[ksr985]

I thought of something while studying eqbm, havent managed to locate the answer yet. please see if you can help:

If i have a reversible reaction of the type:

A(s) gives B(g)

The eqbm constt. Kp for this reaction will be equal to the partial pressure of B at eqbm.

Now if i take 100 atm of A, and at eqbm, i have 5 atm of B, then Kp is 5.

Suppose i take only 2 atm of A. then the maximum amount of B that can ever be present is will have pressure 2 atms. So, will the eqbm never be reached? and if it is, what is the correct value of Kp for the rxn? If eqbm is not reached, but A gets exhausted, how will i know that this is not the eqbm point in the lab, considering the amount of B will be constant thereafter?

[Borek]

To be at equilibrium all substances involved must be present. If there is only one, there is no equilibrium.

[ksr985]

My point exactly. BUT, all reversible reactions continue till eqbm is reached. How will I reach eqbm? How can the terminate before eqbm. In general all components must be present, because the Keq equation invoves their concs/ pressures, but what about this example?

[Borek]

If there is not enough reagents reaction ends before the equilibrium state is reached.

[ksr985]

BUT, the reaction cant end until eqbm is reached. there is no end to a reversible reaction. there is simply eqbm., the minimum in the free energy curve.the eqbm is dynamic, with the backward and forward reaction happening simultaneously. do you see why this is so weird?

[GCT]

I thought of something while studying eqbm, havent managed to locate the answer yet. please see if you can help:

If i have a reversible reaction of the type:

A(s) gives B(g)

The eqbm constt. Kp for this reaction will be equal to the partial pressure of B at eqbm.

Now if i take 100 moles of A, and at eqbm, i have 5 moles of B, then Kp is 5.

Suppose i take only 2 moles of A. then the maximum number of moles of B that can ever be present is 2. So, will the eqbm never be reached? and if it is, what is the correct value of Kp for the rxn? If eqbm is not reached, but A gets exhausted, how will i know that this is not the eqbm point in the lab, considering the amount of B will be constant thereafter?

Kp is in pressure, atm, you made the proposal with moles which is incorrect.  Your query should now be resolved.

[ksr985]

oops, stupid mistake, change all the moles to atmospheres of partial pressure. query remains.

[GCT]

query remains

really?  With any adequate container (not too large), you'll have a partial pressure due to decomposition, doesn't this seem practical?

[GCT]

oops, stupid mistake, change all the moles to atmospheres of partial pressure. query remains.

Alright, I think I see where your confusion is...with adequate amounts of A, there will the pressure will always be, say...5atm and this does not depend on the original amount of A.  If you don't have enough A, all of the A will simply decompose and you'll have less than 5 atm at equilibrium.

[ksr985]

ahh.. exactly. i dont even know if all the A will decompose, that is also uncertain. all i know is that eqbm cannot be acieved. who's to say what position the reaction will occupy at any given time.

[GCT]

ahh.. exactly. i dont even know if all the A will decompose, that is also uncertain. all i know is that eqbm cannot be acieved. who's to say what position the reaction will occupy at any given time.

Well, it'll depend on the temperature and the size of the container.  You can calculate whether all of A will decompose simply by using PV=nRT.  If some solid were to remain, then at equilibrium you would know that the pressure is 5 atm.

[Borek]

BUT, the reaction cant end until eqbm is reached. there is no end to a reversible reaction. there is simply eqbm., the minimum in the free energy curve.the eqbm is dynamic, with the backward and forward reaction happening simultaneously. do you see why this is so weird?

The answer is: reaction CAN end before the equilibrium is reached. Is there a solid in the equilibrium with the non-saturated solution? No. In the solution cotaining EDTA but no Mg(2+) is there an equilibrium with MgEDTA complex? No.

Other example, without a chemical reaction, but the idea remains the same: liquid water is in equilibrium with gaseus water. Get a drop of water and put into small container - there is liquid, there is gas, there is equilibrium. Make the container very large. All water is in the form of gas. Is there an equilibrium between liquid and gas? No. If the given equilibrium can't be reached due to lack of reagents system will be in other equilibrium, period.

[ksr985]

yup. but the question remains unanswered, since keeping the volume of the container constant for the two experiments,ie, with 100 atm of A, and with 2 atm of A, we cannot reach eqbm in the latter, which should not happen. all reversible reactions must attain eqbm.

[ksr985]

very, very good! i think i got it, finally. thanks a lot, man.

[GCT]

yup. but the question remains unanswered, since keeping the volume of the container constant for the two experiments,ie, with 100 atm of A, and with 2 atm of A, we cannot reach eqbm in the latter, which should not happen. all reversible reactions must attain eqbm.

Yes, equilibrium will not be reached in the latter...because it does not apply in this case.  It's simple logic, you'll need to have adequate amounts of A in order for equilibrium to be relevant.  I don't exactly understand why you're saying "it should not happen."

You need to think outside the box, it similar to Ksp, where Ksp pertains to saturation of the ionic compound.  If You provide too little of the ionic compound, all of it will dissolve in a large volume of water.  

In effect, equilibrium is not like maxwell's equations, it is a practical mathematical tool, not a sacred principle.