chem

Equilibrium Partial Pressures

• Focus on solving for the new equilibrium partial pressures of species in a reaction.
  • It is important to specifically find the partial pressure of iodine after the equilibrium has been established.

Equilibrium Constant (K)

• The equilibrium constant (K) remains the same for the reaction as long as the temperature does not change.
  • For the given case, K remains at 0.23.
• Changes in pressure affect the concentrations of reactants and products but do not affect the value of K directly.

Calculating Partial Pressures

• The expression for K is derived from concentrations of reactants and products at equilibrium.
• Example calculation:
  • Left-hand side expression proceeds as follows:
  • K = $(0.23) = (0.041 - 0.4x)^n * (2x)^m$ where n and m represent the stoichiometric coefficients of the species in the reaction.
• The result for x, obtained from this expression, is either 0.018 or -9.6, but only the positive value is meaningful in a physical context.

Formation Constant

• Large K values imply a tendency toward product formation in reactions, indicating a direction favoring complex ion formation.
• When converting nickel (Ni) to its complex ion by ethylenediamine, 67% of nickel is converted into the complex.
  • Here we determine the concentration of ethylenediamine required for this conversion.

Percent Concentration Relationships

• If C represents the initial concentration of nickel before reaction, then after conversion:
  • 67% of C is the concentration of the nickel complex ion.
  • Nickel ion remaining would then constitute 33% of C.

Reaction Categories

• Formation reactions involve the generation of a solid complex from cations and anions, prompting a balanced reaction representation:
  • Example: For lead(II) chloride ($PbCl_2$)
  • Dissolution reaction is: $PbCl_2 (s) \leftrightarrow Pb^{2+} (aq) + 2Cl^{-} (aq)$
• Ksp expressions omit solids and liquids:
  • Ksp = $[Pb^{2+}][Cl^{-}]^2$

Types of Ksp Problems

1. Finding ion concentration at equilibrium
2. Determining if precipitation will occur given a certain concentration (using Q and Ksp comparison)
3. Identifying the states of saturation (supersaturated, saturated, unsaturated) for a solution
   • Supersaturated solutions may precipitate if additional solute is added or disturbed.
4. Relating the concentration of a solution to the total mass in a given volume.
   • Example: 0.5 mg of sodium sulfate in 1 L relates to molarity calculations for sulfate ions specifically.

Conclusion on K and Product Formation

• When the equilibrium constant K is larger, this indicates a higher propensity for products (ions) to form in solution, affecting our calculations and predictions regarding the reaction's outcome.
  • Key definitions and relationships include Ksp and the concept of precipitation based on reaction conditions.