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11.43 6th edition

Posted: Tue Jan 15, 2019 5:12 pm
by Tyra Nguyen 4H
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?

Re: 11.43 6th edition

Posted: Tue Jan 15, 2019 5:20 pm
by Ray Guo 4C
Do you know the volume? I think an ICE box can be used for partial pressure.

Re: 11.43 6th edition

Posted: Tue Jan 15, 2019 5:25 pm
by Emilia z
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.

Re: 11.43 6th edition

Posted: Wed Jan 16, 2019 12:34 pm
by Samantha Chang 2K
When using the ICE box, you can use partial pressure and molarities/concentrations.

Re: 11.43 6th edition

Posted: Thu Jan 17, 2019 10:16 am
by Stevin1H
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