Chemical Equilibrium, Equilibrium Constants, and Le Chatelier's Principle

Fundamentals of Reversible Reactions and Chemical Equilibrium

• Reversible Versus Irreversible Reactions:

  • In introductory chemistry, chemical changes are frequently taught as one-way processes where reactants convert permanently into products (e.g., cooking an egg, which cannot be uncooked).
  • Many chemical systems are reversible, meaning products can react together to reform the original reactants.
  • Reversible reactions are represented with equilibrium double arrows ($\rightleftharpoons$), indicating simultaneous forward and reverse pathways:
  • Forward reaction: Reads from left to right ($\text{reactants} \rightarrow \text{products}$).
  • Reverse reaction: Reads from right to left ($\text{products} \rightarrow \text{reactants}$).
  • Example: The interconversion of nitrogen dioxide and dinitrogen tetroxide:
  • Forward process: $2\,\text{NO}_2(g) \rightarrow \text{N}_2\text{O}_4(g)$
  • Reverse process: $\text{N}_2\text{O}_4(g) \rightarrow 2\,\text{NO}_2(g)$
  • Overall equilibrium equation: $2\,\text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g)$

• Reaction Rates Over Time:

  • When reactants are initially mixed, the rate of the forward reaction is at its maximum because reactant concentrations are at their highest.
  • As reactants are consumed over time, the rate of the forward reaction ($A + B \rightarrow C + D$) continuously decreases.
  • Initially, product concentrations are zero, so the reverse reaction rate is zero; as products accumulate, the rate of the reverse reaction ($C + D \rightarrow A + B$) increases.
  • Eventually, the forward rate decreases and the reverse rate increases until they become exactly equal. At this point, chemical equilibrium is reached.

• Concentrations Approaching Equilibrium:

  • Consider the reversible synthesis of methanol: $\text{CO}(g) + 2\,\text{H}_2(g) \rightleftharpoons \text{CH}_3\text{OH}(g)$.
  • As the system approaches equilibrium, the concentrations of $\text{CO}(g)$ and $\text{H}_2(g)$ decrease, while the concentration of $\text{CH}_3\text{OH}(g)$ increases until both level off.
  • Critical Distinction: Chemical equilibrium does not mean that the concentrations of reactants and products are equal to each other.
  • Chemical equilibrium does mean that reactants are consumed at the exact same rate that they are reformed, resulting in constant, unchanging concentrations of all species over time.

• Nature of Dynamic Equilibrium:

  • At dynamic equilibrium, no macroscopic (observable) changes occur, giving the outward appearance that the chemical reaction has stopped.
  • At the atomic and molecular level, chemical change is continuous: molecules are perpetually colliding and reacting in both directions at identical speeds.
  • Example: The aqueous equilibrium between iron(III) ions, thiocyanate ions, and iron(III) thiocyanate complex ions:     $\text{Fe}^{3+}(aq) + \text{CNS}^-(aq) \rightleftharpoons \text{FeCNS}^{2+}(aq)$
  • At dynamic equilibrium, $\text{Fe}^{3+}$ and $\text{CNS}^-$ constantly combine to form $\text{FeCNS}^{2+}$, while $\text{FeCNS}^{2+}$ simultaneously dissociates back into $\text{Fe}^{3+}$ and $\text{CNS}^-$ at the same rate.
  • The Escalator Metaphor:
  • Walking or running up an escalator that is moving downward at the exact same speed illustrates dynamic equilibrium.
  • The individual is continuously expending effort and running, yet remains at a stationary macroscopic position.

• Criteria for Chemical Equilibrium:

  • Closed System: The reaction must occur in a closed system where no matter can enter or exit (e.g., in a closed container, liquid evaporation rate equals vapor condensation rate; in an open system, vapor escapes and equilibrium cannot establish).
  • Constant Macroscopic Properties: All observable properties (such as color intensity, temperature, pressure, and concentration) remain completely constant over time.
  • Reversibility and Response to Perturbation: The reaction must be fully reversible, such that when a factor affecting reaction rate is altered and then restored to its original value, the system returns to its original equilibrium state.

The Equilibrium Constant ($K_{eq}$) Expression

• Definition and Law of Mass Action:

  • The equilibrium constant ($K_{eq}$) quantitatively describes the relative concentrations of reactants and products present at equilibrium for a reversible reaction.
  • For any generic reversible reaction with balanced stoichiometric coefficients $a$, $b$, $c$, and $d$:     $a\,A + b\,B \rightleftharpoons c\,C + d\,D$
  • The equilibrium constant expression is formulated as:     $K_{eq} = \frac{[C]^c [D]^d}{[A]^a [B]^b} = \frac{[\text{products}]}{[\text{reactants}]}$
  • Square brackets ($[\;]$) denote molar concentrations expressed in units of $\text{M}$ or $\text{mol}\,dm^{-3}$.

• Rules for Writing $K_{eq}$ Expressions:

  • Every reversible reaction obeys this mathematical relationship and possesses a specific numerical $K_{eq}$ value at a given temperature.
  • Phase Inclusions and Exclusions:
  • Only species in the gaseous ($g$) and aqueous ($aq$) states are included in the $K_{eq}$ expression.
  • Substances in the pure liquid ($l$) or pure solid ($s$) states are never included in the expression because their concentrations/densities are fixed and constant throughout the reaction.
  • Example: Heterogeneous decomposition of calcium carbonate:     $\text{CaO}(s) + \text{CO}_2(g) \rightleftharpoons \text{CaCO}_3(s)$
  • Because $\text{CaO}$ and $\text{CaCO}_3$ are solids, their concentrations are constant and omitted from the equilibrium expression:     $K_{eq} = \frac{1}{[\text{CO}_2]}$

• Independence from Initial Concentrations:

  • The ratio defining $K_{eq}$ is a constant value for a specific reaction at a set temperature, regardless of the starting concentrations of reactants and products.
  • Whether high initial reactant concentrations or high initial product concentrations are supplied, the system will adjust to ensure the ratio of equilibrium concentrations satisfies $K_{eq}$.

Interpreting the Magnitude of $K_{eq}$

• Very Large Equilibrium Constant ($K_{eq} > 10^{10}$):

  • The equilibrium mixture contains substantially more products than reactants.
  • The forward reaction goes essentially to completion, leaving virtually unmeasurable or negligible amounts of reactants at equilibrium.
  • Example: Decomposition of ozone into molecular oxygen:     $2\,\text{O}_3(g) \rightleftharpoons 3\,\text{O}_2(g)$$K_{eq} = 2.0 \times 10^{57}$

• Very Small Equilibrium Constant ($K_{eq} < 10^{-10}$):

  • The concentration of reactants at equilibrium is vastly larger than that of the products.
  • The forward reaction barely occurs; most reactants remain completely unchanged and only trace amounts of products are generated.
  • Example: Formation of nitrogen monoxide from atmospheric gases:     $\text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2\,\text{NO}(g)$$K_{eq} = 1.0 \times 10^{-25}$

• Intermediate Equilibrium Constant ($10^{-10} \le K_{eq} \le 10^{10}$):

  • The system maintains significant, comparable concentrations of both reactants and products at equilibrium.
  • Example: Water-gas shift reaction:     $\text{CO}(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CO}_2(g) + \text{H}_2(g)$$K_{eq} = 5.09$

The Reaction Quotient ($Q$) and Predicting Reaction Direction

• Definition and Function of $Q$:

  • Because an equilibrium state cannot be identified visually, the reaction quotient ($Q$) is used to determine whether a reaction has achieved equilibrium or in which direction it must shift to reach it.
  • $Q$ is calculated using the exact same algebraic formula as $K_{eq}$, except that the concentrations inserted into the expression are the instantaneous, non-equilibrium concentrations measured at a specific point in time.

• Comparison Criteria Between $Q$ and $K_{eq}$:

  • $Q = K_{eq}$:
  • The system is currently at dynamic chemical equilibrium.
  • Forward and reverse reaction rates are equal, and concentrations remain unchanged.
  • $Q < K_{eq}$:
  • The ratio of products to reactants is currently smaller than the equilibrium ratio.
  • To increase products and decrease reactants, the reaction proceeds in the forward direction (equilibrium shifts to the RIGHT).
  • $Q > K_{eq}$:
  • The ratio of products to reactants is currently larger than the equilibrium ratio.
  • To consume excess products and replenish reactants, the reaction proceeds in the reverse direction (equilibrium shifts to the LEFT).

• Worked Example: Synthesis of Hydrogen Iodide:

  • Reaction: $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\,\text{HI}(g)$ at $448\,^\circ\text{C}$, with $K_{eq} = 50.5$.
  • Measured concentrations at a given time: $[\text{H}_2] = 0.113\,\text{M}$, $[\text{I}_2] = 0.132\,\text{M}$, and $[\text{HI}] = 0.750\,\text{M}$.
  • Step 1: Formulate the reaction quotient expression:     $Q = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}$
  • Step 2: Substitute instantaneous measured values:     $Q = \frac{(0.750\,\text{M})^2}{(0.113\,\text{M})(0.132\,\text{M})} = \frac{0.5625}{0.014916} \approx 37.7$
  • Step 3: Compare $Q$ to $K_{eq}$:     $Q = 37.7 < K_{eq} = 50.5$
  • Conclusion: Because $Q < K_{eq}$, the reaction is not at equilibrium; it must proceed forward (shift to the right) to produce more $\text{HI}$.

• Additional Applications of $Q$:

  • The production of ammonia: $\text{N}_2(g) + 3\,\text{H}_2(g) \rightleftharpoons 2\,\text{NH}_3(g)$ at $473\,^\circ\text{C}$.
  • Measurements taken during reaction progress (e.g., with measured $[\text{N}_2] = 0.15\,\text{M}$ alongside corresponding $[\text{H}_2]$ and $[\text{NH}_3]$) allow computation of $Q$ to evaluate whether synthesis is complete or continuing forward.

Quantitative Equilibrium Calculations

• Problem Setup and Algebraic Determination of Equilibrium Concentrations:

  • Consider the gas-phase reaction at $448\,^\circ\text{C}$ where $K_{eq} = 50.5$:     $\text{H}_2(g) + \text{Cl}_2(g) \rightleftharpoons 2\,\text{HCl}(g)$
  • A reaction mixture initially contains $2.1\,\text{mol}$ of $\text{H}_2(g)$ and $2.1\,\text{mol}$ of $\text{Cl}_2(g)$ in a vessel such that their initial concentrations are each $1.68\,\text{M}$, with zero initial $[\text{HCl}]$.

• Step-by-Step Algebraic Solution:

  • Step 1: Define concentration changes using a change variable $x$:
  • Let the concentration of $\text{H}_2$ consumed be $x$.
  • From stoichiometric ratios ($1:1:2$), $\text{Cl}_2$ consumed is $x$, and $\text{HCl}$ produced is $2x$.
  • Equilibrium concentrations:       $[\text{H}_2]_{eq} = 1.68 - x$$[\text{Cl}_2]_{eq} = 1.68 - x$$[\text{HCl}]_{eq} = 2x$
  • Step 2: Set up the $K_{eq}$ expression:     $K_{eq} = \frac{[\text{HCl}]^2}{[\text{H}_2][\text{Cl}_2]} = 50.5$$\frac{(2x)^2}{(1.68 - x)(1.68 - x)} = \frac{(2x)^2}{(1.68 - x)^2} = 50.5$
  • Step 3: Take the square root of both sides:     $\sqrt{\frac{(2x)^2}{(1.68 - x)^2}} = \sqrt{50.5}$$\frac{2x}{1.68 - x} = 7.1063$
  • Step 4: Solve algebraically for $x$:     $2x = 7.1063(1.68 - x)$$2x = 11.9386 - 7.1063x$$2x + 7.1063x = 11.9386$$9.1063x = 11.9386$$x = \frac{11.9386}{9.1063} \approx 1.31\,\text{M}$
  • Step 5: Compute final equilibrium concentrations:     $[\text{HCl}]_{eq} = 2x = 2(1.31\,\text{M}) = 2.62\,\text{M}$$[\text{H}_2]_{eq} = [\text{Cl}_2]_{eq} = 1.68\,\text{M} - 1.31\,\text{M} = 0.37\,\text{M}$

Le Chatelier's Principle and System Stresses

• Principle Definition:

  • "A stress that is applied to a reaction that is at equilibrium will shift the equilibrium position in a direction that tends to reduce this stress."
  • Guiding heuristic: "Whatever we do, nature tries to undo."
  • The adjustment is executed by temporarily modifying the relative rate of either the forward or reverse reaction:
  • If forward reaction rate becomes greater than reverse reaction rate, the equilibrium shifts to the RIGHT.
  • If reverse reaction rate becomes greater than forward reaction rate, the equilibrium shifts to the LEFT.

• Disturbances and the Reaction Quotient Mechanism:

  • An applied disturbance shifts instantaneous concentrations or conditions such that $Q \neq K_{eq}$.
  • The system shifts dynamically until the rates re-equalize and $Q = K_{eq}$ once again.
  • Escalator Disturbance Analogy:
  • If a person running upward on a downward escalator is in balance (equilibrium), and the escalator suddenly speeds up (stress), the runner must actively increase running speed to match the increased downward velocity and regain balance.

• Classification of Stresses and Impact on $K_{eq}$:

  • Concentration Change: Perturbs equilibrium position; does NOT change $K_{eq}$.
  • Pressure or Volume Change: Perturbs equilibrium position in gas-phase systems; does NOT change $K_{eq}$.
  • Catalyst Addition: Accelerates equilibration rates; does NOT change $K_{eq}$.
  • Temperature Change: Shifts equilibrium position and DOES change the numerical value of $K_{eq}$.

Detailed Mechanisms of Equilibrium Shifts

• Effect of Concentration Changes:

  • When an equilibrium reaction is subjected to an increase in the concentration of a chemical species, it shifts in the direction that consumes that added substance.
  • When the concentration of a chemical species is decreased (removed), the reaction shifts in the direction that produces that substance.
  • Illustrative reaction: $2\,\text{NO}(g) + \text{Cl}_2(g) \rightleftharpoons 2\,\text{NOCl}(g) + 76\,\text{kJ}$
  • Adding $\text{NO}(g)$ or $\text{Cl}_2(g)$: Stresses the reactant side; the reaction shifts to the right to consume the added reactants and generate more $\text{NOCl}(g)$.
  • Removing $\text{NOCl}(g)$: Shifts the system to the right to replenish $\text{NOCl}(g)$.
  • Throughout all concentration shifts, the numerical ratio at final re-equilibration yields the identical $K_{eq}$ value.

• Effect of Pressure and Volume Changes:

  • Pressure and volume changes impact equilibrium only when gases are involved in the chemical system.
  • Boyle's Law / Inverse Relationship:
  • An increase in pressure corresponds directly to a decrease in system volume.
  • A decrease in pressure corresponds directly to an increase in system volume.
  • Avogadro's Hypothesis and Particle Density:
  • Equal volumes of gases under identical conditions contain equal numbers of particles ($1\,\text{mol}$ of any gas occupies $22.4\,\text{L}$ at standard temperature and pressure, STP).
  • Reducing container volume compresses all gas molecules, simultaneously spiking the total pressure and the molar concentrations of all gaseous species present.
  • Shift Direction Rule:
  • When volume is decreased (pressure increased), the equilibrium shifts toward the side of the reaction with the fewer moles of gas to relieve pressure and occupy less space.
  • When volume is increased (pressure decreased), the equilibrium shifts toward the side with the greater moles of gas to restore pressure.
  • If both sides contain an identical number of gaseous moles ($\Delta n_{gas} = 0$), pressure and volume variations produce no directional shift.
  • Pressure and volume changes do not alter $K_{eq}$.

• Effect of Temperature Changes:

  • Temperature is the only variable that alters the value of the equilibrium constant $K_{eq}$.
  • Thermochemical framework treats thermal energy as a chemical participant:
  • In an exothermic reaction (e.g., $2\,\text{NO}(g) + \text{Cl}_2(g) \rightleftharpoons 2\,\text{NOCl}(g) + 76\,\text{kJ}$), heat is released and functions as a product.
  • In an endothermic reaction, heat is absorbed and functions as a reactant.
  • Shift and $K_{eq}$ Response:
  • Adding heat (increasing temperature) causes the system to shift in the direction that consumes heat (endothermic direction).
  • Removing heat (decreasing temperature) causes the system to shift in the direction that generates heat (exothermic direction).
  • If the equilibrium shifts to the RIGHT:       $\text{More products formed, fewer reactants remaining} \implies K_{eq} \text{ increases}$
  • If the equilibrium shifts to the LEFT:       $\text{Fewer products present, more reactants formed} \implies K_{eq} \text{ decreases}$

• Effect of Adding a Catalyst:

  • Definition: A catalyst is a substance that increases the rate of a chemical reaction without itself undergoing any permanent chemical change.
  • Kinetic Mechanism: A catalyst lowers the activation energy (presentation energy) equally for both the forward and reverse reaction pathways.
  • Equilibrium Consequence:
  • Because forward and reverse reaction rates are accelerated by the exact same proportion, adding a catalyst causes the system to reach dynamic equilibrium more rapidly.
  • A catalyst does not alter the equilibrium position, does not change the equilibrium concentrations of any species, and does not change the numerical value of $K_{eq}$.