Consider the equilibrium $\mathrm{N_2(g)} + 3\mathrm{H_2(g)} \rightleftharpoons 2\mathrm{NH_3(g)}$. The value of $K_p$ at $298\,\text{K}$ is $6.8\times 10^5$ (bar$^{-2}$). At this temperature, ammonia synthesis is

1. NCERT §6.6.1 (Predicting the extent of a reaction): if $K \gg 10^3$, the equilibrium lies STRONGLY to the right (products dominate). 2. Here $K_p = 6.8\times 10^5$, which is much larger than $10^3$. So the equilibrium STRONGLY favours $\mathrm{NH_3}$. 3. In thermodynamic terms, a large $K$ corresponds to a very negative $\Delta G^\circ$ — the products are far more stable than the reactants at this temperature. 4. BUT the reaction is kinetically slow at $298\,\text{K}$, which is why industrial Haber synthesis uses elevated T (lower yield) with a catalyst and high pressure (forces yield back up). 5. Option A reverses the meaning of $K$. Option C would require $K \approx 1$. Option D contradicts the existence of a measurable $K_p$. _Source: NCERT Class 11 Chemistry, Ch 6, §6.6.1 (Predicting Extent) + Table 6.5 ($K_p$ values for ammonia synthesis), p. 14–15._