Exhaustive Guide to Chemical Equilibrium: Law of Mass Action and Constants

Introduction to Chemical Equilibrium

• Conceptual Overview: In previous chemistry studies, focus was often placed on reactions that proceed in one direction until all reactants are converted into products (completion). Chemical equilibrium introduces a more complex reality.

• Fundamental Questions:

  • Do ALL chemical reactions always go to completion?

  • Do ALL chemical reactions proceed in only one direction?

• The Briggs-Rauscher Oscillating Reaction: This specific reaction is cited as an example of complex chemical behavior where the system does not simply settle in one state immediately, highlighting the dynamic nature of chemical processes.

• Visualizing Molecules: All molecular visualizations in this study material were generated using the program HyperChem by HYPERCUBE, INC.

The Dissolution of Sodium Chloride and the State of Equilibrium

• Process of Dissolution: When table salt ($NaCl_{(s)}$) is dissolved in water ($H_2O$), both the solid reactant and the aqueous products exist in the solution simultaneously.

• Reaction Equation:

  • $NaCl_{(s)} \xrightarrow{H_2O} Na^{+}_{(aq)} + Cl^{-}_{(aq)}$

• Saturated Solutions: A saturated solution occurs when so much salt has been added that it can no longer dissolve. At this point, the system has reached its equilibrium state.

• Bidirectional Nature: The reaction can proceed in both the forward and backward directions:

  • Forward (Dissolving): $NaCl_{(s)} \rightarrow Na^{+}_{(aq)} + Cl^{-}_{(aq)}$

  • Backward (Precipitation): $Na^{+}_{(aq)} + Cl^{-}_{(aq)} \rightarrow NaCl_{(s)}$

Defining Equilibrium: Rates and Dynamics

• Definition of Equilibrium: Equilibrium is specifically defined as the state where the Rate of the Forward Reaction is equal to the Rate of the Backward Reaction.

  • $\text{Rate}_{fwd} = \text{Rate}_{bckwd}$

  • $\text{Rate of Precipitation} = \text{Rate of Dissolving}$

• The Equilibrium Symbol: The double arrow ($\rightleftharpoons$) acknowledges that the process is occurring in both directions:

  • $NaCl_{(s)} \rightleftharpoons Na^{+}_{(aq)} + Cl^{-}_{(aq)}$

• Dynamic Equilibrium:

  • Macroscopically: To the naked eye, nothing appears to be happening. The concentrations and amounts of solids/liquids remain constant.

  • Microscopically: Both the forward and backward reactions are occurring "furiously" at the same time. The substance is constantly moving between the solid and aqueous states at equal rates.

Equilibrium Concentration vs. Time Plots

• General Behavior: Graphs plotting Concentration vs. Time show that as a reaction approaches equilibrium, the concentrations of reactants and products change until they flatten out into horizontal lines (indicating a constant value).

• Example Reaction: $CO_{(g)} + H_2O_{(g)} \rightleftharpoons CO_2{(g)} + H_2{(g)}$

  • Reactants ($[CO]$ and $[H_2O]$) decrease over time until they reach equilibrium levels.

  • Products ($[CO_2]$ and $[H_2]$) increase over time until they reach equilibrium levels.

• Calculating Equilibrium Concentrations via Stoichiometry:

  • Reaction: $2NO_{2(g)} \rightleftharpoons N_2O_{4(g)}$

  • Scenario A: Starting with $4\,\text{M}$ $NO_2$ and $0\,\text{M}$ $N_2O_4$. If equilibrium concentration of $N_2O_4$ is $1\,\text{M}$, we calculate the needed $NO_2$ consumption:

    • To make $1\,\text{mol}$ of $N_2O_4$ , you need $2\,\text{mol}$ of $NO_2$ (based on the $2:1$ stoichiometry).

    • Change in $NO_2 = 4\,\text{mol} - 2\,\text{mol} = 2\,\text{mol}$.

    • The equilibrium concentration of $NO_2$ is $2\,\text{M}$.

  • Scenario B: Starting Concentration of both $NO_2$ and $N_2O_4$ is $8\,\text{M}$. Equilibrium concentration of $NO_2$ is observed at $6\,\text{M}$.

    • This indicates a loss of $2\,\text{moles}$ of $NO_2$.

    • Calculation for product yield: $2\,\text{mol NO}_2 \times \frac{1\,\text{mol N}_2O_4}{2\,\text{mol NO}_2} = 1\,\text{mol of N}_2O_4$.

    • Change in $N_2O_4 = 8\,\text{mol} + 1\,\text{mol} = 9\,\text{moles}$.

    • The equilibrium concentration of $N_2O_4$ is $9\,\text{M}$.

The Law of Mass Action and Activity

• Defining the Law: The Law of Mass Action states that the equilibrium constant ($K$) is a ratio of the activities of the products to the activities of the reactants, each raised to the power of their stoichiometric coefficients.

• General Reaction: $aA_{(aq)} + bB_{(s)} \rightleftharpoons cC_{(g)} + dD_{(l)}$

  • $K = \frac{(aC)^c (aD)^d}{(aA)^a (aB)^b}$

• Understanding Activity ($a$): Activity represents how active a species is in the chemical equation once equilibrium is reached. It is defined as current parameters divided by reference parameters.

  • Aqueous Solutions: Activity ($a$) = $\frac{[A]}{1\,\text{M}}$ (where $[A]$ is Molarity). Unitless concentration.

  • Gases: Activity ($a$) = $\frac{pC}{1\,\text{atm}}$ (where $pC$ is partial pressure). Unitless pressure.

  • Pure Solids: Activity ($a$) = $\frac{\text{Density}}{\text{Density}} = 1$.

  • Pure Liquids: Activity ($a$) = $\frac{\text{Density}}{\text{Density}} = 1$.

• Simplifying the K Expression: Because the activities of pure solids and pure liquids are always $1$, they are omitted from the final equilibrium constant expression.

  • $K = \frac{[C]^c (1)^d}{[A]^a (1)^b} = \frac{[C]^c}{[A]^a}$

• Properties of $K$:

  • Unitless: $K$ is a ratio of activities and has no units.

  • Temperature Dependent: For a given reaction at constant temperature, $K$ is constant.

  • Path Independence: Equilibrium is reached whether starting with only reactants or only products.

Interpreting the Magnitude of the Equilibrium Constant ($K$)

• The Scale of K: $K$ can be any positive number, ranging from very small to very large.

• Product-Favored Reactions ($K \gg 1$):

  • The reaction proceeds significantly toward the forward direction.

  • At equilibrium, there are high concentrations of products and very little reactant.

  • Example: $CH_3O_2 + NO_2 \rightleftharpoons CH_3O_2NO_2$ where $K = 1.2 \times 10^{33}$.

  • Completion Convention: A reaction is generally said to go to "completion" if $K$ is on the order of $10^8$ or higher.

• Reactant-Favored Reactions ($K \ll 1$):

  • The reaction proceeds only slightly toward the products.

  • At equilibrium, there are high concentrations of reactants and very little product.

  • Example: $O_3 + NO \rightleftharpoons NO_2 + O_2$ (Small $K$ value example, though the transcript also lists $5.8 \times 10^{-34}$ for specific conditions).

  • Non-Reaction Convention: A reaction is effectively said not to happen if $K$ is on the order of $10^{-12}$ or lower, although it actually occurs to a virtually undetectable extent.

• Significance of K: It is a measure of how far a reaction will proceed in the written direction. It does NOT indicate how fast (kinetics) the equilibrium is reached.

Rules for Manipulating Equilibrium Expressions

• Rule 1: Reversing Reactions: When a reaction is reversed, the new equilibrium constant ($K_{rev}$) is the reciprocal of the original constant ($1/K_{orig}$).

  • If $A + B \rightleftharpoons C + D$ has $K_1$, then $C + D \rightleftharpoons A + B$ has $K_2 = \frac{1}{K_1}$.

• Rule 2: Stoichiometric Coefficients: When a reaction is multiplied by a factor $N$, the equilibrium constant is raised to the power of $N$.

  • $K_{new} = (K_{orig})^{N}$

  • Example: $I_{2(g)} + H_{2(g)} \rightleftharpoons 2HI_{(g)}$ ($K_1 = 0.5$).

  • If multiplied by $0.5$: $K_2 = (0.5)^{1/2} = 0.71$.

  • If multiplied by $2$: $K_3 = (0.5)^{2} = 0.25$.

• Rule 3: Adding Reactions: When two reactions are added together to create a net equation, the new equilibrium constant is the product of the constants for the individual reactions.

  • $K_{net} = K_1 \times K_2$

  • Example Derivation:

    • Rxn 1: $I_{2(g)} + H_{2(g)} \rightleftharpoons 2HI_{(g)}$ ($K_1$)

    • Rxn 2: $I_2CO_{(g)} \rightleftharpoons I_{2(g)} + CO_{(g)}$ ($K_2$)

    • Sum: $I_2CO_{(g)} + H_{2(g)} \rightleftharpoons 2HI_{(g)} + CO_{(g)}$ ($K_{new} = K_1 \times K_2$)

Practice Problems and Solutions

• K Calculation Exercise:

  • Given: $2N_{2(g)} + O_{2(g)} \rightleftharpoons 2N_2O_{(g)}$

  • Equilibrium concentrations: $[N_2] = 0.048\,\text{M}$, $[O_2] = 0.093\,\text{M}$, $[N_2O] = 6.55 \times 10^{-21}\,\text{M}$.

  • Calculation: $K = \frac{[N_2O]^2}{[N_2]^2[O_2]} = \frac{(6.55 \times 10^{-21})^2}{(0.048)^2(0.093)} = 2.00 \times 10^{-37}$.

• NO and O2 Manipulations:

  • Base: $NO_{(g)} + 0.5O_{2(g)} \rightleftharpoons NO_{2(g)}$ ($K = 1.23$)

  • 1. $NO_{2(g)} \rightleftharpoons NO_{(g)} + 0.5O_{2(g)}$ -> $K = \frac{1}{1.23} = 0.81$

  • 2. $2NO_{(g)} + O_{2(g)} \rightleftharpoons 2NO_{2(g)}$ -> $K = (1.23)^2 = 1.51$

  • 3. $2NO_{2(g)} \rightleftharpoons 2NO_{(g)} + O_{2(g)}$ -> $K = \frac{1}{(1.23)^2} = 0.66$

• Complex Addition Problem:

  • Target: $2I_{2(g)} + 2H_2O_{(g)} \rightleftharpoons 4HI_{(g)} + O_{2(g)}$

  • Given:

    • $O_{2(g)} + 2H_{2(g)} \rightleftharpoons 2H_2O_{(g)}$ ($K_1 = 1.4 \times 10^{11}$)

    • $I_{2(g)} + H_{2(g)} \rightleftharpoons 2HI_{(g)}$ ($K_2 = 0.5$)

  • Strategy: Multiply Rxn 2 by $2$ (Squaring $K_2$) and add to the inverse of Rxn 1 (Inv of $K_1$).

  • Result: $K_{new} = (K_2)^2 \times \frac{1}{K_1} = (0.5)^2 \times \frac{1}{1.4 \times 10^{11}} = 1.8 \times 10^{-12}$.