Chemical Kinetics, Chemical Reactions, and Precipitation Notes

Chemical Kinetics, Chemical Reactions, and Precipitation

Chemical Reactions: The Rate Concept

• Reactions are reversible and can achieve chemical equilibrium.
• At equilibrium, the rates of forward and backward reactions are equal.
• Guldberg and Waage defined equilibrium as the point where forward and reverse reaction rates are equal, leading to the equilibrium constant.
• Consider the chemical reaction: $aA + bB = cC + dD$
• Rate of forward reaction: $Rate<em>{fwd} = k</em>{fwd}[A]^a[B]^b$
• Rate of reverse reaction: $Rate<em>{rev} = k</em>{rev}[C]^c[D]^d$
• At equilibrium: $k<em>{fwd}[A]^a[B]^b = k</em>{rev}[C]^c[D]^d$
• Molar equilibrium constant: $\frac{[C]^c[D]^d}{[A]^a[B]^b} = \frac{k<em>{fwd}}{k</em>{rev}} = K$
• Large K indicates that the equilibrium favors the products.
• Small K indicates that the equilibrium favors the reactants.

Types of Equilibria

• Acid-base dissociation:
  • Reaction: $HA + H<em>2O = H</em>3O^+ + A^-$
  • Equilibrium Constant: $K_a$, acid dissociation constant
• Solubility:
  • Reaction: $MA = M^{n+} + A^{n-}$
  • Equilibrium Constant: $K_{sp}$, solubility product
• Complex formation:
  • Reaction: $M^{n+} + aL^{b-} = ML_{(n-ab)+}$
  • Equilibrium Constant: $K_f$, formation constant
• Reduction-oxidation:
  • Reaction: $A<em>{red} + B</em>{ox} = A<em>{ox} + B</em>{red}$
  • Equilibrium Constant: $K_{eq}$, reaction equilibrium constant
• Phase distribution:
  • Reaction: $A<em>{H2O} = A</em>{organic}$
  • Equilibrium Constant: $K_D$, distribution coefficient
• Equilibrium constants can be written for dissociations, associations, reactions, or distributions.

Collision Theory

• Collision theory explains the varying rates of different reactions and suggests ways to alter these rates.
• For a chemical reaction to occur, reacting particles must collide.
• The reaction rate depends on the frequency of collisions.
• Reacting particles may collide without reacting.
• Successful collisions must:
  • Occur with sufficient energy.
  • Have the proper orientation.

Activation Energy

• Activation energy is the minimum energy needed to activate molecules for a chemical reaction.

Transition State

• Reaction coordinate: $A + B \rightarrow A…B \rightarrow A - B$
• Transition state (A…B) is the highest potential energy point along the reaction coordinate.
  • It's a short-lived intermediate that decomposes into products.
• Activation energy (Ea) is the minimum energy required for a reaction to occur.
  • Reactants need activation energy to form an intermediate complex before products are formed.

Le Chatelier's Principle

• When stress is applied to a system at chemical equilibrium, the equilibrium shifts to relieve that stress.
• Devised by Henry-Louise Le Châtelier in 1884.
• Stress can include changes in pressure, temperature, or concentration.

Effect of Temperature

• Exothermic Reaction:
  • $[starting materials] = [products] + HEAT$
  • Increasing temperature favors starting materials.
  • Decreasing temperature favors products.
• Endothermic Reaction:
  • $[starting materials] + HEAT = [products]$
  • Increasing temperature favors products.
  • Decreasing temperature favors starting materials.

Effect of Pressure

• Based on the amount of gaseous phase in equilibrium reactions.
  • $N<em>2(g) + 3H</em>2(g) = 2NH_3(g)$
  • $PCI<em>5(g) = PCI</em>3(g) + Cl_2(g)$
  • $N<em>2(g) + O</em>2(g) = 2NO(g)$

Effect of Concentration

• Adding or removing a component will cause the equilibrium to re-establish itself.
  • $3I^- + 2Fe^{3+} = I_3^- + 2Fe^{2+}$

Catalysts

• Catalysts affect the rate at which equilibrium is attained by influencing both forward and backward reaction rates.
• Catalysts change the activation energy of a reaction.
• Positive Catalysts:
  • Decrease activation energy, providing an alternative smaller path.
• Negative Catalysts:
  • Increase activation energy, providing an alternate larger path.

Gravimetry

• Gravimetry is the quantitative measurement of an analyte by weighing a pure, solid form of the analyte.
• Pure solids are obtained from solutions containing metal ions through precipitation.
• Accurate gravimetric analysis requires careful handling during precipitate formation and treatment.

Solubility Rules

1. Salts containing Group I elements (Li+, Na+, K+, Cs+, Rb+) are soluble.
   • Salts containing the ammonium ion (NH4+) are also soluble.
2. Salts containing nitrate ion (NO3-) are generally soluble.
3. Salts containing Cl-, Br-, or I- are generally soluble.
   • Exceptions: halide salts of Ag+, Pb2+, and (Hg2)2+ (AgCl, PbBr2, and Hg2Cl2 are insoluble).
4. Most silver salts are insoluble.
   • Exceptions: AgNO3 and Ag(C2H3O2).
5. Most sulfate salts are soluble.
   • Exceptions: CaSO4, BaSO4, PbSO4, Ag2SO4, and SrSO4.
6. Most hydroxide salts are only slightly soluble.
   • Group I hydroxides are soluble.
   • Group II hydroxides (Ca, Sr, and Ba) are slightly soluble.
   • Transition metal and Al3+ hydroxides are insoluble (e.g., Fe(OH)3, Al(OH)3, Co(OH)2).
7. Most sulfides of transition metals are highly insoluble (e.g., CdS, FeS, ZnS, Ag2S).
   • Arsenic, antimony, bismuth, and lead sulfides are also insoluble.
8. Carbonates are frequently insoluble.
   • Group II carbonates (CaCO3, SrCO3, and BaCO3) are insoluble, as are FeCO3 and PbCO3.
9. Chromates are frequently insoluble.
   • Examples include PbCrO4 and BaCrO4.
10. Phosphates such as Ca3(PO4)2 and Ag3PO4 are frequently insoluble.
11. Fluorides such as BaF2, MgF2, and PbF2 are frequently insoluble.

Steps in Gravimetric Analysis

1. Preparation of the solution.
2. Precipitation.
3. Digestion.
4. Filtration.
5. Washing.
6. Drying or igniting.
7. Weighing.
8. Calculation.

Precipitation Process

• Supersaturation: Solution contains more dissolved salt than it can carry at equilibrium (metastable condition).
• Nucleation: Minimum number of particles come together to produce microscopic nuclei of the solid phase.
• Precipitation: Initial nucleus grows by depositing other precipitate particles to form a crystal of a geometric shape.

Impurities in Precipitates

• Precipitates tend to carry down other soluble constituents, causing contamination (coprecipitation).
• Inclusion and Occlusion:
  • Occlusion: material not part of the crystal structure is trapped within a crystal.
  • Inclusion: ions of similar size and charge are trapped within the crystal lattice.
• Surface Adsorption:
  • The precipitate surface has a primary adsorbed layer of excess lattice ions, leading to contamination.
• Isomorphous Replacement:
  • Compounds with similar formula types and crystal structures can have one ion replace another in a crystal, forming a mixed crystal.
• Postprecipitation:
  • A second substance slowly forms a precipitate with the precipitating reagent when the precipitate stands in contact with the mother liquor.

Precipitation Equilibria

• "Insoluble" compounds are actually slightly soluble.
  • $AgCl = (AgCl)_{aq} = Ag^+ + Cl^-$
• The precipitate has a definite solubility (g/L or mol/L) at a given temperature (saturated solution).

Solubility Constant

• $AgCl = (AgCl)_{aq} = Ag^+ + Cl^-$
• $K_{sp} = [Ag^+][Cl^-]$
• The solid does not appear in the $K_{sp}$ expression.

Example: Solubility Constant

• $Ag<em>2CrO</em>4 = 2Ag^+ + CrO_4^{2-}$
• $K<em>{sp} = [Ag^+]^2[CrO</em>4^{2-}]$

Example: Solubility Constant

• The $K_{sp}$ of AgCl at 25°C is $1.0 × 10^{-10}$. Calculate the concentrations of $Ag^+$ and $Cl^-$ in a saturated solution of AgCl, and the molar solubility of AgCl.
• Solution:
  • $AgCl = Ag^+ + Cl^-$
  • $K_{sp} = [Ag^+][Cl^-]$
  • Let s represent the molar solubility of AgCl. Therefore,$[Ag^+] = [Cl^-] = s$
  • $s^2 = 1.0 × 10^{-10}$
  • $s = 1.0 × 10^{-5} M$
  • The solubility of AgCl is $1.0 × 10^{-5} M$.

Example: Solubility Constant

• What must be the concentration of added $Ag^+$ to just start precipitation of AgCl in a $1.0 × 10^{-3} M$ solution of NaCl?
• Solution:
  • $[Ag^+](1.0 × 10^{-3}) = 1.0 × 10^{-10}$
  • $[Ag^+] = 1.0 × 10^{-7} M$
• The concentration of $Ag^+$ must just exceed $10^{-7} M$ to begin precipitation.
• Caveat: Supersaturation is often needed before precipitation begins.

Example: Solubility Constant

• What is the solubility of $PbI_2$, in g/L, if the solubility product is $7.1 × 10^{-9}$?
• Solution:
  • The equilibrium is $PbI<em>2 = Pb^{2+} + 2I^-$, and $K</em>{sp} = [Pb^{2+}][I^-]^2 = 7.1 × 10^{-9}$. Let s represent the molar solubility of $Pbl_2$. Then
  • $[Pb^{2+}] = s$
  • and $[I^-] = 2s$
  • $(s) (2s)^2 = 7.1 × 10^{-9}$
  • $s=\sqrt[3]{\frac{7.1 \times 10^{-9}}{4}} = 1.2 × 10^{-3} M$
  • Therefore, the solubility, in g/L, is
  • $1.2 × 10^{-3} mol/L \times 461.0 g/mol = 0.55 g/L$