What is the equilibrium expression for the reaction $\mathrm{P}_{4(\mathrm{~s})}+5 \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{P}_{4} \mathrm{O}_{10(\mathrm{~s})}$ ?

(B) $\mathrm{Kc}=1 /[\mathrm{O}_{2}]^{5}$

(A) $\mathrm{Kc}=\left[\mathrm{P}_{4} \mathrm{O}_{10}\right] / [\mathrm{P}_{4}][\mathrm{O}_{2}]^{5}$

(C) $\mathrm{Kc}=[\mathrm{O}_{2}]^{5}$

(D) $\mathrm{Kc}=\left[\mathrm{P}_{4} \mathrm{O}_{10}\right] / 5\left[\mathrm{P}_{4}\right]\left[\mathrm{O}_{2}\right]$

JEE Main · 2004 · 24 Apr 2004 · Offline · Equilibrium

What is the equilibrium expression for the reaction

P_{4 (s)} + 5 O_{2 (g)} ⇌ P_{4} O_{10 (s)}

Answer: (B)

For a heterogeneous equilibrium the pure solids are dropped from the expression, their activity taken as one. In

P X_{4} (s) + 5 O X_{2} (g) P X_{4} O X_{10} (s)

both the phosphorus and its oxide are solids, so only the oxygen gas remains, giving

K_{c} = \frac{1}{[ O X _{2} ] ^{5}}

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option