Wednesday
March 29, 2017

Post a New Question

Posted by 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
    initial...0.0129......0.....0
    change......-x........x.....x
    equil....0.0129-x.....x.....x

    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.

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question