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?
AIt shifts FORWARD (more $\mathrm{NH_3}$ forms)
BIt shifts BACKWARD (less $\mathrm{NH_3}$)
CNo shift — reacting partial pressures are unchanged
DBoth forward and reverse rates drop to zero
Answer & Solution
Correct answer: C. No shift — reacting partial pressures are unchanged
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._
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