If the equilibrium $\mathrm{N_2(g)} + 3\mathrm{H_2(g)} \rightleftharpoons 2\mathrm{NH_3(g)}$ has equilibrium constant $K_c$, what is $K'_c$ for the REVERSE reaction $2\mathrm{NH_3(g)} \rightleftharpoons \mathrm{N_2(g)} + 3\mathrm{H_2(g)}$?

1. For the forward reaction: $K_c = \dfrac{[\mathrm{NH_3}]^2}{[\mathrm{N_2}][\mathrm{H_2}]^3}$. 2. For the reverse reaction, swap reactants and products: $K'_c = \dfrac{[\mathrm{N_2}][\mathrm{H_2}]^3}{[\mathrm{NH_3}]^2}$. 3. The reverse expression is the RECIPROCAL of the forward: $K'_c = 1/K_c$. 4. This is a general result for any reversible reaction (NCERT §6.4, Table 6.4 row 2): reversing the reaction takes the reciprocal of $K$. 5. Option A would only be true for $K_c = 1$ (a special case). Option C and D apply to multiplied-by-$n$ or squared reactions, not reversal. _Source: NCERT Class 11 Chemistry, Ch 6, §6.4 + Table 6.4 (reversal row), p. 9–10._