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$\ce{N2 (g)}$ and $\ce{H2(g)}$ are allowed to react in closed vessel in given temperature and pressure for the formation of $\ce{NH3 (g)}$ according to,

$$\ce{N2 (g) + 3H2 (g) <=> 2NH3 (g) + \pu{22.4 kcal}}$$ If $\ce{He(g)}$ is added at equilibrium at constant pressure, then which is/are correct?

a) Concentration of $\ce{N2(g)}$, $\ce{H2(g)}$ and $\ce{NH3(g)}$ decrease.

b) Moles of $\ce{NH3(g)}$ decreases.

c) The extent of cooling depends upon the amount of $\ce{He(g)}$ added.

d) Concentration of $\ce{N2(g)}$ and $\ce{H2(g)}$ increases and concentration of $\ce{NH3(g)}$ decreases.

I know that addition of inert gas at constant pressure increases the volume and forces the reaction to proceed towards the side with more gaseous moles (backward in this case). So, option b) and c) are obviously right.

Also, the concentration of all gases decreases suddenly after the inert gas addition due to increase in volume. But I'm unsure of the final state after equilibrium and I can't figure out why option a) is given right.

How can I make a prediction about the concentration of all gases at equilibrium?

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  • $\begingroup$ Can someone help me out with this? $\endgroup$
    – Rishi Shekher
    Oct 6, 2022 at 12:46
  • $\begingroup$ "Also, the concentration of all gases decreases suddenly after the inert gas addition due to increase in volume." I don't understand your question, haven't you answered it already? $\endgroup$
    – Safdar Faisal
    Oct 6, 2022 at 12:51
  • $\begingroup$ "Also, the concentration of all gases decreases suddenly after the inert gas addition due to increase in volume." Here, I meant that concentration of all gases decreases "just" after inert gas addition, but the concentrations of N2 and H2 will also increase by a certain amount, as the reaction shifts in the backward direction to reach a new equilibrium position. So, my question is, what is the net effect of these changes on the "final equilibrium" concentrations? $\endgroup$
    – Rishi Shekher
    Oct 6, 2022 at 13:48
  • $\begingroup$ See Haber Process. $\endgroup$
    – Robert DiGiovanni
    Oct 7, 2022 at 1:25
  • $\begingroup$ Reacting in a closed vessel. At constant pressure one is simply diluting the mixture. But you might try raising the pressure enough for NH3 to liquefy while N2 and H2 remain gaseous. The NH3 could then be removed, driving the reaction towards the product. $\endgroup$
    – Robert DiGiovanni
    Oct 7, 2022 at 1:31

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