Calculating the Equilibrium Constant (HSC SSCE Chemistry): Revision Notes

Different Types of Chemical Reactions

Chemical reactions reach equilibrium in various systems, and the equilibrium constant takes different forms depending on the type of reaction involved. Understanding these different types helps us predict and control chemical processes in solutions, acids and bases, and gas mixtures.

Solubility equilibria

Introduction to precipitation and dissolution

When ionic compounds dissolve in water, they can establish dynamic equilibrium between the solid and its dissolved ions. This type of equilibrium is particularly important for understanding precipitation reactions and predicting whether a solid will form when solutions are mixed.

Consider what happens when you mix potassium iodide with lead nitrate solutions. A precipitation reaction occurs:

$2\text{KI}(\text{aq}) + \text{Pb}(\text{NO}_3)_2(\text{aq}) \rightarrow \text{PbI}_2(\text{s}) + 2\text{KNO}_3(\text{aq})$

In solution, aqueous substances exist as free-moving ions. We can rewrite this equation showing all ions:

$2\text{K}^+(\text{aq}) + 2\text{I}^-(\text{aq}) + \text{Pb}^{2+}(\text{aq}) + 2\text{NO}_3^-(\text{aq}) \rightarrow \text{PbI}_2(\text{s}) + 2\text{K}^+(\text{aq}) + 2\text{NO}_3^-(\text{aq})$

Notice that $\text{K}^+$ and $\text{NO}_3^-$ ions appear on both sides unchanged. These are called spectator ions because they don't participate in the actual chemical reaction. Removing spectator ions gives us the simplified ionic equation:

$2\text{I}^-(\text{aq}) + \text{Pb}^{2+}(\text{aq}) \rightarrow \text{PbI}_2(\text{s})$

The solubility product (Ksp)

Although lead iodide forms a precipitate, it still maintains dynamic equilibrium with a small amount of dissolved ions in the saturated solution. By convention, we write the solid on the left and its ions on the right, using equilibrium arrows:

$\text{PbI}_2(\text{s}) \rightleftharpoons 2\text{I}^-(\text{aq}) + \text{Pb}^{2+}(\text{aq})$

When we write the equilibrium expression for this dissolution process, we call the equilibrium constant the solubility product ($K_{sp}$). The expression is:

$K_{sp} = [\text{Pb}^{2+}] \times [\text{I}^-]^2$

Notice that this is a heterogeneous system (containing both solid and aqueous phases), so the solid is excluded from the equilibrium expression. Only the concentrations of dissolved ions appear in the expression.

Ionic product versus solubility product

The equilibrium constant can only be used when concentrations are measured at equilibrium. When we measure ion concentrations in a system that hasn't reached equilibrium yet, we calculate the ionic product instead of the solubility product.

Comparing the ionic product with $K_{sp}$ allows us to predict what will happen in the solution:

Acid-base dissociation equilibria

Equilibrium constants for acids

Acids behave differently in water. Strong acids completely ionise to form ions, while weak acids only partially ionise. Weak acids are those in which only some molecules break apart to form ions, while others remain as intact molecules.

Acetic acid ($\text{CH}_3\text{COOH}$) provides a good example. In water, some acetic acid molecules ionise to form hydrogen ions ($\text{H}^+$) and acetate ions ($\text{CH}_3\text{COO}^-$), while other molecules remain unchanged. Since ionisation only occurs in water, we correctly write the equation including water and the hydronium ion ($\text{H}_3\text{O}^+$):

$\text{CH}_3\text{COOH}(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightleftharpoons \text{H}_3\text{O}^+(\text{aq}) + \text{CH}_3\text{COO}^-(\text{aq})$

This is an equilibrium reaction, so we can write an equilibrium expression. The equilibrium constant for the ionisation of an acid is called the acid dissociation constant ($K_a$):

$K_a = K_{eq} \times [\text{H}_2\text{O}]$

In dilute aqueous solutions, the water concentration remains essentially constant, so we incorporate it into the constant. The expression for $K_a$ becomes:

$K_a = \frac{[\text{H}_3\text{O}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}$

For any weak acid, we can write a general equation:

$\text{HA}(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightleftharpoons \text{H}_3\text{O}^+(\text{aq}) + \text{A}^-(\text{aq})$

Therefore, the general expression for the acid dissociation constant is:

$K_a = \frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]}$

The magnitude of $K_a$ provides valuable information about the acid's strength. A larger $K_a$ value means the numerator (ion concentrations) is larger relative to the denominator (molecular acid concentration), indicating greater ionisation. In other words, the larger the $K_a$, the stronger the weak acid.

Dissociation constants for bases

A weak base is one where only some molecules react with water to produce ions. Ammonia ($\text{NH}_3$) is a common example of a weak base. The equation for ammonia reacting with water is:

$\text{NH}_3(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightleftharpoons \text{NH}_4^+(\text{aq}) + \text{OH}^-(\text{aq})$

Since this establishes equilibrium, we can write an equilibrium expression. Because water's concentration remains approximately constant in dilute solution, the equilibrium constant is denoted as $K_b$:

$K_b = K_{eq} \times [\text{H}_2\text{O}]$

The equilibrium expression for $K_b$ is:

$K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]}$

The general equation for a base reacting with water is:

$\text{B}(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightleftharpoons \text{BH}^+(\text{aq}) + \text{OH}^-(\text{aq})$

Therefore, the general expression for the base dissociation constant is:

$K_b = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]}$

Just as with acids, larger $K_b$ values indicate stronger bases (greater degree of ionisation).

Gaseous equilibria expressed in terms of pressure

Introduction to pressure-based equilibrium constants

All previous examples expressed equilibrium constants using concentration values. However, for gaseous systems, we can also express equilibrium constants in terms of pressure. The symbol for this pressure-based equilibrium constant is $K_p$.

Pressure can be measured using various units including atmospheres (atm), kilopascals (kPa), or millimetres of mercury (mmHg). When using $K_p$, you must use consistent pressure units for all gaseous species in the reaction, but you can choose which unit system to use.

Two essential concepts underpin pressure-based equilibrium calculations:

• Mole fraction
• Partial pressure

Mole fraction

When multiple gases mix in the same container, we determine the mole fraction for each species. This represents the proportion of the total moles that each gas contributes.

For example, imagine a container holding $1$ mole of chlorine gas ($\text{Cl}_2$) and $5$ moles of phosphorus trichloride ($\text{PCl}_3$). The total is $6$ moles of gas.

$\text{Mole fraction of Cl}_2 = \frac{\text{Number of moles of Cl}_2}{\text{Total number of moles of gas present}} = \frac{1}{6}$

$\text{Mole fraction of PCl}_3 = \frac{\text{Number of moles of PCl}_3}{\text{Total number of moles of gas present}} = \frac{5}{6}$

The general equation for mole fraction is:

$\text{Mole fraction of gas A} = \frac{\text{Number of moles of gas A}}{\text{Total number of moles of gas present}}$

Partial pressure

Gases exert pressure by colliding with container walls. When a mixture of gases occupies a container, the total pressure results from collisions of all gas particles. The partial pressure is the proportion of total pressure contributed by collisions of one particular gas.

Using our chlorine and phosphorus trichloride example with a total pressure of $120$ atm: the phosphorus trichloride molecules are five times more abundant, so they're five times more likely to collide with the walls.

$\text{Partial pressure of Cl}_2 = \text{Mole fraction of Cl}_2 \times \text{Total pressure}$ $= \frac{1}{6} \times 120 = 20 \text{ atm}$

$\text{Partial pressure of PCl}_3 = \text{Mole fraction of PCl}_3 \times \text{Total pressure}$ $= \frac{5}{6} \times 120 = 100 \text{ atm}$

The general equation for partial pressure is:

$\text{Partial pressure of gas A} = \text{Mole fraction of gas A} \times \text{Total pressure of system}$

Writing Kp expressions

To write the expression for equilibrium constant in terms of pressure, we substitute partial pressures for concentrations. For a generalised gaseous reaction:

$a\text{A}(\text{g}) + b\text{B}(\text{g}) \rightleftharpoons c\text{C}(\text{g}) + d\text{D}(\text{g})$

The expression for $K_p$ is:

$K_p = \frac{P_C^c \times P_D^d}{P_A^a \times P_B^b}$

The value of $K_p$ differs from $K_c$ (the concentration-based equilibrium constant):

$K_p \neq K_c$

Always specify clearly which equilibrium constant you're using. While these two constants are related through the ideal gas equation, that relationship is beyond this course's scope.