Class 11 NEET Chemistry Equilibrium Questions Part 2

Concept:The relation between $K_P$ and $K_C$ is $K_P = K_C (RT)^{\Delta n_g}$, where $\Delta n_g$ = (moles of gaseous products) – (moles of gaseous reactants).When $\Delta n_g = 0$, $K_P = K_C$; when $\Delta n_g \neq 0$, $K_P \neq K_C$.Chemical Equation / Formula:The general relation: $K_P = K_C (RT)^{\Delta n_g}$.Explanation:For each reaction, calculate $\Delta n_g$ and check if it is zero.Option A: $\mathrm{H_2O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{H_2}(g)+\mathrm{CO_2}(g)$Reactants: 2 moles gas; Products: 2 moles gas → $\Delta n_g = 0$ → $K_P = K_C$.Option B: $\mathrm{N_2}(g)+3\mathrm{H_2}(g) \rightleftharpoons 2\mathrm{NH_3}(g)$Reactants: 4 moles gas; Products: 2 moles gas → $\Delta n_g = -2$ (not zero) → $K_P \neq K_C$.Option C: $\mathrm{H_2}(g)+\mathrm{I_2}(g) \rightleftharpoons 2\mathrm{HI}(g)$Reactants: 2 moles gas; Products: 2 moles gas → $\Delta n_g = 0$ → $K_P = K_C$.Option D: $\mathrm{N_2}(g)+\mathrm{O_2}(g) \rightleftharpoons 2\mathrm{NO}(g)$Reactants: 2 moles gas; Products: 2 moles gas → $\Delta n_g = 0$ → $K_P = K_C$.Only Option B has $\Delta n_g \neq 0$.Shortcut:Quickly compute $\Delta n_g$ for each reaction: if total moles of products equals total moles of reactants (all gases), then $K_P = K_C$; otherwise $K_P \neq K_C$.Only $\mathrm{N_2}+3\mathrm{H_2} \rightleftharpoons 2\mathrm{NH_3}$ has unequal moles (4 → 2).Answer:Option B: $\mathrm{N_2}(g)+3\mathrm{H_2}(g) \rightleftharpoons 2\mathrm{NH_3}(g)$.