Questions: Choose the correct equilibrium constant expression for the following reaction: I2(aq) + I^-(aq) <=> I3^-(aq) - [I3^-] / [I^-]^2[I2] - [I^-]^2[I2] - 1 / [I3^-] - [I^-]^2[I2] / [I3^-]

Solution Steps

The given reaction is:

\[ \mathrm{I}_{2}(a q) + \mathrm{I}^{-}(a q) \leftrightharpoons \mathrm{I}_{3}^{-}(a q) \]

This is a reversible reaction involving the formation of the triiodide ion (\(\mathrm{I}_{3}^{-}\)) from iodine (\(\mathrm{I}_{2}\)) and iodide ions (\(\mathrm{I}^{-}\)).

For a general reaction of the form:

\[ aA + bB \leftrightharpoons cC + dD \]

The equilibrium constant expression (\(K\)) is given by:

\[ K = \frac{[C]^c [D]^d}{[A]^a [B]^b} \]

For the reaction:

\[ \mathrm{I}_{2}(a q) + \mathrm{I}^{-}(a q) \leftrightharpoons \mathrm{I}_{3}^{-}(a q) \]

The equilibrium constant expression is:

\[ K = \frac{[\mathrm{I}_{3}^{-}]}{[\mathrm{I}^{-}] [\mathrm{I}_{2}]} \]

Compare the derived expression with the given options:

1. \(\frac{\left[I_{3}^{-}\right]}{\left[I^{-}\right]^{2}\left[I_{2}\right]}\)
2. \({\left[I^{-}\right]^{2}\left[I_{2}\right]}\)
3. \(\frac{1}{\left[I_{3}{ }^{-}\right]}\)
4. \(\frac{\left[I^{-}\right]^{2}\left[I_{2}\right]}{\left[I_{3}{ }^{-}\right]}\)

The correct expression is:

\[ \frac{[\mathrm{I}_{3}^{-}]}{[\mathrm{I}^{-}] [\mathrm{I}_{2}]} \]

This matches the first option, except for the power of \([\mathrm{I}^{-}]\). The correct expression should not have \([\mathrm{I}^{-}]\) squared, so none of the options exactly match the derived expression. However, the closest match is the first option if we assume a typo in the power of \([\mathrm{I}^{-}]\).

Final Answer