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It would actually have taken just a tad longer due to the fact that human lungs branch down to aveoli sacs < 1 mm in diameter, but the simpler calculation puts the meaning of "explosive decompression" on a human scale.
n = pV/RT = 1 atm × 4.8 L / [(0.08206 atm L /mol K) × (273+37)] = 0.19 mol
N = n NAv = 0.19 × 6.022×1023 = 1.14×1023 molecules A = pR2 = ¼pD2 = ¼p(2.5×10-2 m)2 = 4.9×10-4 m2 dN/dt = Z×A Z = p / (2pmkT)½ Z = 1.01×105 Pa / (2p × [29×10-3 / NAv] kg × 1.38×10-23 J/K × 310 K)½ Z = 2.81×1027 molecules m-2 s-1 dN/dt = 2.81×1027 m-2 s-1 × 4.9×10-4 m2 = 1.38×1024 molecules/s
T ~ N / (dN/dt) = 1.14×1023 / 1.38×1024 = 0.083 s
= 83 ms
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is conveniently analyzed by collecting the gas evolved during the reaction. During such an experiment, 50.0 mg of nitramide was allowed to decompose at 15°C. The volume of (dry) gas evolved after 70.0 min. was measured to be 6.59 cm3 at 1 bar pressure. Find the rate constant and the half-life for the nitramide decomposition.
MW = 2(14.01)+2(16.00)+2(1.008) = 62.04 g/mol
n0 = m/MW = 5.00×10-2 g / 62.04 g/mol = 8.06×10-4 mol ng = pV/RT ng = 1 bar (1 atm / 1.01 bar) 6.59×10-3 L / [(0.08206 atm L/mol K) × 288 K ] ng = 2.76×10-4 moles gas therefore nt = (8.06-2.76)×10-4 = 5.30×10-4 mol solid
nt = n0 e-kt or
k = - (1/t) ln(nt / n0) = [ ln(n0 / nt) ] / t
T½ = 0.693/k = 6930 s = 115.5 min |
Derive d(ln k)/dT for the last expression, and use estimates of its values at the midpoints between the temperatures in the table below to estimate n for this (hydrolysis of ATP) reaction.
It'll be simplest to multiply both sides of your expression by T 2. Then if we call that derivative D, the expression you derive should read T 2 D = nT + E/R where n and E have been presumed constants. Solve that for n given the two midpoint estimates. Remember that T is in K.
T ( °C ) | 39.9 | 43.8 | 47.1 |
k ( s-1 ) | 4.67×10-6 | 7.22×10-6 | 10.0×10-6 |
ln k = ln(c) + n ln(T) - E/RT
d(ln k)/dT = 0 + n/T + E/RT2 T2 d(ln k)/dT = nT + E/R
Two equations and two unknowns: Ti2 Di = nTi + E/R Taking the difference at the two midpoint temperatures, the E/R term cancels out, leaving
n = [ T22D2 - T12D1 ]
/ [ T2 - T1 ]
Although we've manipulated it correctly, we've just applied a 2nd order formula to a first order process! The hydrolysis of ATP is first order. We should have applied it instead to a second order process since both the collision and activated complex theory k's are k2. |
Derive the rate equation for the loss of N2O5 by assuming that the intermediate NO3 is in steady state.
- d[N2O5]/dt = - d[NO]/dt = k2 [NO] [NO3]
0 ~ d[NO3]/dt = k1[N2O5] -
k-1[NO2][NO3]ss -
k2[NO][NO3]ss
-d[N2O5]/dt ~ k1k2[NO][N2O5] / ( k2[NO] + k-1[NO2] ) |
In other words, why does the "reactant-like" A-BC as a complex geometry give higher AB vibration than does the "product-like" AB-C as a complex geometry (even if their reaction exothermicities were the same)?
[HINT: use the shape of the potential energy surface for the A+BC AB+C reaction.]
At the activated complex, the exothermicity has not yet been felt. So a "reactant-like" complex,
A-BC, feels the enormous forces of the exothermic falloff as B is jumping toward A; that's
AB vibration. If the complex were a "product-like" one, AB-C, those same forces would be felt as
C jumps away from AB; that's translation of the separating products instead.
In terms of the "bent gutter" view of the ABC potential surface, A-BC has not yet "turned the bend," so the exothermic falloff pulls the system forcibly in the -RAB direction. But AB-C has already "turned the bend," and those falloff forces push on +RBC instead. So high vibration in highly exothermic reactions is consistent with Hammond's Postulate that such reactions are associated with "early" activated complexes (or "reactant-like"). |
The cross-section for N2 differs with the process considered for it. For example, its hard-sphere collision cross-section, s = 0.43 nm2, whereas its BET monolayer footprint cross-section, s = 0.16 nm2. What does that suggest to you about the orientation of N2 molecules in a surface monolayer?
Aside from the fact that we're mixing apples and oranges since the hard sphere
s deals with long-range van der Waals forces which deflect gas molecules
even during near misses, and BET's s deals with liquid packing dimensions,
the
(sBET /
sHS )½ value of 0.61 appears to represent
the ratio of diameters vaguely consistent with the notion that N2 "thickness"
(across the N atom) is perhaps
2/3 of its "length"
(along the internuclear direction).
That would suggest that N2 may be standing up on the surface rather than
"lying down" with both atoms at the same average distance from the surface. Such a posture would
maximize the number of N2 molecules (and N atoms) in contact with the surface.
(There are molecules know not to stand up; H2 is catalytically decomposed on many metal surfaces, resulting in twice as many H atoms skittering around.) |