Homework Problem #3 7

[805377003]

At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.100 M and [NO]=0.600 M.
N2(g)+O2(g)↽−−⇀2NO(g)

If more NO is added, bringing its concentration to 0.900 M, what will the final concentration of NO be after equilibrium is re‑established?

I've attempted this problem but I keep getting it wrong. Can someone walk my through the steps?

[IsaacLaw1E]

You need to make an ICE table. Because you added [NO], there will be more [N2] and [O2] formed, so the reverse reaction occurs.
2NO <-> N2 + O2, and the equilibrium values are [NO] = 0.9 - 2x, [N2] = [O2] = 0.1 + x. The Ka is found with the given equilibrium concentrations, so Ka = (0.1)(0.1)/(0.6^2) = 0.028

[ALee_1J]

Using [N2]=[O2]=0.100 M and [NO]=0.600 M, calculate the Ka value. This value will allow you to calculate the values at equilibrium should the reaction be disturbed. Then solve an ICE table with the values [N2]=[O2]=0.100 M and [NO]=0.900 M for the initial concentration. Reminder: solving for x doesn't always mean that that's the final concentration of a substance at equilibrium.

[George Hernandez 3I]

Like the others said, solving for x means finding the unknown changes to the concentrations. Remember that K will always stay the same, with the exception of temperature change. So by adding more product, some of that product will decompose/form to more reactants until it reaches the product/reactant ratio again.