Question

Write the expression for the equilibrium constant, `K_(c)` for each of the following reactions :
` (i) 2 NOCl (g) hArr 2 NO (g) + Cl_(2)`
(ii) ` 2 Cu ( NO_(3)) _(2) (s) hArr 2 CuO (s) + 4 NO_(2) (g) + O _ (2) (g) `
(iii) ` CH_(3) COOHC_(2) H_(5) (aq) + H_(2)O (l) hArr CH_93) COOH (aq) + C_(2) H_(5) OH(aq) `
(iv) ` Fe^(3+) (aq) + 3 OH ^(-) (aq) hArr Fe (OH)_(3) (s)`
(v) ` I_(2) (s) + 5 _(2) hArr 2 IF_(5)`

Answer 1

`K_(c) = ([NO(g)]^(2) [Cl_(2) (g) ])/([NOCl(g) ]^(2))`
` K_(c) = ([CuO(s)]^(2) [NO_(2) (g)]^(4) [O_(2)(g)])/([Cu (NO_(3))_(2) (s)]^(2))= [NO_(2)(g)]^(4) [O_(2)(g)]`
`K_(c) = ([CH_(3) COOH (aq)] [C_(2)H_(5)OH(aq)])/([CH_(3) COOC_(2) H_(5) (aq) ] [H_(2)O(l) ])`
` K_(c) = [Fe(OH)_(3)(s)]/([Fe^(3+) (aq) ][OH^(-)(aq)]^(3) )= 1/([Fe^(3) (aq) ][OH^(-)(aq)]^(3)) " "(v)K_(c) = ([IF_(5)]^(2))/[I_(2) (s) [F_(2)]^(5))= ([IF_(5)]^(2))/([F_(2)]^(5))`