Download Solutions - Illinois State Chemistry

Chemistry 362
Spring 2013
Dr. Jean M. Standard
March 8, 2013
Name __________KEY________________
Physical Chemistry II – Exam 2 Solutions
1.) (14 points) The harmonic vibrational frequency of NaF is 536.1 cm–1. Determine the
ratio of populations of the v=1 and v=0 levels of the NaF molecule at 300 K. Repeat
your calculation to determine the ratio of populations of the v=2 and v=0 levels of the
NaF molecule at 300 K. Explain how these populations play a role in dictating which
transitions will be observed in the infrared spectrum: v=0→1, v=1→2, or v=2→3.
The population ratio of the v=1 to v=0 levels is given by
n1
g − E −E / k T
= 1 e ( 1 0) B ,
n0
g0
where n j is the population, g j is the degeneracy factor (=1 for harmonic oscillators),
E j is the energy, kB is the
€ Boltzmann constant, and T is the temperature.
€
€ between the harmonic oscillator levels is
€ The energy difference
€
E 1 − E 0 = 23 hν 0 − 12 hν 0 = hν 0 .
The harmonic frequency ν 0 in s–1 may be determined from the harmonic frequency ω e in
cm–1 using the relation
€
€
ν0 = ωe⋅ c
(
= 536.1 cm
€
−1
)(2.99793 × 10
10
)
cm/s
ν 0 = 1.6072 × 1013 s−1 .
Substituting, the population ratio at 300 K becomes
€
n1
= exp{−hν 0 /kB T}
n0
% 6.62618 × 10 −34 Js 1.6072 × 1013 s-1
'
= exp& −
1.38066 × 10 −23 J/K ( 300K)
'(
n1
= 0.076.
n0
(
)(
(
€
)
) )'*
'+
1.
2
continued
The population ratio of the v=2 to v=0 levels is given by
n2
− E −E / k T
= e ( 2 0) B .
n0
The energy difference is
€
E 2 − E 0 = 2hν 0 .
Substituting, the population ratio at 300 K becomes
€
n2
= exp{−2hν 0 /k B T}
n0
% 2 6.62618 × 10 −34 Js 1.6072 × 1013 s-1
'
= exp& −
1.38066 × 10 −23 J/K ( 300K)
'(
n2
= 0.0058.
n0
(
)(
(
)
) )'*
'+
From these results, we see that the population of the v=1 level is only about 8% that of the
v=0 level,
€ and the population of the v=2 level is only 0.6% of the v=0 level. The intensity of
a transition from an initial state to a final state is directly proportional to the population of the
initial state. Thus, even though all the transitions mentioned, v=0→1, v=1→2, and v=2→3,
are allowed by the selection rules for harmonic oscillators ( Δv = ±1), only the v=0 level has
any significant population; therefore, only the v=0→1 transition will have any significant
intensity in the spectrum. The v=1→2 transition may make a small contribution to the
spectrum since the population of v=1 is about 8%.€
(Of course, in the harmonic oscillator approximation, the v=0→1, v=1→2, and v=2→3
transitions all would occur at the same frequency in the spectrum; however, only the v=0→1
transition would contribute significantly to the intensity. For a more realistic anharmonic
oscillator, the v=0→1, v=1→2, and v=2→3 transitions would occur at slightly different
frequencies in the spectrum, and in that case, the v=0→1 transition would be far more intense
than the others.)
2.) (14 points) Compare and contrast the quantum mechanical view of our ability to know
simultaneously the position and momentum of a particle to that of the classical view.
Make sure that you include some comments about the way in which the Heisenberg
Uncertainty Principle plays a role in the quantum mechanical view.
The classical view of the position and momentum of a particle is a deterministic view. That
is, if we start with initial conditions for the position and momentum (or velocity) at time=0,
then using Newton’s equations, we can solve for the position and momentum of the particle
at any future time. So, from classical mechanics, the position and momentum of the particle
can both be known exactly at any time.
The quantum mechanical view is drastically different. First, the quantum mechanical view is
based upon probabilities; it is not deterministic. Second, unlike classical mechanics, the
position and momentum cannot both be known simultaneously. The Heisenberg Uncertainty
!
Principle, Δpx Δx ≥ , does not allow for exact knowledge of both the position and
2
momentum. For example, if the position is exactly known so that Δx = 0 , this implies that
we know nothing about the momentum, Δpx = ∞ (and vice versa).
€
€
€
3
3.) (15 points) The first excited state wavefunction of the harmonic oscillator is given by
% α (1/ 4
" µk % 1/ 2
−α x 2 / 2
ψ1 ( x ) = ' * ⋅ 2α ⋅ x e
, where α = $ 2 ' . For the first excited state,
#! &
&π )
determine the average value of the linear momentum, px, by explicit integration.
€
The expectation value of the linear momentum is defined as
∞
px
∫ ψ * ( x ) pˆ x ψ ( x ) dx .
=
−∞
The momentum operator pˆ x is defined as
€
pˆ x = − i!
€
d
.
dx
Substituting, and using the wavefunction for the first excited state of the harmonic oscillator
given above, the expectation value
becomes
€
∞
px
=
∫ ψ1 * ( x ) pˆ x ψ1( x ) dx
−∞
( α +1/ 2
= − 2i!α * )π ,
∞
∫ xe−α x
2
/2
−∞
2
d
xe−α x / 2 dx .
dx
Evaluating the first derivative gives
€
2
2
2
d
xe−α x / 2 = e−α x / 2 – αx 2e−α x / 2 .
dx
(
)
Substituting, and splitting the expression into two integrals yields
€
px
=
∞
∫ ψ1 * ( x ) pˆ x ψ1( x ) dx
−∞
( α +1/ 2
= − 2i!α * )π ,
px
€
( α +1/ 2
= − 2i!α * )π ,
∞
∫ xe−α x
2
−∞
/2
∞
∫
−∞
xe
−α x 2
(e
dx
−α x 2 / 2
+
– αx 2e−α x
2
/2
( α +1/ 2
2i!α * )π ,
2
) dx
∞
∫ x 3e−α x
−∞
2
dx .
4
5
3.) continued
The integrals that must be evaluated are
∞
∫
2
∞
x e−αx dx
2
x 3 e−αx dx .
∫
and
−∞
−∞
Note that both the functions in the integrals are odd functions. For odd functions, the integral
over the range [ −∞,∞
€ ] equals zero. Therefore,
∞
∫
€
∞
2
x e−αx dx = 0 and
2
x 3 e−αx dx = 0 .
∫
−∞
−∞
Substituting,
€
px
% α (1/ 2
= − 2i!α ' *
&π )
1/ 2
%α(
= − 2i!α ' *
&π )
px
€
= 0.
∞
∫
2
xe−α x dx
% α (1/ 2
2i!α 2 ' *
&π )
+
−∞
1/ 2
%α(
⋅ 0 + 2i!α 2 ' *
&π )
⋅0
∞
∫ x 3e−α x
−∞
2
dx
4.) (14 points) Consider the molecule XeF4, which has four equivalent Xe-F bonds in the
square planar structure shown below.
F
F
Xe
F
F
a.) How many vibrational modes does XeF4 possess?
Since XeF4 is nonlinear, the number of vibrational modes is 3N–6, where N is the number
of atoms.
Thus, # modes = 3*5–6 = 9 modes.
b.) One of the vibrational modes of XeF4 is the out-of-plane deformation, in which the
F-Xe-F bonds bend so that the Xe atom is out of the plane of the fluorine atoms, as
shown below, and then it returns to equilibrium.
Xe
F
F
F
F
Is this mode IR active? Explain.
YES, the mode is IR active. Initially, at equilibrium the dipole moment of the molecule
is zero. When the fluorine atoms bend out of the plane, the center of negative charge of
the fluorines shifts out of the plane, creating a net dipole moment. This change in the
dipole moment during the course of the vibration causes the mode to be IR active.
c.) Another of the vibrational modes of XeF4 is the symmetric stretch (or breathing
mode), which corresponds all the Xe-F bonds increasing at the same time, followed
by all the Xe-F bonds decreasing at the same time. Is this mode IR active? Explain.
NO, this mode is not IR active. When the Xe-F bonds all stretch or compress in a
symmetric fashion, the dipole moment of the molecule remains zero. With no change in
dipole moment during the course of the vibration, the mode is IR inactive.
6
5.) (14 points) The harmonic vibrational frequency in wavenumbers of the DCl molecule
is 2144.7 cm–1. The anharmonicity constant is 0.01251.
a.) Treat the DCl molecule as a harmonic oscillator and determine the fundamental
vibrational transition in wavenumbers.
For a harmonic oscillator, the vibrational energy in wavenumbers is given by
ωv =
Ev
= ω e ( v + 12 ) .
hc
The fundamental transition corresponds to v = 0 →v = 1. This transition is
€
ω 0→1 = ω 1 − ω 0 =
ω 0→1 €= ω e
3
2ωe
−
1
2ωe
ω 0→1 = 2144.7cm−1 .
€
b.) Treat the DCl molecule as an anharmonic oscillator and determine the fundamental
vibrational transition in wavenumbers.
For an anharmonic oscillator, the vibrational energy in wavenumbers is given by
ωv =
2
Ev
= ω e ( v + 12 ) − ω e x e ( v + 12 ) .
hc
The fundamental transition corresponds to v = 0 →v = 1. This transition is
€
2(
2(
%
%
ω 0→1 = ω 1 − ω 0 = 'ω e ( 23 ) − ω e x e ( 23 ) * − 'ω e ( 12 ) − ω e x e ( 12 ) *
&
) &
)
€
ω 0→1 = ω e − 2ω e x e
(
)
ω 0→1 = 2144.7cm−1 − 2 2144.7cm−1 (0.01251)
ω 0→1 = 2091.0cm−1
€
7
6.) (14 points) A buckyball (C60, with mass 1.197×10–24 kg) is confined to move in two
dimensions on a surface. The confined region is a square area 5.0 nm on each side. The
buckyball is excited with electromagnetic radiation with a wavelength of 13.54 m. It
undergoes a transition from its ground state ( n x = n y = 1) to an excited state. [1 nm =
10–9 m]
a.) Determine the energy (in Joules)€of the excited state of the buckyball.
A photon with an energy corresponding to ΔE would have a frequency given by
λ
E photon = ΔE = hν . Since, for light, λν = c , we can substitute ν = , and calculate the
c
energy difference ΔE ,
€
€ hc
ΔE =
λ
€
€
€
(6.62607 × 10
−34
=
−26
ΔE = 1.4671 × 10
)(
)
J s 2.99793 × 10 8 m/s
13.54 m
J.
The two-dimensional particle in a box energies are given by
€
E nx ny =
(
n x2
+
n y2
(
)
h2
.
8mL2
)
For a transition from the ground state with n x ,n y = (1,1) to some unknown final state
€
n x ,n y , the energy difference
ΔE is
(
)
ΔE €= E n x n y − E 11
€
€
(
= E n x n y − 12 + 12
= E nx ny
h2
−
.
4mL2
Solving for the excited state energy E n x n y yields,
€
€
€
E nx ny
2
h
) 8mL
h2
= ΔE +
.
4mL2
2
8
9
6 a.) continued
Substituting,
E n x n y = ΔE +
h2
4mL2
2
(6.62607 × 10 J s)
4 (1.197 × 10 kg)(5.0 × 10
−34
= 1.4671 × 10−26 J +
−24
−9
)
m
2
E n x n y = 1.8339 × 10−26 J .
€
b.) Determine the excited state quantum numbers n x and n y of the buckyball after it
absorbs the radiation.
The two-dimensional particle in a box energies
are given by
€
€
(
E n x n y = n x2 + n y2
)
h2
.
8mL2
Solving for the unknown quantum number squared sum n x2 + n y2 yields,
€
n x2
+
n y2
8mL2 E n x n y
=
€ h2
.
Substituting,
€
n x2
+
n y2
=
=
8mL2 E n x n y
h2
(
)(
(6.62607 × 10
−34
2
) (1.8339 × 10 J)
J s)
8 1.197 × 10−24 kg 5.0 × 10−9 m
−26
2
n x2 + n y2 = 10 .
Therefore, the quantum numbers of the excited state must be
€
(n ,n ) = (1,3)
x
€
y
or
(n ,n ) = (3,1) .
x
y
7.) (14 points) The harmonic vibrational frequency (in wavenumbers) of the OH radical is
3650 cm–1. Use this information to predict the harmonic vibrational frequency of the
OD radical in wavenumbers.
By definition, the harmonic frequency
ν 0 is given by
1/ 2
1 $k'
ν0 =
& )
2π % µ (
.
The harmonic frequency in wavenumbers is
€
ωe
1/ 2
ν0
1 %k(
=
=
' * .
c
2πc & µ )
Therefore, the ratio of frequencies (in wavenumbers) can be written
€
1/ 2
1 $ kOD '
&
)
2πc % µOD (
ω e (OD)
=
1/ 2 .
ω e (OH)
1 $ kOH '
&
)
2πc % µOH (
Assuming that kOD = kOH , this equation simplifies to
€
€
# µOH &1/ 2
ω e (OD)
= %
( .
ω e (OH)
$ µOD '
Calculating the reduced masses, we have
µOH
€
€
€
(15.9949amu)(1.0078amu) = 0.94806 amu
mO mH
=
=
mO + mH
(15.9949amu +1.0078amu)
µOD =
(15.9949amu)(2.1040amu) = 1.78877 amu .
mO mD
=
mO + mD
(15.9949amu + 2.0140amu)
10
11
7.) continued
Substituting the reduced masses into the frequency ratio yields,
# µOH &1/ 2
# 0.94806 amu &1/ 2
ω e (OD)
= %
= %
= 0.7280.
(
(
$ 1.78877 amu '
ω e (OH)
$ µOD '
Solving for the OD frequency gives
€
ω e (OD) = ω e (OH)⋅ 0.7280
(
)
= 3650 cm–1 (0.7280)
ω e (OD) = 2657 cm–1 .
€