Chemical Equilibrium: find concentration from 2 solutions (ICE table)

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Discussion Overview

The discussion revolves around a chemical equilibrium problem involving the concentrations of NO2 and N2O4, specifically focusing on the application of an ICE table and the equilibrium constant expression. Participants are exploring the correct approach to substituting values into the equilibrium expression and the implications of their choices.

Discussion Character

  • Debate/contested

Main Points Raised

  • One participant questions why the values 0.39 and 0.65 are plugged into (1-2x) instead of (1-2x)^2 in the context of the equilibrium expression.
  • Another participant points out that while both solutions appear correct, using x=0.65 leads to a negative concentration of NO2, which is not physically possible.
  • A participant reiterates their confusion about why the equilibrium constant K is not calculated using (1-2x)^2, given that 2 moles of NO2 were provided.
  • One response suggests trying the values in the equilibrium expression to see the results, implying that the calculations may clarify the situation.

Areas of Agreement / Disagreement

Participants express differing views on the correct method for substituting values into the equilibrium expression, indicating that there is no consensus on the approach to take.

Contextual Notes

There is uncertainty regarding the correct application of the equilibrium constant expression and the implications of the values chosen for x. The discussion does not resolve these uncertainties.

Who May Find This Useful

Students and individuals studying chemical equilibrium, particularly those interested in the application of ICE tables and equilibrium constant calculations.

doridoridori
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Homework Statement
I've recently encountered this problem https://www.physicsforums.com/threads/need-major-help.117630/
In case the link doesn't open, here's the problem itself:
9. At a certain temperature, T, K for the reaction below is 7.5 liters/mole.

2NO2 <===> N2O4

If 2.0 moles of NO2 are placed in a 2.0-liter container and permitted to react at this temperature, what will be the concentration of N2O4 at equilibrium?

a) 0.39 moles/liter

b) 0.65 moles/liter

c) 0.82 moles/liter

d) 7.5 moles/liter

e) none of these
Relevant Equations
Super stuck on this. All of the deets are given in the question I linked. I'm still getting used to this forum so don't mind please if I accidentally mess something.
Also this is the equation itself:

7.5=x/(1-4x-4x^2)
30x^2-31x+7.5=0
x=0.39(approx.)
x=0.65(also approx.)
Hello! I've recently encountered this problem https://www.physicsforums.com/threads/need-major-help.117630/ and solved it and I'm stuck at choosing between 2 solutions. I don't understand why do we need to plug in 0.39 and 0.65 to (1-2x) and NOT to (1-2x)^2. I mean, we were given 2NO2, not just NO2. I see that in ICE table the result is [1-2x] but then why Keq=N2O4/(NO2)^2 and not just NO2?
 
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Those are both correct solutions to your equation, but x=0.65 corresponds to a negative concentration of NO2, which is physically impossible.
 
Chestermiller said:
Those are both correct solutions to your equation, but x=0.65 corresponds to a negative concentration of NO2, which is physically impossible.
Hello! Yes, that's essentially what I was asking. Sorry if I made it all unclear! So to know the right answer, we are plugging 0.65 and 0.39 into (1-2x). What I don't understand is why don't we plug it into K=x/(1-2x)^2 instead? I mean, we were given 2 moles of NO2, wouldn't it be reasonable then to calculate for 1-2x squared?
 
doridoridori said:
Hello! Yes, that's essentially what I was asking. Sorry if I made it all unclear! So to know the right answer, we are plugging 0.65 and 0.39 into (1-2x). What I don't understand is why don't we plug it into K=x/(1-2x)^2 instead? I mean, we were given 2 moles of NO2, wouldn't it be reasonable then to calculate for 1-2x squared?
That is what you do plug into. Try those values and see what you get.
 

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