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School of Science and Engineering (SSE)
Spring 2017
General Chemistry II (CHE 1402)
Homework 2 - CORRECTION
I
1) Based on the figure above, what is the boiling point of diethyl ether under an external
pressure of 1.32 atm
1 atm = 760 torr  1.32 atm = 1.32 x 760 = 1003 torr
The boiling point of diethyl ether under this pressure would be 40 oC.
2) Elements A and B have identical densities and unit cell volumes for their crystalline
lattice. Element A crystallizes in a body-centered cubic lattice and element B crystallizes
in a face-centered cubic lattice. What can be concluded concerning the molar mass of A
and B.
Let’s call:
dA the density of element A
VA the volume occupied by element A
mA the mass of atoms of A in the unit cell
MA the molar mass of element A
XA the number of atoms of A in the unit cell
dB the density of element B
VB the volume occupied by element B
mB the mass of of atoms of B in the unit cell
MB the molar mass of element B
XB the number of atoms of B in the unit cell
m
 m  d V
m A  d A  V A m B  d B  VB
V
Because densities are equal, dA = dB
mA = mB
Because unit cell volumes are equal, VA = VB
d
1
X
M 
M  M
m A  mB  X A   A   X B   B   A  B
 Na 
 Na  M B X A
A
B
In the lattice occupied by A, there are 2 atoms (8x1/8 + 1). In the lattice occupied by B,
there are 4 atoms (8x1/8 + 6x1/2). Therefore, XA = 2 and XB = 4.
MA XB 4

 2
MB XA 2
 M A  2 M B
Conclusion: the molar mass of element A is twice the molar mass of element B.
II
1. Explain why the orders of the boiling points of the compounds, shown in the following
graph, are as follows:
HF (19.5 °C) > HI (-35.4 °C) > HBr (-66.8 °C) > HCl (-.85 °C)
H2O (100 °C) > H2Te (-1.0 °C) > H2Se (-42.0 °C) > H2S (-60.7 °C)
BiH3 (16.88 °C > SbH3 (-17.0 °C) > NH3 (-33.4 °C) > AsH3 (-63.0 °C) > PH3 (-87.7 °C)
2
Dispersion forces tend to increase in strength with increasing molecular weight. If we
compare the hydrogen compounds of the halogens (group 7A), the molecular weight
increases down a column (MCl<MBr<MI), which explains why bp(HCl) < bp(HBr) <
bp(HI). HF has an abnormally high boiling point because it has strong intermolecular
forces resulting from hydrogen bonding. Because hydrogen bonds are stronger than
dipole-dipole and dispersion forces, HF has a higher boiling point than HI: bp(HI) <
bp(HF).
Hydrogen bonding is the same reason that explains why H2O has the highest boiling point
of the hydrogen compounds of the chalcogens (group 6A).
Finally, the comparison of the boiling points of the hydrogen compounds of group 5A (N,
P, As, Sb, Bi), bp(NH3) > bp(PH3) > bp(AsH3) because NH3 does have hydrogen
bonding. In the case of SbH3 and BiH3, the molecular weights are sufficiently high for the
dispersion forces to overcome hydrogen bonding, therefore bp(BiH3) > bp(SbH3) >
bp(NH3).
2. Indicate the main type of intermolecular forces that operates in each of the following
substances: IF, BF3, NF3, XeF4.
IF: dipole-dipole forces
BF3: dispersion forces
NF3: dipole-dipole forces
XeF4: dispersion forces
Note: BF3 and XeF4 have a trigonal planar and square planar molecular geometries
respectively. They therefore are symmetric molecule and have no dipole.
trigonal planar
square planar
III
1. Place the following substances in order of decreasing volatility assuming that their
polarities are similar: CH4, CBr4, CH2Cl2, CH3Cl, CHBr3, CH2Br2
Volatility is a measure of the tendency of a liquid to vaporize. If we assume that the
compounds have similar polarities, we therefore assume that they have similar
intermolecular forces and that volatility is dependent solely on the molecular weight. The
lighter the substance, the more volatile.
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MCH4 = 16 g/mol
MCH3Cl = 50 g/mol
MCH2Cl2 = 85 g/mol
MCH2Br2 = 174 g/mol
MCHBr3 = 253 g/mol
MCBr4 = 332 g/mol
Conclusion:
The order of decreasing volatility is the following: CH4 > CH3Cl > CH2Cl2 > CH2Br2 >
CHBr3 > CBr4
2. What are the major attractive forces that exist in each of the following molecules.
CO2, H2, IF, BF3, and CCl4
CO2: dispersion forces
H2: dispersion forces
IF: dipole-dipole forces
BF3: dispersion forces
CCl4: dispersion forces
Note: a symmetric molecule does not have a dipole.
IV
Ethanol, C2H5OH, melts at –114°C and boils at 78 °C. The enthalpy of fusion of ethanol
is 5.02 KJ/mol, and its enthalpy of vaporization is 38.56 KJ/mol. The specific heats of
solid and liquid ethanol are 0.97 J/g-K and 2.3 J/ g-K respectively.
What is the enthalpy change required to convert 10.0 g of ethanol at –150 °C to the vapor
at 78°C
We need to calculate the enthalpy change for each segment of the heating curve and then
sum them to get the total enthalpy change (Hess’s law).
There are 4 segments.
ΔH1 = m x Cs x ΔT1
ΔH2 = ΔHfusion x n
ΔH3 = m x Cliq x ΔT2
ΔH4 = ΔHvap x n
ΔT1 = -114+150 = 36 oC = 36 K
ΔT2 = 78+114 = 192 oC= 192 K
H TOTAL = H1 + H 2 + H 3 + H 4
= (m  Cs x T1 ) + (H fusion  n) + (m x C liq  T2 ) + (H vap  n)

 H vap 
 H fusion 

= m   (C s  T1 ) + 
 + ( C liq  T2 ) + 
M
 M 



4

 H fusion  H vap 

= m   (C s  T1 ) + ( C liq  T2 ) + 
M




 5.02  38.56

= 10.0   (0.97  36) + (2.3  192) + 
 10 3 
46.07



3
H TOTAL = 14.2  10 J  14.2 kJ
Conclusion: the enthalpy change required for this transformation is 14.2 kJ.
V
1. The phase diagram of a substance is given above. What is the physical state of this
substance at 25°C and 1.05 atm.
The physical state of the substance is a liquid.
On the phase diagram, w corresponds to the solid phase, x to the liquid phase and y to the
gas phase.
2. Shown here is a portion of the phase diagrams for H2O and CO2.
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a. Explain why the melting curve of H2O deviates to the left and that of CO2 deviates to
the right.
b. A sample of CO2 cannot be melted by heating it in a container open to the
atmosphere. Explain why this is so.
c. The critical temperature and pressure of H2O are 374°C and 218 atm, respectively. At
critical pressure and temperatures above 374°C, what is the physical state H2O?
a. The melting curve of CO2, like most of substances, slopes to the right because as
pressure increases, the more compact solid form is favored and a higher temperature is
required to melt the solid at higher pressure.
In contrast, the melting point line of H2O is atypical, slanting to the left with increasing
pressure. This is explained by the fact that water is among the very few substances whose
liquid form is more compact than its solid form.
b. As shown on the phase diagram of CO2, CO2(s) sublimes at atmospheric pressure. CO2
can be melted at the appropriate temperature only above a pressure of 5.11 atm.
c. At critical pressure and temperatures above 374°C, water is in the state of a
supercritical fluid.
VI
1. What fraction of the volume of each corner atom is actually within the volume of, a
primitive cubic, body centered cubic, and a face-centered cubic unit cells, ?
2. At room temperature copper (Cu) crystallizes in a face-centered cubic unit cell that has
an edge length of 3.61 A° Assuming that atoms touch along the diagonal of the face of
the unit cell, determine the density and radius of copper metal (MCu=63.5 g/mol).
3. Determine the percentage of empty space in the unit cell.
1.
The fraction of the volume of each corner atom in the unit cell of all the lattices of the
cubic crystal system is 1/8.
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2.
Let’s call: a, the length of the edge of the cube
r, the radius of the atom
X, the number of copper atoms occupying one unit cell
M 
X 

m
4  63.5
Na  X  M

d



 8.96 g / cm 3
3
3
23
10
2 3
V
a
Na  a
6.023  10  (3.61 10  10 )
Copper metal has a density of 8.96 g/cm3.
The length of the diagonal of a square is a 2
a 2  4r  r  a
2
2
 3.61
 1.28 A°
4
4
The radius of copper metal is 1.28 A°
3.
The volume of a sphere (ie a Cu atom) is
4
  r3 .
3
4
4
X    r 3 X    r 3
3
3
Portion of occupied space =



3
Vcell
Vcell
 4r 
 43 r 3 




2
2
 2


3
X  4  r  2 2 X   2


 0.74  74%
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3  43 r 3
VCu atoms
X  Vsphere
Portion of empty space = 100-74 = 26%
The portion of empty space in the unit cell is 26%.
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