Keq equilibration

[Rucap09]

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When the starting conditions of a reaction are out of balance with Keq, Le Chet explains how products and reactants will adjust to reflect the Keq value. How would this work in a scenario where you have A+B <-> C+D when one product/reactant is too high and the other is too low?

For ex, if Keq=4 and you have 12 moles of gas, ideally at equilibrium you'd have [C]=[D]=4 and [A]=2 giving (4x4)/(2x2)=16/4=4.

What if instead you start with [C]=8 and [D]=2, would they both move towards an equal value? Same thing for an uneven distribution of reactants?

Hope this is clear...?

 

[justadream]

There is something called "Q" which is calculated the exact same way as Keq except that it is done under conditions when you are necessarily at equilibrium.

Obviously, if Q = Keq, then you are at equilibrium.

If Q<Keq, that means you have too much reactants (remember reactants are in the denominator) so the reaction will proceed in the forward direction to reach Keq.

If Q>Keq, that means you have to much products (products are in the numerator) so the reaction will proceed in the backwards direction to reach Keq.

 

[Rucap09]

I know how the reaction behaves in regards to Qrx being greater/less than Keq. My question is what happens when two reactants are unbalanced in relation to each other like in the example I gave. Would [C] and [D] equilibrate to become equal if their coefficients in the balanced reaction are equal.

 

[techfan]

You can only react as many Ds as you have Cs (limiting reagent), you can't shift over from A+B to make only C, you'd make both C AND D. So you'd just have excess D that wouldn't react and it would react the same as if you only had 2 D and 2 C.

Here's a pretty good tutorial if you want a step-by-step http://chemcollective.org/activities/tutorials/equil/quant_limit

 

[Rucap09]

Haha no I understand that. Sorry should have clarified, this is under the assumption that you've created the system this way...as in you put this mix of gasses into a closed container.

Obviously 8 D's won't be able to react completely with 2 C's but if you increase single reactants or products in an A+B <-> C+D mechanism you will have a restorative effect from Le Chatelier's.

Are you saying that for a reaction with Keq=4, if you started with a system having [8][2]/[2][2] or [8][2]/[1][4] that they'd behave identically as a system with [4][4]/[2][2] and no rebalancing of the products would occur due to the forward and reverse rates being equal? That doesn't seem correct to me...

 

[techfan]

As I understand it, the reaction would be running 4x in the forward direction for every 1x in the reverse direction. So by giving 8 C and 2D you still are running the reaction forward 4 times and reverse 1x. Keq is simply a ratio of forward/reverse. Which you seem to understand.

Le Chatlier deals with reactions that start in equilibrium. By defining Keq as 4 and giving the molar concentrations that you did the system is in equilibrium, now if you do something to disturb it (add A, or remove B) AFTER the system is in equilibrium, it'll shift, but it won't shift to "make the numbers pretty". That's why you add or subtract x from each side in an ICE table, to see (calculate) how your shifting of the equilibrium will affect the system.

In the C=8 D=2 A=1 B=2 reaction as soon as C combines with D to make an A there is already a B waiting to gobble it up and vice versa as soon as A combines with B to make a D there are 4 Cs waiting to gobble it up (8/2=4). So it won't go back to the nice 4,4 2,2 ratio

 

[Rucap09]

See that's where I'm confused.

You're saying that a reaction with Keq=4 will interpret [8][2] in the denominator as equilibrium and will remain unchanged, but previously you said the excess product would effectively be ignored and the reaction would run identically to if it was [2][2]. If the latter was the case, then Qrx would be 4/4 and the reaction would run in the forward direction to establish equilibrium. Can you clarify how those two sentences aren't contradictory?

For example, lets instead take it out of equilibrium, so that now you have [9][2]/[1][4]. If you're Qrx is now 4.5 and your Keq is 4, which reactant/products change to establish equilibrium?

 

[techfan]

Sorry, when I made my first post I didn't understand the concept behind Keq as well as I should've. I should not have said limiting reagent since the reaction is in equilibrium with C=8 and D=2 A=1 B=2. My second post was written better.

As for having Qrx=4.5 and Keq=4; Keq<Qrx and the reaction shifts left to the reactant side(TBR had a cool little trick where if you put K and Q in alphabetical order and draw the correct inequality sign with a line in the middle the inequality sign points to the direction the reaction will shift).

I think the problem that you're having is trying to fit common sense to theoretical concepts. By using A, B, C, D and using Keq=4 it's hard to appreciate what that means because we (or at least I) tend to think of reactions in terms of individual atoms instead of molecules that can be composed of multiple atoms. There aren't any stoichiometrical constants so you're seeing A combining with B at a rate that is 4x faster than C combining with D and then seeing that if you have 8x more of C than you do of B it somehow makes the same amount of A and B. Yes, that is true in theory, but by not using real molecules it is hard to appreciate that mass is still being conserved and that this example is more easily imagined if C was a smaller component of A and/or B.