Posted by laura on .
A sample of PCl5 weighing 2.69 grams is placed in a 1.000 liter flask and vaporized completely at 250 o C.
The final pressure observed at this temperature is 1.000 atmospheres. Note that some, but not all of the PCl5 decomposes according to the equation: PCl5 (g) ---------> PCl3 (g) + Cl2 (g)
What are the pressures ( in atm) of PCl5 , PCl3 and Cl2 once the reaction is complete and at equilibrium ?
chem sorry the last one did not have the question -
I assume you don't need a Kp but if you have one listed, please let me know. I believe this is the way to solve the problem although it seems the long way around to me.
moles PCl5 initially = 2.69/molar mass PCl5 = about 0.0129 but you need to confirm ALL of these numbers.
Total moles AT EQUILIBRIUM from PV = nRT. Plug in P, V, R, and T, and I get something like 0.0233 for total moles AT EQUILIBRIUM (not starting.
...........PCl5 ==> PCl3 + Cl2
total moles at equilibrium = 0.0129 - x + x + x = 0.0233
x = 0.0104 and 0.0129-x = 0.0025.
Now solve PV = nRT three times, once for PPCl5, once for PPCl3 and once for PCl2 using the value for T, n (x), R, and total P of 1 atm above. The values you get will be the partial pressure of each of the gases at equilibrium and at the conditions listed. Check my thinking.