Why don't pressure or volume affect K?

[n00b]

I know that temperature is the only factor that changes K (the equilibrium constant). What I want to know is, why don't pressure and volume affect K? Say, for example, that an increase in pressure moves equilibrium to the right. Wouldn't this then increase the concentration of products, reduce concentration of reactants and therefore increase K?

[modest]

I know that temperature is the only factor that changes K (the equilibrium constant).

 

Boltzmann's constant should not change (so long as the units stay the same). It is constant.

 

What I want to know is, why don't pressure and volume affect K? Say, for example, that an increase in pressure moves equilibrium to the right. Wouldn't this then increase the concentration of products, reduce concentration of reactants and therefore increase K?

 

It will increase the temperature, not the constant k. If you increase the pressure then the temp will go up. If you look at the P-V equation:

[math]p V = N k T [/math]

N is the number of molecules and it stays constant. k is also constant. If you increase the pressure, p, then you will need to increase the temperature, T, for the equality to be true. Plug in some arbitrary numbers to see how the equations behaves:

 

p V = N k T

10*1=1*2*5

double the pressure:

20*1=1*2*?

you'll have to double the temp as well:

20*1=1*2*10

By doubling the pressure 10 -> 20, the temp doubled 5 -> 10.

 

~modest

[Gordon Freeman]

I know that temperature is the only factor that changes K (the equilibrium constant). What I want to know is, why don't pressure and volume affect K?

 

Well hold on there cowboy, I'm no chemist, but I know my math properties, and the original poster was onto something.

 

It will increase the temperature, not the constant k. If you increase the pressure then the temp will go up.

 

Well if pressure goes up, and pressure directly relates to temperature, and the temperature goes up, and temperature directly relates to K, then you have yourself the transitive property in affect. So by law of transitive property, pressure does affect the equilibrium constant, but only in variance of temperature, which was also misused in the equation in the post before.

 

[math]

p V = N k T

[/math]

 

That only applies to a singular state of matter because the more pressure you have, the temperature will eventually drop drastically to condense the matter into a state of lower energy. The ideal gas laws will say that pressure and volume are unrelated because gases don't rely on those factors. However, pressure and volume are significant factors in finite environments because again, there will eventually be a state change.

 

So literally no, pressure and volume aren't the immediately calculable variables that determine the constant. But theoretically, they are variables that change the constant, which is proven because you can derive their value given the other variables.

[modest]

Howdy, Gordon. WElcome to Hypography :)

 

Well if pressure goes up, and pressure directly relates to temperature, and the temperature goes up, and temperature directly relates to K, then you have yourself the transitive property in affect.

 

As I said in my last post, temperature is not related to k. k is constant under any system of units.

 

So by law of transitive property, pressure does affect the equilibrium constant, but only in variance of temperature, which was also misused in the equation in the post before.

 

[math]

p V = N k T

[/math]

 

That only applies to a singular state of matter because the more pressure you have, the temperature will eventually drop drastically to condense the matter into a state of lower energy. The ideal gas laws...

 

The ideal gas law applies only to a gas. It should be clear we are talking about a gas.

 

The ideal gas laws will say that pressure and volume are unrelated because gases don't rely on those factors.

 

Let's look at the question in the opening post from the standpoint of a thought experiment rather than a physical law. If someone were to hand you a cylinder filled with carbon dioxide gas and sealed so that no molecules could go in the cylinder and none out. The volume of the cylinder is constant because you cannot compress the cylinder. If the person asked you to increase the pressure of the gas in the cylinder, how could it be done?

 

~modest

[lawcat]

Modest, can I call you a space cowboy? :(

 

Let me take this conversation in a slightly different direction with intent to explain.

 

Gas constant k, is not Boltzman constant. Boltzman constant is used in kinetic gas theory to calculate the movement (speed etc.) of molecules. The gas constant k is used in static gas theory (pressure, heat etc.).

 

The gas constant k is what OP is talking about. The gas constant is a ratio of specific heats c (Cp for constant pressure specific heat, and Cv for constant volume specific heat.) So, k= Cp/Cv. (p/v subscript denotes constant process)

 

Specific heat c, or Cp, Cv, is a ratio of heat Q required to raise the mas m of a substance by T degrees. So, c = Q/m*T. This turns out to be constant. In other words, for a constant volume of mass m, or cosntant pressure in mass m, as heat Q goes up so does temperature T, and vice versa. The constant remains a constant.

 

Since gas constant is defined as a ratio k = Cp/Cv, and c's are constant, the k is constant and does not depend on temperature; or pressure and volume as long as those are constants in the process.

 

But if things are not constant, then you have a polytropic siituation and k can not be used. Then, the polytropic exponent n should be used. Hope this gives the OP some material to start research.

 

EDIT: Thank you modest. I misused gas constant terminology. Anyway, the k that OP is talking about is the ratio of specific heats k, not the BOltzman constant k (kappa).

[modest]

Modest, can I call you a space cowboy? :smile:

 

:hyper: :hihi:

 

Gas constant k, is not Boltzman constant.

 

:confused:

 

I think, perhaps, you mean that the gas constant, R, is not Boltzmann's constant. That would be true, but you can use either in the ideal gas law.

 

The Boltzmann constant (

k

or

k

_B

) is the physical constant relating energy at the particle level with temperature observed at the bulk level. It is the gas constant

R

divided by the Avogadro constant

N

_A

:

 

[math]k = \frac{R}{N_{\rm A}}\,[/math]

Boltzmann constant - Wikipedia, the free encyclopedia

 

Boltzmann's constant

k

is a bridge between Macroscopic scale and microscopic physics. Macroscopically, the ideal gas law states that, for an ideal gas, the product of pressure

p

and volume

V

is proportional to the product of amount of substance

n

and absolute temperature

T

:

[math] pV = nRT \,[/math]

where

R

is the gas constant (8.314 472 J K

⁻¹

 mol

⁻¹

). Introducing the Boltzmann constant transforms the ideal gas law into an equation about the microscopic properties of molecules,

[math]p V = N k T \,[/math]

where ''N'' is the number of molecules of gas.

Boltzmann constant - Wikipedia, the free encyclopedia

 

The gas constant k is what OP is talking about.

 

In terms of pressure and temperature.

 

~modest

[lawcat]

You are correct. Sorry about misuse of "gas constant" term. the Boltzman constant and ratio of specific heats are certainly two different things.

 

But both are constant for constant volume and pressure processes. The ratio of specific heats k is constant because by definition k=Cp/Cv, we are dealing with constant pressure and volume parameters. And Boltzman constant k is constant because k = R*/Na; by definition universal gas constant R* is constant, and Avogadro's number Na is constant for equal volumes of gases for constant temp and pressure.

 

So to answer the original question, the k is constant because underlying assumption of pressure, volume, and temperature make it so. But if you are dealing with polytropic processes, then the k's do not apply.

 

Thank you modest.

[n00b]

Thanks for all your answers (even though they're all way over my head). Um, actually the K I was referring to in the original post was the equilibrium constant as per:

 

 

According to Wikipedia:

All equilibrium constants depend on temperature and pressure (or volume).

However, in our high school Chemistry class we were told explicitly that, although pressure and volume could effect the position of equilibrium (shift it left or right), only changes in pressure would effect K. This is the bit I don't understand, as per my original post.

 

Sorry for being daft if all your answers above did actually answer my question.

[Qfwfq]

Howdy folks. :Guns:

 

Sorry for being daft if all your answers above did actually answer my question.

Actually, no they didn't. :doh:

 

I regret not having come to see this thread the past few days because, reading your OP just now, it seemed obvious to me what you're talking about; I could have pointed out the mistake. Unfortunately I'm no great chemist and so I would have to think the matter over and shake plenty of cobwebs out of my faded memories before I could say anything certain, but I suspect it is a matter of how equilibrium is computed (depending on pressure and volume for a same given K value). Try working it out in a case of a single compound vs. more than one element, so that the total number of moles differs between the two sides of the formula.

[modest]

I'm terribly sorry, n00b. When I read the OP I skimmed over it. Very sorry.

 

I know that temperature is the only factor that changes K (the equilibrium constant). What I want to know is, why don't pressure and volume affect K? Say, for example, that an increase in pressure moves equilibrium to the right. Wouldn't this then increase the concentration of products, reduce concentration of reactants and therefore increase K?

 

If you increase the concentration of the products A and B:

[math]K_C = \frac{C^c D^d}{A^a B^b}[/math]

then the equilibrium will shift to the right,

[math]aA + bB \rightarrow cC + dD[/math]

this happens because there are more of A and B to react (because their concentration was increased). There are more collisions going on between them, so they will react more than C and D. When the equilibrium shifts right the denominator in the equation will shrink,

[math]K_C = \frac{C^c D^d}{A^a B^b}[/math]

and the numerator will get larger back to their original values keeping K_C constant.

 

A good google search might be "le Chatelier's principle".

 

~modest

Edited August 7, 2010 by modest
got left and right mixed up

[Qfwfq]

As usual, it is simpler in the case of the ideal gas approximation but then it can be generalized for other relations between pressure, temperature and specific volume (which is the inverse of concentration), so let's write the gas equation as the proportionality:

 

[math]P\propto CT=\frac{n}{V}T[/math]

 

and let's consider the reaction [imath]2A + 3B \rightarrow C[/imath] so there are 5 moles total on one side and 1 on the other. If all three are gasses and we push the piston down to half the volume, without letting temperature change.

 

Without the reaction taking place all three concentrations would double and this would mean doubling the pressure, but it is easy to see that this would change K:

 

[math]\frac{2\{C\}}{(2\{A\})^2 (2\{B\})^3}=\frac{1}{2^4}\frac{\{C\}}{\{A\}^2 \{B\}^3}[/math]

 

Instead what happens is that the equilibrium changes, with more moles of [imath]C[/imath] and less of [imath]A[/imath] and [imath]B[/imath], for the same value of K.

 

In general that concentration could be some function [imath]C(P, T)[/imath] of pressure and temperature, this can make it more complicated but the crux is that if there is some interdependence of pressure and concentratoin at a given temperature, the same value of K can give a different equilibrium.