# Calculating Kp for gas mixture from picture

1. Mar 7, 2015

### CroSinus

1. The problem statement, all variables and given/known data
A chemical reaction is described by the equation N2O4 <> 2NO2. The pressure equilibrium constant for the given reaction is equal to 8 Pa. Which picture describes the equilibrium? (Please, see the attached file!).

2. Relevant equations
N2O4 <> 2NO2

3. The attempt at a solution
First I wrote the expression for the equilibrium constant of the given reaction. Kp = p^2(NO2) / p(N2O4). Then I tried to substitute partial pressures with the number of molecules for each type of the gass. It is not clear to me how can I use only number of molecules to replace partial pressures since I have the partial pressure squared in numerator. The pressure squared prevents for some things to be canceled in the expression for Kp if I apply Dalton's law for gas mixtures. Can somebody explain this to me, please.

Last edited: Mar 7, 2015
2. Mar 8, 2015

### Staff: Mentor

Agreed, there is not enough data to answer the problem.

For the first picture, assuming the total pressure is Pt, partial pressure of NO2 is $\frac 1 3 P_t$ and partial pressure of N2O4 is $\frac 2 3 P_t$. Then

$$K_p = \frac {P^2_{NO_2}}{P_{N_2O_4}} = \frac {(\frac 1 3 P_t)^2}{\frac 2 3 P_t} = \frac 1 6 P_t$$

Assume Pt=48 Pa and the first image is the correct answer. Problem is, the same can be done for every other image.

3. Mar 8, 2015

### CroSinus

My textbook says the correct answer is C. But how did they get that result? Here's the original text of the question in Croatian.

Last edited: Mar 8, 2015
4. Mar 9, 2015

### Staff: Mentor

Making an error.

C is NOT the correct answer. There is no correct answer to this problem, as it is incomplete.

As I have shown you above - you can "prove" each answer to be the correct one, just assuming different values of the total pressure.

5. Mar 9, 2015

### CroSinus

Right. I agree with you. Unless you know the total pressure of the mixture, it is impossible to solve the problem. Funny textbook.