Chemistry I
Final Exam
January 16, 2009
Total Score: 100 points
Time: 3 hours (18:00-21:00)
Constants
R = 8.314 J / mol K
= 0.08206 L atm / K mol
= 8.314 L kPa / K mol
h = 6.63×10-34 J⋅s
h = 1.05×10-34 J⋅s
k = 1.3806504 × 10−23 J / K
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1. The Haber process for the synthesis of ammonia (NH3) by using nitrogen and
hydrogen is one of the most important industrial processes for the well-being of
the humanity. It is used extensively in the production of fertilizers as well as
polymers and other products.
(A) What volume of hydrogen at 15 atm and 350 ℃ must be supplied to
produce 1 tonne (1t = 1000 kg) of NH3? (6%)
(B) What volume of hydrogen is needed in part (a) if it is supplied at 376 atm
and 250 ℃? (4%)
Answer: (this is virtually a copy of the exercise 4.51)
(A) The number of moles of H 2 needed will be 1.5 times the amount of NH3
produced, as seen from the balanced equation:
1
2
N 2 (g) + 32 H 2 (g) → NH 3 (g)
(3%)
or
N 2 (g) + 3 H 2 (g) → 2 NH 3 (g).
Once the number of moles is known, the volume can be obtained from the
ideal gas law:
  3 mol H 2  (1.0 × 103 kg)(103 g ⋅ kg −1 ) 

 RT


3
17.03 g ⋅ mol−1
nH 2 RT
2 nNH3 RT
 2 mol NH 3 


V=
=
=
P
P
P
6

10 g
3
(0.082 06 L ⋅ atm ⋅ K −1 ⋅ mol−1 )(623 K)
 
−1 
 2   17.03 g ⋅ mol 
=
15.00 atm
(
)
= 3.0 × 105 L.
(3%)
(B) The ideal gas equation
PV
PV
PV
PV
1 1
= 2 2 simplifies to 1 1 = 2 2
n1 RT1 n2 RT2
T1
T2
(2%)
because R and n are constant for this problem.
2
(15.00 atm)(3.0 × 105 L) (376 atm)V2
=
623 K
(523 K)
(2%)
V2 = 1.0 × 104 L
or
(4%)
3
2. Consider the following van der Waals coefficients of 8 different gases:
L2atm/mol2
b,
L/mol
molecular mass,
Gas
a,
A
0.034
0.0237
4
B
0.224
0.0266
2
C
0.211
0.0171
20
D
2.32
0.0398
84
E
4.19
0.0511
131
F
6.49
0.0562
71
G
3.59
0.0427
44
H
4.17
0.0371
6~30
g/mol
(A) Which of the above gases has the smallest attractive forces? (2%)
(B) Which gas is least dense at 1 atm and 25°C? (2%)
(C) Which gas may have the smallest compression factor, Z? (2%) Please
explain it. (2%)
(D) Which gas may provide the largest Joule-Thomson effect (cooling during
expansion)? (2%) Please explain it. (2%)
(E) With the large difference of the molecular mass between gas E and H, they
have the similar value of van der Waals coefficient a.
may be gas H： CH4, NH3, F2? (3%)
Which molecule
Answer:
(A) A
(B) B
(C) F
The stronger the attraction between molecules is, the smaller the compression
factor, Z, is.
Gas F has the strongest attraction among the molecules, so it has
the smallest compression factor, Z.
(D) F
The stronger the attraction between molecules is, the larger the Joule-Thomson
effect is.
Gas F has the strongest attraction among the molecules, so it has
largest Joule-Thomson effect.
(E) NH3.
4
3.
The unit cell of diamond is shown in figure 1. Here some carbon atoms arranged
as an expanded ccp (fcc) arrangement, and others occupy half the tetrahedral
holes. (hint: it looks like the zinc-blende structure, but all atoms are carbon
atoms) Each carbon atom is covalently bonded to four neighbors through sp3
hybrid bonds.
(A) Is diamond a (A) molecular, (B) network, or (C) metallic solid?
(2%)
(B) Is this structure a close-packed structure?
(2%)
(C) What is the coordination number of a carbon atom in diamond?
(2%)
(D) How many carbon atoms in an unit cell?
(2%)
(E) Calculate the density of diamond from the data of carbon atom given:
covalent radius = 77 pm & atomic weight = 12.0 g/mol.
(5%)
Figure 1
Ans: (A) B.
(B) No, not closed packed
(C) 4
(E) let the side of a unit cell = a 2a / 2
a/2
a/2
2×77 pm = (31/2 ×a/2)/2 a = 356 pm
5
(D) 8 atoms
8 C atoms in a unitcell density = 8×12.0 / ((6×1023) × (356×10-10)3) = 3.55 g/cm3
)ote: There are 2 correct answers for Questions 4, 5, 6: each
correct answer gets +2%, and each incorrect answer gets -2%.
4.
Which of the following molecules are likely to form hydrogen bonds: (A)
CH3COOH ; (B) CH3CHO ; (C) C2H5OC2H5 ; (D) (CH3)2CHOH .
(4%)
Ans: (A)(D)
5.
Which of the following statements are not correct:
(4%)
(A) CH3(CH2)8CH3 molecule has no multiple bond its shape cannot remain
rodlike cannot form a liquid crystal.
(B) Atoms in glass do not lie in an orderly array, but glass is hard and
transparent like quartz Glass is a crystalline solid.
(C) Two metals with very similar atomic radii can form a substitutional alloy.
(D) Ionic radius: Mg2+ = 72 pm & O2− = 140 pm MgO has a cesium chloride
structure.
Ans: (B)(D)
6.
Which of the following statements are correct:
(4%)
(A) The surface tension of a liquid is independent of the temperature.
(B) Molecules with more electrons are more polarizable and have greater
London interaction.
(C) NaCl can be dissolved in ethanol (alcohol). There are ion-dipole
interactions between ethanol molecules and ions.
(D) Ozone (O3) is gaseous at room temperature. There is no dipole-dipole
interaction between O3 molecules.
6
Ans: (B)(C)
7.
(5%) A student rides a bicycle to class every day, a 10 km round trip that takes 20
minutes in each direction. The students burns 880 kJ/h cycling. The same trip in
a car would require 1.2 L of gasoline. Assume that the student goes to class 150
days per year and that the enthalpy of combustion of gasoline can be
approximated by that of octane. What is the yearly energy requirement of this
journey by (a) bicycle and (b) car.
Molecular and thermodynamic data for octane C8H18(l): molar mass
M=114.22 g/mol, density d = 0.702 g/cm3, standard enthalpy of combustion ∆Hcº
= -5471 kJ/mol, standard enthalpy of formation ∆Hfº = -249.9 kJ/mol, standard
Gibbs free energy of formation ∆Hfº = -5471 kJ/mol, and standard molar entropy
Smº = -249.9 J/K·mol.
Solution: (this is virtually a copy of the exercise 6.49 with somewhat modified data)
(a) Energy requirement for the bicycle journey per year: 150 days × 2 times each day
× 20/60 hour cycling × 880 kJ/h = 88000 kJ = 8.80·107 J
(b) Energy requirement for the bicycle journey per year: 150 days × 1.2 L of gasoline
each day × 702 g/L (density of octane) ÷ 114.22 g/mol (molar mass of octane)
× 5471 kJ/mol = 6052491 kJ = 6.05·109 J
8.
(10 × 2%) Answer the following short questions:
(A) Give two types of non-expansion work.
7
(B) What is P and T for the STP conditions?
(C) The constant R/NA is associated with the name of a famous scientist. Who?
(D) What is ∆H for an exothermic process, ∆H=0, ∆H>0 or ∆H<0?
(E) Give a relationship between CV and ∆H for an ideal gas.
(F) Can you estimate Cp,m for a linear triatomic molecule at 100 K?
(G) Which of the following thermodynamic functions are not state functions:
pressure, work, entropy, enthalpy, heat, volume?
(H) Is the average translational energy of a molecule equal to its average
rotational energy?
(I)
Which of the thermodynamic functions measures the heat transfer at
constant pressure?
(J) Standard enthalpies of formation are +49.0 kJ/mol for benzene and −74.8
for methane. Which of these two compounds is more stable?
Solutions.
A. e.g., extension work, gravitational work, surface expansion work, electrical work,
magnetic work.
B. T=Oº C, p=1 atm.
C. Boltzmann.
D. ∆H<0.
E. CV=∆U/∆T at constant volume, but ∆U=∆H−∆(pV)=∆H−V·∆p at constant volume;
then CV=∆H/∆T −V·∆p/∆T; for an ideal gas it further reduces to CV=∆H/∆T −n·R.
F. 7/2 R.
G. Heat and work.
H. Not, it is true only for nonlinear molecules.
I. ∆H.
8
J. It is impossible to answer this question on the base of the given data.
9.
(6%) Assume the following molecules adopt disordered arrangements in their
crystal form, what would their residual molar entropy be? (a) BF3, (b) COF2, and
(c) SO2F2.
N
Answers: (a) W = 1, S = 0. (b) W = 3 A, S = k ln 3
NA
,
NA
= R ln 3 = 9.1 J/K. (c) W = 6
S = R ln 6 = 14.9 J/K.
10. (6%) (a) Why are so many exothermic reactions spontaneous? (b) Explain how
an endothermic reaction can be spontaneous.
Answers:
(a) Exothermic reactions tend to be spontaneous because the result is an
increase in the entropy of the surroundings. Using the mathematical
relationship ∆Gr = ∆H r − T ∆S r , it is clear that if ∆H r is large and negative
compared to ∆S r , then the reaction will generally be spontaneous.
(b) For an endothermic reaction ∆Hr > 0, but if ∆Sr < 0 and T∆Sr > ∆Hr  ,
then ∆Gr < 0, the reaction will be spontaneous.
11.
(7%) Suppose that 50.0 g of H2O(l) at 20.0°C is mixed with 65.0 g of H2O(l) at
50.0°C at constant atmospheric pressure in a thermally insulated vessel.
Calculate ∆S and ∆Stot for the process.
Cp,m for H2O(l) is 75.3 J / K / mole.
Answers:
To calculate the change in entropy for the hot and cold water, the amount of
energy which flows from one to the other must first be calculated:
9
qc = − qh
= nc ⋅ C p,m ( H 2 O) ⋅ (T f − Ti ,c ) = −nh ⋅ C p,m ( H 2 O) ⋅ (T f − Ti ,h )
where nc and nh are the moles of cold and hot water, respectively, and
Ti , c and Ti , h are the initial temperatures of the cold and hot water, respectively.
Dividing
both
sides
by
we
C p,m ( H 2 O)
obtain:
nc ⋅ (T f − Ti ,c ) = − n h ⋅ (T f − Ti ,h ).
The moles of hot and cold water are:
nc =
and
Tf =
50 g
65 g
= 2.78 mol and n h =
= 3.61 mol
−1
18.015 g ⋅ mol
18.015 g ⋅ mol −1
the
final
temperature
Tf
is
therefore:
(2.78 mol ⋅ 293.15 K ) + (3.61 mol ⋅ 323.15 K ) = 310.1 K.
(2.78 mol + 3.61 mol)
With Tf, we can calculate ΔS for the hot and cold water, and the total ΔS for
the entire system:
 Tf
∆S c = nc ⋅ C p,m ln
 Ti ,c
= +11.9 J ⋅ K −1



 = (2.78 mol ) 75.3 J ⋅ K −1 ⋅ mol −1 ln 310.1 K 
 293 K 




 Tf
∆S h = nh ⋅ C p,m ln
 Ti ,h
= −11.1 J ⋅ K −1
Here, ∆S is :



 = (3.61 mol ) 75.3 J ⋅ K −1 ⋅ mol −1 ln 310.1 K 
 323 K 




(
(
)
)
∆S = ∆S c + ∆S h = +0.8 J ⋅ K −1 .
Because it is a insulated system,
there is no heat transfer between system and surrounding and ∆Ssurr = 0
Therefore, ∆S tot is :
∆S tot = ∆S + ∆Ssurr = +0.8 + 0 = + 0.8 J ⋅ K −1 .
12.
(6%) Explain why each of the following statements is false. (a) Reactions with
10
negative Gibbs free energies of reaction occur spontaneously and rapidly. (b) An
exothermic reaction producing more moles of gas molecules than are consumed
has a positive standard reaction Gibbs free energy.
Answers:
(a) With ∆Gr < 0, the reaction will occur spontaneously. However, it does not tell how
fast a process takes place.
(b) ∆nr > 0, ∆Sr > 0 and with ∆Hr < 0, ∆Gr = ∆Hr – T ∆Sr < 0.
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