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Answer 1 · 2014-09-14T13:26:15

These are equilibrium constants for various equilibrium reactions.

If you have an equilibrium of the type

aA + bB ⇌ cC + dD

you can write an equilibrium constant expression:

$K_{eq} = \frac{{[C]}^{c} {[D]}^{d}}{{[A]}^{a} {[B]}^{b}}$ $K_"eq" = (["C"]^c["D"]^d)/(["A"]^a["B"]^b)$ , where [ ] indicates concentration

You do NOT include the concentrations of solids or liquids in the $K_{eq}$ $K_"eq"$ expression, because their concentrations do not change.

For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

$K_{eq} = \frac{{[{NH}_{3}]}_{3}^{2}}{[N_{2}] {[H_{2}]}_{2}^{3}}$ $K_"eq" = ["NH"_3]^2/(["N"_2]["H"_2]^3)$

For the reaction, C(s) + H₂O(g) ⇌ CO(g) + H₂(g)

$K_{eq} = \frac{[CO] [H_{2}]}{[H_{2} O]}$ $K_"eq" = (["CO"]["H"_2])/(["H"_2"O"])$

If one of the reactants is an acid, the equation for the equilibrium is

HA + H₂O ⇌ H₃O⁺ + A⁻

The equilibrium constant expression is

$K_{eq} = \frac{[H_{3} O^{+}] [A ⁻]}{[HA]}$ $K_"eq" = (["H"_3"O"^+]["A"⁻])/"[HA]"$

To show explicitly that this equilibrium involves an acid, we call the equilibrium constant $K_{a}$ $K_a$ .

So $K_{a}$ $K_a$ is the equilibrium constant for the dissociation of an acid.

$K_{a}$ $K_a$ is usually called the acid dissociation constant.

For the dissociation of HF in water, we write

HF + H₂O ⇌ H₃O⁺ + A⁻

and

$K_{a} = \frac{[H_{3} O^{+}] [F ⁻]}{[HF]}$ $K_a = (["H"_3"O"^+]["F"⁻])/"[HF]"$