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Adding an inert gas will not affect Q if that reaction is held at a constant volume. But, if the reaction is held at constant pressure, then adding inert gas will change Q bc the partial pressures of the gases will change.

I don't understand this. If the reaction is held at constant pressure, this means that you changed the volume when adding the inert gas. Wouldn't the partial pressures of all the reactants change by the same factor, so Q wouldn't change?

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Wouldn't the partial pressures of all the reactants change by the same factor, so Q wouldn't change?

If the number of gas particles is the same for reactants and products, then Q would not change. However, if you have a reaction like

$$\ce{2N2O(g) <=> N2O4(g)}$$

even though the reactant and product partial pressure changes by the same factor, the changes don't cancel out because the coefficients are different.

$$ Q = \frac{[\ce{N2O4}]}{[\ce{N2O}]^2}$$

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  • $\begingroup$ Say PN2O = 2 and PN2O4=4. Say the volume doubles as a result of adding an inert gas at constant pressure. So now PN2O = 1 and PN2O4=2. Won't Q be the same? $\endgroup$ – Jon Apr 3 at 1:03
  • $\begingroup$ No because $P_\ce{N2O}$ gets squared. $\endgroup$ – Karsten Theis Apr 3 at 1:13
  • $\begingroup$ Couldn't you make the same argument for adding an inert gas at constant volume? Wouldn't any change be magnified for the squared reactant? $\endgroup$ – Jon Apr 3 at 1:17
  • $\begingroup$ At constant volume, the partial pressures don't change at all. $\endgroup$ – Karsten Theis Apr 3 at 2:04
  • $\begingroup$ But, if you add an inert gas, won't that increase the total pressure in the system, so the reactant and product partial pressures will increase, and due to the squared reactant in the Q expression, the N2O partial pressure will increase more than that of N2O2? $\endgroup$ – Jon Apr 3 at 2:25

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