
14. Without performing any calculations, predict whether there is an increase or a decrease in
entropy for each of the following process: (treat all gases as ideal).
(2% each, total 8%)
(a) The pressure of 1 mole of oxygen gas is allowed to double isothermally.
(b) Carbon dioxide is allowed to expand isothermally to 10 times its original volume.
(c) The temperature of 1 mol of helium is increased 25C at constant pressure.
(d) SO2(g)＋Br2(g)＋2 H2O(l) → H2SO4(aq)＋2 HBr(aq).
15. Ozone (O3) in the atmosphere can reaction with nitric oxide (NO):
O3(g) + NO(g)  NO2(g) + O2(g)
Calculate the G for this reaction at 25C using data listed below:
O3(g): Hf° = 142.7 kJ/mol , S° = 238.93 J/mol∙K
O2(g): Hf° = 0 kJ/mol , S° = 205.14 J/mol∙K
NO(g): Hf° = 90.25 kJ/mol , S° = 210.76 J/mol∙K
NO2(g): Hf° = 33.18 kJ/mol , S° = 240.06 J/mol∙K
(6%)
16. Estimate the temperature at which it is thermodynamically possible for carbon to reduce
iron(III) oxide to iron under standard conditions by endothermic reaction
2 Fe2O3(s) + 3 C(s) → 4 Fe(s) + 3 CO2(g) (7%)
CO2(g): Hf° = -393.5 kJ/mol , S° = 213.7 J/mol∙K
C(s): S° = 5.7 J/mol∙K
Fe(s): S° = 27.3 J/mol∙K
Fe2O3(s): Hf° = -824.2 kJ/mol , S° = 87.4 J/mol∙K
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102B Chemistry (II) final exam
Answer
1.
(a) Dispersion interactions (or London force) (3%)
(b) (i) Hydrogen bonding, (ii) dipole-dipole interactions and (iii) dispersion interactions (or
London interactions) (3%)
(c) NaCl is an ionic solid. To dissolve NaCl in any solvent, (i) ion-ion interactions and (ii)
dispersion interactions (or London interactions) must be overcome. (2%)
To dissolve something in H2O, (iii) hydrogen bonding, (iv) dipole-dipole and (v) dispersion
interactions between H2O molecules must also be overcome. (3%)
2.
(a) Number of O atoms in a unit cell = 1⁄4×8+1=3
Number of Nb atoms in a unit cell = 1⁄8×8+1⁄2×4=3
(Number of Nb) : (Number of O) = 3 : 3  NbO (3%)
(b) The coordination number of Nb is 4 (2%)
The coordination number of O is 4 (2%)
3.
(a) n = 8000/28 = 285.7 moles
𝑃=
(b)
4.
𝑛𝑅𝑇
𝑉
𝑃=
𝑛 2
𝑛𝑅𝑇
𝑉−𝑛𝑏
= 134.3 (atm) (4%)
100
− 𝑎( ) =
𝑉
285.7×0.082×(300+273)
100−285.7×0.03913
285.7 2
− 1.39 (
100
) = 139.8 (atm) (6%)
The length a of the unit cell :
a=
(128×10−10 +138×10−10 )×2
√3
density =
5.
285.7×0.082×(300+273)
=
= 3.07 × 10−8 (cm)
63.5+65.4
6×1023 ×(3.07×10−8 )3
= 7.428 (g/cm3 ) (6%)
(a) vrms = (3RT/M)1/2
𝑇=
2
M𝑣𝑟𝑚𝑠
3𝑅
或 𝑇=
=
2×10−3 ×2522.42
3×8.314
2
M𝑣𝑟𝑚𝑠
3𝑅
=
= 510 𝐾
40×10−3 ×5642
3×8.314
= 510 𝐾 (4%)
1
(b) 𝑃𝑉 = 𝑛𝑀𝑣𝑟𝑚𝑠 2 = 𝑛𝑅𝑇
3
𝑃𝐻2 =
2
𝑛𝑀𝐻2 𝑣𝐻
2
3𝑉
=
1×2×2522.42 .
3𝑉
=
12725004
3𝑉
5
𝑃𝐴𝑟 =
2
𝑛𝑀𝐴𝑟 𝑣𝐴𝑟
3𝑉
𝑃𝐻2 ≅ 𝑃𝐴𝑟
=
1×40×5642 .
3𝑉
=
12723840
3𝑉
(4%)
(c)
𝑇𝐻 2
=
𝑇𝐴𝑟
(d)
𝑓(𝑣𝐻2 )
𝑓(𝑣𝐴𝑟 )
𝑀𝐻 (2𝑣𝐻 )2
2
2
3𝑅
𝑀𝐴𝑟 (2𝑣𝐴𝑟 )2
3𝑅
=
𝑀𝐻2 (2𝑣𝐻2 )2
𝑀𝐴𝑟 (2𝑣𝐴𝑟 )2
3
𝑀𝑣2
𝐻2
2
−
2 ) 𝑣 2 𝑒 2𝑅𝑇
4𝜋(2𝜋𝑅𝑇
𝐻2
3
𝑀𝑣2
𝐴𝑟
𝑀𝐴𝑟 2 2 −
4𝜋(
𝑣𝐴𝑟 𝑒 2𝑅𝑇
)
2𝜋𝑅𝑇
𝑀𝐻
=
3
2
2
𝑓(𝑣𝐻2 )
2
2522.4
= ( ) (
)𝑒
𝑓(𝑣𝐴𝑟 )
40
5642
2×(2×2522.4)2
=
40×(2×564)2
3
𝑀𝐻2 2
= (
𝑀𝐴𝑟
) (
2
𝑀𝑣 2 𝑀𝑣𝐻2
( 𝐴𝑟 −
)
2𝑅𝑇
2𝑅𝑇
2
𝑣𝐻
2
2
𝑣𝐴𝑟
=1
𝑀𝑣2
)𝑒
(4%)
𝑀𝑣2
𝐻
2)
𝐴𝑟 −
( 2𝑅𝑇
2𝑅𝑇
3
2 2 2522.42 0
=( ) (
) 𝑒 = 0.224
40
5642
(4%)
6.
(a) H of freezing = -6.01×103 + (38.1 – 75.3)×(-8) = -5.71×103 J∙mol-1 = -5.71 kJ∙mol-1
(b) w < 0 (because the molar volume of ice > that of water)
7.
(3%)
(3%)
(a) C3H8 + 5 O2  3 CO2 + 4 H2O
Hc = (2×348 +8×412 + 5×496) – (3×2×743 + 4×2×463) = -1690 kJ∙mol-1 (單位或為 kJ)
(3%)
(b) H2 + 1/2 O2  H2O, Hc = (436 + 1/2×496) – (2×463) = -242 kJ∙mol-1 (單位或為 kJ) (3%)
(c) C3H8 = -1690/44 = -38.4 kJ∙g-1
H2 = -242/2 = -121 kJ∙g-1  H2 is more efficient
8~12
(3%)
每題 3%
8. B ;
9. E ;
10. C
;
11. E ;
12. A
13. (a) Suniv = Ssys + Ssurr > 0
The entropy of any isolated system increases in any spontaneous process.
(b)
(2%)
lim 𝑆 = 0
𝑇→0𝐾
The entropy of a pure substance in its thermodynamically most stable form is zero at the
absolute zero of temperature. (2%)
14. (a) a decrease in entropy
(2%)
6
(b) an increase in entropy
(2%)
(c) an increase in entropy
(d) a decrease in entropy
(2%)
(2%)
15. H = Hf(O2(g)) + Hf(NO2(g)) – Hf(O3(g)) – Hf(NO(g))
= 0 + 33.18 – 142.7 – 90.25 = -199.77 kJ/mol (單位或為 kJ)
(分段給分 2%)
S = S(O2(g)) + S(NO2(g)) – S(O3(g)) – S(NO(g))
= 205.14 + 240.06 – 238.93 – 210.76 = -4.49 J/mol∙K (單位或為 J/K)
(分段給分 2%)
G = H – TS = -199.77 – 298×(-4.49×10-3) = -198 kJ/mol (單位或為 kJ)
(分段給分 2%)
16. Hr = 3×Hf(CO2(g)) – 2×Hf(Fe2O3(s))
= 3×(-393.5) – 2×(-824.2) = +467.9 kJ/mol (單位或為 kJ)
(分段給分 2%)
Sr = 3×S(CO2(g)) + 4×S(Fe(s)) – 2×S(Fe2O3(s)) – 3×S(C(s))
= 3×213.7 + 4×27.3 – 2×87.4 – 3×5.7 = +558.4 J/mol∙K (單位或為 J/K)
(分段給分 2%)
T = Hr / Sr = 467.9×10 ÷ 558.4 = 838 K (分段給分 2%)
The minimum temperature at which reduction occurs at 1 bar is 838 K (565°C).
(分段給分 1%)
3
7