 My
mother-in-law, a former Ohio farm girl, loans me an iron horseshoe
weighing 2916.3 g (big draft animal!) which I give
to the smithy in the picture. He heats it to 870°C
(red heat) in his forge and drops it into a
bucket of room temperature (25°C) water. It sizzles and steams the
water a bit then finally comes to equilibrium at 59.5°C. There was
originally 10 kg of cool water, but we lost 17 g to steam as the shoe
entered the bucket; so only 9.983 kg of water came to that equilibrium.
What's the apparent heat capacity (J/mol°C) of iron?
[ CP(H2O) = 4.18 J/g°C and
DHvap = 40.657 kJ/mol ]
|
| There are two kinds of water to consider: the 9.983 kg
that arrived at 59.5°C and the 17 g that arrived at 100°C then
evaporated. Thus sums of the heats involved in both represent the
change in enthalpy of the iron, DH.
DH = 9,983 g × 4.18 J/g°C ×
(59.5°C - 25°C)
That enthalpy was supplied as heat by 2,913.6 g of iron dropping from 870°C to 59.5°C; so iron's specific heat is CP = 1.483×106 J / [ 2916.3 g × (870-59.5)°C ] = 0.628 J/g°C But we wanted not the specific heat but the molar heat capacity, hence CP = 0.628 J/g°C × 55.85 g/mol = 35 J/mol°C |
Imagine a winter day with an outside temperature of 5°C and 50% humidity; that governs the actual water vapor pressure in our homes (unless we take hot showers or boil water for a cup of tea). But we keep our homes at 20°C, where the vapor pressure would have to be higher to maintain that comfortable 50% humidity. So what is the indoor relative humidity on such a day? DHvap = 44 kJ/mol (different from problem #1 since that previous vaporization occurred at 100°C as the water near the surface boiled).
| The water vapor pressure outside and inside is
PH2O = 0.5 × PH2Osat'd
Psat'd(5°C) = 1 atm e-(DHvap/R)×[(1/278 K)-(1/373 K)]
So PH2O = 2.98 torr. But saturated water vapor at the temperature of the house would be:
Psat'd(20°C) = 1 atm e-(DHvap/R)×[(1/293 K)-(1/373 K)]
So the % humidity is (2.98/15.8)×100% = 19%; that's pretty dry. |
OK,
resolve this paradox. The triple point is said to be unique because all
3 phases of a pure compound are in equilibrium at that pressure and
temperature. What's the big deal? Everywhere along the solid-liquid
equilibrium line in the compound's phase diagram there exists solid,
liquid, and their equivalent vapor pressures! And that
equilibrium extends over an enormous pressure range and a modest
temperature range as well. What then makes the triple point unique?
(Ignore "D," the supercooled water.)
| But the vapor pressure above the ice/water equilibrium isn't the same as the applied pressure except at the triple point. You know that because when boiling, the pressure of the steam over the water/steam equilibrium is 1 atm. But we just calculated the vapor pressure of water at 5°C to be only 6 torr, and it would be smaller still at 0°C! So the triple point has the vapor too at the external pressure. |
| A liter of 0.60 M salt weighs 1020 g and contains
0.60×(22.99+35.45)g = 35.06 g salt hence 985 g water. Hence 1000 g
of water would be holding (1000/985)×0.60 = 0.61 moles of salt,
and the molality of the solution is 0.61 m. However, the salt
dissociates (presumably) completely, giving 1.22 m of ions; that's the
vant Hoff factor of 2 entering here.
So DTf = - 2(1.86 °C kg/mol)×0.61 m
= - 2.3°C
|
C2H4O can be acetaldehyde (ethanal),
H O
| //
H--C--C
| \
H H
or oxirane (ethylene oxide),
O
/ \
H--C---C--H
/ \
H H
which forms a 3-membered ring. Both isomers have the same molecular
formula (hence the same molecular weight), but oxirane boils at 11°C
while acetaldehyde boils at 20°C. Why?
| There are no -O-H bonds here; so the classical
hydrogen bond seems not to be in effect. Or does it?
Oxygen exerts a powerful electron-withdrawing effect on the species to which it's bonded. So in both compounds, the hydrogens on a carbon to which the oxygen is bonded lose some of their electron density and become partially positive. That makes them attractive to the oxygens from neighboring molecules. In acetaldehyde (ethanal), only one hydrogen gets ripped off this way, and so it becomes much more positive than would the four hydrogens in oxirane, all supplying ¼ of the extra electron density the single oxygen demands. So ethanal's aldehyde hydrogen is more attractive to a neighbor's oxygen, and the intermolecular force is expected to be stronger. But wait! Although it may have the stronger attraction, the hydrogens in oxirane are four times as numerous! Why don't those numbers win? Look at the geometry of those molecules. Ethanal is perfectly constructed to double "hydrogen bond" to a neighbor like:
H O ... H H
| // \ |
H--C--C C--C--H
| \ // |
H H ... O H
making a pair harder to separate. Oxirane can't do that with all four of its much weaker hydrogens. So ethanal (acetaldehyde) has the higher boiling point. |
| Henry's constant measures the equilibrium between
the vapor and its dissolved component. If KH is large, it
means that a high vapor pressure is required to dissolve the species.
If KH is low, the gas dissolves readily from even low
pressures!
So looking at the Periodic Table, we'd expect those molecules to react very like their buddies above them. What do we know about the solubilities then of CH4, NH3, H2S, and HF? Clearly, making such an argument, SiH4 should have the poorest solubility, hence the highest KH. HCl can be made up in extremely concentrated solution; so it should have the highest solubility and hence the lowest KH. PH3 should behave like NH3 and dissolve well to make the weak base, PH4OH. And H2S dissolves to make the far weaker acid, H2S(aq). So the extremes are easy, HCl and SiH4. But will the water "like" PH3 or H2S best? Water is a polar molecule, and it will form better intermolecular attractions with the more polar of those two. The sulfur, being the more electronegative, wins, and the list of increasing KH (hence decreasing solubility) should be:
|