# Thermodynamics: Equilibrium HELP

1. Dec 16, 2004

### apchemstudent

Thermodynamics: Equilibrium HELP!!!

Consider the following equilibrium:

N2O4(g) <--> 2NO2(g)

We can assume the heat of formation and standard change in entropy do not vary with temperature.

At what temperature will an equilibrium mixture contain equal amounts of the two gases?

where G = standard free energy change

-G/(R*ln(K(eq))) = T

delta G = delta H - T*(delta S)

There's two unknowns K(eq) and T. How do i figure out the value of K(eq) to solve this problem? Like, is there some sort of relationship of the K(eq) when both the gases are at the same amount?

K(eq) = X^2/X = X <---- this will become a variable, so how can i determine the K(eq).

Last edited: Dec 16, 2004
2. Dec 18, 2004

### GCT

A particular temperature of a reaction will be associated with a specific free energy value. A free energy value will have a relationship with the ratio of product over relationship, that is the equilibrium K of the reaction.

The solution to this problem lies in a simple substitution method, you'll need to substitute the free energy value (G). Try and figure it out. I believe this is all you'll need.

Feel free to ask further questions.

Last edited: Dec 18, 2004
3. Dec 18, 2004

### Gokul43201

Staff Emeritus
Just fixing a typo in GenChem's post :

"A particular temperature of a reaction will be associated with a specific free energy value. A free energy value will have a relationship with the ratio of products over reactants , that is the equilibrium K of the reaction."

4. Dec 20, 2004

### apchemstudent

I don't see this substitution going anywhere. First of all if we try to remove one variable such as T, we can solve for:

T = (G - H)/S. However We'll end up having 2 unknowns still with G and
K(eq).

If I substitute G = H - T*S into -G/(RlnK(eq)), I'll still have 2 unknowns T and K(eq).

5. Dec 21, 2004

### Gokul43201

Staff Emeritus
$$N_2O_4(g) \equiv 2NO_2(g)$$
If you start with 1 mole of $N_2O_4(g)$, then at equilibrium, you have some $1-x$ moles, where $x$ is the dissociation constant. As a result, you will have produced $2x$ moles of $NO_2(g)$.

$$K(eq) = \frac{[NO_2]^2}{[N_2O_4]} = \frac{4x^2}{1-x}$$

But, you are given that, at equilibrium,

$$[NO_2] = [N_2O_4] [/itex] [tex]=> 2x = 1-x => x = 1/3 moles$$

Plugging this into the above expression gives K(eq) = 2/3.

6. Dec 22, 2004

### apchemstudent

I really think there's something definitely wrong with this problem. I took a look at the answer book at school and they assumed the gases were at a partial pressure of 1 atm each. That's how they figured out the problem. Is there some reason they might use this?

7. Dec 23, 2004

### Gokul43201

Staff Emeritus
Could you write down the question exactly as it appears in your text/homework ?

8. Dec 23, 2004

### apchemstudent

Consider the following equilibrium:

N2O4(g) <--> 2NO2(g)

At what temperature will an equilibrium mixture contain equal amounts of the two gases? Assume the heat of formation and standard change in entropy do not vary with temperature.

9. Dec 23, 2004

### GCT

Of course, that makes sense. Rememeber when you devised the formula K(eq) = X^2/X in your first post? Instead of concetrations it is in terms of pressure, first find the relationship, the corresponding way you should write it in terms of pressure and that should solve your problem.

1 atm should be divided equally since the molar ratios are equivalent (same amount of each gases refers to #, and this relates directly to moles). Each gas should take up .5 atm (deduced according to PV=nRT), remember that equilibrium also relates to Keq. You need to find this relationship.