Chemistry 3/4 2013 Thread

[psyxwar]

Hey guys, when does Le Chateliers principle apply? Help!!

Well, think of it this way. The equilibrium constant is well, constant at a given temperature. So when you're doing something that would change one of the values of the equation and thus change the equilibrium constant for that temperature, Le Chatelier's principle kicks in to ensure that the K is remains constant.

For example, look at heterogenous equilibrium. As solids and liquids do not contribute to our equilibrium constant, then changes in their amount are not subject to Le Chatelier's principle. Similarly, adding an inert gas in order to change the pressure of a mixture won't affect the equilibrium position either, because nothing's really changed -- the partial pressures of each of the products and reactants (assuming they're all gaseous) are still the same.

[thushan]

Hey guys, when does Le Chateliers principle apply? Help!!

Psyxwar has almost gotten it. But there's a couple of things that we need to iron out:

The equilibrium constant, K, is just that, a constant. It is constant at a given temperature, whether the system is in equilibrium or not. To define K roughly (say for A (g) + E (g) <--> C (g) + D (g))

K = [C]eq [D]eq / [A]eq [E]eq

These are EQUILIBRIUM concentrations. You can only use concentrations of each of the species to calculate K if the system is in equilibrium.

Now, there's another quantity called Q or CF (I prefer Q) - reaction quotient or concentration fraction. This is very similar to K, but it's defined as

Q = [C] [D] / [A] [E].

Note that you can use these concentrations to calculate Q whether the system is in equilibrium or not. If the system were in equilibrium, Q = K. When the system is not in equilibrium, K will remain the same (since its defined as a reaction quotient at equilibrium), but Q will not be equal to K. The system will change such that Q approaches K.

I say this because  "So when you're doing something that would change one of the values of the equation and thus change the equilibrium constant" is not entirely correct. It's Q that you are changing, not K. The system will then act as to oppose the initial change in Q, so that it will equal K again.

As for Le Chatelier's Principle, it is merely a shortcut, a trick, not an explanation. Like the left hand gun rule or right hand slap rule (if you do Physics). More on this:

http://www.chemguide.co.uk/physical/equilibria/lechatelier.html

[Homer]

I am having difficulties understanding the effect of temperature, volume and pressure on the equilibrium position :/
Carbon monoxide is used as a fuel in many industries. It reacts according to the equation:

2CO(g) + O2(g) (forth and back arrow) 2CO2(g)

In a study of this exothermic reaction, an equilibrium system is established in a closed vessel of constant volume at 1000°C.
a   Predict what will happen to the equilibrium position as a result of the following changes:
i   decrease in temperature

[brightsky]

since the reaction is a combustion reaction, it must be exothermic. for an exothermic reaction, a decrease in temperature will result in an increase in Ka value. so the equilibrium position will move to the right. you can predict this using le chatelier's principle. suppose the system were in equilibrium. suppose you now decreased the temperature. then the system would adjust so as to partially increase the temperature. how does it do this? well exothermic reactions release heat into the surroundings, so if you want to heat the surroundings up, thereby increasing the temperature, you have to bring about a nett forward reaction; the system will favour the exothermic reaction.

[scribble]

yep, you can use le chatlier to work out the effects of a change in temp/volume on the equilibrium position
So le chatlier says that "if an equilibrium system is subjected to a change, the system will adjust itself to partially oppose the effects of that change"
brightsky already explained temperature. Just you're using K not Ka, Ka is for acids!
for volume, if you decrease the volume of a gas, youre increasing the pressure right? so whats the system gonna do? it'll adjust to decrease the pressure again. it does that by favouring the side of the reaction with more gas particles, because we're assuming that the more gas particles in a given volume, the higher the pressure will be. so with 2CO(g) + O2(g) <=> 2CO(2), theres 3 gas particles on the left and 2 on the right, so a decrease in volume will increase the pressure, causing the equilibrium to shift forwards to decrease the pressure.

be careful though, changing the pressure by adding an unreactive gas to a system at equlibrium will do nothing! this has to do with partial pressures. think of your K. it's made of concentrations of gases. and its always constant so long as you don't change the temp. At equilibrium, Q=K. If you add an unreactive gas, the concentration of your gases don't change since concentration is n/V, and neither n or V is changing. this means that Q does not change, and is still equal to K and no changes in the equilibrium position will happen

[zvezda]

Hey,
so how does changing the pressure of an equilibrium mixture not change the K constant?
Im struggling to imagine it, because if youre changing the pressure by, say, decreasing the volume, there'll be a net forward reaction assuming there's less particles on the right side of the equation. Therefore, the ratio of concentration of product to concentration of reactants will be higher, thus changing the value of K.
Am I missing something?

[brightsky]

yes but just because a nett forward reaction occurs, doesn't mean the Kc value is changed. when you change the pressure of an equilibrium mixture, you make it so that the system momentarily goes out of equilibrium. so you are changing the Qc value. suppose the Qc value goes up, then Qc > Kc, and so the system will adjust so to regain equilibrium and a nett back reaction will occur.

the Kc value never changes. doing stuff to the system does not change the Kc value. you're just throwing the system out of equilibrium. but the Kc value still stays the same; the system still needs to fight to attain the same Kc value.

the exception to this is of course temperature. temperature, as we all know, is the only factor that changes the Kc value. nothing else does. the reason for this is beyond me. i can refer you to a couple of complicated formulas that may help elucidate why, but for the purposes of vce, just commit that to memory.

[scribble]

mmm, you're thinking the right thing, its just le chatlier doesnt always work; it happens to work with temperature and volume changes, but not pressure changes due to addition of a gas. and when one model doesn't work, you go to a more advanced model. which in this case is looking at K.
So K by definition is equal to [A][ B] / [C] for the reaction C(g) <=> A(g) + B(g) when its at equilibrium right? We also know that K at a specific temperature is a constant, which means it will NEVER change, unless you change the temperature.
So lets take the system at equilibrium. (that is Q = K) If you add a gas that won't react with the system, theres no change [A] [B ] or [C] since they're all equal to n/V and adding an unreactive gas changes neither the number of mol of A,B or C or the volume of the container. This means that Q is still equal to K. so theres no equilibrium shift.

We can use this idea to think about volume changes as well. at V1, at equilibrium K = (n(A)/V1 * n(B)/V1) / n(C)/V1. This simplifies to n(A)*n(B) / n(C)*V1. Now if you increase the volume of the system and make it V2. Q =  n(A)*n(B) / n(C)*V2. V2 is bigger than V1, so the denominator of Q is bigger than the denominator of K. to make Q=K, you need to make the numerator bigger. which means that n(A) and n(B) must increase. hence the reaction will move forwards with an increase in volume. and this agrees with le chatlier, who says that if you increase volume, you decrease pressure, and the system tried to increase the pressure by moving forwards. and because le chatlier requires less calculations, you use his principle instead.

edit: awks i wrote [B ] without the space, and half my post became bolded;; also i should really learn how to use latex;;;;

[zvezda]

Ahhh yes, thats clear now. Thanks guys.
Btw scribble i like that bit of reasoning at the end lol

[scribble]

la-Z way is always the best way. C:

[lzxnl]

Chatelier does work with adding inert gases. The inert gases do not disrupt the equilibrium, so nothing changes

[scribble]

thats true, since chatlier does say partially oppose the *effects* of the change, and adding an inert gas has no effects. OTL
its just that people (inc me last year) get really thrown by the idea we use pressure to explain volume changes and then we say that changing the pressure with an inert gas does nothing.

[lzxnl]

If you go from the perspective of concentration and how partial pressures are directly related to concentrations, adding an inert gas doesn't change either partial pressures or the concentration. I don't see how that's confusing.

[scribble]

but partial pressures arent on the vce course. hence Kc and not Kp

[lzxnl]

I know. Death to VCAA.