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Question:

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Decreasing the product's concentration seemed to follow the logic of Le Chatelier's principle, so a decrease in volume will force the reaction to the left to increase the pressure (increasing the number of molecules). However, decreasing nitrogen gas concentration also has the same effect. I chose volume compression over the other because I assumed. I am probably stretching what it means by "sealed" though. Apparently I got the question incorrect.

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  • $\begingroup$ Don't hold me on this, but I'd say that $N_{2}$ was removed from the system. Removing reactant will shift the equilibrium to the left to oppose the change, so more $NH_3$ would have to decompose into $N_{2}$ in order to do this. However, I'm intrigued by the volume increasing. If we're abiding by the ideal gas laws, pressure is inversely proportional to volume, so increasing the volume decreases the pressure. There are less gaseous moles of product, so decreasing the pressure would also shift the equilibrium to the left. $\endgroup$ – user60221 Mar 18 '18 at 0:52
  • $\begingroup$ By the way, the 'sealed container' is used to provide a closed system, so that an equilibrium can be achieved. $\endgroup$ – user60221 Mar 18 '18 at 0:58
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The given graph indicates that $\ce{[NH3]}$ remains constant till the 15 minute mark which implies that the equilibrium was established till then. Now, after the 15 minute mark, $\ce{[NH3]}$ decreases and reaches a constant value which means, the equilibrium was disturbed and re-established. $$\ce{N2 + 3H2 <=> 2NH3 + heat}$$ Since we observe a decrease in $\ce{[NH3]}$, the reaction must have shifted in the backward direction according to the given equilibrium.
Out of the five options, only match for this is (C) removing some amount of $\ce{N2}$. Let us analyse the other four options,
(A) Decrease in temperature: Note that the above reaction is exothermic and hence according to Le Chatlier’s principle, the reaction shifts in the forward direction for a decrease in temperature.
(B) Decrease in volume: Since we are dealing with a gaseous(ideal) equilibrium, a decrease in volume implies an increase in pressure and hence, according to Le Chatlier’s principle, the reaction shifts in the direction where there is less no. of moles ie., forward direction here.
(D) Removal of $\ce{NH3}$: As Gaurang pointed out in the comments, removal of $\ce{NH3}$ will cause a sharp fall along a vertical line, but the graph indicates a gradual fall in $\ce{[NH3]}$. Hence this is also not valid.
(E) Increase in volume: Assuming a sudden increase in volume, say n times, $\ce{[NH3]}$ should also fall by n-times. Clearly, the graph doesn’t match with our condition. Hence this is also incorrect.

So (C) must be the correct answer, as removing some $\ce{N2}$, shifts the reaction in the backward direction, thereby decreasing $\ce{NH3}$ gradually till equilibrium is re-attained.

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