How much energy? How fast do the molecules/atoms move?
Translation
Thermodynamics, Maxwell-Boltzmann, equipartition theorem, translational kinetic energy of gas = (3/2) k T per molecule. (CM2004)
At room temperature, T = 300 K, Etrans = 6.21 x10-21 J molecule-1 = m v2 / 2
For He atom, m = 4 amu, v = 1.36 x103 ms-1 (~3000 mph!!!)
For C6H6m = 78 amu v = 3.05 x102 ms-1
This is the velocity on average, not of any particular molecule.
Rotation
Similarly the equipartition theorem states that the average rotation energy of a diatomic molecule is (1/2) k T per molecule per rotational direction.
At T = 300 K, Erot,x = μRe2ωx2 /2 = (1/2) k T = 2.07 x10-21 J molecule-1
for H – 35Cl: Re = 1.2746x10-10 m (spectroscopic data)
μ = mHmCl/ (mH + mCl) ~ (1 x 35)/(1+35) amu ~ 1.614x10-27 kg
ωx ~ 6.288x1012 radians s-1 = 1.000x1012 rev s-1
a revolution every 1.0 ps on average.
Vibration
If we assume a vibrational energy of 10-19 J molecule-1 for NO which has a force constant k = 1.54x103 Nm-1 and Re = 1.1508x10-10 m, can estimate (classically) the extent to which the bond stretches.
At maximum (or minimum) R all the energy is Potential Energy (the atoms come to rest at the limits of the vibration, no Kinetic Energy) therefore
(k/2)(R – Re)2 = 10-19 J, from which (R – Re) = 1.14x10-11 m
about 10% of Re
When R = Re all the energy is kinetic energy, therefore
μvR2/2 = 10-19 J , from which vR = 4016 ms-1
and from the classical equations of motion, estimate that a vibration period is 2.26x10-13 s or the vibrational frequency is 4.40x1014 s-1.
Quantum Mechanical Molecular Energy of theNuclei.
The motion of nuclei in molecules does not obey classical mechanics. The energies of translational, rotational and vibrations motion of nuclei obey Quantum Mechanics.
In all three cases quantum mechanics finds the energy to be quantized. Only certain values of energy are allowed. The equations for the energy involve quantum numbers (1 quantum number for every independent coordinate).
Translation. The ‘Particle in the box’ problem. The energy of motion of a particle of mass m in 1 D (say along x) and restricted to a path length of L (e.g. a molecule contained in a box; an electron traveling along a copper wire) is found to be:
En= n2 h 2 / (8mL2) Joules n = 1, 2, 3, … all positive integer values >0
h = 6.626x10-34 JsPlanck’s constant
mmass, kg
Llength of ‘box’, m
Units (Js)2/(kg m2) J
In 3 D, a box of lengths A, B, C in the x, y, z directions, the energy is specified by three quantum numbers.
En,p,q= n2 h 2 /(8m A2) + p2 h 2 /(8m B2) + q2 h 2 /(8m C2) Joules
n = 1, 2, 3, …; p = 1, 2, 3, …;q = 1, 2, 3, …
The sum of the three 1 D energies.
The 3D case demonstrates ‘degeneracy’ if the ‘box’ is a cube, A = B = C:
En,p,q=(n2+ p2 + q2 )h 2 /(8m A2) J
n = 1, 2, 3, …; p = 1, 2, 3, …;q = 1, 2, 3, …
n = 1, p = 2, q = 3 has energy E1,2,3=(1+ 4+ 9)h 2 /(8m A2) = 14 h 2 /(8m A2) J
as do: E1,3,2E2,1,3E2,3,1E3,1,2E3,2,1
6 different energy levels with the same energy – 6 degenerate levels.
Consider a molecule in a (1D) room: O2(g), m = 32 amu = 5.314x 10-26 kg, L = 10 m
En = 1.03283x10-44n2Joules
How this relates to the Classical translational energy is a matter for CM3109/CM4113 (but it does suggest that, on average, n ~ 1011)
Points to note about the ‘Particle in the Box’ problem:
1)n ≠ 0. The particle cannot have zero translational energy. This has to do with the Heisenberg uncertainty principle.
2)Energy levels ~ n2, get further apart with increasing n.
3)1D box is non-degenerate (no 2 energy levels have the same energy), 3D box has degenerate levels under special circumstances.
Quantized Rotational Energy for a diatomic molecule
Rotation. The ‘Rigid Rotator’ problem. The energy of two masses, mA and mB at a fixed distance Re rotating about their centre of mass, is found to be:
Erot = (ħ 2 / 2 I) J (J + 1) Joules J = 0, 1, 2, 3 …
I = μRe2momentum of inertia, kg m2
ħ = h / 2π Js
μ = mAmB/ (mA + mB)reduced mass, kg
Each energy level is degenerate. There are (2 J + 1) levels with the same energy. There is a second quantum number mJ which has integer values between – J and + J , that is (2 J + 1) values all with the same energy determined by J. 2 quantum numbers as there are 2 separate rotational axes/motions.
For H – 35Cl (data given previously)
Erot = (ħ 2 / 2 I) J (J + 1) = 2.12x10-22J (J + 1) Joules
(again the connection with the Classical estimate is complicated and is covered in CM3109/CM4113 but shows that J ~ 3 on average)
In spectroscopy of diatomic molecules, it is usual to write
Erot = BJ (J + 1)
B = (ħ 2 / 2 I) = (ħ 2 / 2 μRe2)Joules is called the rotational constant and in spectroscopy it is usual to give the value in wavenumbers, cm-1
E (J) = h c E(cm-1)
c velocity of light, cm s-1 !! For B = 2.12x10-22 ≡ 10.7 cm-1
Points to note:
1) Energy levels ~(J2 + J), get further apart with increasing J
2) J = 0 is allowed.
3) Degeneracy (2 J + 1)
4) Excellent agreement with spectroscopy in the microwave region.