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One such observer watched a ball lighting for a couple of minutes until it happened to float over to a metal bucket filled with 2 gallons of water. Its internal fields could not support themselves in the dead electrical short of the metal, and the ball exploded with a terrible bang. The observer reported that the water in the bucket, which had been at room temperature (25°C), was now boiling!
What can we say about the minimum energy content (kJ) of that lightning ball? We can safely ignore the heat capacity of the bucket's metal, since it's swamped by water's 1.0 cal g-1. Water's density is 1.0 g cm-3.
From the Periodic Table handed out in class, 1 gallon = 3.7854 L. So the volume of
water raised to the boiling point was 7.57 L = 7.57×103 cc or (since
water is the definition of unit specific gravity) 7.57×103 g.
DE = C DT = 1.0 cal g-1 °C-1 (7.57×103 g) (75 °C) = 5.68×105 cal
From the Periodic Table, 1 cal = 4.184 J, so
DE = 5.68×105 cal (4.184 J/cal) = 2.38×103 kJ
Clearly the ball lightning would have had more energy than that. However much water boiled (unknown) would have consumed its heat of vaporization, and the metal bucket itself had a heat capacity we've not taken into account. Still 2 MJ in a basketball sized glowing sphere is impressive!
Watch your units! And it doesn't matter how much the bullet weighs, so pick a convenient weight. Ignore the immovable object's heat absorption; it happens so fast, the lead hasn't an opportunity to share the wealth. Now you know why most meteors (50,000 m/s) burn up even in the atmosphere.
Choose a 1 gram bullet; it doesn't matter because a 2 gram bullet would have twice the intial
kinetic energy at the same speed and require twice the energy to melt...so the problem
is intensive rather than extensive.
DE = C DT + DHfus = 0.0327 cal/°C (327-25)°C + 5.86 cal = 15.74 cal = 65.84 J
½ m v2 = ½ 10-3 kg v2 = 65.84 J
= 65.84 kg m2 s-2
DHcomb = - 1,112.3 kcal/g (4.184 kJ/kcal) (132.15 g/mol) = - 6.15×105 kJ/mol (!?!)
This is clearly way too high, but it's what you'd have to believe from the data given. So the data's wrong. It should have read "11,123 cal/g" instead. But we're stuck with the typo:
4 H2(g) + 2 O2(g) 4 H2O(liq) DH2 = 4 (-285.9) kJ/mol
9 CO2(g) + 4 H2O(liq) C6H5CHCHCHO + ½ (21) O2(g) DH = + 6.15×105 kJ/mol
If we add the last 3 equations (and their enthalpies) together, we get
DHf = 6,150,000 - 3,541 - 1,144 kJ/mol = 6.14×105 kJ/mol
Had we used the correct value for the combustion,
DHf = 1,470 - 3,541 - 1,144 kJ/mol = - 3,215 kJ/mol (makes more sense)
Energy of combustion because pressure wasn't fixed; volume was!|
DE = qV meaning heat transferred at fixed volume.
DE = DH - DPV
= DH - RTDngas|
DE = 615,000 kJ/mol -
8.314×10-3 kJ/mol K (298 K) (9 - ½)
Even with the corrected DHcomb = 1,470 kJ/mol, DE = 1,474 kJ/mol not different from DH within the significant figures.
EDTA + Fe3+ Fe(EDTA)3+
(sure, EDTA complexes as an anion, EDTA-4, but we'll ignore that here)
Clearly entropy favors the latter since only two free species get tied up in the complex rather than four! That means that the EDTA complex would have a larger stability (equilibrium) constant than the ethylene diamine one.
* means no water of hydration in the solid
Assuming the dissolution was just barely spontaneous, then DG
~ 0 and DS = DH/T or
DSsolution = 53,090 J/mol / 298 K = 178 J/mol K
Now DG might be much lower than 0, in which case, DS = DH/T - DG/T might be much larger than 178 J/mol K. So that value is a minimum.
|Molecule||Ethanol, C2H5OH||Acetic Acid, CH3CO2H||H2O||CO2|
nATP = 454/30.5 = 14 and change
At the melting point, DGfus = 0 =
So T = DHfus/DSfus = 3050 J/mol / 8.25 J/mol K = 370 K
|If they're perfect, S0 = 0.00 by definition. There's only one way, W, to build a perfect crystal, so R ln(W) = 0.|
Now there's 6 ways to find the molecule, so
S0 = R ln(6) = 14.90 J/mol K.
You could also arrive at that by the entropy of mixing formula:
where the sum is over 6 terms each one having Xi = 1/6.
because it vindicates what he said about temperature effects on equilibrium.
|Well last semester anyway, he said, "Increasing temperature favors products of endothermic reactions but reactants of exothermic ones."|
We want to see what sign we'd expect from the differential d ln(K) / dT.
Fortunately, we know that dT -1/dT =
- T -2 dT; so if
DH° and DS°
are sufficiently insensitive to T (over the range of dT of interest), then
we can treat them both as constants, and
and if dT increases, a positive DH° will correlate with a positive d ln(K), favoring products. Vice versa for a negative (exothermic) DH°.
[Of course, if ln(K) increases, then K must increase, favoring products. And if ln(K) decreases, K decreases, favoring reactants.]
|Comments or QUESTIONS to Chris Parr||Return to CHM 1316 HomePage||Last modified 6 April 2000.|