Consider this equilibrium: I2(s) + H2O(l) ⇌ H+(aq) + I-(aq) + HOI(aq). What is the Keq expression for this equilibrium?

Answer 1

The expression is #K_(eq)# = [H⁺][I⁻][HOI]

We often consider a generic reaction of the form

aA +bB ⇌ cC +dD

The equilibrium constant expression for this reaction is

#K_(eq) = ([C]^c[D]^d)/([A]^a[B]^b)#

Notice that it is always "products" over "reactants". Each component A, B, C, and D is raised to the power of its coefficient a, b, c, and d in the balanced equation.

The concentration of a solid or of a liquid does not change during a reaction. We omit it from the equilibrium constant expression.

Consider the equation

I₂(s) + H₂O(l) ⇌ H⁺(aq) + I⁻(aq) + HOI(aq)

I₂ is a solid, and H₂O is a liquid. Their concentrations do not change during the reaction. We omit them from the #K_(eq)# expression.

All the products are in solution. They can change their concentrations. We keep them in the #K_(eq)# expression.

Thus, the #K_(eq)# expression is

#K_(eq)# = [H⁺][I⁻][HOI].

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Answer 2

[ K_{eq} = \frac{[H^+][I^-][HOI]}{[I_2][H_2O]} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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