[Meter]

The exact wording of the problem:

NO₂(g) can decompose to give NO(g) and O₂(g) according to the following equilibrium

2NO₂ ⇌ O₂(g) + 2NO(g)

When a mixture of the three gases was introduced into an empty vessel and the system was allowed to come to equilibrium, it was found that the NO concentration had decreased by 0.00050 mol L^-1. How had the concentrations of O₂ and NO₂ changed?

I am new to ICE tables (they don't teach this in high schools where I'm from). Since I am given no initial concentrations, and no dimensions of the vessel, am I supposed to calculate the initial concentrations using an arbitrary volume (moles per 1 liter, for example)?

[Borek]

You can assume a volume of 1 L, sure.

Not that it is necessary. You are told what was the change in the concentration, as long as the volume is constant changes in concentrations will follow stoichiometry (volume cancels out). So whether you assume 1 L, or 2 L, or just V, the result will be the same.

[Meter]

You can assume a volume of 1 L, sure.

Not that it is necessary. You are told what was the change in the concentration, as long as the volume is constant changes in concentrations will follow stoichiometry (volume cancels out). So whether you assume 1 L, or 2 L, or just V, the result will be the same.

So, something like the change in O₂ = twice the change in NO and the change in NO₂ = change in NO?

[Borek]

Yes.

And ICE table is just a way of making keeping track of such changes in an easily manageable way.

[Meter]

Yes.

And ICE table is just a way of making keeping track of such changes in an easily manageable way.

I see now.

Having done some problems now, the problem itself is finding the change (that is where the algebra comes in). If the change is given, a lot can be told from the stoichiometrical equation alone.

[mjc123]

So, something like the change in O2 = twice the change in NO and the change in NO2 = change in NO?

Change in O₂ = half the change in NO.

[Meter]

So, something like the change in O2 = twice the change in NO and the change in NO2 = change in NO?

Change in O₂ = half the change in NO.

Yes, that's what I meant. Thank you.

[MNIO]

ICE... Initial... Change... Equilibrium

in this case.. your reaction is
   2 NO₂  1 O₂ + 2 NO

and your ice table looks like this with the info you were given
                             [NO₂]          [O₂]            [NO]
  initial                     
  change                                               -0.00050M
  equilibrium

and you were asked to fill in this information.
                             [NO₂]             [O₂]              [NO]
  initial                     
  change           +0.00050M      -0.00025M      -0.00050M
  equilibrium

******
Those values in red were calculated via the coefficients of the balanced equation.  Do you see that?
You were NOT asked to fill in initial nor equilibrium values. just the info in red.  The assumption here is constant volume.