Advanced Thermodynamics   Quiz #3

1. Predict the symmetry factors for
   
   
   1. PCl₅

• Non-linear
• No more than one C₃ axis
• But 3 C₂ axes perpendicular to the C₃
• And a horizontal plane...
• Hence D_3h by the scheme
• Containing E, 2C₃, and 3C₂ for a symmetry factor of 6

H₂O₂

• Non-linear
• No more than one C₂ axis (perpendicular to the plane)
• And the plane is a symmetry element
• So it's C_2h by the scheme
• Containing E and C₂ for a symmetry factor of 2

C₂H₄

• Non-linear
• No axis higher than C₂ (perpendicular to the plane)
• But two other C₂ axes in the plane
• And the plane is a symmetry element
• Hence it's D_2h by the scheme
• Having E, C₂^x, C₂^y, and C₂^z for a symmetry factor of 4

C₆H₆ , benzene

• Non-linear
• Only one C₆ axis (perpendicular to the plane)
• But six C₂ axes in the plane (through atoms and through bonds)
• And the plane is a symmetry element
• So it's D_6h by the scheme
• With E, (C₂, 2C₃, 2C₆)_{principal axis}, 3C₂', and 3C₂" for a symmetry factor of 12

IF₄^-

• Non-linear
• Only one C₄ axis (perpendicular to the plane)
• Four C₂ axes in the plane (through and between bonds)
• And the plane is still a symmetry element
• Makes it D_4h by the scheme
• With E, (C₂, 2C₄)_{principal axis}, 2C₂', and 2C₂" for a symmetry factor of 8

Predict the K_P at 1000 K for the abstraction reaction

·NH₂(g) + H₂(g) NH₃(g) + ·H(g)

Molecule	Symmetry	Term	Vibrations (cm-1)	Rotations (cm-1)	1st Excited (cm-1)	D0 (1st H) (kJ/mol)
·NH2(g)	C2v	2B1	3219 1497 3301	23.728 12.942 8.16	10249 (2A1)	-
H2(g)	Dh	1g+	4395	60.80	91690 (1u+)	436.0
NH3(g)	C3v	1A1 *	3336 (A1) 955 (A1) 3444 (E) 1626 (E)	9.444 6.196 6.196	46136 (1A2")	449.4
·H(g)	-	2S½	-	-	82259 (2S½)	-

* NH₃ can "invert" (pass N through the plane of the H atoms) to yield a "second ground state" only 0.793 cm^-1 above the ¹A₁. Since they share virtually the same frequencies, we can "inversion double" the otherwise expected electronic degeneracy.

(Cancel everything cancellable. Estimate everything estimable. Use the handouts.)

- ₀ / kT

_i

+D₀/RT

1. The largest term is usually e^+D₀/RT and so we can get a quick feel for the magnitude of the answer if we calculate that first. We'll need the conversion factor from our Periodic Table cheat sheet which tells us that 1 cm^-1 = 11.9620 J/mol (but we're unlikely to need all that accuracy!). That's 1 cm^-1 = 0.0120 kJ/mol for convenience. So...
   
   e^+D₀/RT = e^{+(449.4-436.0 kJ/mol)÷(0.0120 kJ/mol cm-1) / 695 cm-1} = e^+1.607 = 5
   
   So the gross thermochemistry says breaking a weaker H₂ bond to make a stronger H-NH₂ bond gives you an exothermicity which would suggest more products than reactants...but not by much here...so the partition function factors can make a difference!
   
   
2. K_TRANS = [(M_NH3×M_H) / (M_NH2×M_H2)]^3/₂ = [(17×1)/(16×2)]^3/₂ = 0.39
   
   In other words, tying up more of the mass in one of the molecules was an unfavorable thing to do from an entropic point of view. This alone drops what we had begun as a K_P of 5 down to 2... amazing!
   
   
3. As both NH₂ and NH₃ each have 3 rotational constants (ABC), the constants and temperature factors in their partition functions will cancel out, but, of course, their symmetry factors won't. It doesn't take a Rocket Scientist (my brother's one of those for a fact) to guess or determine the symmetry factors of NH₂ and NH₃ to be 2 and 3, respectively.
   
   The H atom has no rotation (or vibration for that matter), so H₂'s rotational partition function won't be cancelled by any other linear molecule's z_ROT. So we have to include it, constants, temperature, B and symmetry factors and all.
   
   We should worry a moment about H₂. That 61 cm^-1 rotational constant is huge. If kT had been its 209 cm^-1 as at room temperature, we couldn't get away with using the simple integral approximation to H₂'s z_ROT. But with kT here 695 cm^-1 (over an order of magnitude larger than B!), we don't have a thing to worry about.
   
   K_ROT = [ (ABC)^½_NH2]×[ B_H2 / 695 cm^-1] / [ (ABC)^½_NH3]
   = [2 (23.728×12.942×8.16)^½]×[2 (60.80)/695] / [3 (9.444×6.196²)^½] = 0.31
   
   So the loss of a rotor (H₂) wasn't made up by the sloppiness of the ammonia rotations, and we've now reduced our composite K_P to 0.59! Partition functions have taken their toll and actually reversed the direction of spontaneity over simple energy considerations! What a neat question!
   
   
4. Since kT is 695 cm^-1, we don't need to worry much about molecular frequencies above about 2100 cm^-1. Ignoring one there will miss out on a factor of 1.05, as we showed on the blackboard 1/(1-e^-3) = 1.05. Now if there were 20 vibrations like that, we'd be off by a factor of (1.05²⁰) almost 3, but, fortunately, the high frequency vibrations in these molecules are all in excess of 3200 cm^-1, and each of those will contribute a completely negligible factor of 1.01.
   
   So that leaves us with worrying only about NH₂'s 1497 and NH₃'s 955 and doubly degenerate 1626 cm^-1 vibrations.
   
   K_VIB = (1-e^-1497/695) / [ (1-e^-955/695)×(1-e^-1626/695)² ] is all that matters
   = 0.88 / [ (0.75)(0.90)² ] = 1.44
   
   ...which raises our accumulated K_P to 0.87 since ammonia's vibrations are sloppier than the amino radical's.
   
   
5. The easiest is the electronic partition function since the excited states are so high in energy that only the ground state degeneracy contributes. The radicals are both doublets (g=2) and the molecules are both singlets (g=1), but ammonia molecule suffers from "inversion doubling," so it's g=2.
   
   K_EL = g_NH3 g_H / [ g_NH2 g_H2 ] = 2×2 / 2×1 = 2
   
   
6. So the product totals now to 1.74, saved by inversion doubling! (But still only about ½ what you'd expect from energy considerations.)
   
   K_P = (0.39)(0.31)(1.44)(2)(5) = 1.74