Table of Contents
Scope and Focus of This Chapter
This chapter deals with how the law of mass action is actually used in typical chemical systems. You will apply the general concepts of equilibrium and equilibrium constants (introduced earlier) to specific, practically important equilibria. The aim is not to derive the law again, but to see how it helps to:
- Predict the direction in which a reaction will proceed.
- Calculate equilibrium compositions.
- Understand and control technological and environmental processes.
We focus on a few representative classes of systems:
- Gas-phase reactions.
- Acid–base and solubility equilibria (as previews of later, more detailed chapters).
- Simple examples relevant to industrial and biological chemistry.
Throughout, the central tools are:
- The equilibrium constant $K$ in its appropriate form.
- The reaction quotient $Q$ and the $Q$ vs. $K$ comparison.
- Mass balances (material balances) and ICE tables (Initial–Change–Equilibrium).
Using the Law of Mass Action to Predict Reaction Direction
The law of mass action provides an expression for the equilibrium constant $K$ for a reaction written in a specific stoichiometric form, for example:
$$
\alpha A + \beta B \rightleftharpoons \gamma C + \delta D
$$
The equilibrium constant (for concentrations) is:
$$
K_\mathrm{c} = \frac{[C]^\gamma [D]^\delta}{[A]^\alpha [B]^\beta}
$$
To decide whether a given mixture will move toward products or reactants, you compute the reaction quotient $Q$ by the same expression, but using the current concentrations (or partial pressures), not necessarily equilibrium ones.
- If $Q < K$: the system has “too many reactants,” so it will shift toward products.
- If $Q > K$: the system has “too many products,” so it will shift toward reactants.
- If $Q = K$: the system is already at equilibrium.
This diagnostic is used constantly in practical applications.
Example: Gas-Phase Synthesis of Ammonia (Conceptual Use of $Q$ vs. $K$)
Consider the reaction (balanced):
$$
\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)}
$$
Assume at a certain temperature, $K_\mathrm{p}$ is known from data. In a reactor you may measure partial pressures $p_{\ce{N2}}$, $p_{\ce{H2}}$, and $p_{\ce{NH3}}$ and compute:
$$
Q_\mathrm{p} = \frac{(p_{\ce{NH3}})^2}{p_{\ce{N2}} (p_{\ce{H2}})^3}
$$
- If a fresh feed has mostly $\ce{N2}$ and $\ce{H2}$ and little $\ce{NH3}$, then $Q_\mathrm{p} \ll K_\mathrm{p}$, and the reaction proceeds strongly toward $\ce{NH3}$.
- During operation, if $Q_\mathrm{p}$ becomes close to $K_\mathrm{p}$, the forward reaction slows and net production ceases; process engineers then remove $\ce{NH3}$ or change pressure/temperature to keep the process economical.
This type of reasoning (without doing full numerical calculations) is a key application of the law of mass action in industrial chemistry.
Calculating Equilibrium Compositions
In many practical problems you are given:
- Initial amounts (or concentrations) of reactants and products.
- The equilibrium constant $K$.
The task is to calculate the equilibrium composition. A common systematic approach is to set up an ICE table.
General ICE Table Strategy
For a reaction:
$$
\alpha A + \beta B \rightleftharpoons \gamma C + \delta D
$$
- Initial: Write the initial concentrations (or partial pressures).
- Change: Assume the reaction proceeds by an amount $x$ (using stoichiometric coefficients).
- Equilibrium: Express all equilibrium concentrations in terms of $x$.
- Insert these expressions into the $K$ expression and solve for $x$.
- Back-substitute to get all equilibrium concentrations.
Example: Equilibrium Between $\ce{NO2}$ and $\ce{N2O4}$ (Gas-Phase Dimerization)
Reaction:
$$
\ce{2 NO2(g) <=> N2O4(g)}
$$
Suppose:
- Initially, only $\ce{NO2}$ is present at concentration $c_0$.
- At a given temperature, $K_\mathrm{c}$ is known.
Set up:
- Initial:
- $[\ce{NO2}] = c_0$
- $[\ce{N2O4}] = 0$
Let the reaction proceed to the right by an amount $x$ (in mol/L):
- Change:
- $[\ce{NO2}]$: $-2x$ (two moles of $\ce{NO2}$ are consumed per mole of $\ce{N2O4}$ formed)
- $[\ce{N2O4}]$: $+x$
- Equilibrium:
- $[\ce{NO2}] = c_0 - 2x$
- $[\ce{N2O4}] = x$
Equilibrium constant:
$$
K_\mathrm{c} = \frac{[\ce{N2O4}]}{[\ce{NO2}]^2} = \frac{x}{(c_0 - 2x)^2}
$$
You solve this equation for $x$. Once $x$ is known, all equilibrium concentrations are determined. In practice:
- If $K_\mathrm{c}$ is very large, you expect most $\ce{NO2}$ to convert to $\ce{N2O4}$.
- If $K_\mathrm{c}$ is very small, the reaction hardly proceeds to the right.
This method is typical for gas-phase and solution-phase reactions.
Working with $K_\mathrm{c}$ and $K_\mathrm{p}$ in Gas Reactions
Gas reactions often use partial pressures and $K_\mathrm{p}$:
$$
K_\mathrm{p} = \frac{(p_C)^\gamma (p_D)^\delta}{(p_A)^\alpha (p_B)^\beta}
$$
If needed, $K_\mathrm{p}$ and $K_\mathrm{c}$ are related (for ideal gases) by:
$$
K_\mathrm{p} = K_\mathrm{c} (RT)^{\Delta n}
$$
where $\Delta n$ is the change in the amount of gas (products minus reactants), and $R$ is the gas constant. In applications:
- Use $K_\mathrm{p}$ when your data are in pressures (common in industrial reactors).
- Use $K_\mathrm{c}$ when your data are in concentrations (common in laboratory solutions).
Example: $\ce{H2}$–$\ce{I2}$–$\ce{HI}$ Equilibrium
Reaction:
$$
\ce{H2(g) + I2(g) <=> 2 HI(g)}
$$
Given initial partial pressures and $K_\mathrm{p}$, you proceed analogously to the $\ce{NO2}$/$\ce{N2O4}$ example:
- Express equilibrium partial pressures in terms of an unknown extent $x$.
- Insert into the $K_\mathrm{p}$ expression.
- Solve for $x$.
- Interpret which side of the equilibrium is favored based on the magnitude of $K_\mathrm{p}$.
Acid–Base Equilibria as Applications (Preview)
In acid–base chemistry, the law of mass action is applied to proton-transfer equilibria. Even though these are treated in detail later, a few simple applications illustrate the method.
Weak Acid Equilibria and $K_\mathrm{a}$
Consider a weak monoprotic acid $\ce{HA}$ in water:
$$
\ce{HA(aq) + H2O(l) <=> H3O+(aq) + A-(aq)}
$$
Water appears in large excess and is treated as constant, so the equilibrium constant simplifies to the acid dissociation constant:
$$
K_\mathrm{a} = \frac{[\ce{H3O^+}][\ce{A^-}]}{[\ce{HA}]}
$$
Typical application: calculate the pH of a weak acid solution.
Example setup (symbolic):
- Initial:
- $[\ce{HA}] = c_0$
- $[\ce{H3O^+}] = [\ce{A^-}] = 0$ (ignoring water autoprotolysis)
- Change:
- $\ce{HA}$: $-x$
- $\ce{H3O^+}$: $+x$
- $\ce{A^-}$: $+x$
- Equilibrium:
- $[\ce{HA}] = c_0 - x$
- $[\ce{H3O^+}] = x$
- $[\ce{A^-}] = x$
Insert in:
$$
K_\mathrm{a} = \frac{x \cdot x}{c_0 - x} = \frac{x^2}{c_0 - x}
$$
Solving for $x$ gives $[\ce{H3O^+}]$ and thus the pH via:
$$
\mathrm{pH} = -\log_{10} [\ce{H3O^+}]
$$
This is a direct application of the law of mass action to a specific but very important class of reactions.
Weak Base and Conjugate Pairs
Similarly, for a weak base $\ce{B}$:
$$
\ce{B(aq) + H2O(l) <=> BH+(aq) + OH-(aq)}
$$
The law of mass action yields the base dissociation constant $K_\mathrm{b}$:
$$
K_\mathrm{b} = \frac{[\ce{BH^+}][\ce{OH^-}]}{[\ce{B}]}
$$
Again, ICE tables are used to compute equilibrium concentrations and pH (through $[\ce{OH^-}]$).
These applications will be extended and specialized in the separate chapter on acids and bases; here the focus is that they are straightforward uses of equilibrium concepts.
Solubility Equilibria and the Solubility Product (Preview)
Dissolution of sparingly soluble ionic solids is another major application. The law of mass action leads to solubility product constants, $K_\mathrm{sp}$.
Consider a slightly soluble salt:
$$
\ce{MX(s) <=> M+(aq) + X-(aq)}
$$
The general equilibrium constant, including the solid, would be:
$$
K = \frac{[\ce{M^+}][\ce{X^-}]}{a_{\ce{MX(s)}}}
$$
Because the activity (effective concentration) of a pure solid is taken as 1, this simplifies to:
$$
K_\mathrm{sp} = [\ce{M^+}][\ce{X^-}]
$$
This is called the solubility product.
Using $K_\mathrm{sp}$ to Calculate Solubility
If one mole of $\ce{MX}$ dissolves to give one mole of $\ce{M^+}$ and one mole of $\ce{X^-}$, then in pure water:
- Let $s$ be the molar solubility (mol/L).
Equilibrium:
- $[\ce{M^+}] = s$
- $[\ce{X^-}] = s$
Thus:
$$
K_\mathrm{sp} = s \cdot s = s^2 \quad \Rightarrow \quad s = \sqrt{K_\mathrm{sp}}
$$
For a salt like $\ce{M(OH)2}$:
$$
\ce{M(OH)2(s) <=> M^{2+}(aq) + 2 OH^-(aq)}
$$
- If $s$ is the solubility:
- $[\ce{M^{2+}}] = s$
- $[\ce{OH^-}] = 2s$
Equilibrium expression:
$$
K_\mathrm{sp} = [\ce{M^{2+}}][\ce{OH^-}]^2 = s (2s)^2 = 4 s^3
$$
Thus you can compute $s$ from $K_\mathrm{sp}$, or vice versa.
Precipitation and Dissolution: Using $Q_\mathrm{sp}$
Analogous to $Q$ and $K$, you define a reaction quotient for solubility, $Q_\mathrm{sp}$:
- For $\ce{MX}$:
$$
Q_\mathrm{sp} = [\ce{M^+}][\ce{X^-}]
$$
Comparison:
- If $Q_\mathrm{sp} < K_\mathrm{sp}$: the solution is unsaturated; more solid can dissolve.
- If $Q_\mathrm{sp} = K_\mathrm{sp}$: the solution is saturated; it is in equilibrium with the solid.
- If $Q_\mathrm{sp} > K_\mathrm{sp}$: the solution is supersaturated; precipitation is thermodynamically favored.
This is the basis for:
- Predicting whether a precipitate forms when solutions are mixed.
- Designing qualitative analysis schemes for separating ions by selective precipitation.
These topics are further elaborated in the later chapters on solubility equilibria and qualitative analysis, but the underlying tool is again the law of mass action.
Competing and Coupled Equilibria
In many real systems, several equilibria occur simultaneously. Applying the law of mass action then means:
- Writing an equilibrium expression for each reaction.
- Combining them with mass balances and charge balance (in electrolyte solutions).
Example: Complex Formation and Solubility
Consider the dissolution of a metal salt $\ce{AgCl}$ in the presence of $\ce{NH3}$, which forms complexes with $\ce{Ag^+}$:
- Solubility equilibrium:
$$
\ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)}
$$
with $K_\mathrm{sp} = [\ce{Ag^+}][\ce{Cl^-}]$. - Complex formation equilibrium:
$$
\ce{Ag+(aq) + 2 NH3(aq) <=> [Ag(NH3)2]+(aq)}
$$
with the formation constant:
$$
\beta_2 = \frac{[\ce{[Ag(NH3)2]+}]}{[\ce{Ag^+}][\ce{NH3}]^2}
$$
Because complex formation removes free $\ce{Ag^+}$ from solution, the first equilibrium is “pulled” to the right, increasing the apparent solubility of $\ce{AgCl}$. Quantitatively, you would combine the equilibrium expressions with:
- Mass balance for silver: total silver $=$ free $\ce{Ag^+}$ $+$ complexed silver.
- Mass balance for chloride and ammonia.
- Charge balance if needed.
This is a typical applied problem in coordination and analytical chemistry, where the law of mass action is the main mathematical framework.
Biological and Environmental Applications (Qualitative)
In biological and environmental systems, equilibria are often too complex to solve exactly in introductory courses, but the same principles apply:
- Blood buffer systems: The $\ce{CO2}$–$\ce{H2CO3}$–$\ce{HCO3^-}$ system is governed by acid–base equilibria described by the law of mass action. The body controls pH by adjusting $\ce{CO2}$ and $\ce{HCO3^-}$ concentrations.
- Carbonate equilibria in natural waters: Dissolved $\ce{CO2}$, carbonic acid, bicarbonate, and carbonate are linked by multiple equilibrium constants. These govern hardness of water, dissolution/precipitation of $\ce{CaCO3}$, and buffering capacity.
- Nutrient availability in soils and waters: The solubility of phosphates, iron, and other nutrients is determined by $K_\mathrm{sp}$, complex formation constants, and acid–base equilibria.
In all these cases, the law of mass action:
- Links species concentrations at equilibrium.
- Predicts how changes in conditions (e.g., $\ce{CO2}$ level, pH, ionic strength) shift equilibria.
Practical Problem-Solving Tips
When applying the law of mass action in various contexts, a few systematic steps help:
- Write the balanced reaction in the direction you want to define $K$.
- Write the correct $K$ expression ($K_\mathrm{c}$, $K_\mathrm{p}$, $K_\mathrm{a}$, $K_\mathrm{b}$, $K_\mathrm{sp}$, or a formation constant), following stoichiometric exponents.
- Identify what is constant and can be omitted:
- Pure solids and liquids (activity set to 1).
- Solvent (usually water) in dilute solutions.
- Set up an ICE table when asked for equilibrium compositions.
- Use $Q$ vs. $K$ when asked about direction of change or precipitation.
- Estimate first, calculate second:
- Check magnitude of $K$ to anticipate which side is favored.
- Use approximations (e.g., $c_0 - x \approx c_0$) sensibly when $x$ is small; verify at the end.
- For multiple equilibria, write all equilibrium expressions and combine them with:
- Mass balances on key elements.
- Charge balance in ionic solutions (when relevant).
These techniques generalize across all the applied topics where the law of mass action appears in the remainder of the course: acids and bases, redox equilibria, solubility, coordination chemistry, environmental chemistry, and biochemical systems.