The Chemportal

I've picked this problem from some general chemistry files that were retrieved from some tutoring sessions during my undergraduate years in chemistry (where I was the tutor).  Let me know if there's anything erroneous in the content.

When a mixture of xenon and flourine is sealed in a glass bulb and placed in the sunlight, beautiful crystals of XeF2 are slowly formed and deposited on the walls of the container. (XeF2 is a volatile solid with a vapor pressure of 4.5 torr at 25 degrees C).

In one such experiment, a 10.0 L glass bulb is filled with 1.00 atm Xe and 0.100 atm F2 at 25 degrees C. After sitting in the sunlight for a period of time, 4.60 g of SOLID XeF2 is formed. The bulb is then removed from the light and the reaction stops. Determine the partial pressure of each gas in the bulb at this point in the reaction.

Since XeF2 solid was formed, the container is saturated with XeF2 gas.  The quantity of Xe and F2 which together formed XeF2 (g) can be ascertained by considering how much of XeF2 (g) as well as XeF2 (s) was produced.

Due to saturation, the vapor pressure of XeF2 is 4.5 torr, to find the number of moles of XeF2 gas that exists, the vapor pressure value is incorporated into the ideal gas equation.

4.5 torr XeF2(1 atm/760 torr)=0.0059211 atm XeF2

n=PV/(RT)=0.0059211 atm XeF2(10.0 L)/(0.08206 Latm/molK(298 K))= 0.0024213 moles XeF2

The problem states the amount of XeF2 that was formed as a solid, this amount is converted to the mole unit. 

4.60 g XeF2(1 mole XeF2/(131 g + 2(19 g))=0.027219 moles XeF2 solid

The total number of moles of XeF2 that was formed as a gas and solid is thus,

.0024213 + 0.027219 = 0.029640 moles XeF2 has formed

This means that we need to subtract this amount from the initial mole values for Xe and F2.  The stoichiometric equation, Xe (g) + F2 (g) ----> XeF2 (g), indicates that the ratio between the reagents and products are unity; the number of Xe (g) and F2 (g) are equivalent with the number quantity of XeF2 (g) that is subsequently produced.

Moles of Xe (g) consumed = Moles of F2 (g) consumed = Moles of XeF2 (g) and XeF2 (s) 

PV=nRT, n=PV/(RT)=0.100 atm F2 (10.0 L)/(.08206 Latm/molK(298 K))=0.040893 moles of F2

The amount is 10 times greater for Xe since its initial vapor pressure is 10 times higher

1.00 atm/0.100 atm = 10, 0.40893 moles Xe

0.40893 initial moles Xe - 0.029640 Xe reacted = 0.37929 moles Xe remain
0.040893 initial moles F2 - 0.029640 F2 reacted = 0.011253 moles F2 remain

Incorporate these values into the ideal gas equation to calculate the pressure value.

P=nRT/V=0.37929 moles Xe(0.08206 Latm/molK)298 K/10.0 L=0.92751 atm Xe

P=nRT/V=0.011253 moles F2(0.08206 Latm/molK)298 K/10.0 L=0.027518 atm F2

0.0059211 atm is the vapor pressure for XeF2 as was stated already at the introduction.

Edited 02/27/2007 for simplification.