CAcT HomePage Chemical Equilibrium: Previous page: Mass action law

# Special Chemical Equilibria

### Skills to develop

• Write the equilibrium constant expression when the solvent is one of the products or reactants.

• Write the equilibrium constant expression for heterogeneous equilibria.

# Special Chemical Equilibria

We treat chemical equilibria in which the solvents is a reactant or product, and heterogeneous equilibria as special chemical equilibria. The expressions for equilibrium contants of these systems are discussed.

The mass action law is valid for the case when the solvent is a reactant or product. However, due to the large amount of solvent present, the equilibrium constant expression can be simplified.

Special consideration also apply to heterogeneous equilibria, in which, solids or liquids are involved. Phases such as liquid and solid are not sensitive to pressure, and their "concentration" are constant as long as these phases are present. Their "concentrations" are not defined, and that is why we use a " " to mark the special meaning of these phases. Perhaps "activity" or "property" is a better term to use than "concentration" in these cases.

## Equilibrium Constant for Reactions Involving Solvents

As a solvent in the system, the concentration of the solvent is so large that its concentration rarely change during the course of the reaction.

For example, when water is used as a solvent, the concentration of water is almost always a constant. Its concentration can be calculated to be 55.6 M. (1 L of water has 1000 g, and molecular weight 18 g/mol; and thus [H2O] = 1000/18 = 55.6 mol/L)

For example, consider the following reaction,

CH3COOH + C2H5OH = CH3COOC2H5 + H2O According to the Mass action law, we can use either
```       [CH3COOC2H5] [H2O]
K' =  -------------------
[CH3COOH] [C2H5OH]
or
[CH3COOC2H5]
K  =  -------------------
[CH3COOH] [C2H5OH]
```
Of course, K = K' / [H2O] = K' /55.6

## Heterogeneous Equilibria

Heterogeneous reactions involve at least two phases. That is two of gas, liquid, and solid are present as reactants or products.

In heterogeneous equilibria, the activities (or concentrations) of solid and liquid but not gas are always a constant. In these cases, their activities or concentrations are omitted in the expression of Q or K.

Let us look at these expressions for the equilibria:

CaCO3(s) = CaO(s) + CO2(g),       Kp = P(CO2)
2 H2O(l) = 2 H2O(g),       Kp = P2(H2O)
2 H2O(l) = 2 H2(g) + O2(g),       Kp = P2(H2) P(O2)
AgCl(s) = Ag+(aq) + Cl-(aq),       K = [Ag+] [[Cl-]
Heterogeneous equilibria discusses this type of system in more details.

### Confidence Building Questions

• What is the concentration of water in pure water?

Hint 55.6
Consider...
For most solutions, [H2O] = 1000 / 18 = ?

• Give the Kp expression for the reaction HgO(s) = Hg(s) + O2(g)

Hint P(O2)
Consider...
This reaction is used to produce O2 from HgO.

• For the phase transition, H2O(l) = H2O(g), the vapor pressure of water depends on temperature. At 298 K, the saturated vapor pressure is 23.8 torr. What is the equilibrium constant based on pressure Kp in atm? (1 atm = 760 torr)

Hintr 0.031 atm
Consider...
Kp = P(H2O) in this case. Just want you to consider one of the common equilibrium state. The amount of water present does not affect the vapor pressure.

• At some low temperature, the reaction of ethyl acetate with water, CH3COOC2H5(aq) + H2O(l) = CH3COOH(aq) + C2H5OH(aq) has an equilibrium constant of 0.10. If [CH3COOC2H5] = 0.90 M in a system that is at equilibrium, what is [C2H5OH]?

Consider...

```CH3COOC2H5(aq) + H2O(l) = CH3COOH(aq) + C2H5OH(aq)
0.9                 x             x
```
x2 / 0.90 = K = 0.1;       x = ?
What was the original [CH3COOC2H5] before any hydration reaction takes place? (Answer, 1.2)

• The value of Kp is 81 Pa at some temperature for the reaction, NH4Cl(s) = NH3(g) + HCl(g). What is the partial pressure (in Pa) of HCl in an enclosed system containing solid NH4Cl and its vapor?

```   NH4Cl(s) = NH3(g) + HCl(g).