SME
Physical Chemistry I: Thermodynamics & Kinetics
Statistical thermodynamics
Use this for “HCl vibrates at 2886 cm⁻¹. Find its vibrational partition function, the fraction in v = 0 and the molar vibrational energy at 1000 K.”
Chart
This is an example. Type your own values to start a new problem.
ν̃
Vibrational wavenumber
example · cm⁻¹
θᵥ
Vibrational temperature
calculated · K
T
Temperature
example · K
qvib
Vibrational partition function
calculated
p₀
Fraction in level v = 0
calculated
p₁
Fraction in level v = 1
worked out (not typed)
p₂
Fraction in level v = 2
worked out (not typed)
p₃
Fraction in level v = 3
worked out (not typed)
U
Molar vibrational energy
worked out (not typed) · J/mol
Formulas
θᵥ
=
1.4388
×
ν̃
3,000 = 1.4388 × 2,085
qvib
=
1
1 −
e
−θᵥ ÷
T
1.157
=
1
1 −
e
−3,000 ÷ 1,500
p₀
=
1
qvib
0.8647
=
1
1.157
p₁
=
p₀
×
e
−θᵥ ÷
T
0.117
=
0.8647
×
e
−3,000 ÷ 1,500
p₂
=
p₁
×
e
−θᵥ ÷
T
0.01584
=
0.117
×
e
−3,000 ÷ 1,500
p₃
=
p₂
×
e
−θᵥ ÷
T
0.002144
=
0.01584
×
e
−3,000 ÷ 1,500
U
=
8.314
×
θᵥ
e
θᵥ ÷
T
− 1
3,904
=
8.314
×
3,000
e
3,000 ÷ 1,500
− 1
Assumptions
•
One vibration, treated as a harmonic oscillator: its levels are evenly spaced, hcν̃ apart.
•
Energies count from the zero-point level, so q = 1 ÷ (1 − e−θᵥ/T) and level v holds the fraction p₀e−vθᵥ/T.
•
θᵥ = hcν̃ ÷ k = 1.4388ν̃ with hc ÷ k = 1.4388 cm·K; R = 8.314 J/(mol·K).
Step-by-step
WE KNOW
ν̃ = 2,085 cm⁻¹, T = 1,500 K
FIND
vibrational temperature (θᵥ), vibrational partition function (qvib), fraction in level v = 0 (p₀), fraction in level v = 1 (p₁), fraction in level v = 2 (p₂), fraction in level v = 3 (p₃), molar vibrational energy (U)
Step 1 · Find vibrational temperature (θᵥ)
Use θᵥ = 1.4388 × ν̃
hc
k
=
1.4388
cm·K
turns
the
wavenumber
into
a
temperature.
θᵥ = 1.4388 × 2,085
θᵥ = 3,000 K
Step 2 · Find vibrational partition function (qvib)
Use qvib = 1 ÷ (1 − e−θᵥ ÷ T)
Sum
e
−vθᵥ/
T
over
every
level:
a
geometric
series,
1
1 −
e
−θᵥ/
T
.
qvib
=
1
1 −
e
−3,000 ÷ 1,500
qvib
=
1
1 − 0.13534
qvib
=
1
0.86466
qvib = 1.157 → q is above 1 by a clear margin: v = 1 and higher hold a real share of the molecules.
Step 3 · Find fraction in level v = 0 (p₀)
Use p₀ = 1 ÷ qvib
The ground level’s Boltzmann factor is 1, so its share is 1 over the sum.
p₀
=
1
1.15653
p₀
=
0.8647
Step 4 · Find fraction in level v = 1 (p₁)
Use p₁ = p₀ × e−θᵥ ÷ T
Each
level
up
multiplies
the
share
by
e
−θᵥ/
T
.
p₁
=
0.8647
×
e
−3,000 ÷ 1,500
p₁
=
0.8647
×
0.13534
p₁
=
0.117
Step 5 · Find fraction in level v = 2 (p₂)
Use p₂ = p₁ × e−θᵥ ÷ T
One more level up: the same factor again.
p₂
=
0.11702636
×
e
−2,999.898 ÷ 1,500
p₂
=
0.117
×
e
−1.9999
p₂
=
0.117
×
0.13535
p₂
=
0.01584
Step 6 · Find fraction in level v = 3 (p₃)
Use p₃ = p₂ × e−θᵥ ÷ T
And once more for v = 3.
p₃
=
0.01584
×
e
−3,000 ÷ 1,500
p₃
=
0.01584
×
0.13534
p₃
=
0.002144
Step 7 · Find molar vibrational energy (U)
Use U = 8.314 × θᵥ ÷ (eθᵥ ÷ T − 1)
U
=
RT²
d(ln
q)/dT
works
out
to
Rθᵥ
e
θᵥ/
T
− 1
,
counted
from
the
zero-point
level.
U
=
8.314
×
3,000
e
3,000 ÷ 1,500
− 1
U
=
8.314
×
3,000
7.389 − 1
U
=
8.314
×
3,000
6.389
U
=
24,942
6.389
U
=
3,904
J/mol
Check
Put the numbers back in. Both sides should match.
3,000 = 1.4388 × 2,085 ✓
1.157
=
1
1 −
e
−3,000 ÷ 1,500
✓
0.8647
=
1
1.15653
✓
0.117
=
0.8647
×
e
−3,000 ÷ 1,500
✓
0.01584
=
0.11702636
×
e
−2,999.898 ÷ 1,500
✓
0.002144
=
0.01584
×
e
−3,000 ÷ 1,500
✓
3,904
=
8.314
×
3,000
e
3,000 ÷ 1,500
− 1
✓
← Previous lesson
Unimolecular rate against pressure (Lindemann)
Next lesson →
Rotational partition function and the most populated level
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.