Medium
Write the expression for the equilibrium constant,$K_{c}$ for each of the following reactions:
$(i)$ $2 NOCl_{(g)} \longleftrightarrow 2 NO_{(g)} + Cl_{2(g)}$
$(ii)$ $2 Cu(NO_{3})_{2(s)} \longleftrightarrow 2 CuO_{(s)} + 4 NO_{2(g)} + O_{2(g)}$
$(iii)$ $CH_{3}COOC_{2}H_{5(aq)} + H_{2}O_{(l)} \longleftrightarrow CH_{3}COOH_{(aq)} + C_{2}H_{5}OH_{(aq)}$
$(iv)$ $Fe^{3+}_{(aq)} + 3 OH^{-}_{(aq)} \longleftrightarrow Fe(OH)_{3(s)}$
$(v)$ $I_{2(s)} + 5 F_{2(g)} \longleftrightarrow 2 IF_{5(g)}$
$(i)$ $2 NOCl_{(g)} \longleftrightarrow 2 NO_{(g)} + Cl_{2(g)}$
$(ii)$ $2 Cu(NO_{3})_{2(s)} \longleftrightarrow 2 CuO_{(s)} + 4 NO_{2(g)} + O_{2(g)}$
$(iii)$ $CH_{3}COOC_{2}H_{5(aq)} + H_{2}O_{(l)} \longleftrightarrow CH_{3}COOH_{(aq)} + C_{2}H_{5}OH_{(aq)}$
$(iv)$ $Fe^{3+}_{(aq)} + 3 OH^{-}_{(aq)} \longleftrightarrow Fe(OH)_{3(s)}$
$(v)$ $I_{2(s)} + 5 F_{2(g)} \longleftrightarrow 2 IF_{5(g)}$
(N/A) The equilibrium constant $K_{c}$ is defined as the ratio of the product of molar concentrations of products to that of reactants,each raised to the power of their stoichiometric coefficients. Pure solids and liquids are taken as unity $(1)$.
$(i)$ $K_{c} = \frac{[NO]^{2} [Cl_{2}]}{[NOCl]^{2}}$
$(ii)$ Since $Cu(NO_{3})_{2(s)}$ and $CuO_{(s)}$ are solids,their concentrations are taken as $1$.
$K_{c} = [NO_{2}]^{4} [O_{2}]$
$(iii)$ Since $H_{2}O_{(l)}$ is a pure liquid,its concentration is taken as $1$.
$K_{c} = \frac{[CH_{3}COOH] [C_{2}H_{5}OH]}{[CH_{3}COOC_{2}H_{5}]}$
$(iv)$ Since $Fe(OH)_{3(s)}$ is a solid,its concentration is taken as $1$.
$K_{c} = \frac{1}{[Fe^{3+}] [OH^{-}]^{3}}$
$(v)$ Since $I_{2(s)}$ is a solid,its concentration is taken as $1$.
$K_{c} = \frac{[IF_{5}]^{2}}{[F_{2}]^{5}}$
$(i)$ $K_{c} = \frac{[NO]^{2} [Cl_{2}]}{[NOCl]^{2}}$
$(ii)$ Since $Cu(NO_{3})_{2(s)}$ and $CuO_{(s)}$ are solids,their concentrations are taken as $1$.
$K_{c} = [NO_{2}]^{4} [O_{2}]$
$(iii)$ Since $H_{2}O_{(l)}$ is a pure liquid,its concentration is taken as $1$.
$K_{c} = \frac{[CH_{3}COOH] [C_{2}H_{5}OH]}{[CH_{3}COOC_{2}H_{5}]}$
$(iv)$ Since $Fe(OH)_{3(s)}$ is a solid,its concentration is taken as $1$.
$K_{c} = \frac{1}{[Fe^{3+}] [OH^{-}]^{3}}$
$(v)$ Since $I_{2(s)}$ is a solid,its concentration is taken as $1$.
$K_{c} = \frac{[IF_{5}]^{2}}{[F_{2}]^{5}}$
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