Chemical Equilibrium

Core Concept

• Chemical equilibrium occurs when forward and reverse reaction rates are equal
• Concentrations of reactants and products remain constant (not necessarily equal)
• Denoted by $\rightleftharpoons$ in chemical equations

Equilibrium Constant ($K$)

For a general reaction: $aA+bB\rightleftharpoons cC+dD$

The equilibrium constant is:

$K=\frac{[C{]}^c[D{]}^d}{[A{]}^a[B{]}^b}$

• $K>1$: Products are favored
• $K<1$: Reactants are favored
• $K\approx 1$: Neither is strongly favored

Types of Equilibrium Constants

• $K_c$: Based on concentrations (mol/L)
• $K_p$: Based on partial pressures (atm)
  • Relationship: $K_p=K_c(RT{)}^{\Delta n}$
  • Where $\Delta n=$ moles of gaseous products - moles of gaseous reactants

Le Châtelier’s Principle

System at equilibrium will shift to counteract any imposed change:

1. Concentration changes:

   • Adding reactant: Shifts right
   • Adding product: Shifts left
   • Removing reactant: Shifts left
   • Removing product: Shifts right

2. Pressure/Volume changes (gases only):

   • Increasing pressure: Shifts toward fewer gas molecules
   • Decreasing pressure: Shifts toward more gas molecules

3. Temperature changes:

   • Endothermic reaction ($\Delta H>0$):
     • Increasing temperature shifts right
     • Decreasing temperature shifts left
   • Exothermic reaction ($\Delta H<0$):
     • Increasing temperature shifts left
     • Decreasing temperature shifts right

4. Catalysts:

   • No effect on equilibrium position
   • Only speeds up rate of reaching equilibrium

Calculating Equilibrium Concentrations

1. Write balanced equation
2. Set up ICE table (Initial, Change, Equilibrium)
3. Express $K$ in terms of equilibrium concentrations
4. Solve for unknown values

Reaction Quotient ($Q$)

• Same form as $K$ but uses non-equilibrium concentrations
• Comparing $Q$ and $K$:
  • $Q<K$: Reaction shifts right
  • $Q>K$: Reaction shifts left
  • $Q=K$: System at equilibrium

Common Ion Effect

• Adding an ion already in solution shifts equilibrium
• Example: Adding $C{l}^{-}$ to $AgCl\rightleftharpoons A{g}^{+}+C{l}^{-}$ shifts left

Acid-Base Equilibrium

• $K_a$: Acid dissociation constant
• $K_b$: Base dissociation constant
• Relationship: $K_a\times K_b=K_w=1.0\times 10^{-14}$ at 25°C

Solubility

$$\begin{array}{rl}& K_{sp}=\text{solubility product}\\ & \end{array}$$

value /ksp >>> 1000 ⇒ approx rule todo (u can ignore x with this )

x from the ICE tables also represents solubility of a compound

When coverting from pH to [$H3O^{+}$], you want to use the n decimals (not the leading num) as the sig figs for your answer

Strong acid: HCl, HBr, HI, HNO3, H2SO4, HClO. Ionizes 100% in water.

• Technically still at an equilibrium. .But the foward reaction is much much faster

Weak acids:

• Don’t ionize 100% in water.
• reach equilvrium with their anion and hydronium and therefore do nor ionize more

$$K_{eq}=\frac{[H_3{O}^{+}][C{H}_{3}O{O}^{-}]}{[C{H}_{3}COOH]}=K_a$$

Group 1 bases are strong

Weak Base

• Reacts with water to form hydroxide ion and a cation.
• Reaches equilibrium

$$\begin{array}{rl}& K_{eq}=K_b=\frac{[H_{2}C{O}_3][O{H}^{-}]}{[HC{O}_3-]}\end{array}$$