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

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SUMMARY

The discussion centers on the application of the ICE table method to determine concentrations in a chemical equilibrium problem involving the reaction of NO2 and N2O4. Participants clarify the correct approach to substituting values into the equilibrium constant expression, specifically why the concentrations of NO2 should be calculated using (1-2x) rather than (1-2x)^2. The confusion arises from the initial conditions of the reaction, where 2 moles of NO2 are present, leading to the conclusion that using (1-2x) is necessary to avoid negative concentrations, which are physically impossible.

PREREQUISITES
  • Understanding of chemical equilibrium concepts
  • Familiarity with ICE tables for equilibrium calculations
  • Knowledge of equilibrium constant expressions (Keq)
  • Basic algebra skills for solving equations
NEXT STEPS
  • Study the derivation of the equilibrium constant expression for reactions involving multiple species
  • Learn how to construct and interpret ICE tables for various chemical reactions
  • Explore the implications of negative concentrations in chemical equilibrium
  • Practice solving equilibrium problems with varying initial conditions and concentrations
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Chemistry students, educators, and professionals involved in chemical equilibrium analysis, particularly those working with gas-phase reactions and equilibrium 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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