Advanced Thermodynamics   Final

1. Since you don't have STANJAN at your desks this evening, we'll have to assume that
   CH₄(g) + 4 NO(g) CO₂(g) + 2 N₂(g) + 2 H₂O(g)

   really represents all the important species in that explosion. (It doesn't.) And the table below should give us all the data we need to estimate the stoichiometric explosion overpressure (i.e., how much in excess of 1 atm does the adiabatic explosion generate?).

   Species	CH4(g)	NO(g)	CO2(g)	N2(g)	H2O(g)
   CP, J/mol K	35.31	29.84	37.1	29.12	33.58
   Hf°, kJ/mol	-74.81	90.25	-393.5	0.00	-241.82

   Remember that explosions occur at fixed volume (so fast that products have not yet expanded). Since this is an estimate, we'll presume all gases are ideal and (foolish us) that C_P doesn't change with T.

T_expl = - E°_expl / vC_V(products only), but that's not what we have from the tables. So we have to convert H to E and C_P to C_V. Fortunately, everything being an ideal gas, C_V = C_P - R. (Where R=8.314 J/mol K.)

It's a little trickier with E, but since H°=E°+P°V, H°=E°+(P°V). And as everything is an ideal gas, P°V=nRT°, H°=E°+RT°n. Aha! n=0. (5 moles in; 5 moles out.) So E°=H°!

Life is good. Especially at °.

E° = H° = (-393.5) + 2(0) + 2(-241.82) - (-74.81) - 4(90.25) = - 1163.3 kJ/mol as written.

vC_V(products only) =
(37.1-8.314) + 2(29.12-8.314) + 2(33.58-8.314) = 120.9 J/mol K = 0.1209 kJ/mol K. Although we have C_P for reactants, they aren't present after the stoichiometric combustion, so they don't soak up E°.

Therefore, T_expl = 1163.3 kJ/mol / 0.1209 kJ/mol K = 9620 K.

T_expl = T+T₀ = 9620 + 298 = 9918 K

By our modified Charles' Law,
P_final=P_initial(T_final/T_initial), and P_expl = 1 atm (9918 K/298 K) = 33.3 atm.
So the overpressure is 32.3 atm. STANJAN would find it to be much less.

On the right is Shaw's Fig. 5.13. It is a BET plot for the adsorption of N₂ on porous silica gel at 77 K.

Given that N₂ molecules occupy a surface area of 16.2×10^-20 m² each at this temperature, estimate the surface area per gram of this sample.

Note that the ordinate values have been multiplied by 1,000 to give the labelled values (thus the "10³") and the volumes have already been scaled on a per gram basis. Remember that "stp" means 1 atm and 273 K not 298 K.

Given the size of the graph, let's just pretend that it goes through the origin, i.e., the intercept is zero. We'll incur only a trivial error thereby.

So if the intercept is close enough to zero, the slope is any y on the line divided by its x. Your choice actually. Mine is (0.16,0.001) or slope = 0.00625 g sample per cc N₂ at STP. Or V_mono=1/0.00625=160 cc N₂ @ STP per gram sample.

Now all we need to know is how much area 160 cc of N₂ occupies when it adsorbs. That we can get from the 16.2×10^-20 m² per molecule by determining the number of molecules in 160 cc of gas at STP. Assume an ideal gas, then n=PV/RT or
n = 1 atm × 0.160 L / (0.08206 L atm/mol K × 273 K) = 0.0071 moles
N = n N_Av = 0.0071 × 6.022×10²³ = 4.3×10²¹ molecules.
A_mono = 16.2×10^-20 m²/molecule × 4.3×10²¹ molecules = 700 m² / g sample

All microstates are equally likely. Knowing that, determine the ratio of B to A versions of a system at a fixed energy from the following information: Version A consists of 3 oscillators and 4 quanta, but due to structural differences, B consists of 4 oscillators and only 3 quanta. (Clearly B's quanta are 4/3 the size of A's, but that's of no consequence here. All that matters here is that for A, N=3 and Q=4, but for B, N=4 and Q=3.)

Since all microstates are equally likely, the chances of seeing the system as B vs. A is just the ratio of the microstates available to those two, viz., W_B/W_A.

So we must either remember that for the easy case of the harmonic oscillator,
W = (Q+N-1)!/[Q!(N-1)!],
or we have to draw out the energy levels and populate the legitimate configurations and add up their weights.

I choose the formula. :)

Since Q+N-1 is the same for both A and B, the ratio becomes even simpler:

[B]/[A] = [Q_A!(N_A-1)!] / [Q_B!(N_B-1)!] = [4!2!]/[3!3!] = 1.33

And indeed any [(n+1)!(n-1)!] is > [n!n!] means that oscillators are worth more than quanta in creating microstates. Interesting.

Problem #3 determined a fixed energy equilibrium constant for a microscopic system. But we're more familiar with a fixed temperature K_P for a macroscopic (~N_Av) system. Let's estimate that to one significant figure from statistical mechanics for the reaction:

⁷⁹Br₂(g) + ¹²C⁸¹Br₄(g) ⁷⁹Br⁸¹Br(g) + ⁷⁹Br¹²C⁸¹Br₃(g)

We can't really assume a temperature such that vibrations are inactive (v_Br2 = 323 cm^-1 !), but we do know that they'll all be almost the same for either substituted version of each molecule. And T needn't be above carbon tetrabromide's boiling point since CBr₄ has quite a reasonable vapor pressure. So let's say T=298 K. (Why am I so cavalier about that?)

Even the big term e^D₀/RT, will be uninteresting since D₀ depends on those same small differences in the zero-point energies of much too similar bromine isotopes.

So what's left?

Electronic? Not on a bet! Different isotopes have the same electronic structure.

Rotational symmetries? BINGO. While there'll be no T dependence in K_ROT, that doesn't mean K_ROT itself vanishes to 1. It is true that the moments of inertia change by some really trivial amount on substitution of 79 for 81 and vice versa, but we're looking for a first significant figure effect. And that's symmetry factors, .

K_ROT = [_{⁷⁹Br2 C⁸¹Br4}] / [_{⁷⁹Br⁸¹Br ⁷⁹BrC⁸¹Br3}] = [ 2 × 12 ] / [ 1 × 3 ] = 8

where we've recognized that z_ROT is inversely proportional to and that ⁷⁹BrC⁸¹Br₃ has only the one C₃ axis for rotation and the tetrahedral C⁸¹Br₄ can be oriented by grabbing any one of the 4 Br's then can be rotated about that Br-C axis to any of 3 orientations, 4×3=12. Or we could've gone to the symmetry tables for C_3v and T_d, respectively.

So our 1 significant figure guess for K = 8.

Below 216.58 K, its triple point, CO₂ has no liquid phase and so must sublime to create its vapor pressure. Estimate H_subl at 205 K from the data below.

T (K)	194.7	205	216.58
PCO2 (atm)	1.000	2.242	5.114

ln(P₂/P₁) = - (H_subl/R) [(1/T₂) - (1/T₁)]

H_subl = R ln(P₂/P₁) / [(1/T₁) - (1/T₂)]

Now our only question is what to choose for "1" and "2" from among the 3 data points. The value at 205 could be estimated by letting T₁=194.7 K and T₂=216.58 K. The resulting H_subl would be a best estimate for the middle of the range or <T> = ½(194.7+216.58) ~ 206 K, close enough. But we wouldn't know how H_subl might have changed over the interval; so we'd have no built-in error estimate.

Better would be to calculate H_subl between the 1st two T's and again between the 2nd two T's and take THAT average. After all, the formula above varies non-linearly in T (and P) but to the extent that C_P's don't change very much, we'd expect H_subl to vary linearly with T. So

H_A = 8.314 ln(2.242/1.000) / [(1/194.7) - (1/205)] = 26,011 J/mol = 26.01 kJ/mol

H_B = 8.314 ln(5.114/2.224) / [(1/205) - (1/216.58)] = 26,543 J/mol = 26.54 kJ/mol

H_subl(205 K) = ½ (26.01+26.54) = 26.28 kJ/mol +/- 0.27 kJ/mol

Just for jollies:

H_C = 8.314 ln(5.114/1.000) / [(1/194.7) - (1/216.58)] = 26,149 J/mol = 26.15 kJ/mol, not as good as the answer above.

10 pt. BONUS:

CV is often approximated as a+bT+cT2 where a,b,c are experimentally fitted parameters. Use those parameters to find A,B,C where Z~DeAlnT+BT+CT2 and D is just an undetermined constant independent of T.

C_V = 2kT [d lnZ / dT]_V + kT² [d²lnZ / dT²]_V]

ln Z = ln D + ln e^AlnT+BT+CT² = ln D + AlnT + BT + CT²

[d lnZ / dT]_V = A/T + B + 2CT

[d²lnZ / dT²] = -A/T² + 2C

Substitute:

C_V = k [ 2A + 2BT + 4CT² ] + k [ -A + 2CT² ]

C_V = k [ A + 2BT + 6CT² ] = a + bT + cT²

so

A = a/k, B = b/2k, and C = c/6k.