An INERT GAS (e.g. argon) is added to an equilibrium mixture of $\mathrm{N_2(g)} + 3\mathrm{H_2(g)} \rightleftharpoons 2\mathrm{NH_3(g)}$ at CONSTANT VOLUME (and constant temperature). How does the equilibrium position change?

1. At CONSTANT VOLUME, adding an inert gas increases the TOTAL pressure of the mixture. 2. However, it does NOT change the PARTIAL PRESSURES of N$_2$, H$_2$, NH$_3$ — each is still $n_i RT/V$ with the same $n_i$ and $V$. 3. The equilibrium expression $K_p = (p_{\mathrm{NH_3}})^2 / (p_{\mathrm{N_2}}\,(p_{\mathrm{H_2}})^3)$ depends ONLY on partial pressures of the reacting species. Since those are unchanged, $Q = K$ still — no shift. 4. Contrast: at CONSTANT TOTAL PRESSURE, adding inert gas would force the volume to increase, lowering reactant partial pressures — and the equilibrium WOULD shift toward more moles of gas (here, reverse). That's a different scenario. 5. Option D is wrong: rates do not drop to zero just because inert gas was added; the reaction is still dynamic. _Source: NCERT Class 11 Chemistry, Ch 6, §6.8.2 (Effect of Inert Gas on equilibrium), p. 21._