Q.The equilibrium for the dissociation of $XY_2$ is given as,

[!FORMULA] $2XY_2(g) \rightleftharpoons 2XY(g) + Y_2(g)$

if the degree of dissociation x is so small compared to one, show that

[!FORMULA] $2K_P = PX^3$

where P is the total pressure and $K_P$ is the dissociation equilibrium constant of $XY_2$.

Given a reaction's starting amounts and either its equilibrium constant or one measured equilibrium quantity, the full equilibrium composition can be worked out systematically using an Initial - Change - Equilibrium (ICE) table: list the initial moles (or concentrations) of every species, the change each undergoes (in terms of a single unknown extent of reaction, x, scaled by each species' own stoichiometric coefficient), and add them to get the equilibrium amount.

For $H_2(g)+I_2(g)\rightleftharpoons2HI(g)$, starting with a mol $H_2$ and b mol $I_2$ in volume V, and x mol of each reacting to give 2x mol HI, the equilibrium concentrations are $\tfrac{a-x}{V}$, $\tfrac{b-x}{V}$, $\tfrac{2x}{V}$, giving $K_C=\dfrac{4x^2}{(a-x)(b-x)}$ (and, since $\Delta n_g=0$, $K_P=K_C$ directly).

For the dissociation $PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)$, starting with a mol $PCl_5$ and x mol dissociating, the equilibrium concentrations are $\tfrac{a-x}{V}$, $\tfrac xV$, $\tfrac xV$, giving $K_C=\dfrac{x^2}{(a-x)V}$; converting to $K_P$ (since $\Delta n_g=+1$) via total moles $n=a+x$ at equilibrium gives $K_P=\dfrac{x^2P}{a^2-x^2}$.

For ammonia synthesis $N_2(g)+3H_2(g)\rightleftharpoons2NH_3(g)$, starting with a mol $N_2$ and b mol $H_2$, x mol $N_2$ reacting with 3x mol $H_2$ to form 2x mol $NH_3$, the equilibrium concentrations are $\tfrac{a-x}{V}$, $\tfrac{b-3x}{V}$, $\tfrac{2x}{V}$, giving $K_C=\dfrac{4x^2V^2}{(a-x)(b-3x)^3}$; since $\Delta n_g=-2$, $K_P=K_C(RT)^{-2}$. …