Medium
At a definite temperature,the equilibrium constant $K_{c}$ is given by the following equation: $K_{c} = \frac{[I_{2}][H_{5}IO_{6}]^{5}}{[IO_{3}^{-}]^{7}[H_{2}O]^{9}[H^{+}]^{7}}$. Write the balanced chemical equilibrium equation.
(N/A) The equilibrium constant expression is defined as the ratio of the product of concentrations of products to the product of concentrations of reactants,each raised to the power of their stoichiometric coefficients.
From the given expression $K_{c} = \frac{[I_{2}][H_{5}IO_{6}]^{5}}{[IO_{3}^{-}]^{7}[H_{2}O]^{9}[H^{+}]^{7}}$,the products are $I_{2}$ and $H_{5}IO_{6}$ with coefficients $1$ and $5$ respectively.
The reactants are $IO_{3}^{-}$,$H_{2}O$,and $H^{+}$ with coefficients $7$,$9$,and $7$ respectively.
Thus,the balanced chemical equilibrium equation is:
$7IO_{3(aq)}^{-} + 9H_{2}O_{(l)} + 7H_{(aq)}^{+} \rightleftharpoons I_{2(aq)} + 5H_{5}IO_{6(aq)}$
From the given expression $K_{c} = \frac{[I_{2}][H_{5}IO_{6}]^{5}}{[IO_{3}^{-}]^{7}[H_{2}O]^{9}[H^{+}]^{7}}$,the products are $I_{2}$ and $H_{5}IO_{6}$ with coefficients $1$ and $5$ respectively.
The reactants are $IO_{3}^{-}$,$H_{2}O$,and $H^{+}$ with coefficients $7$,$9$,and $7$ respectively.
Thus,the balanced chemical equilibrium equation is:
$7IO_{3(aq)}^{-} + 9H_{2}O_{(l)} + 7H_{(aq)}^{+} \rightleftharpoons I_{2(aq)} + 5H_{5}IO_{6(aq)}$
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View SolutionThe rate of forward reaction is two times that of reverse reaction at a given temperature and identical concentration. $K_{equilibrium}$ is
Medium
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