Question

What is the equilibrium constant expression for the reaction,
${{\mathbf{P}}_4}(\;{\text{s}}) + 5{{\text{O}}_2}(\;{\text{g}}) \rightleftharpoons {{\mathbf{P}}_4}{{\text{O}}_{10}}(\;{\text{s}})$
(A) ${{\text{K}}_{\text{c}}} = \dfrac{{\left[ {{{\text{P}}_4}{{\text{O}}_{10}}} \right]}}{{\left[ {{{\text{P}}_4}} \right]{{\left[ {{{\text{O}}_2}} \right]}^5}}}$
(B) ${K_c} = \dfrac{{\left[ {{P_4}{O_{10}}} \right]}}{{5\left[ {{P_4}} \right]\left[ {{O_2}} \right]}}$
(C) ${{\text{K}}_{\text{c}}} = {\left[ {{{\text{O}}_2}} \right]^5}$
(D) ${{\text{K}}_{\text{c}}} = \dfrac{1}{{{{\left[ {{{\text{O}}_2}} \right]}^5}}}$

Answer 1

Many chemical systems, as well as physiological processes like oxygen transport by haemoglobin in blood and acid–base balance in the human body, need an understanding of equilibrium constants. Equilibrium constants include stability constants, formation constants, binding constants, association constants, and dissociation constants.

Complete answer:
A chemical reaction's equilibrium constant is the value of its reaction quotient at chemical equilibrium, a condition reached by a dynamic chemical system after a period of time has passed in which its composition shows no discernible tendency to change. The equilibrium constant is independent of the initial analytical concentrations of the reactant and product species in the mixture for a particular set of reaction circumstances. As a result, established equilibrium constant values may be used to calculate the composition of a system at equilibrium, given its beginning composition.
The generic chemical equation describes a system that is experiencing a reversible reaction.
$\alpha {\mkern 1mu} {\text{A}} + \beta {\mkern 1mu} {\text{B}} + \cdots \rightleftharpoons \rho {\mkern 1mu} {\text{R}} + \sigma {\mkern 1mu} {\text{S}} + \cdots$
When forward and reverse reactions occur at the same rate, a thermodynamic equilibrium constant, represented by ${K^ \ominus }$ , is defined as the value of the reaction quotient ${Q_t}\;$ . The chemical composition of the mixture does not vary with time in chemical equilibrium, and the Gibbs free energy change $\Delta G$ for the reaction is zero. Given enough time, if the composition of an equilibrium mixture is altered by adding a reagent, a new equilibrium position will be achieved. The composition of the mixture at equilibrium is linked to an equilibrium constant by
${K^ \ominus } = \dfrac{{{{\\{ {\text{R}}\\} }^\rho }{{\\{ {\text{S}}\\} }^\sigma }...}}{{{{\\{ {\text{A}}\\} }^\alpha }{{\\{ {\text{B}}\\} }^\beta }...}} = \dfrac{{{{[{\text{R}}]}^\rho }{{[{\text{S}}]}^\sigma }...}}{{{{[{\text{A}}]}^\alpha }{{[{\text{B}}]}^\beta }...}} \times \Gamma ,$
For ${{\mathbf{P}}_4}(\;{\text{s}}) + 5{{\text{O}}_2}(\;{\text{g}}) \rightleftharpoons {{\mathbf{P}}_4}{{\text{O}}_{10}}(\;{\text{s}})$
The reaction's equilibrium constant expression is given by ${{\text{K}}_{\text{c}}} = \dfrac{1}{{{{\left[ {{{\text{O}}_2}} \right]}^5}}}$
Solids (and pure liquids) have a concentration of one and do not appear in the equilibrium constant formula.
Hence option d is correct.

Note:
Chemical equilibrium is the condition of a chemical process in which both the reactants and products are present in concentrations that have no further tendency to vary with time, resulting in no visible change in the system's characteristics. When the forward reaction and the reverse reaction both proceed at the same pace, this condition occurs. The forward and backward response rates are usually not zero, although they are roughly equal. As a result, there are no net changes in the reactant and product concentrations. Dynamic equilibrium is the term for such a situation.