[imanooblar]

So given a standard combustion reaction:
CH4 + 2O2 --> CO2 + 2H2O

Hypothetically speaking, if I remove water from the products, then the reaction will shift towards the products to create more CO2 and H2O to relive that stress.

My question is, if you remove water, and the system shifts to the right to produce more CO2 and H2O, then doesn't the equilibrium change?

For example, in the original equation, there is 1 mole of carbon dioxide to every 2 moles of water. Let's say I remove 2 moles of water, but keep the carbon dioxide there. Now, as the reaction proceeds to replenish the water removed, it will do so while producing 1 mole of CO2 for every 2 moles of water. So now, I effectively have 2 moles of water (which was "replenished" due to the shift") and 2 moles of CO2 (1 new one create, and one left over from before). Is that even possible? If it is, then doesn't the equilibrium expression change now, because instead of 1 mole of CO2 to 2 moles of water, I have 2 moles of CO2 and 2 moles of water.

[MOTOBALL]

My recollection of Le Chatelier's principle is that " if a constraint is placed on a system in equilibrium, the system reacts to oppose the constraint".

If CO2 + 2H2O  CH4 + 2O2

then you have an equilibrium, and removing H2O from your original eqn. will take the eqlbm. to the right to produce more CO2 and H2O,
 
but does that reverse reaction actually happen ??

[imanooblar]

sorry, I don't really understand what you mean by if the reverse reaction will happen or not?

If it's in equilibrium, then both the forward and reverse reaction will be occurring in order to maintain equilibrium.

I guess to rephrase my question..

Will the equilibrium expression change, once I remove a product and the reaction shifts to the products side to replenish it.

So before I had [CO2][H2O]²/[CH4][O2]²

After I remove some of the product (water in this case) then the reaction will shift to give me a new equilibrium expression, if that makes any sense.

This is what I'm thinking it will look like, even though I know it's wrong, just want to know why it's wrong and what happens to the original product so the equilibrium expression doesn't change.

[H2O]²[CO2²/[CH4][O2]²

[Borek]

That's the beauty of the equilbrium. Write expression for K. This expression is called reaction quotient, Q, and it doesn't have to equal K before equilibrium is reached. However, if Q doesn't equal K, mixture is not at equilibrium and it will react, shifting towards products (or reactants), according to LeChatelier's principle, till Q=K. If you remove some of the substances - situation repeats itself, and whatever is left reacts till Q=K. Repeat ad nauseam

[imanooblar]

Hey Borek,

I understand that if Q<K then reaction will shift toward products, and if Q>K then the reaction will shift toward the reactants. I guess my question is how can Q eventually equal K if we are removing one thing from the reaction.

Like the example I gave, lets say we have one mole of each. I take away one more of CO2 from the products. Now to reestablish equilibrium, the reaction will shift creating more products, in this case CO2 and H2O in a 1:2 ratio. Now I have 1 mole of CO2 and effectively 4 moles of H2O because 2 was left over from the original. Now the reaction will shift back to the reactants, using up CO2 and H2O in a 1:2 ratio, so I'm back to having 0 moles of CO2 and 2 moles of H2O. So, from what it seems like, equilibrium can never be reestablished, it can only keep trying to reestablish equilibrium but never reach it cause it's in a never ending cycle, or a new equilibrium will be reached. Hope you understand my confusion.

Here is my confusion in reaction form:

Original:
CH4 + 2O2 --> CO2 + 2H2O

Take out 1 mole of CO2, so reaction will shift to the products side. Then you get this equation:
CH4 + 2O2 --> CO2 + 2H2O + 2H2O (Red color is what is added to replenish lost CO2)

Now we have too much H2O so reaction will shift back to reactants. Then you get this equation:
CH4 + 2O2 <--  2H2O

So now we're back to where we started when we removed the CO2, and thus the never ending cycle continues.

[Borek]

This is not a 1 mole this way or other way, you are allowed mole fractions... Calculation of the final concentrations is sometimes difficult, but given initial concentrations of all substances, reaction stoichiometry and K value always possible.


Let's assume we started with equilibrium mixture containing four substnces - A, B, C, and D. let's assume they react

A + 2B <-> C + 2D

(this is your example, I just don't want to use original substances, as values we will use are not correct).

Further let's assume - like you stated - that initially we had 1 mole of each substance, and mixture was at equilibrium. Finally, to make things simple, let's assume it all happens in 1L, and we are dealing with concentrations, not partial pressures of gases.

Initially all concentrations equalled 1, so

K = [A][ B]² / [C][D]² = 1

Now, we remove all C. According to LeChatelier's principle A and B reacts, producing C and D. How much reacted? Let's say x A reacted. If so, 2x of B reacted and x C was produced together with 2x of D (that's all from the reaction stoichiometry). So after equilbrium has been reached we have 1-x of A, 1-2x B, x C and 1+2x D. And we know that

K = [A][ B]² / [C][D]² = [1-x][1-2x]² / [ x][1+2x]² = 1

Solve for x - and you know new concentration of C and you can calculate new concentrations of everything...

[imanooblar]

Alright thanks Borek!!

I only used 1 mole this way or the other way just to simplify things, but I do understand that mole fractions can be used. Mathematically wise, it makes complete sense and thanks for clearing up that confusion. Completely forgot about solving for it in that sense.

Just having a hard time visualizing it, not that it's really important but just bothers me. Would it be correct to visualize it as as long as Keq remains the same then equilibrium is satisfied, not necessarily that the reactants and products have to go back to the same ratios.

For example, assuming Keq is equal to 1, as long as [C][D]^2 = [A][ B]^2 then the equilibrium is satisfied. So therefore, even though you remove C and the reaction shifts towards the products to produce more C and D, because the D value is squared so in the end more D will be produced to make up for the initial removal of C. Does that make sense?

Let's say before we had 3 of C and 2 of D. Putting into the equation will give us a [C][D]^2 = 12, now if we remove C, in order to make up for it, we could have 1.3 of C and 3 of D to make 12 also. Of course I would solve for X like you demonstrated, but just to get the concept, is this sort of how it works?

[Borek]

Honestly - I have no idea what you mean 

[imanooblar]

Haha alright let me try to explain this differently.

Basically, from your example, as long as Keq = 1 then equilibrium is satisfied.

In our case, A + 2b -> C + 2D, we had the equation K = [A][ B]2 / [C][D]2 = 1.

Now my only question is that before I thought that C and D had to always be in a 1:2 ratio because that's how the equation is written out. But now I'm thinking that as long as [C][D]2 = [A][ B]2 then it is okay.

So say in your example, you had 2 mole of A, 4 moles of B, 2 mole of C, and 4 moles of D. From this, you can see that [C][D]2 = [2][2x4]2 = 32. Now once I remove C, and the reaction shifts to the right, as long as [C][D]2 = 32 collectively then it's okay. So it doesn't necessarily have to be a C:D ratio 1:2. It could be say... 1.28:5 which still gives [C][D]2 = 32. Does that make more sense? Hard to explain by typing it out 

[Borek]

Basically, from your example, as long as Keq = 1 then equilibrium is satisfied.

By definition.

Now my only question is that before I thought that C and D had to always be in a 1:2 ratio because that's how the equation is written out.

They can be in any ratio.

But now I'm thinking that as long as [C][D]2 = [A][ B]2 then it is okay.

In this particular case - yes.

So say in your example, you had 2 mole of A, 4 moles of B, 2 mole of C, and 4 moles of D. From this, you can see that [C][D]2 = [2][2x4]2 = 32. Now once I remove C, and the reaction shifts to the right, as long as [C][D]2 = 32 collectively then it's okay. So it doesn't necessarily have to be a C:D ratio 1:2. It could be say... 1.28:5 which still gives [C][D]2 = 32. Does that make more sense? Hard to explain by typing it out 

No, it doesn't make sense

[C][D]² doesn't HAVE to equal 32. When reaction proceeds not only C & D are being produced, concentrations of A and B change as well, so both sides of the equation chnage - till Q=K.

[imanooblar]

Alright thanks Borek. I think I got it now 

[MOTOBALL]

Now I'm confused !!!

You wrote      CH4 + 2O2    CO2 + 2H2O

which is NOT presented as an eqilbm reaction.

If it is an eqlbm rxn, then the reverse

                    CO2 + 2H2O  CH4 + 2O2

will occur.  I questioned whether your combustion IS an eqlbm rxn.  Is it ?

[Borek]

Now I'm confused !!!

You wrote      CH4 + 2O2   CO2 + 2H2O

which is NOT presented as an eqilbm reaction.

If it is an eqlbm rxn, then the reverse

                    CO2 + 2H2O   CH4 + 2O2

will occur.  I questioned whether your combustion IS an eqlbm rxn.  Is it ?

Now you know why I have selected to use A + 2B <-> C + 2D as a better example.

In short: in all reactions there is some equilibrium involved, but there are cases where taking it into account is a moot.

In the case of combustion that is because there are many parallel/alternate reactions taking place, there are many stages that lead to the final products, and while each step is technically reversible, chances of getting back to the same reactants are dim. There is also an obvious thermodynamical component - products are much more stable, so the equilibrium at each stage lies far to the right, which makes reverse reactions theromodynamically unlikely, but not at all impossible. If you think about it, most of the chemical synthesis methods leading to products we use in everydays life, are thermodynamically (or kinetically) unlikely, but we force them by clever choice of catalysts and conditions.

[imanooblar]

Hey sorry Motoball for using a combustion reaction when trying to explain something at equilibrium. Should have picked arbitrary letters like Borek.

From what I understand, it can be reach equilibrium (thermodynamics), but how long it will take to get there is a different story (kinetics).  It needs a high level of heat to overcome it's activation energy , but it is thermodynamically favorable because it is an exothermic reaction.

Anyways, hope Borek's explanation was clear enough.