## 11.43 6th edition

Tyra Nguyen 4H
Posts: 74
Joined: Fri Sep 28, 2018 12:25 am

### 11.43 6th edition

The question states:

Consider the reaction 2NO(g)N2(g)+O2(g). If the initial partial pressure of NO(g) is 1.0 bar, and x is equilibrium concentration of N2(g), what is the correct equilibrium relation?
(a) K=x2/(1.0-x)
(b) K=x2
(c) K=x2/(1.0-2x)2
(d) K=4x2/(1.0-2x)2
(e) K=2x/(1.0-x)2

I'm not sure how to solve this if there are 4 variables and 2 of them, the pressure of O2 and K, are unknown. Would this require an ICE box? Does an ICE box only apply to molarities or can it be applied to partial pressures as well?

Ray Guo 4C
Posts: 90
Joined: Fri Sep 28, 2018 12:15 am

### Re: 11.43 6th edition

Do you know the volume? I think an ICE box can be used for partial pressure.

Emilia z
Posts: 28
Joined: Fri Sep 28, 2018 12:21 am

### Re: 11.43 6th edition

It looks like this question is just testing you on your ability to understand the equation for K, and not necessarily solve for any value. Therefore the solution is just K = pressure of the products / pressure of the reactants, since some of this is unknown you substitute 'x' for those pressures and that should be your answer.

Samantha Chang 2K
Posts: 69
Joined: Fri Sep 28, 2018 12:17 am

### Re: 11.43 6th edition

When using the ICE box, you can use partial pressure and molarities/concentrations.

Stevin1H
Posts: 89
Joined: Fri Sep 28, 2018 12:17 am

### Re: 11.43 6th edition

This question is asking for the equilibrium constant, K, expression. This question requires an ice box and I believe the ice box can be applied to both concentration and partial pressure. For the ice box, the initial value of NO would be 1.0 bar (given) and the initial values of N2 and O2 are 0. The change in molarity of NO would be -2x because of the coefficients and the change for N2 and O2 would be +x. So the equilibrium expression for NO would be (1-2x) and N2 and O2 would be +x.

And the equilibrium constant expression would be (N2)(O2)/(1-2x)^2 which is equal to (x^2)/(1-2x)^2