Kc Values - Temperature and Pressure - Equilibrium

FaoiSin

Would anybody be able to explain to me why Temperature does not affect the value of Kc but Pressure does? Can't get my head around it.

ChemHickey

YeaaahFaoiSin wrote: »

Would anybody be able to explain to me why Temperature does not affect the value of Kc but Pressure does? Can't get my head around it.

Temperature does... oh it does.

For the haber process at like 127K its around a few thousand, around 500K its around a few hundred and at a few thousandK its .04 or something to those effects. Don't know the exact figures but temp does affect it.

FaoiSin

Oh wait I meant the other way around. Pressure doesn't but Temperature does? I'm not sure of either but that's what the person who corrected my mock wrote beside the Equilibrium question.

ChemHickey

YeaaahFaoiSin wrote: »

Oh wait I meant the other way around. Pressure doesn't but Temperature does? I'm not sure of either but that's what the person who corrected my mock wrote beside the Equilibrium question.

I'm not 100% certain but it does doesn't it? Providing it is a gaseous reaction. If pressure is applied to a gaseous reaction the system will shift to the side with the least number of molecules. This would mean more either product or reactant would be formed, changing their concentration at equilibrium and therefore kc. This doesn't occur in a liquid phase reaction as equilibrium won't shift to the side with lower molecules (it's not in the gaseous phase)

Hope it helps a bit. Its just some abstract thinking on it by me so I might be wrong.

SuchGold

Change in pressure will have an effect on equilibrium only if all of the substances are gases and the total number of molecules on each side is different.

Liveforrugby

Le Chatelier's Principle is a very useful tool to predict the direction of the equilibrium. There are mainly 3 factors which disturbs the equilibrium.
1) Change in concentration
2) Change in pressure (by changing volume)
3) Change in temperature

I will explain the effects of these changes by using Le Chatelier's Principle.
The application is rather simple. If something increases, try to decrease it. If something decreases, try to increase it.

I will use the following reaction to explain all these effects.

2SO2(g) + O2(g) ↔ 2SO3(g) .......... ΔH = −197 kJ

1) Change in concentration:
- If more O2 is added to the system at equilibrium, the system will try to decrease the added O2. Therefore the equilibrium shifts to right. That is forward reaction will be effective. SO2 and O2 will react and will produce more SO3.
- Same thing occurs when more SO2 is added.
- If more SO3 is added. the system will try to decrease the added SO3. Therefore the equilibrium shifts to left. That is reverse reaction will be effective. SO3 will decompose into SO2 and O2.
- If some SO3 is removed from the system at equilibrium, the system will try to increase the removed SO3. Therefore the equilibrium shifts to right. That is forward reaction will be effective. SO2 and O2 will react to form SO3.
- If some SO2 is removed from the system at equilibrium, the system will try to increase the removed SO2. Therefore the equilibrium shifts to left. That is reverse reaction will be effective. SO3 will decompose into SO2 and O2.

2) Change in pressure (by changing volume)
- If the pressure of the system is increased by decreasing the volume, the system will try to decrease the pressure. Since the partial pressures of gases are directly proportional to their number of moles, the equilibrium shifts to right. Why? Because, at the right there are 2 moles, at the left there are 4 moles. As the number of moles decreases, the pressure also decreases.
- If the pressure of the system is decreased by enlarging the volume, the system will try to increase the pressure. The reverse of the first case will happen.
- Same thing occurs when some O2 is removed.

NOW YOUR QUESTION;
3) Change in temperature
- If the temperature of the system is increased, the system will try to decrease this increase by consuming heat. Since the forward reaction is exothermic (ΔH = −197 kJ), the reverse reaction will be effective. The equilibrium will shift to left. Endothermic reverse reaction will consume heat.
- If the temperature of the system is decreased, the system will try to increase this decrease by poducing heat. Since the forward reaction is exothermic (ΔH = −197 kJ), the forward reaction will be effective. The equilibrium will shift to right. Exothermic forward reaction will produce heat.

This is IMPORTANT for you;
In cases of concentration or pressure changes, the system will readjust itself to satisfy Kc or Kp value.
But, in case of temperature changes, the value of Kc or Kp changes. Whenever equilibrium shifts to right by the effect of temperature change, Kc and Kp will be greater and whenever equilibrium shifts to left by the effect of temperature change, Kc and Kp will be smaller.

4) Effect of Catalyst: Catalysts increase the rates of both the forward and reverse reactions equally. Thus, they reduce the time to reach the equilibrium. They have no effect on either the yield of the reaction or the equilibrium constants.

*stolen not taking credit.

Liveforrugby

By the way the above won't really answer your question as nobody really knows =P But it should clear some things up =P

Mista

ChemHickey wrote: »

I'm not 100% certain but it does doesn't it? Providing it is a gaseous reaction. If pressure is applied to a gaseous reaction the system will shift to the side with the least number of molecules. This would mean more either product or reactant would be formed, changing their concentration at equilibrium and therefore kc. This doesn't occur in a liquid phase reaction as equilibrium won't shift to the side with lower molecules (it's not in the gaseous phase)

Hope it helps a bit. Its just some abstract thinking on it by me so I might be wrong.

I thought this too, but my teacher and book both say that pressure has no effect on the Kc value.. I don't really get why either, but ye just gotta accept it.



As for temperature, it depends on whether the forward reaction is exothermic or endothermic. :pac:

Liveforrugby

Actually I just found the answer but I don't think any of us will understand it.

http://en.wikipedia.org/wiki/Van_'t_Hoff_equation

RedTexan

Kc is only temperature dependent, pressure does not affect the value of Kc, don't ask why, this is just what our teacher always insisted! It does of course affect the position of equilibrium but not Kc. And for the equal numbers of moles on each side, that's only significant when you are given the volume of the container (i.e. volume doesn't matter if number of moles are equal).

reznov

In the case of a pressure stress, the equilibrium will shift to the side with the least molecules being produced. If less molecules are produced, there will be less pressure on the system, relieving it.

finality

http://www.chemicalforums.com/index.php?topic=20112.0

Like, in this example
C > 2D (equilibrium system, arrows should be going both ways :L )


Kc= [D]^2/[C]

when pressure is doubled both their concentrations double, (pV=k so volume is halved, that means concentration is doubled)

But to maintain the value of Kc, the reaction is pushed towards the left. So the concentration of C increases further.

Kc= 4[D]^2/2x[C]
where x is the amount [C] increases according to Le Chatelier's principle.

But this is equal to the original Kc. Basically an increase in pressure will cause the reaction to shift in a case like this, but only to maintain the value of Kc.

Changes in temperature affect the value of Kc because you've just got one side of the system increasing in concentration and one side decreasing. Like, say the forward reaction is endothermic here and temperature is increased. We could end up with, for example,

Kc= 2[D]^2/([C]/2)
=4[D]^2/[C]

so Kc is quadrupled :P

bladet1delta

Be careful not to get mixed up between yield and your Kc.

While, as per the Haber Process, a change in pressure and a change in temperature will both affect the yield of ammonia.
However, a change in pressure won't affect the Kc value in your calculation.

Mista

If more product is formed, does the Kc value increase or decrease? Or does it depend?

reznov

Higher the Kc, the more the system is geared towards producing products.

DepoProvera

Think about it: Kc is gotten by putting product concentration over reactant concentration so the more product you form(relative to reactants) the more topheavy the fraction becomes.

Mista

DepoProvera wrote: »

Think about it: Kc is gotten by putting product concentration over reactant concentration so the more product you form(relative to reactants) the more topheavy the fraction becomes.

Oh yea.. makes sense :pac: cheers