SPC Ramuel Mendoza Raagas

Teaching Fellow: David Perlmutter

Chemistry E-1a
Problem Set 4
October 10, 2003
Looks like we'll be doing a lot of PV = nRT for this thing.
Pressure x Volume = number of moles * Real Gas Constant times Temperature.
Pressure x Volume = number of moles * Real Gas Constant times Temperature.
[Pressure] = atmospheres (necessarily so--- torricelli must be converted into atmospheres!)
[Volume] = Liters
[Temperature] = units Kelvin
1. Is oxygen oxidized in this reaction as the three moles of reactant O with oxidation number typical 2- charge in the chlorate ion becomes zero oxidation number as product?

1. What is the vapor pressure of water at 25 degrees Celsius (298.15 Kelvin, that is)? The Logan Notes tell us that it's 23.76 torricelli. How much is that in atmoshpheres? 0.031263 = 0.0313 atmospheres! Let's subtract this value from the total pressure. 0.98947 atmospheres is the total pressure (0.989 atm), which is 0.0313 atmospheres H2O(g) plus 0.95281 atmospheres = .963 atm of other gas stuff. Oxygen is the only gas of the featured reaction.

PO2 = .953 atmospheres.

 The partial pressure of O2(g) is 0.953 atmospheres.
2. We do a PV = nRT thing. Algebraically, n = PV/RT. Then we'll multiply the molar mass of Oxygen by n, that is the number of moles. The molecular mass of Oxygen gas is 32 grams per mole. n = (.95281 atnmospherres) * .423 Liters / .08206 L.atm/mople/Kelvin times 298.15 Kelvin.

0.0946 moles of Oxygen gas was what I got around to calculating.

Mathematica tells me that I've got 1.5316 times two grams of oxygen atoms. This would give me 1.5316 grams of Oxygen gas, or 3.06 grams of O2(g)
 A mass of 3.06 grams of O2(g) has been collected.

3. I've got one gram of KClO3, considering the stoichiometric relation 3/2 O2 per one mole KClO3. That's 2.04 grams KClO3
 2.04 grams KClO3 have decomposed.
4. PV = nRT again. P = nRT / Volume. Pressure inside bulb = 0.0946 moles times .08206 L-atm/mol-Kelvin times 318.15 Kelvin divided by 0.2 Liter. 12.35 atmospheres would be the pressure. Nine thousand three hundred and eighty five torricelli that would be
1. What is n? It would be the same for both of our diatomic gases hydrogen and oxygen. n = PV/RT encore. n = 2 atm * 1 Liter / [(.08206 L-atm/mol/Kelv) * (298.15 Kelvin)]
.0817(455) = .0817 moles of gas is what we've got all in all.
2. 0.0409 moles is what we've got of hydrogen gas. Ditto, 0.0409 moles of oxygen gas. Obviously, hydrogen would be our limiting reactant. Only .0204 moles of Oxygen would get consumed, whereas all 0.0408 moles of hydrogen gas would get consumed. 0.0204 moles of gaseous water would get porduced. 0.368 grams of water would be produced. That is, 367 milligrams of water would be produced.
3. Let's express the concentration of water in vapour phase as a mole fraction. 23.76 torricelli divided by 1520 toricelli equals what? n 1520 torricelli pressure tehres is to begin with? But what about after the reaction (condensation -heh-heh reaction in a different sense? The water-making reaction decreases entropy. One point five moles of gas becomes only mole of gas. Thus the pressure after reactions is not two atmospheres, but only 1.5 atmoshphseres. What would that be in torricelli? 1140 torricelli, I'd say. What's 23.76 torricelli divided by 1140? 0.0208 mole fraction what ewe end u with . Like 2.08 per cdnet of the waters is n vapour 7.65(44) milligrans of waater vapour there would be whwenentemperatueree cools bcaakc down aafterr thtree spsark.; three hundred sixssxixty grmilligrams of liquid water there would be.
4. Do we have to do a Maxwell distribution? Anyway, the easy plug-in turn-key thing is that vrms = square root of the quotient of triple the Kelvin-format temperature times the Boltzman's constant divded by the gas' mass. We get our mass from our answer for 2. b). Boltzman's constant k = 1.38 x 10-23 Joules per unit Kelvin.

Whereas the molar gas constant may be expressed in two forms, one of which is 0.08206 Liter-atmospheres/ mol-Kelvin--- k, Boltzman's gas constat = 1.38 x 10-23 Joules per unit Kelvin. Note that the k Boltzmann's constant unlike the molar ideal gas R constant does not bother carrying Liters or atmospheres expressly as dimensional factors. Bolztman's k has a dimensional configuration of [Work / Temperature]. R has Liter-atmospheres in its numerator, but we may appreciate the fact that Liter-atmospheres is itself one way of dimensionaly8 exprersssing a quantity of work done.

2. Should we like get the equilibrium constant for this thing? Keq? Kc? Kp? Kc = Kp(RT)Dn?
five hundred degres Celsius is like 773.15 Kelvin. 840 degreecs Celksisuss is like 1113.15 Kelvin. The diatkmoic gas becomes a monatomic gas! ANywssaazya in this problem we'll end up with a mixtur e fo F2 molecules and F atoms which need no covalent interactions amongts eac othe r to keep a gaseodus sstsatase/. Additional;a pressur e fo 0.384 atmoshphers coems with partial fddisssociasiasteion of Flurororrirne gas. .984 atmospheres makes for n = RT/PV = (.08206 * 1114.15 Kelvin) /
3. Considering that Xenon fluoride has got a vapor pressure of 4.5 torricelli (which is much lower than that for water)... 0.005921 atmospheres is Xenon fluoride's vapour pressure.

.0410 moles
 Xe(g) + F2(g) XeF2(s) .410 moles

PV = nrt
1.1 atmospheres * ten Liters = n RT
11.0 Liter-atmospheres / (.08206 * 298.15 Kelv) = number of moles
.450 moles gas is what we started out with.
.410 moles Xenon gas is what we started out with.
.041 moles Fluorine gas is what we started out with.

What's the molar mass of Xenon difluoride? 169.3 grams mole?
.027128 moles of XeF2(s) is what we end up with (4.60 grams formed divided by molecular mass of 169.287 grams per mole = .02722 moles). Stoichiometrically this means that we've used up ..02722 moles moles of Xenon gas and .02722 molesof F2 gas.
We end up with three gases in the ten liter glass bulb. There's .383 moles Xenon gas left. 0.014 moles (or twenty-four millimoles) Fluorine gas left.
0.005921 atmospheres is Xenon fluoride's vapour pressure. We can use this value to get a mole fraction.
What's the overall final total Pressure? It's .397 moles of gas, at least.
Pf = Ptot = .407
The total final pressure is less than the 1.1 atm we started out with. IOt turs out being .971 atmospheres plus the .00592
.977 atmospheres is what the final total pressure is
http://members.aol.com/profchm/dalton.html http://www.cm.utexas.edu/academic/courses/Summer2002/CH301/Labrake/chpt12-notes.pdf http://www.stfx.ca/people/gmarango/chem100/Course-Notes/Gases.doc http://ist-socrates.berkeley.edu/~phy7b/spitzer/1-3.pdf http://www.webelements.com/webelements/compounds/text/Xe/F2Xe1-13709369.html