Константа рівноваги для реакцій чистих твердих тіл та рідин

We've been learning about equilibrium. We've seen that some chemical reactions are reversible. The enthalpy and entropy balance. So, the free energy change equals zero. We've also studied equilibrium constants and calculated them from concentrations and pressures. But, we've also been saying that you don't include pure liquids and solids. Why?! Let's investigate.

What are Equilibrium Constants?

What are equilibrium constants? A reaction is at equilibrium when the forward and reverse reactions have equal rates. Remember, the reaction rate is the change in concentration over time. Let's write a generic elementary reaction. Its balanced equation is $aA+bB\rightleftharpoons cC+dD$ . The rate laws are $k_{f}[A]^{a}[B]^{b}=k_{r}[C]^{c}[D]^{d}$ . We defined the equilibrium constant as $K_{eq}=\frac{k_{f}}{k_{r}}=\frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}$ .

So, why do we not use solids and liquids in equilibrium expressions? Their concentrations don't change! If their concentrations don't change, they don't affect the rates.

Why Liquid Concentrations Don't Change

Imagine a beaker with 50.0 milliliters of water. Do you remember its molarity? It's 55.5 $\frac{mol}{L}$ . What's the molarity if we had 25.0 milliliters? It's still 55.5 $\frac{mol}{L}$ . So, let's say we had a chemical reaction with water as a reactant. When the reaction reaches equilibrium, we'll still have water. Its volume will be less. But, its molar concentration is approximately the same.

Why is its concentration "approximately the same?" Because molar concentrations are temperature-dependent. What if we have a very exothermic or endothermic reaction? Our temperature would change. And technically, so would the water's concentration. Thankfully, in the lab, we can assume the concentrations are constant. That isn't true for chemical engineers in industrial production. They work at very high temperatures and pressures.

Why Solid Concentrations Don't Change

Now let's think about concentrations of solids. Let's say we have a cubic chunk of gold. The volume is 1.0 liter or 1000.0 $cm^{3}$ . Gold's density is $19.3 \frac{g}{cm^{3}}$ . Let's do some quick dimensional analysis.

$19.3 \frac{g}{cm^{3}}\times1000.0cm^{3}\times\frac{1.0mol}{196.97g}=97.98mols$ .
We can write its molarity as $97.98\frac{mols}{L}$ .

What if we had a chunk of gold that was exactly half the volume. We would also have half the number of moles. So, our molarity is constant.

So, let's say we had an equilibrium reaction using solid gold. Some of the gold reacts. But, some gold is still in its solid elemental state. The concentration of the gold did not change. So, we don't need to include it in the equilibrium constant calculation.

Example Problems

Write the equilibrium constant expression for each reaction.

$PbCl_{2(s)} \rightleftharpoons Pb_{(aq)}^{2+} + 2Cl_{(aq)}^{-}$

$K_{eq}=\frac{[Pb^{2+}][Cl^{-}]^{2}}{\cancel{[PbCl_{2(s)}]}}$

$H_{2}O_{(l)} \rightleftharpoons H_{(aq)}^{+} + OH_{(aq)}^{-}$

$K_{a}=\frac{[H^{+}][OH^{-}]}{\cancel{[H_{2}O_{(l)}]}}$

This reaction is called the self-ionization of water.

Weak acids and bases that are in equilibrium have an acid dissociation constant $K_{a}$ . It's the equivalent of the equilibrium constant.

$CaCO_{3(s)}\rightleftharpoons CaO_{(s)} + CO_{2(g)}$

$K_{p}=\frac{[CO_{2(g)}]\cancel{[CaO_{s}]}}{\cancel{[CaCO_{3(s)}]}}=[CO_{2(g)}]$

This reaction has a solid on both the reactant and product sides. We canceled both because they are solids.

For this reaction, the equilibrium constant equals the partial pressure of carbon dioxide gas.