[madscientist]

Im reposting this because im not so sure of my answer any help would be much appreciated. ( i have deleted the original post)

hi everyone,

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

SO_2(g) + 1/2 O_2(g) <=====> SO_3(l)

my working is:

K = a _SO3  / (a _SO2 )(a _O2 ) ^1/2

Where "a" is the activity of the species in the reaction, is this right? ive only used "a" because one of the species is in its condensed phase and didnt think you could use concentration terms in the equilibrium expression


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

so:

                            -RT lnK = dG_rxn

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

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

                                  lnK = 28.612690166

                                    K = e^28.612690166

                                       = 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   :albert:

[madscientist]

OMG! now it makes sense.. thanks for the help people.. :ranting:

[Moonshyne]

Not sure if you were kidding or serious about understanding it... but it will reply anyway!

I am pretty sure that you are not supposed to put species with states of matter other than gas and aqueous into an equilibruim expression. So the first expression should read
K = 1/(O₂)^1/2(SO₂)
I am unclear as to why the "a" is necessary, and even relevant in that matter...
You calculated K correctly...
I don't like the units bar, but I'm sure that 1 bar equals 0.986923 atmospheres. (irrelevant information)
The reason that there is so much SO₃ compared to reactants is because the system is trying to convert the gas in the system to equilibrium, and since SO₃ keeps getting condensed to liquid form, it is taking away from products. Using Le Chatlier's Principle, when products are removed, the system reacts in a way to minimize the change, so more products are formed. The vapor pressure of SO_3(g) is very low above the liquid SO₃, so the reaction continues to drive toward products. This also makes sense because your K value is rediculously large, so the system makes more products.

-hope this helps

[madscientist]

Thanks for the reply m8 that confirms it for me,

regarding the use of the activity "a", i think they just wanted us to know why species in states other than gas, liquid solid...(a=1), arnt included...? but who knows how these creatures known as "professers" think?  :alien1:lol

I seriously wasnt serious in my last post, toungue was firmly planted in my cheek (ausy slang) , that smily cracks me up though so i had to add it.

cheers

madscientist :albert:

[Hunt]

I can't seem to understand why activities are used here. I thought that the activity is the effective concentration of an ion, it is due to the electrostatic interaction between charged particles, correct? When working with gases, should we take into account the effective concetration? I really doubt so ...

In the problem above, I think we can deduce that the rxn shifts forward because Q ? 1 << K , but why is it always that we take the activity or conc of solids / liquids to be 1 ?