Write the expression for the equilibrium constant, `K_c` for each of the following reactions
(1)`2NOCl_((g)) hArr 2NO_((g)) + Cl_(2(g))`
(2) `2Cu(NO_3)_(2(s)) hArr 2CuO_((s)) + 4NO_(2(g)) + O_(2(g))`
(3)`CH_3COOC_2H_(5(aq)) + H_2O_((l)) hArr CH_3COOH_((aq)) + C_2H_5OH_((aq))`
(4)` Fe_((aq))^(3+) + 3OH_((aq))^(-) hArr Fe(OH)_(3(s))`
(5)`I_(2(s)) + 5F_(2(s)) hArr 2IF_(5(g))`

Explanation

(1)`K_c=([NO_((g))]^2[Cl_(2(g))])/([NOCl_((g))]^2)`
(2)`K_c=([CuO_((s))]^2[NO_(2(g))]^4 [O_(2(g))])/([Cu(NO_3)_(2(s))]^2)`
In this equilibrium CuO and `Cu(NO_3)_2` are solid. So, at equilibrium its concentration are not changed and constants. So, ignore in expression, `K_c=[NO_(2(g))]^4 [O_(2(g))]`
(3)`K_c=([CH_3COOH_((aq))][C_2H_5OH_((aq))])/([CH_3COOC_2H_(5_(aq))][H_2O_((l))])`
Here , `H_2O` is liquid its concentration is constant during reaction, so ignore it.
`K_c=([CH_3COOH_((aq))][C_2H_5OH_((aq))])/([CH_3COOC_2H_(5(aq))])`
(4)`K_c=([Fe(OH)_(3(s))])/([Fe_((aq))^(3+) ][OH_((aq))^(-)]^3)`
In this equilibrium `Fe(OH)_3` is solid and amount of solid i.e. its concentration is equal to density. Density is constant at temperature, so its not affect the equilibrium constant. `(Fe(OH)_(3(s)))` is ignore
`therefore K_c=1/([Fe^(3+)][OH^-]^3)`
(5)`K_c` for the product `IF_5` of the reaction of `I_(2(s))` and `F_(2(g))` is ,
`K_c=([IF_(5(g))]^2)/([I_(2(s))] [F_(2(g))]^5)`
`I_(2(s))` is solid so its concentration remain constant so it is ignore in expression.
`therefore K_c=([IF_(5(g))]^2)/([F_(2(g))]^5)`