Q. What is the equilibrium expression for the reaction $P_{4}(s) + 5O_{2}(g) \rightleftharpoons P_{4}O_{10}(s)$ ?
A) $K_{c} = \frac{\left\lbrack P_{4}O_{10} \right\rbrack}{\left\lbrack P_{4} \right\rbrack\left\lbrack O_{5} \right\rbrack^{5}}$
B) $K_{c} = \frac{\lbrack P_{4}O_{10}\rbrack}{5\left\lbrack P_{4} \right\rbrack\lbrack O_{2}\rbrack}$
C) $K_{c} = \left\lbrack O_{2} \right\rbrack^{5}$
D) $K_{c} = \frac{1}{\left\lbrack O_{2} \right\rbrack^{5}}$
Explanation
(d)
In the expression for equilibrium constant $(K_{p}\ or\ K_{c})$ species in solid state are not written ($i.e$., their molar concentrations are taken as 1)
$P_{4}(s) + 5O_{2}(g) \rightleftharpoons P_{4}O_{10}(s)$
Thus, $K_{c} = \frac{1}{\left\lbrack O_{2} \right\rbrack^{5}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.