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CHEMISTRY 120 (2) DR. J. F. C. TURNER
3RD EXAMINATION
MONDAY, 18TH NOVEMBER, 2002
2 HOURS
ANSWER ALL QUESTIONS
YOU MAY NOT USE PROGRAMMABLE CALCULATORS, NOTE CARDS OR ANY OTHER SIMILAR WRITTEN MATERIALS
SHOW ALL YOUR WORK RELATIVE ATOMIC MASS AND CONVERSION FACTORS ARE GIVEN AT THE END OF THE
EXAM
NAME
SECTION
1(a)
Aluminum (Al) and phosphorous (P4) react at elevated temperatures to form aluminum phosphide.
Write a balanced chemical equation for this reaction.
Al(s) + 1/4P4(s)
AlP(s)
or
4AlP(s)
4Al(s) + P4(s)
3
Write down the oxidation states for both Al and P in both the reactants and the products.
Reactants
Products
Al
0
+3
P
0
-3
4
(c)
Is aluminum phosphide ionic or covalent?
Ionic
1
(d)
Aluminum phosphide reacts with water to form phosphine, PH3(g) and aluminum oxide, Al2O3(s).
Write a balanced equation for this reaction.
(b)
2AlP(s) + 3H2O(l)
Al2O3(s) + 2PH3(g)
3
(e)
The enthalpies of formation of AlP(s), PH3(g), H2O(l) and Al2O3(s)
are given in the table. Use these values and Hess’ Law to calculate
the enthalpy of reaction, ∆H r° , for the reaction between aluminum
phosphide and water.
2AlP(s) + 3H2O(l)
∆Hr
∆H1
Al2O3(s) + 2PH3(g)
Compound
∆H °f /kJmol-1
AlP(s)
H2O(l)
PH3(g)
Al2O3(s)
-166.5
-285.8
5.4
-1675.7
∆H2
2Al(s) + 1/2P4(s) + 3H2(g) +3/2 O2(g)
∆Hr = ∆H1 + ∆H2
∆H1 = [-3(∆Hf(H2O(l))] + [-2(∆Hf(AlP(s))]
∆H2 = [(∆Hf(Al2O3(s))] + [2(∆Hf(PH3(g))]
∆Hr = [-3(∆Hf(H2O(l))] + [-2(∆Hf(AlP(s))] + [(∆Hf(Al2O3(s))] + [2(∆Hf(PH3(g))]
= [-3(-285.8)] + [-2(-166.5)] + [-1675.7] + [2(5.4)]
= [857.4] + [333] + [-1675.7] + [10.8]
= -474.5 kJmol-1
6
Is this reaction endothermic or exothermic?
exothermic
1
The enthalpy of a reaction can be written as
H = U + ∆( PV )
and when gas is generated or absorbed in a reaction, this equation can be written as
H = U + (∆n).RT
where ∆n is the change in the number of moles of gas. Assuming that T = 298 K and R = 8.314
Jmol-1K-1, calculate the change in internal energy for the reaction between aluminum phosphide and water
when 2 moles of aluminum phosphide and 3 moles of water are used.
(f)
(g)
Al2O3(s) + 2PH3(g)
2AlP(s) + 3H2O(l)
∆n = 2
∆Hr = ∆Ur + ∆n.RT
∆Hr = ∆Ur + 2.(8.314.298)
∆Ur = ∆Hr- 2.(8.314.298)
= -474.5 - 4.995
= -479.5 kJmol-1
(h)
°
f
(
+
(aq)
Given that ∆H H
)
-1
°
f
(
= 0 kJmol , ∆H H2O(l)
)
-1
= -285.8 kJmol
and that ∆H
°
r
6
for the
neutralization reaction
H+(aq) + OH-(aq)
(
)
∆Ηr = -55.8 kJmol-1
H2O(l)
−
is -55.8 kJmol-1, calculate ∆H °f OH(aq)
.
∆Ηr = -55.8 kJmol-1
H+(aq) + OH-(aq)
H2O(l)
∆Η1 = [−∆Ηf (H+(aq))] + [−∆Ηf (OH-(aq))]
∆Η2 = [∆Ηf (H2O(l))]
H2(g) + 1/2O2(g)
∆Ηr = ∆Η1 + ∆Η2 = -55.8 kJmol-1
∆Η1 = [−∆Ηf (H+(aq))] + [−∆Ηf (OH-(aq))]
∆Η2 = [∆Ηf (H2O(l))]
+
∆Ηr = [−∆Ηf (H
(aq))]
+ [−∆Ηf (OH-(aq))] + [∆Ηf (H2O(l))] = -55.8 kJmol-1
∆Ηr = [0] + [−∆Ηf (OH-(aq))] + [−285.8] = -55.8 kJmol-1
∆Ηf (OH-(aq)) = -285.8 + 55.8 = -230 kJmol-1
6
30
3(a)
Write down the equation for the de Broglie relationship between momentum and the associated
wavelength for a particle
h
h
λ=
or λ =
p
mv
2
(b)
Calculate the momentum of an electron with a wavelength of 0.1 Å, given that Planck’s constant =
6.63 x 10-34 Js. With what velocity does this electron move if the electron mass is 9.1 x 10-31 kg?
h
λ=
p
λ = 0.1 × 10 −10 =
6.63 × 10 −34
p
6.63 × 10 −34
= 6.63 × 10 −23 kgms −1
0.1 × 10 −10
p = mv
p=
p 6.63 × 10 −23
=
= 7.29 × 10 7 ms −1
m
9.1 × 10 −31
v=
4
(c)
(i)
Write down the equation for
The relationship between energy and frequency for electromagnetic radiation
E = hν
and
(ii)
The relationship between the velocity, the frequency and the wavelength of electromagnetic
radiation
c = λν
4
(d)
Calculate the frequency and wavelength of a photon which has an energy of 3.798 x 10-15 J
E = hν
∴ν =
E 3.798 × 10 −15
=
= 5.728 × 1018 s −1
h
6.63 × 10 −34
c = λν
o
3 × 108
= 5.247 × 10 −11 m = (0.5247 Α)
18
5.728 × 10
(e)
6
The Bohr relation for the energy of an electron with a principle quantum number n is given by
−B
, where B = 2.179 × 10 −18 J . An electron in a hydrogen atom moves from a state where n = 8 to a
n2
state where n = 1 . Calculate
(i)
the energy of this transition
− 2.179 × 10 −18 − 2.179 × 10 −18
E8 =
=
= −3.405 × 10 − 20 J
64
82
− 2.179 × 10 −18 − 2.179 × 10 −18
E1 =
=
= −2.179 × 10 −18 J
2
1
1
∆E = E1 − E 8 = − 2.179 × 10 −18 − − 3.405 × 10 − 20 = −2.145 × 10 −18 J
6
(ii)
The wavelength and frequency of the photon emitted.
E=
(
) (
)
E = hν
∴ν =
E 2.145 × 10 −18
= 3.235 × 1015 s −1
=
− 34
h
6.63 × 10
c = λν
3 × 10 8
= 9.273 × 10 −8 m
3.235 × 10 15
6
(iii)
If 1 mole of hydrogen atoms in a state where n = 8 move to a state where n = 1 , calculate the
quantity of energy released.
− 2.179 × 10 −18 − 2.179 × 10 −18
E8 =
=
= −3.405 × 10 − 20 J
64
82
− 2.179 × 10 −18 − 2.179 × 10 −18
E1 =
=
= −2.179 × 10 −18 J
2
1
1
∆E = E1 − E 8 = − 2.179 × 10 −18 − − 3.405 × 10 − 20 = −2.145 × 10 −18 J per atom
(
) (
)
Per mole, E = (2.145 × 10 )× (6.02 × 10 ) = 1.29 kJmol
−18
23
-1
6
34
3(a)
Write down and name the three atomic quantum numbers that define the motion of an electron in
an atom
l
orbital quantum number
n
Principle quantum number
ml
Magnetic quantum number
6
(b)
What is the fourth quantum number that is necessary for the complete description of an electron
in an atom?
s
spin quantum number
2
(c)
Write down a statement of the Pauli Exclusion Principle
(d)
No pair of electrons can have identical quantum numbers
2
(d)
How many values are allowed for
ml for a given l
2l + 1
l for a given n
n −1
2
(e)
For the elements in bold and underlined in the partial Periodic Table, write out the full electronic
configuration for the atom in the form Be: 1s22s2
O
1s22s22p4
Ne
1s22s22p6
Na
1s22s22p63s1
Al
1s22s22p63s23p1
P
1s22s22p63s23p3
Ca
1s22s22p63s23p64s2
Mn
1s22s22p63s23p64s23d5
14
H
He
Li
Be
B
C
N
O
F
Ne
Na
Mg
Al
Si
P
S
Cl
Ar
K
Ca
Sc
Ti
V
Cr
Fe
Mn
s
Co
Ni
Cu
Zn
d
p
(f)
Write the l value that corresponds to each block in the partial Periodic Table in the boxes below
each block.
3
(g)
Cr exhibits common oxidation states of (II), (III) and (VI). What are the electronic configurations
for these oxidation states? Write the configurations either in the form 1s22s2, or using the noble gas core
configurations.
Cr(II) 1s22s22p63s23p63d4
Cr(III)
1s22s22p63s23p63d3
Cr(VI) 1s22s22p63s23p6
(h)
For the following orbitals, write down the l value and the letter associated with each one
l=0
s
l=1
p
l=2
d
l=1
p
6
8
43
4(a)
Curve 1 and curve 2 represent
the variation of energy of two atoms with
distance.
Which curve shows the energy change
with respect to distance for
(i)
a bonding orbital
Curve 1
Curve 1
Curve 2
9
Curve 2
(ii)
an antibonding orbital
Curve 2
Energy
Curve 1
9
Distance
Mark your choice with a check in the
appropriate box.
2
(b)
State the reason why the energy
rises rapidly, in region A for both curves.
B
Internuclear repulsion raises the
energy as the nuclei become closer
together
(c)
A
4
What is special about point B on curve 1? What distance in the molecule does it correspond to?
Point B is the equilibrium distance – the point of lowest energy and represents the bond length in
the molecule
4
(d)
Sketch the two orbitals that correspond to curves 1 and 2, using s orbitals to illustrate the phase
relationships between the two orbitals.
+
σ bonding orbital
+
σ* antibonding orbital
4
(e)
VSEPR theory is a method of determining the shapes of main group molecules. Name and draw
the basic structure for four five and six stereochemical units (lone pairs or bond pairs) around a central
atom
Four
Five
Six
9
(f)
atom.
Sketch the structures of SF6 and SF4, taking care to show the stereochemistry at the central sulfur
F
S
F
F
F
F
F
F
S
F
F
F
8
31