[sameeralord]

Hello everyone,

It is funny that I thought I understood dynamic equilibrium really well but now I have gone blank again.  I can do all the calculations and I understand
Le chatelier principle well. Here are the problems I have.

2A+B  <----> C

* If forward reaction equals backward reaction wouldn't those two balance out and be zero. Is the reason why this doesn't happen is because both reactions are continuosly occuring. But please elaborate if you can.

*I checked so many sites on this topic and none of them really talks much about what happens to the moles in dynamic equilibrium. Can anyone tell me what happens to the moles in dynamic equilibrium. I understand that if the reaction occurs according to stoichiometric ratios there would be one way reaction. So roughly when we react 2 moles of A and 1 mole of B do we get like 0.8 moles of C.I'm just asking does something like this happen.

*If in equilibrium the reactants and products don't act according to their stoichiometric ratios why do we assume so in calculations. Do they always provide us with ICL data in equilibrium questions.

If anyone can take their time and at least answer one of these questions I would be very greatful. Thank you

[Astrokel]

hey sam,

*If in equilibrium the reactants and products don't act according to their stoichiometric ratios why do we assume so in calculations. Do they always provide us with ICL data in equilibrium questions.

ICE table? In equilibrium, the stoichiometric ratio might be difference. However, the change is always the same according to their mole ratio. Usually, in equilibrium question, initial [reactants] will be given and the % of product at equilibrium, so you could work out the unknown change(x value).

*I checked so many sites on this topic and none of them really talks much about what happens to the moles in dynamic equilibrium. Can anyone tell me what happens to the moles in dynamic equilibrium. I understand that if the reaction occurs according to stoichiometric ratios there would be one way reaction. So roughly when we react 2 moles of A and 1 mole of B do we get like 0.8 moles of C.I'm just asking does something like this happen.

No, it doesn't happen that way. In a reversible reaction, they still react according to their ratio but there is one major difference.

Let's say you have [reactant A] = 0.1 and [reactant B] = 0.2

A+ 2B ----> 2C

If this is a one-way reaction, ALL A and B will all be used up and react according to the ratio.

However, if it is

A + 2B <---> 2C

If it is a reversible reaction, A and B will not be used up totally but still react according to their ratio. Might be 0.05 in A and 0.1 of B. That is why at equilibrium the mole ratio is not the same in reversible case.

* If forward reaction equals backward reaction wouldn't those two balance out and be zero. Is the reason why this doesn't happen is because both reactions are continuosly occuring. But please elaborate if you can.

What do you mean by cancel out? Become a one way reaction?

[sameeralord]

What do you mean by cancel out? Become a one way reaction?

Thanks once again for helping me out Astrokel . What I mean by this is if the forward rate equals the back rate. For every mole of B produced another mole would be consumed. Doesn't this mean nothing happens and everything cancel out.
A ----- B

[Borek]

It doesn't mean nothing happens, but it does mean that effects of forward and back reactions cancel out.

[Astrokel]

Consider A <---> B, for example at equilibrium, there is 60% of B. Since the rate for both forward and backward reaction is the same, at any instance there will stll be 60% of B and 40% of A, therefore it's called equilibrium, like borek had mentioned, ONLY the NET exchange between products and reactants is zero, but the amounts at equilibrium will not be zero.

[Yggdrasil]

As with many concepts in chemistry, it is useful to consider both a macroscopic and microscopic view of the system.  Consider a reaction A<-->B at equilibrium.  At the macroscopic level, you will see a beaker containing A and B which looks fairly static.  Over time, the concentrations of A and B do not change, so you think that nothing is going on.  However, now consider looking at the microscopic level.  If you could follow a single molecule of A, you will see it float around as A for a bit, then suddenly change to B, then change back to A after some time, and so on.  So, even though it appears that nothing is going on at the macroscopic level, there are still plenty of dynamics at microscopic level.

[sameeralord]

First of all I should thank everyone who helped me.  Astrokel , Borek  and Yggdrasil  . I think we can finally put this to bed now (Hopefully). Here is my final understanding in terms of an anology. Tell me if I'm right.
 
There are two buckets X and Y. There are two people. One pouring water from X to Y and other vise versa.

Intially  bucket X was filled with water (reactants). First the water was poured to bucket Y very fast (example 3 cups of water per second). Then as the water in the other bucket increases the reverse reaction takes place (Y to X). The reverse reaction is intially slow (Half a cup of water per second). Then it comes to a stage where both people scoop only one cup of water at a second and this is equilibrium. There is no apparent change in concentration. Am I right?

[Borek]

Yes, although from your example it is not clear why the speed of water transfer changes.

But if you will imagine they are using not buckets but huge flat tanks, and to move water they are using bakers, amount of water that you can take with a baker will be directly proportional to the water level. So, there is a reason to the speed at which water is transferred, thus equilbrium state is not somebodys decision, but it is just the effect of the way the system works.

[cliverlong]

Sameeralord

If I modify your idea slightly, although a bit more complicated, it may explain the behaviour of dynamic equilibrium better.

Most of the effort is in understanding the setup

I am going to model

A <-> B

Say we have two large cylinders, capacity 1000 litres each labelled A and B

Initially, A is full of water and B is empty.

The cylinders are connected by two sets of horizontal pipes running between the cylinders at heights 0cm, 10cm, 20cm etc to a height of 100cm

The pipes allow water to flow in only one direction, so A empties into B and B empties into A

The pipes pump liquid at the same rate of 1 litre / second between the cylinders

Initially, A is full and is pumping at 11 litres per second into B (11 pipes).

Initially, B is empty and there is nothing to pump into A.

However as A empties, the water level in A drops and as B fills the water level in B rises.

So ...

Some of the pipes from A to B are now above the level of water in A and there is nothing to pump

Some of the previously empty pipes from B to A are now below the level of water in B and can now start pumping 

So ...

the flow from A decreases and the flow from B increases << This is the key idea

 until a dynamic equilibrium is established where the flow between both is equal.

Note that the liquid is still moving between each cylinder (microscopic behaviour), and at a fair rate, but the overall level in each cylinder is constant at equilibrium  (macroscopic behaviour).


Also notice

If the rate of flow increases in the same proportion on both sides, the exchange will happen quicker yet the same equilibrium will be reached quicker - so an analogy to rate of reaction there

Conversely, flow decreased proportionally, equilibrium reached more slowly

Further

If the rates of flow are changed not in proportion, say liquid flows in each pipe from A to B at 1 litre / second but those from B to A at 1.5 litre / second then a new equilibrium position  is reached where a greater volume of A is maintained.

And

If we started to drain liquid out of B without returning it to A, then there would be less B to pump back into A and the level of A would drop to replenish the lost B until a new equilibrium was established.

All the processes and initial and final states above have analogy with a chemical reaction where A converts to B and reaches an equilibrium whose position and rate depends on various factors of the reaction in each direction.

The model can be made more sophisticated, and complicated, to model say

2R₁ + R₂  <-> 2P₁

We would take three large cylinders,  labelled R₁, R₂ and P₁ all interconnected so liquid flowed from R₁ and R₂ into P₁ and from P₁ into R₁ and R₂.

The initial state would be R₁ and R₂ are full of water and P₁ is empty.

Again, liquid flows initially fast from R₁ and R₂ into P₁.

However as R₁ and R₂ empty, the flow from P₁ back into R₁ and R₂  increases until a dynamic equilibrium is established

By adjusting the flow rates:

The equilibrium would be reached faster by faster flow rates
If the ratio between the flow rates changed, the position of equilibrium would change

The scenario could be where R₁ and P₁ pumped at 1 litre per second through each pipe but R₂ pumped at 2 litre / second. This would model a reaction where the stoichiometry showed a greater molar ratio of R₂ reacting with R₁

There are other scenarios that could be investigated where the amount of one of the reactants , say R₁ was in excess, and see how that affected the outcome and equilibrium position.

Clive

[sameeralord]

WOW that was quite a reply clive long .Thanks a lot for your lengthy response . Unquestionably snack .You covered lots of aspects chemical equilibrium. I understand the concept well now . lol I'm overflowing with it . I think this topic covers equilibrium better than many sites in the internet. Good job guys now I have a question on acid base equilibrium    Thanks again Clive  and everyone else who helped