For the equilibrium of the reaction, HgO(s) $\stackrel{}{⇌}$ Hg(g) + $\frac{1}{2}$O2(g), kP for the reaction at total pressure of P is:

(1) ${\mathrm{K}}_{\mathrm{P}}\mathrm{=}\frac{\mathrm{2}}{{\mathrm{3}}^{3/2}}{\mathrm{p}}^{3/2}$

(2) ${\mathrm{K}}_{\mathrm{P}}\mathrm{=}\frac{\mathrm{2}}{{\mathrm{3}}^{1/2}}{\mathrm{p}}^{1/2}$

(3) ${\mathrm{K}}_{\mathrm{P}}\mathrm{=}\frac{1}{{\mathrm{3}}^{2/3}}{\mathrm{p}}^{3/2}$

(4) ${\mathrm{K}}_{\mathrm{P}}\mathrm{=}\frac{1}{{\mathrm{3}}^{2/3}}\mathrm{p}$

Concept Videos :-

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Concept Questions :-

KP/KC - Factors

(1) HgO(s) $\stackrel{}{⇌}$ Hg (g) + $\frac{1}{2}$O2 (g)

Kp = PHg(g) × (PO2) $\frac{1}{2}$

Total moles at equilibrium = $\frac{3\mathrm{x}}{\mathrm{2}}$

PHg = $\frac{\mathrm{x}}{3\mathrm{x} / 2}\mathrm{ }= \mathrm{P} =\mathrm{ }\frac{\mathrm{2}}{\mathrm{3}}\mathrm{P}$

PO2 = $\frac{\mathrm{x}/2}{3\mathrm{x}/2}\mathrm{ }\mathrm{P}\mathrm{ }=\mathrm{ }\frac{1}{3}\mathrm{P}$

Kp = $\frac{2}{3}\mathrm{P}{\left(\frac{1}{3}\mathrm{P}\right)}^{1/2}=\frac{\mathrm{2}}{{\mathrm{3}}^{3/2}} {\mathrm{P}}^{3/2}$

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