Chemical equilibrium is all about balance. When reactions reach a steady state, the concentrations of reactants and products stabilize. Understanding this balance is key to predicting how reactions behave and calculating important values.

The equilibrium constant (K) is the star of the show. It tells us the ratio of products to reactants at equilibrium. By using K and some clever math tricks, we can figure out concentrations, pressures, and even pH levels for all sorts of chemical systems.

Equilibrium Concentrations and Pressures

Calculations with equilibrium constant

• Equilibrium constant (K) represents ratio of product of concentrations of products raised to their stoichiometric coefficients divided by product of concentrations of reactants raised to their stoichiometric coefficients
  • For general reaction $aA + bB \rightleftharpoons cC + dD$, equilibrium constant expression is $K = \frac{[C]^c[D]^d}{[A]^a[B]^b}$
    • $[A]$, $[B]$, $[C]$, and $[D]$ represent molar concentrations of species A, B, C, and D respectively
    • $a$, $b$, $c$, and $d$ represent stoichiometric coefficients of species A, B, C, and D respectively
• If equilibrium concentrations or pressures are known, they can be substituted into equilibrium constant expression to calculate K
  • For example, if $[A] = 0.1 M$, $[B] = 0.2 M$, $[C] = 0.3 M$, and $[D] = 0.4 M$ at equilibrium for reaction $A + 2B \rightleftharpoons C + D$, then $K = \frac{(0.3)(0.4)}{(0.1)(0.2)^2} = 3$
• If equilibrium constant and all but one of equilibrium concentrations or pressures are known, unknown concentration or pressure can be calculated using equilibrium constant expression
  • For example, if $K = 3$, $[A] = 0.1 M$, $[B] = 0.2 M$, and $[D] = 0.4 M$ at equilibrium for reaction $A + 2B \rightleftharpoons C + D$, then $[C]$ can be calculated as $[C] = \frac{K[A][B]^2}{[D]} = \frac{(3)(0.1)(0.2)^2}{0.4} = 0.3 M$

ICE table method

• ICE table method used to solve for equilibrium concentrations or pressures when initial concentrations or pressures and equilibrium constant are known
• ICE table consists of three rows:
  1. Initial (I): concentrations or pressures of reactants and products before reaction starts
  2. Change (C): changes in concentrations or pressures of reactants and products as reaction proceeds to equilibrium
  3. Equilibrium (E): concentrations or pressures of reactants and products at equilibrium
• Change in concentration or pressure for each species determined by stoichiometry of reaction and extent of reaction (x)
  • For example, for reaction $A + 2B \rightleftharpoons C + D$, if initial concentration of A is $[A]_0$ and extent of reaction is x, then change in concentration of A is -x, change in concentration of B is -2x, change in concentration of C is +x, and change in concentration of D is +x
• Equilibrium concentrations or pressures are sum of initial concentrations or pressures and changes in concentrations or pressures
  • For example, for reaction $A + 2B \rightleftharpoons C + D$, if initial concentration of A is $[A]_0$ and extent of reaction is x, then equilibrium concentration of A is $[A] = [A]_0 - x$, equilibrium concentration of B is $[B] = [B]_0 - 2x$, equilibrium concentration of C is $[C] = x$, and equilibrium concentration of D is $[D] = x$
• Equilibrium concentrations or pressures substituted into equilibrium constant expression, and resulting equation solved for x
  • For example, for reaction $A + 2B \rightleftharpoons C + D$ with $K = 3$, $[A]_0 = 0.1 M$, and $[B]_0 = 0.2 M$, equilibrium constant expression is $3 = \frac{(x)(x)}{(0.1-x)(0.2-2x)^2}$, which can be solved for x using quadratic formula or other methods
• Value of x then used to calculate equilibrium concentrations or pressures of all species
  • For example, if x = 0.0577 M for reaction $A + 2B \rightleftharpoons C + D$ with $[A]_0 = 0.1 M$ and $[B]_0 = 0.2 M$, then equilibrium concentrations are $[A] = 0.1 - 0.0577 = 0.0423 M$, $[B] = 0.2 - 2(0.0577) = 0.0846 M$, $[C] = 0.0577 M$, and $[D] = 0.0577 M$

Multiple equilibria systems

• In system with multiple equilibria, concentrations or pressures of species involved in each equilibrium affected by other equilibria
• To solve for equilibrium concentrations or pressures in system with multiple equilibria:
  1. Write equilibrium constant expressions for each equilibrium
  2. Use ICE table method to set up equations for equilibrium concentrations or pressures of each species in terms of extent of reaction variables (x, y, etc.)
  3. Substitute equilibrium concentrations or pressures into equilibrium constant expressions
  4. Solve resulting system of equations for extent of reaction variables
  5. Use values of extent of reaction variables to calculate equilibrium concentrations or pressures of all species
• For example, consider system with two equilibria: $A \rightleftharpoons B$ with $K_1 = 2$ and $B \rightleftharpoons C$ with $K_2 = 3$. If initial concentration of A is $[A]_0 = 0.1 M$ and initial concentrations of B and C are zero, then ICE tables for each equilibrium are:

Equilibrium in acid-base reactions

• Acid-base reactions are type of equilibrium reaction in which acid and base react to form salt and water
• Equilibrium constant for acid-base reaction called acid dissociation constant (Ka) for forward reaction or base dissociation constant (Kb) for reverse reaction
  • For general acid dissociation reaction $HA \rightleftharpoons H^+ + A^-$, acid dissociation constant expression is $K_a = \frac{[H^+][A^-]}{[HA]}$
    • $[HA]$, $[H^+]$, and $[A^-]$ represent molar concentrations of undissociated acid, hydrogen ion, and conjugate base respectively
  • For general base dissociation reaction $B + H_2O \rightleftharpoons BH^+ + OH^-$, base dissociation constant expression is $K_b = \frac{[BH^+][OH^-]}{[B]}$
    • $[B]$, $[BH^+]$, and $[OH^-]$ represent molar concentrations of base, conjugate acid, and hydroxide ion respectively
• pH of solution at equilibrium can be calculated using equilibrium concentrations of $H^+$ and $OH^-$
  • $pH = -\log[H^+]$
    • For example, if $[H^+] = 1 \times 10^{-7} M$, then $pH = -\log(1 \times 10^{-7}) = 7$
  • $pOH = -\log[OH^-]$
    • For example, if $[OH^-] = 1 \times 10^{-7} M$, then $pOH = -\log(1 \times 10^{-7}) = 7$
  • $pH + pOH = 14$ at 25°C
    • For example, if $pH = 7$, then $pOH = 14 - 7 = 7$
• ICE table method can be used to solve for equilibrium concentrations of $H^+$ and $OH^-$ in acid-base reaction, and then pH can be calculated
  • For example, for dissociation of 0.1 M acetic acid ($CH_3COOH$) with $K_a = 1.8 \times 10^{-5}$, ICE table is:

|$CH_3COOH$|$\rightleftharpoons$|$H^+$|+|$CH_3COO^-$| |:--|:--|:--|:--|:--|:--| | Initial | 0.1 | | 0 | | 0 | |Change|-x|+x|+x| |Equilibrium|0.1-x|x|x| Equilibrium constant expression is $K_a = \frac{(x)(x)}{0.1-x} = 1.8 \times 10^{-5}$. Solving for x gives x = $1.34 \times 10^{-3} M$. Therefore, $[H^+] = 1.34 \times 10^{-3} M$ and $pH = -\log(1.34 \times 10^{-3}) = 2.87$.