5. Spectroscopic measurements indicate that the rotational constants of SO2, a non-linear molecule, are 2.02736, 0.34417, and 0.29354 cm ¹. Note: ks = 0.69503476 cm-¹/K. Reference (NIST): https://cccbdb.nist.gov/expgeom2.asp?casno=7446095&charge=0 (a) Compute Zrot at T = 385 K assuming its symmetry number is o= 2. Hint: Using the rotational temperatures for the molecule might make this easier. (b) At what temperature would the partition function equal 2.4 × 104?

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5. Spectroscopic measurements indicate that the rotational constants of SO2, a non-linear
molecule, are 2.02736, 0.34417, and 0.29354 cm ¹. Note: ks = 0.69503476 cm-¹/K.
Reference (NIST): https://cccbdb.nist.gov/expgeom2.asp?casno=7446095&charge=0
(a) Compute Zrot at T = 385 K assuming its symmetry number is o= 2. Hint: Using the
rotational temperatures for the molecule might make this easier.
(b) At what temperature would the partition function equal 2.4 x 104?
Transcribed Image Text:5. Spectroscopic measurements indicate that the rotational constants of SO2, a non-linear molecule, are 2.02736, 0.34417, and 0.29354 cm ¹. Note: ks = 0.69503476 cm-¹/K. Reference (NIST): https://cccbdb.nist.gov/expgeom2.asp?casno=7446095&charge=0 (a) Compute Zrot at T = 385 K assuming its symmetry number is o= 2. Hint: Using the rotational temperatures for the molecule might make this easier. (b) At what temperature would the partition function equal 2.4 x 104?
Label
A
B
с
D
E
F
G
H
J
K
L
M
N
O
P
Q
R
S
T
U
V
W
X
Equation
Etotal= Eetec + Evib + Erot + Etrans
E₂ = hw (v + ²)
Zvib =
Evib =
h²
Erot= 27 J+1)=Bjj+1)
Etrans (nx. ny. ₂) =
3N-6
[₁ hw; (v₁ + ²)
3N-6
(-1
w(n)=
-
E-ENG
E=
N = n₁
Ztrans
P₁ = gjeti/kg
z
τα
z=₁9₁e-B, or z=₁e-/kg
h² (n.2
B=
8m L
-
Z=
1 az
z ap
Zrot
Zrot =
1
kgT
(9₁)",
n₂!
Zvib =
■
N!
+
A₁ e₁/kt
gjet/kat
Σge-B
(2mmk, T., 3/2
h²
B
erot= ks
z = Zelec Zvib Zrot Ztrans
kgT
Ba
erib=
a In Z
aß
√R T3
, Ꮎ Ꮎ Ꮎ
e-Bho/2
1-e-Bhu
ho
kB
2
V
+
3N-6
e-Bhoj/z
=
-ПIG
-e-Bhay
-Oyib./2T
-e-vib/T
Zvibi
P(b) = (1-e-b/T)e-/T
Transcribed Image Text:Label A B с D E F G H J K L M N O P Q R S T U V W X Equation Etotal= Eetec + Evib + Erot + Etrans E₂ = hw (v + ²) Zvib = Evib = h² Erot= 27 J+1)=Bjj+1) Etrans (nx. ny. ₂) = 3N-6 [₁ hw; (v₁ + ²) 3N-6 (-1 w(n)= - E-ENG E= N = n₁ Ztrans P₁ = gjeti/kg z τα z=₁9₁e-B, or z=₁e-/kg h² (n.2 B= 8m L - Z= 1 az z ap Zrot Zrot = 1 kgT (9₁)", n₂! Zvib = ■ N! + A₁ e₁/kt gjet/kat Σge-B (2mmk, T., 3/2 h² B erot= ks z = Zelec Zvib Zrot Ztrans kgT Ba erib= a In Z aß √R T3 , Ꮎ Ꮎ Ꮎ e-Bho/2 1-e-Bhu ho kB 2 V + 3N-6 e-Bhoj/z = -ПIG -e-Bhay -Oyib./2T -e-vib/T Zvibi P(b) = (1-e-b/T)e-/T
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