December 25, 2024, 02:11:34 AM
Forum Rules: Read This Before Posting


Topic: Keq and Equilibrium  (Read 1150 times)

0 Members and 1 Guest are viewing this topic.

Offline GengiMane

  • New Member
  • **
  • Posts: 5
  • Mole Snacks: +0/-0
Keq and Equilibrium
« on: November 07, 2019, 03:56:31 PM »
I was giving a practice question from my chemistry teacher which has stumped me. I have the reaction CO+Br2=CoBr2. Where [COBr2] = 0.12M, [CO] = 0.060M, [Br2] = 0.080M. The equilibrium constant is 25, but how can I verify the equilibrium constant is the same when the volume is doubled and 0.010M of COBr2 turns in CO and Br2?

I've tried subtracting and adding 0.01M of COBr to CO and Br. But I still cannot get the equilibrium constant back 25.
Any help is greatly appreciated thanks.

Offline chenbeier

  • Sr. Member
  • *****
  • Posts: 1337
  • Mole Snacks: +102/-22
  • Gender: Male
Re: Keq and Equilibrium
« Reply #1 on: November 07, 2019, 04:22:53 PM »
Double of Volume means half of the Mole before equlibrium
 COBr2 = 0,06 M, CO = 0,03 M and Br2 = 0,04 M.
New equlibrium change 0.01 M means COBr2 = 0,05 M CO = 0,04 M and Br2 = 0,05 M

K = 0,05 M/(0,04. M * 0.05 M)  = 25 q.e.d.

Offline GengiMane

  • New Member
  • **
  • Posts: 5
  • Mole Snacks: +0/-0
Re: Keq and Equilibrium
« Reply #2 on: November 08, 2019, 03:17:03 PM »
Oml thank you so much I see where I went wrong. I was adding just 0.005M to CO and 0.005M Br2.

Sponsored Links