April 26, 2024, 11:21:35 PM
Forum Rules: Read This Before Posting


Topic: equilibrium constant K  (Read 5572 times)

0 Members and 1 Guest are viewing this topic.

Offline madscientist

  • Full Member
  • ****
  • Posts: 225
  • Mole Snacks: +14/-7
  • University of New England Australia
equilibrium constant K
« on: February 21, 2006, 11:04:42 PM »
hi everyone,

Just wondering if this is how you write the expression for the equilibrium constant for the following reaction:

SO2(g) + 1/2 O2(g) <=====> SO3(l)

my working is:

K = [a SO3 ] / [a SO2 ][a O2 ] 1/2

I also need to calculate the value of "K" from dGrxn = -70890.0 J/mol

so:

                            -RT lnK = dGrxn

    -(8.314 J/K.mol)(298K) lnK = -70890.0 J

                                  lnK = -70890.0 J / -(8.314 J/K.mol)(298K)

                                  lnK = 28.612690166

                                    K = e28.612690166

                                       = 2.6689*1012

another part to this question is:

If 1.00 bar of SO2 and 1.00 bar of O2 are enclosed in a system in the presence of some SO3 liquid, in which direction would the equilibrium move?

Now all I could think of is that they dont want a numerical answer and that because the value for the equilibrium constant K is extremely large, this indicates that there is alot more products than reactants at equilibrium (doesnt it), especially when if you look at that equation the product (SO3) is in its liquid state and its activity is approx = 1

therefore:  K = 1 /  [a SO2 ][a O2 ] 1/2

this says that at equilibrium, there is 2.6689*1012 times more SO3 than the multiplication of [a SO2 ][a O2 ] 1/2.

can anyone tell me if my reasoning is correct?

cheers,

madscientist  :drunk:
« Last Edit: February 21, 2006, 11:09:40 PM by madscientist »
The only stupid question is a question not asked.

Offline Mitch

  • General Chemist
  • Administrator
  • Sr. Member
  • *
  • Posts: 5298
  • Mole Snacks: +376/-3
  • Gender: Male
  • "I bring you peace." -Mr. Burns
    • Chemistry Blog
Re:equilibrium constant K
« Reply #1 on: February 22, 2006, 04:49:44 AM »
What is the "a"in your K constant equation. Isn't [] used for concentrations, how can you have a concentration for a gas?
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 Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27664
  • Mole Snacks: +1801/-410
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re:equilibrium constant K
« Reply #2 on: February 22, 2006, 05:57:16 AM »
how can you have a concentration for a gas?

Oxygen concentration in air is about 20%  :P
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Sponsored Links