May 06, 2024, 11:26:56 AM
Forum Rules: Read This Before Posting


Topic: equillibrium constants  (Read 5446 times)

0 Members and 1 Guest are viewing this topic.

Offline Kaleyrvt

  • Regular Member
  • ***
  • Posts: 57
  • Mole Snacks: +1/-0
equillibrium constants
« on: January 15, 2007, 11:40:36 PM »
Consider the equilibrium reaction:
   A (g)
2B (g)  +  C (g)

When 1.00 mol of A is placed in a 4.00 L container at temperature t, the concentration of C at equilibrium is 0.050 mol/L. What is the equilibrium constant for the reaction at temperature t ?

again, having troubles with this equillibrium stuff-If someone could please just point me in the proper direction that would be great. I know the equilibrium constant is:
Ke=[C}c[D]d / [A]ab           but how do I input all of this data?

Sorry guys, I am trying :( Any hints would be greatly appreciated :)

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27675
  • Mole Snacks: +1801/-410
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: equillibrium constants
« Reply #1 on: January 16, 2007, 03:53:57 AM »
First of all - rewrite Ke so that it contains substances and coefficients present in your reaction equation.

Secodd: use reaction stoichiometry to calculate amounts of substances in the equilibrium. How much A was destroyed if there is 0.050 C produced? How much B was produced at the same time?
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline Yggdrasil

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 3215
  • Mole Snacks: +485/-21
  • Gender: Male
  • Physical Biochemist
Re: equillibrium constants
« Reply #2 on: January 16, 2007, 03:57:14 AM »
Also note that since you are working with gases, you will need to use partial pressures instead of concentrations in calculating Ke.

Offline xiankai

  • Chemist
  • Full Member
  • *
  • Posts: 785
  • Mole Snacks: +77/-37
  • Gender: Male
Re: equillibrium constants
« Reply #3 on: January 16, 2007, 07:52:26 AM »
the letters in brackets stand for the partial pressure/concentration of that substance.

the letters outside the brackets are the number of moles.

it should have looked more like [A]a, where a is the power of [A].

also, the products belong to the numerator (at the top), while the reactants belong to the denominator (at the bottom).
one learns best by teaching

Offline Kaleyrvt

  • Regular Member
  • ***
  • Posts: 57
  • Mole Snacks: +1/-0
Re: equillibrium constants
« Reply #4 on: January 16, 2007, 11:11:42 AM »
so then,  Ke= 2+[C]
                              [A]


does [A]=0.25 mol/L   and [C]=0.050 mol/l ???   is that right?? how do I find B???

Offline Kaleyrvt

  • Regular Member
  • ***
  • Posts: 57
  • Mole Snacks: +1/-0
Re: equillibrium constants
« Reply #5 on: January 16, 2007, 07:41:45 PM »
Anybody have any tips??
 :'(

Offline Yggdrasil

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 3215
  • Mole Snacks: +485/-21
  • Gender: Male
  • Physical Biochemist
Re: equillibrium constants
« Reply #6 on: January 16, 2007, 08:14:35 PM »
From the stoichiometry of the reaction, you know that for every molecule of C formed, two molecules of B are formed.

Offline Kaleyrvt

  • Regular Member
  • ***
  • Posts: 57
  • Mole Snacks: +1/-0
Re: equillibrium constants
« Reply #7 on: January 16, 2007, 09:06:22 PM »
A= 0.25 mol/L
B=0.050 mol/L
C=2 x 0.050 =0.1 mol/L

Ke =    [C]   =     [0.1][0.050]   = 0.02   
                   [A]               [0.25]      

Thus, the equilibrium constant is 0.02 at temperature t.

Sponsored Links