July 11, 2020, 11:18:06 AM
Forum Rules: Read This Before Posting


Topic: ANSWER THIS PLEASE  (Read 7049 times)

0 Members and 1 Guest are viewing this topic.

NISHANT

  • Guest
ANSWER THIS PLEASE
« on: January 03, 2005, 11:21:42 AM »
Q:1 mole of N2 and 3 moles of PCl5 are placed in a 100 litre vessel and was heated at 500K.The eq. pressure is 2.05atm.Assuming ideal behaviour,calculate degree of dissocaition of PCl5 and kp of the reaction that occurs.

here it is very clear that N2 will not take part in the reaction.so if we calculate the initial preesure and the partial pressure of N2 and PCl5,partial preesure of N2 or I could say pressure due to the N2 molecules would not change in the course of the reaction.Partial pressure of PCl5 would change and we can easily find the kp of the reaction.
But in the sloution of the book this is not given and even if we calculate all the things by the method which I have supposed Partial pressure of PCl5 AT EQ. COMES OUT TO BE NEGATIVE??

CAN YOU PLEASE SPOT THE MISTAKE

Offline Mitch

  • General Chemist
  • Administrator
  • Sr. Member
  • *
  • Posts: 5294
  • Mole Snacks: +376/-2
  • Gender: Male
  • "I bring you peace." -Mr. Burns
    • Chemistry Blog
Re:ANSWER THIS PLEASE
« Reply #1 on: January 03, 2005, 03:11:23 PM »
please read forum rules. It's the first post of all forums.
Most Common Suggestions I Make on the Forums.
1. Start by writing a balanced chemical equation.
2. Don't confuse thermodynamic stability with chemical reactivity.
3. Forum Supports LaTex

Offline Donaldson Tan

  • Editor, New Asia Republic
  • Retired Staff
  • Sr. Member
  • *
  • Posts: 3178
  • Mole Snacks: +261/-12
  • Gender: Male
    • New Asia Republic
Re:ANSWER THIS PLEASE
« Reply #2 on: January 03, 2005, 08:22:34 PM »
PCl5 <-> PCl3 + Cl2
let x be amount of PCl5 dissociated
amount of PCl5 at equilibrium = 3-x
amount of PCl3 at equilibrium = x
amount of Cl2 at equilibirum = x

V = 100L = 0.1m3
T = 500K

P = nRT/V

partial P of N2 = (1)R(500)/(0.1) = 5000R
partial P of PCl5 = (3-x)R(500)/(0.1) = 5000R(3-x)
partial P of PCl3 = (x)R(500)/(0.1) = 5000R(x)
partial P of Cl2 = (x)R(500)/(0.1) = 5000R(x)

given eqbm P is 2.04atm = 207665Pa
(make sure all your quantities are in SI units)

5000R + 5000R(3-x) + 5000R(x) + 5000R(x) = 207665
5000R(1+3-x+x+x) = 207665
5000R(4+x) = 207665
4+x = 207665/5000R = 4.9955
x = 0.9955 ~ 1 mole
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

NISHANT

  • Guest
Re:ANSWER THIS PLEASE
« Reply #3 on: January 04, 2005, 10:03:24 AM »
but if we calculate the pressure due to N2 in the beginning,it comes out to be different,but N2 has not taken part in the reaction then why so?

Offline AWK

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 7391
  • Mole Snacks: +516/-86
  • Gender: Male
Re:ANSWER THIS PLEASE
« Reply #4 on: January 05, 2005, 06:30:05 AM »
From this data, assuming ideal gas behaviour and reaction:
PCl5 = PCl3 + Cl2
we can calculate the total moles of N2 + PCl5 + PCl3 + CL2 in the mixture (4.9963)
Hence, at the equilibrium we have  0.9963 moles of Cl2,  0.9963 moles of PCl3 and 2.0037 moles of PCl5
 N2 do not take part in reaction but modifies partial pressures of PCl5, PCl3 and Cl2
AWK

Sponsored Links