Table of Contents
In this chapter, we assume you already know what a chemical equilibrium is and how reversible reactions behave in general. We now focus on how to describe equilibria quantitatively using the law of mass action and the related equilibrium constant.
1. The Law of Mass Action: Basic Idea
The law of mass action gives a mathematical relationship between the concentrations (or, more generally, activities) of reactants and products at equilibrium.
Consider a general homogeneous reaction in solution or gas phase:
$$
a\,\mathrm{A} + b\,\mathrm{B} \rightleftharpoons c\,\mathrm{C} + d\,\mathrm{D}
$$
Here, $a, b, c, d$ are the stoichiometric coefficients from the balanced chemical equation.
Law of mass action:
At equilibrium and at a given temperature, the ratio
$$
\frac{[\mathrm{C}]^{c}[\mathrm{D}]^{d}}{[\mathrm{A}]^{a}[\mathrm{B}]^{b}}
$$
has a constant value, called the equilibrium constant $K$ (more precisely $K_c$ when expressed using concentrations):
$$
K_c = \frac{[\mathrm{C}]^{c}[\mathrm{D}]^{d}}{[\mathrm{A}]^{a}[\mathrm{B}]^{b}}
$$
Here, $[\mathrm{X}]$ denotes the equilibrium concentration of species X, usually in $\text{mol·L}^{-1}$.
Key points:
- Only equilibrium concentrations are used.
- Each concentration is raised to the power of its stoichiometric coefficient.
- At a fixed temperature, $K$ is constant for that reaction.
2. Writing the Equilibrium Constant Expression
2.1 From the balanced equation
The steps are:
- Write the balanced chemical equation.
- Put the product species in the numerator, each to the power of its coefficient.
- Put the reactant species in the denominator, each to the power of its coefficient.
Example:
$$
\mathrm{N_2(g)} + 3\,\mathrm{H_2(g)} \rightleftharpoons 2\,\mathrm{NH_3(g)}
$$
Equilibrium constant (in terms of concentrations):
$$
K_c = \frac{[\mathrm{NH_3}]^{2}}{[\mathrm{N_2}]\,[\mathrm{H_2}]^{3}}
$$
Another example in solution:
$$
\mathrm{Fe^{3+}(aq)} + \mathrm{SCN^{-}(aq)} \rightleftharpoons \mathrm{[FeSCN]^{2+}(aq)}
$$
Equilibrium constant:
$$
K_c = \frac{[\mathrm{FeSCN}^{2+}]}{[\mathrm{Fe^{3+}}][\mathrm{SCN^{-}}]}
$$
2.2 Omission of pure solids and pure liquids
For equilibria involving solids and pure liquids, the activities of pure phases are effectively constant and are set equal to 1. As a result, pure solids and pure liquids do not appear in the equilibrium constant expression.
Example:
$$
\mathrm{CaCO_3(s)} \rightleftharpoons \mathrm{CaO(s)} + \mathrm{CO_2(g)}
$$
Pure solids: $\mathrm{CaCO_3(s)}$ and $\mathrm{CaO(s)}$.
Equilibrium constant (in terms of partial pressure, see below):
$$
K_p = p(\mathrm{CO_2})
$$
The solids are omitted; only the gas appears in $K$.
Similarly, for an acid dissociation in water (where water is the pure solvent):
$$
\mathrm{HA(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{H_3O^+(aq)} + \mathrm{A^-(aq)}
$$
The solvent water is a pure liquid (activity ≈ 1), so it does not appear:
$$
K_c = \frac{[\mathrm{H_3O^+}][\mathrm{A^-}]}{[\mathrm{HA}]}
$$
(The special treatment of water and acid/base equilibria is handled in separate chapters; here we only illustrate the omission principle.)
3. Different Forms of the Equilibrium Constant: $K_c$ and $K_p$
The equilibrium constant can be defined in different ways depending on the system:
- $K_c$ using concentrations in solution or gas phase
- $K_p$ using partial pressures for gas equilibria
3.1 $K_c$ (based on concentration)
Use $K_c$ when the species are:
- Dissolved in solution, or
- Gases expressed via molar concentrations.
General form:
$$
K_c = \frac{\prod [\text{products}]^{\nu_{\text{prod}}}}{\prod [\text{reactants}]^{\nu_{\text{reac}}}}
$$
where $\nu$ are the stoichiometric coefficients.
Example (gas phase, but expressed as concentrations):
$$
\mathrm{H_2(g)} + \mathrm{I_2(g)} \rightleftharpoons 2\,\mathrm{HI(g)}
$$
$$
K_c = \frac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]}
$$
3.2 $K_p$ (based on partial pressure)
For gas-phase equilibria, it's often convenient to use partial pressures instead of concentrations. The resulting equilibrium constant is denoted $K_p$.
For a general gas-phase reaction:
$$
a\,\mathrm{A(g)} + b\,\mathrm{B(g)} \rightleftharpoons c\,\mathrm{C(g)} + d\,\mathrm{D(g)}
$$
The equilibrium constant in terms of partial pressures is:
$$
K_p = \frac{p(\mathrm{C})^{c} p(\mathrm{D})^{d}}{p(\mathrm{A})^{a} p(\mathrm{B})^{b}}
$$
where $p(\mathrm{X})$ is the partial pressure of gas X (typically in bar or atm).
Example:
$$
\mathrm{N_2O_4(g)} \rightleftharpoons 2\,\mathrm{NO_2(g)}
$$
$$
K_p = \frac{[p(\mathrm{NO_2})]^2}{p(\mathrm{N_2O_4})}
$$
3.3 Relationship between $K_p$ and $K_c$ (for gases)
For gaseous reactions, $K_p$ and $K_c$ are related through the ideal gas law. Consider:
$$
a\,\mathrm{A(g)} + b\,\mathrm{B(g)} \rightleftharpoons c\,\mathrm{C(g)} + d\,\mathrm{D(g)}
$$
Define:
$$
\Delta n_\text{gas} = (c + d) - (a + b)
$$
i.e., the change in moles of gas (moles of gaseous products minus moles of gaseous reactants, using stoichiometric coefficients).
The relation is:
$$
K_p = K_c (R T)^{\Delta n_\text{gas}}
$$
where:
- $R$ is the gas constant,
- $T$ the absolute temperature in kelvin.
Special case:
- If $\Delta n_\text{gas} = 0$ (same total moles of gas on both sides), then $K_p = K_c$.
Derivation uses $p = cRT$ for each gas (ideal gas behavior), but the details are typically discussed in physical chemistry chapters.
4. Dimensionless Nature of the Equilibrium Constant (Conceptual)
Strictly speaking, the equilibrium constant is defined in terms of activities, which are dimensionless. Commonly, however, we approximate activities by:
- Concentrations in $\text{mol·L}^{-1}$ (for solutions)
- Partial pressures in bar or atm (for gases)
To make $K$ dimensionless, each concentration or pressure should be divided by a standard reference value (e.g., $c^\circ = 1\,\text{mol·L}^{-1}$, $p^\circ = 1\,\text{bar}$). In introductory work, these reference values are usually omitted in notation, and $K$ is treated symbolically as “unitless.”
For practical purposes in this course:
- Write $K_c$ and $K_p$ with the usual concentration or pressure terms.
- Be aware that in more advanced treatments, activities and standard states are used instead.
5. Interpreting the Magnitude of the Equilibrium Constant
The numerical size of $K$ tells you how the equilibrium composition is biased between reactants and products at a given temperature.
Let a generic reaction be written as:
$$
\text{reactants} \rightleftharpoons \text{products}
$$
Then:
- $K \gg 1$:
- Products are strongly favored at equilibrium.
- Equilibrium mixture contains mostly products; the reaction is said to “lie to the right.”
- $K \approx 1$:
- Neither side is strongly favored.
- Significant amounts of both reactants and products are present.
- $K \ll 1$:
- Reactants are strongly favored at equilibrium.
- Only a small fraction of reactants is converted; the reaction is said to “lie to the left.”
Examples (qualitative):
- $K = 10^{15}$: Reaction is essentially “complete” in the forward direction under typical conditions.
- $K = 10^{-10}$: Reaction barely proceeds to products; reverse reaction is strongly favored.
- $K = 0.5$: Comparable amounts of reactants and products at equilibrium, with a slight excess of reactants.
The numerical value of $K$ depends on how the reaction equation is written (direction and coefficients), but its physical meaning is always tied to the equilibrium composition at a given temperature.
6. Dependence of $K$ on the Stoichiometric Equation
The equilibrium constant belongs to a specific chemical equation as written. Modifying the equation changes the value of $K$ in a predictable way.
6.1 Reversing the reaction
If you reverse the chemical equation, the new equilibrium constant is the reciprocal of the original one.
If:
$$
\mathrm{A} \rightleftharpoons \mathrm{B} \quad \text{has} \quad K
$$
then:
$$
\mathrm{B} \rightleftharpoons \mathrm{A} \quad \text{has} \quad K' = \frac{1}{K}
$$
6.2 Multiplying the equation by a factor
If you multiply all stoichiometric coefficients by a factor $n$, the new equilibrium constant is the old one raised to the power $n$.
If:
$$
\mathrm{A} \rightleftharpoons \mathrm{B} \quad \text{has} \quad K
$$
then:
$$
2\,\mathrm{A} \rightleftharpoons 2\,\mathrm{B}
$$
has:
$$
K' = K^{2}
$$
More generally:
- Multiply all coefficients by $n$: $K' = K^{n}$.
- Divide all coefficients by $n$: $K' = K^{1/n}$.
6.3 Adding (combining) reactions
Sometimes overall reactions are built by adding several component reactions. The total equilibrium constant is then the product of the individual equilibrium constants.
Suppose:
- Reaction 1: $\quad \mathrm{A} \rightleftharpoons \mathrm{B}$ with $K_1$
- Reaction 2: $\quad \mathrm{B} \rightleftharpoons \mathrm{C}$ with $K_2$
If we add the two equations, we get:
$$
\mathrm{A} \rightleftharpoons \mathrm{C}
$$
The overall equilibrium constant is:
$$
K_\text{overall} = K_1 \cdot K_2
$$
This rule is very useful, for example, when:
- Building stepwise equilibrium reactions (e.g., complex formation, polyprotic acids, solubility products),
- Or when tabulated equilibrium constants are given for elementary steps.
7. Using the Equilibrium Constant in Calculations (Overview)
Detailed calculation strategies are covered more thoroughly elsewhere, but the core principle is:
- Write the balanced equation and the expression for $K$.
- Express unknown equilibrium concentrations in terms of:
- Known initial amounts, and
- A variable representing the extent of reaction (often denoted $x$).
- Substitute into the $K$ expression to obtain an equation for $x$.
- Solve for $x$, then compute the equilibrium concentrations.
Example outline (without full solution):
For:
$$
\mathrm{H_2(g)} + \mathrm{I_2(g)} \rightleftharpoons 2\,\mathrm{HI(g)}
$$
If initial concentrations of $\mathrm{H_2}$ and $\mathrm{I_2}$ are known and no $\mathrm{HI}$ is present initially, define a reaction progress $x$, set up:
- $[\mathrm{H_2}]_\text{eq} = [\mathrm{H_2}]_0 - x$
- $[\mathrm{I_2}]_\text{eq} = [\mathrm{I_2}]_0 - x$
- $[\mathrm{HI}]_\text{eq} = 2x$
Then insert into:
$$
K_c = \frac{[\mathrm{HI}]_\text{eq}^2}{[\mathrm{H_2}]_\text{eq}[\mathrm{I_2}]_\text{eq}}
$$
and solve for $x$.
(The systematic treatment of such equilibrium calculations, including approximations, appears in associated problem-focused subsections.)
8. Summary of Key Points
- The law of mass action states that for a reaction at equilibrium and fixed temperature, the ratio of product terms to reactant terms (each concentration or pressure raised to its stoichiometric coefficient) is constant.
- This constant is the equilibrium constant $K$ (e.g., $K_c$, $K_p$).
- Pure solids and pure liquids do not appear in the equilibrium constant expression.
- For gas-phase reactions, both concentration-based ($K_c$) and pressure-based ($K_p$) forms are used; they are related by $K_p = K_c (RT)^{\Delta n_\text{gas}}$ for ideal gases.
- The magnitude of $K$ indicates whether products or reactants predominate at equilibrium.
- $K$ depends on how the reaction equation is written:
- Reversing: $K_{\text{reversed}} = 1/K$.
- Scaling coefficients by $n$: $K' = K^{n}$.
- Adding reactions: multiply the corresponding $K$ values.
- In practice, $K$ is treated as dimensionless but is formally based on activities, not raw concentrations or pressures.
These concepts provide the quantitative foundation for understanding how chemical equilibria behave and for predicting equilibrium compositions under various conditions.