CHM 5414 Thermodynamics
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Exam #2 Solutions
14 November 1996
- Figure 25 (page 77 in Chem-Thermo) says the rise of an explosion's temperature
is the curved line. Let us analyze this for cases with ideal gases and
no diluents.
- Describe the shape of that curve if the products are gases with constant
heat capacities but the reactants are solids of negligible heat capacity.
Since we don't have STANJAN with us, ignore the radicals!
With negligible heat capacities, the reactants won't be absorbing any of the heat
of reaction. So all the energy produced goes to heat the product molecules.
But given x% of reaction, we produce x% of the final products immersed in x% of
the energy of reaction. So it scales perfectly!
Thus the temperature at all points in the reaction (except at the beginning when
there are no products) is the same as the final temperature! In other words,
the curve should lie along the line marked Tf .
For the stoichiometric explosion of ethane in pure oxygen,
the thermodynamic properties (CRC) of both reactants and products are given below.
Find the temperature at 0%, 50%, and 100% reaction completion and comment on
Nash's figure.
| Molecules | C2H6(g)
| O2(g) | CO2(g)
| H2O(g)
|
 H°f , kJ/mol | -83.8
| 0.0 | -393.5 | -241.8
|
| CP, J/mol°K | 52.6
| 29.4 | 37.1 | 33.6
|
0% is dead simple. The reaction hasn't started; so we're at the initial temperature
which we'll take to be the standard T=298°K.
100% is a little more involved. We must evaluate
Tf = 298°K - E° / [
products CV ]
and we realize to our horror that the table data is CP and H°
instead! No problem; we know CV = CP - R for ideal gases
of which the problem assured us. And
E = H -(PV)
E = H - RTn
So all we need is a proper stoichiometric reaction and we can convert to energies.
And we need to subtract R=8.315 J/mol°K from each CP in the table
before using it as a CV.
C2H6(g) + 7/2 O2(g) 
2 CO2(g) + 3 H2O(g)
which makes
n = 2 + 3 - (1 + 3.5) = + 0.5
H° = 2(-393.5) + 3(-241.8) - (-83.8) = -1428.6 kJ/mol which makes
E° = - 1428.6 - (0.5)(8.314x10-3 kJ/mol°K)(298°K) = - 1429.8 kJ/mol
Since all of the reactants have been consumed at 100% completion, we need only
consider the CV of products, scaled by their stoichiometric coefficients:
Tf = 298°K - (-1429.8x103 J/mol) / [ 2 (37.1 - 8.314) + 3 (33.6 - 8.314) ] J/mol°K
Tf = 298 + 10716 = 11,014°K (but probably much less)
But the tricky part is 50% completion. Clearly E is only
half as great (715 kJ/mol) but we've only half the moles of products as well; so
the CV of products is only half as large. However, we have reactants lying
around soaking up that (half) energy . . . but only half as many moles as we started
with! All this means that the denominator heat capacity sum will be
(1/2) (52.6 - 8.314) + (7/4) (29.4 - 8.314) + (37.1 - 8.314 ) + (3/2) (33.6 - 8.314) = 125.8
or
T50 = 298°K - (-715x103 J/mol) / [ 125.8 J/mol°K ]
T50 = 298 + 5686 = 5,984°K (also an overestimate)
If Nash's T vs % reaction really has the negative curvature shown, this 50% completion
temperature we've found should lie above the average of Ti and
Tf, which is (298+11014)/2 = 5,656°K < T50 as predicted.
Into an adibatic container is introduced a mole of ice at 0°C and a mole of
steam at 100°C. (a) How much of what remains at what final temperature? (b) What
is the overall S?
Hfus(ice)=6.008 kJ/mol
CP(water)=75.291 J/mol
Hvap(steam)=40.656 kJ/mol
- Clearly, some ice melts and some steam condenses, but the great disparity of those
phase change energies means that it all doesn't come to a nice average 50°C! So
we should proceed in stages to see what happens.
Hvap >> Hfus
so all of the ice will melt by condensing only a fraction of the steam.
Qmelt = + 6.008 kJ requiring that much energy from the condensation of
X(steam)x(40.656 kJ/mol) = - 6.008 kJ or X(steam) = - 0.148 moles lost to melt the ice.
But if all that (one mole) ice melt absorbed enough energy to come to the boiling point, it
would only require
Qheat = 75.291x10-3 kJ/°K (100°K) = 7.53 kJ more,
condensing only
Y(steam)x(40.656 kJ/mol) = - 7.53 kJ or Y(steam) = - 0.185 moles lost to (nearly)
boiling the ice melt.
Now everything's at 100°C; no more changes can take place, and we have
Steam = 1.000 - 0.148 - 0.185 = 0.667 moles
Water = 1.000 + 0.148 + 0.185 = 1.333 moles
all at 100°C. This might help explain why steam burns are so much more
serious that hot water burns. Hvap
is nasty.
- Ahhhh...a plug-in.
Sice = Hfus/Tfus + CP(water) ln(Tvap/Tfus)
Sice = (6.008x103 J/mol)/(273°K) + (75.291 J/mol°K)ln(373°K/273°K)
Sice = 45.5 J/mol = 45.5 J (since one mole of ice was involved)
Ssteam = - (Hvap/Tvap) n(moles condensed)
Ssteam = - (40.656x103 J/mol)/(373°K)x(0.148 + 0.185)
Ssteam = - 36.3 J/°K
Stotal = Sice + Ssteam = + 45.5 - 36.3 = + 9.2 J/°K
spontaneous since the adiabatic walls ensure that Ssurroundings=0.
A water boiler is rated for 5 atm. At what T(°C) will it explode?
Ahhhh...another plug-in.
Clausius-Clapeyron tells us that pressure will raise the boiling point of a liquid.
ln(P2/P1) = - (Hvap/R) [ (1/T2) - (1/T1) ]
or
T2 = 1 / [ (1/T1) - (R/Hvap) ln (P2/P1) ]
T2 = 1 / [ (1/373°K) - (8.314 J/mol°K / 40656 J/mol) ln (5 atm/1 atm) ]
T2 = 425°K = 152°C
Napoleon's cheap white tin buttons spontaneously changed to gray tin in a
Russian winter and fell off, freezing soldiers. For the white to gray tin
transition at 25°C, H°=-2.03 kJ/mol
and S°=-7.1 J/mol°K.
For pure Sn, at what T(°C) would this slow process start?
G° = H° - TS° = - 2030 J/mol - 298°K(-7.1 J/mol°K)
G° = + 86 J/mol
(Well, we're not actually entitled to both significant figures. We only know H to +/-10 J/mol;
so G° = + 90 J/mol is more like it.)
Nevertheless, since S<0,
we ought to decrease the temperature to render G negative
(spontaneous). But by how much?
Hmmmmm.
dG = VdP - SdT
but dP=0 even in Russia. So dG = - SdT, or more to the point,
dG = - S dT
or, since we want dG = -90 J/mol,
dT = - dG/S = + 90 J/mol / (-7.1 J/mol°K) = - 13°K
So the transition temperature is T=25-13=12°C.
That's not a Russian winter, but the buttons weren't pure Sn either!
Napoleon's tailors weren't stupid; they amalgamated the tin to avoid what was
then known as "tin pest" (the crumbly gray form). But nobody thought to test
the amalgam against as cold a winter as one finds in Russia! So the French
uniforms fell open: for want of a button an empire is lost. C'est la guerre.
Last modified 24 November 1996.