Answer :
Answer:
Explanation:
To calculate the fundamental vibration frequency, force constant, and zero-point energy of carbon monoxide (CO), we can use the information provided and apply some fundamental equations from molecular spectroscopy.
1. Fundamental Vibration Frequency
The band at 2144 cm
−
1
−1
corresponds to the vibrational transition of CO. The fundamental vibration frequency
ν in Hz can be calculated using the wavenumber
ˉ
ν
ˉ
:
=
ˉ
×
ν=
ν
ˉ
×c
where:
ˉ
=
2144
cm
−
1
ν
ˉ
=2144cm
−1
c is the speed of light in cm/s (
≈
2.998
×
1
0
10
cm/s
c≈2.998×10
10
cm/s)
=
2144
cm
−
1
×
2.998
×
1
0
10
cm/s
ν=2144cm
−1
×2.998×10
10
cm/s
=
6.435
×
1
0
13
Hz
ν=6.435×10
13
Hz
2. Force Constant
The force constant
k can be calculated using the formula:
=
(
2
)
2
k=μ(2πν)
2
where:
μ is the reduced mass of the CO molecule.
ν is the fundamental vibration frequency in Hz.
The reduced mass
μ is calculated using:
=
1
×
2
1
+
2
μ=
m
1
+m
2
m
1
×m
2
For CO:
Mass of carbon (
m
C
) = 12 amu
Mass of oxygen (
m
O
) = 16 amu
=
12
×
16
12
+
16
=
192
28
≈
6.857
amu
μ=
12+16
12×16
=
28
192
≈6.857amu
Convert this mass into kg (1 amu =
1.660539
×
1
0
−
27
kg
1.660539×10
−27
kg):
=
6.857
×
1.660539
×
1
0
−
27
≈
1.141
×
1
0
−
26
kg
μ=6.857×1.660539×10
−27
≈1.141×10
−26
kg
Now calculate
k:
=
(
2
)
2
k=μ(2πν)
2
=
1.141
×
1
0
−
26
×
(
2
×
6.435
×
1
0
13
)
2
k=1.141×10
−26
×(2π×6.435×10
13
)
2
≈
1.141
×
1
0
−
26
×
(
4.038
×
1
0
14
)
2
k≈1.141×10
−26
×(4.038×10
14
)
2
≈
1.141
×
1
0
−
26
×
1.631
×
1
0
29
k≈1.141×10
−26
×1.631×10
29
≈
1.867
×
1
0
3
N/m
k≈1.867×10
3
N/m
3. Zero-Point Energy
The zero-point energy (ZPE) can be calculated using:
0
=
1
2
ℎ
E
0
=
2
1
hν
where
ℎ
h is Planck’s constant (
6.626
×
1
0
−
34
J s
6.626×10
−34
J s) and
ν is the frequency in Hz.
0
=
1
2
×
6.626
×
1
0
−
34
×
6.435
×
1
0
13
E
0
=
2
1
×6.626×10
−34
×6.435×10
13
0
≈
2.14
×
1
0
−
20
J
E
0
≈2.14×10
−20
J
0
≈
0.134
eV
E
0
≈0.134eV
Summary
Fundamental Vibration Frequency:
6.435
×
1
0
13
Hz
6.435×10
13
Hz
Force Constant:
1.867
×
1
0
3
N/m
1.867×10
3
N/m
Zero-Point Energy:
2.14
×
1
0
−
20
J
2.14×10
−20
J (or
0.134
eV
0.134eV)
