Reactions Chemical Reactions

Stoichiometry of Chemical Reactions Chemical Reactions • A balanced chemical equation reflects the Law of the Conservation of Matter. • It must have the same number of atoms of the same kind on both sides. • A chemical reaction indicates the type of reactants and products and their relative amounts in a reaction. 4 Fe (s) + 3 O2(g) → 2 Fe2O3(s) • The above reaction says: • 4 Fe atoms + 3 O2 molecules → 2 molecules of Fe2O3 • 4 moles of Fe + 3 moles of O2 → 2 moles of Fe2O3 • The numbers in the front are called stoichiometric coefficients. • The letters (s), (g), (l) and (aq) indicate the physical states of reactants and products. Problems: Balancing Reactions – using whole numbers 1. __C3H8(g) + __ O2(g) → __CO2(g) + __ H2O(g) 2. ___B4H10(g) + __ O2(g) → __ B2O3(g) + __ H2O(g) 3. ___Pb (s) + __ O2 (g) + ___H2O (l) → __ Pb(OH)2 (s) 4. ___K3PO4 (aq) + __ Ca(NO3)2 (aq) → __ Ca3(PO4)2 (s) + __KNO3(aq) 5. ___Na3N (s) → __ Na (s) + ___ N2 (g) 6. ___C19H38O2 (l) + ___O2 (g) → ____ CO2 (g) + ___ H2O (l) 7. ___ TiCl4 (s) + ___H2O (g) → ___TiO2 (s) + ___HCl (g) Chemical Reaction Reversibility: Equilibrium • The direction of a chemical reaction is indicated by arrows when writing a chemical equation. • A single arrow (→) denotes the direction of completion (nonreversible). • Example: 4 Al(s) + 3 O2(g) → 2 Al2O3(s) • A double arrow ( ) indicates that the reaction does not go to completion but reaches a point where the forward rate = the reverse rate. • The reaction is at equilibrium. • Example: Ca2+ (aq) + 2 HCO3– (aq) CaCO3(s) + CO2(g) +H2O(l) Net Ionic Equations • To conveys only what is the chemical reaction (i.e., what is producing a liquid or gas or solid or undergoing reduction/oxidation a “net ionic equation” is written. • All other species (e.g., ions) that are not participating in the reaction are not stated in the chemical reaction. • Non participatory ions are referred to as spectator ions. Pb2+(aq) + CrO42-(aq) → PbCrO4(s) • In summary: • Complete formula/molecular chemical equation Pb(NO3)2(aq) + K2CrO4(aq) → PbCrO4(s) + 2 KNO3(aq) • Complete ionic chemical equation. Pb2+(aq) + 2 NO3-(aq) + 2 K+(aq) + CrO42-(aq) → PbCrO4(s) + 2 K+(aq) + 2 NO3-(aq) • Net ionic chemical equation. Pb2+(aq) + CrO42-(aq) → PbCrO4(s) Criteria for Chemical Reactions to Occur • Kinetics & Thermodynamics requirements must be meet. • Given that the kinetics and thermodynamics conditions have been met, the driving forces behind a chemical reaction are: • Making a gas • Making a liquid • Making a precipitate (solid) • Exchange of electrons (redox reaction) • At least one of the four driving forces must have occurred and/or be observed for a chemical reaction to have occurred. Chemical Reactions: Types Two main categories: • Decomposition • A reactant decomposes to two or more products • Synthesis • Two or more reactants combine to make product(s) Types of Synthesis reactions • Double Displacement • Precipitation reactions • Single Displacements • Gas Forming reactions • Neutralization reactions • Acid & Base • Reduction/Oxidation • Combustion reactions • Corrosion Solubility • Solubility is the ability of a solute to dissolve in a solvent to make a solution. • Factors that affect solubility: • Thermodynamics • Types of intermolecular attractive forces. • Environmental conditions • Temperature • Partial pressure (solute) • The ratio of volume of solvent to amount of solute. • Solubility is generally given in units of grams of solute that will dissolve per 100 grams of water (solvent). • Solubility curves are a graphical representation used to predict whether a solution is unsaturated, saturated or supersaturated for a given quantity of solvent (i.e., usually 100 grams) at varies temperatures. Reference: sideplayer.com Precipitation Reactions: Double Displacement (Metathesis) • A precipitate (i.e., solid) may form when two aqueous ionic solutions are combined. • Where or not a precipitate occurs depends on the solubility of the substance and the experimental conditions (e.g., temperature, amount of solute: solvent ratio, pressure, acidity). • Soluble vs. insoluble • Substance that have high solubility tend to remain in solution will substances with low solubility are said to be insoluble (solid). • Solubility Tables • Double displacement reactions result in the formation of a precipitate. • The anions exchange places with the cations – they switch their partners. • Example: Reference credit: pinterest.com Pb(NO3) 2(aq) + 2 KI(aq)--> PbI2(s) + 2 KNO3 (aq) Gas-Forming Reactions: Single Displacement Metal carbonates: • Metal carbonates + acid →CO2 (g) + salt + water • Example: CaCO3(s) + 2 HCl (aq) → CO2 (g) + H2O (l) + CaCl2 (aq) Net ionic: CaCO3(s) + 2 H+ (aq) → CO2 (g) + H2O (l) + Ca+2 (aq) Metal Acid reactions • Metal (s) + acid → Gas + salt • Example: Mg (s) + 2 HCl (aq) → H2 (g) + MgCl2 (aq) Net ionic: Mg(s) + 2 H+ → H2(g) + Mg+2 (aq) Acid & Base Overview • An acid is a substance that donates a H+ ion (H3O+) when in water (BronstedLowry). • Example: HCl (aq) → H+ (aq) + Cl- (aq) HCl (aq) + H2O (l) → H3O+ (aq) + Cl- (aq) • A base is a substance that accepts a H+ ion (H3O+) when in water (BronstedLowry). • Example: NH3 (aq) + H20 (l)  → NH4+ (aq) + OH- (aq) • Acid & Base Classification: • Strong: complete ionization in water. • Examples: HCl (acid) & KOH (base) • Also known as strong electrolytes. • Weak: Partial ionization in water • Example: HF (acid) or NH3 (base) • Also known as weak electrolytes. Acid Base Reactions: Neutralization • Acid-Base Reactions are referred to as NEUTRALIZATION reactions. • NEUTRALIZATIONS • When the number of moles of acid in solution = the moles of base in solution • DOES NOT mean that the pH of the solution is necessarily 7 • The driving force for strong acid- strong base reactions is the formation of water. • Example: KOH(aq) + HNO3 (aq) → KNO3 (aq) + H2O(l) Net ionic equation: OH- (aq) + H+ (aq) → H2O (l) • A common product of many acid-base reactions is the formation of water and a SALT (MX). The salt can be soluble or insoluble. Generic: HX + MOH → MX + H2O Mn+ comes from base & Xn- comes from acid Strong Acids vs. Weak Acids: It’s all about the Ionization Strong Acids Weak Acids An example of Weak Acids There are ONLY six strong acids HCl hydrochloric H2SO4 sulfuric HClO4 perchloric HNO3 nitric HBr hydrobromic HI hydroiodic CH3CO2H acetic acid H2CO3 carbonic acid H3PO4 phosphoric acid HF hydrofluoric acid H2SO3 sulfurous acid HNO2 nitrous acid HCN hydrocyanic acid Electrolytes Nonelectrolyte Weak electrolyte Strong electrolyte • Given similar polarities, when a solute dissolves in a solvent it can: • Dissolve as molecules (miscible). • Dissolve as ions (ionization). • Solutes that dissolve as molecules are called nonelectrolytes. • Don’t form ions • Solution can not conduct a current. • Solutes that dissociate into ions when dissolved in a solvent are known as electrolytes. • If the solute completely dissociates it is called a strong electrolyte. • If the solute partially dissociates, it is referred to as weak electrolyte. • Both solutions will conduct a current. Reference: socratic.org An Overview of Oxidation/Reduction (Redox) Terminology & Reactions Oxidation: Ga(l) → Ga3+(aq) + 3 e• Electrons are LOST by a substance • Oxidation number (state) increases as e- are donated. • Therefore, Ga is OXIDIZED • Ga is the REDUCING AGENT in this balanced half-reaction. Reduction: Br2 (l) + 2e- → 2 Br (aq) • Electrons are GAINED by a substance • Oxidation number (state) decreases as e- are accepted. • Therefore, Br2 is REDUCED • Br2 is the OXIDIZING AGENT in this balanced half-reaction. NOTE: If something is being reduced something else is being oxidized Oxidation States/Oxidation Numbers • Main group metal elements tend to lose electrons to achieve the “noble look” (i.e., have the same number of electrons as the noble gas in the presiding row (period)). • Example: Ca → looses 2 electrons to become Ca2+ • Alkali metals (1A) always have an oxidation state of +1. • Alkaline earth metals (2A) always have an oxidation state of +2. • Nonmetal elements tend to gain electrons to achieve the “noble look” ( i.e., have the same number of electrons as the noble gas at end of their row (period)). • Example: P gains 3 electrons to become → P3• Halogens mostly have an oxidation state of -1. • F always has an oxidation state of -1. • Oxygen most prevalent oxidation state is -2. • Example: Na2O (sodium oxide) • In peroxides: oxygen oxidation state is -1 • Example: NaO (sodium peroxide). • In superoxide: oxygen oxidation state is -1/2. • Example NaO2 (sodium superoxide). Common Oxidation States (Charges) Reference credit: OpenStax2017 Problem: Determine the oxidation state for P in PO4-3 Stoichiometric Basics • Reaction stoichiometry describes what is happening in a reaction. • Stoichiometry uses a balanced chemical reaction to predict yield (theoretical), as well as determine yield (percent/actual). • Strategy Basics: • Required: a balance reaction to determine the stoichiometric relationship between: a. Reactants to reactants or b. Reactants to products or c. Products to products • Go to the mole! a. If mass (g) is given, then divide by molecular or atomic mass (g/mol) to get mole: mole A = 𝐬𝐚𝐦𝐩𝐥𝐞 𝐦𝐚𝐬𝐬 𝐀 (𝐠) 𝐠 ) 𝐦𝐨𝐥 𝐦𝐨𝐥.𝐦𝐚𝐬𝐬 𝐨𝐟 𝐀 ( b. If volume and molarity (M) are given then: Ma x Va = moles A • Use stoichiometric factors to convert from mole A to mole B to solve problem. Problem: Given the following chemical reaction: C3H8 (l) + O2 (g)→ CO2 (g) + H2O(g) 1. Determine the theoretical yield (grams) for H2O if 33.1 grams of C3H8 was completely combusted. Assume that you have plenty of O2 to complete the reaction. 2. How many molecules of H2O would have been produced? Limiting Reactant & Percent Yield • Stoichiometry can be used not only to predict the theoretical yield of a reaction, but also to determine whether or not a reactant is limited. • If a reactant is in short supply then it LIMITS the quantity of product that can be formed. • Limiting reactant. • Knowing the limiting reactant allows you to determine the theoretical yield. • If the theoretical yield is known and the actual amount of product produced is known then a percent yield can be calculated. percent yield = ( 𝒂𝒄𝒕𝒖𝒂𝒍 𝒚𝒊𝒆𝒍𝒅 ) 𝒕𝒉𝒆𝒐𝒓𝒆𝒕𝒊𝒄𝒂𝒍 𝒚𝒊𝒆𝒍𝒅 x 100 Theoretical Yield vs. Percent Yield Theoretical Yield • It is the predicted yield for a chemical reaction based upon the LIMITING reagent. • Verbiage associated with theoretical yield: • Should produce, could produce, would produce. Percent Yield • It is all about the actual yield. • Verbiage associated with actual yield: • Did produce, did make, was produced, was made, was prepared. • Must know the theoretical yield to determine percent yield. • Use stoichiometry to predict theoretical yield. percent yield = ( 𝒂𝒄𝒕𝒖𝒂𝒍 𝒚𝒊𝒆𝒍𝒅 ) 𝒕𝒉𝒆𝒐𝒓𝒆𝒕𝒊𝒄𝒂𝒍 𝒚𝒊𝒆𝒍𝒅 x 100 Problem#1: Silicon nitride (Si3N4) is a ceramic material used to make a component for jet engine turbine blades. Si (s) + N2 (g) → Si3N4 (s) 1. Determine the theoretical yield in grams for silicon nitride if 2.00 g Si and 1.50 g N2 react. 2. Determine the percent yield of the reaction if 2.15 g of silicon nitride was produced. 3. Identify the excess reactant and determine how much of the excess reactant was not consumed during the reaction. Problem#2: Aspirin (C9H8O4), is synthesized by reacting salicylic acid (C7H6O3) with acetic anhydride (C4H6O3). The chemical equation for this reaction is given below: C7H6O3 (s) + C4H6O3 (aq) → C9H8O4 (s) + CH3COOH (aq) The synthesis was carried out by reacting 1.20 g of salicylic acid with 1.80 grams of acetic anhydride. A percent yield of 96.5% was determined. 1. Determine the actual mass of aspirin that was prepared. 2. Acetic anhydride is a liquid. How many milliliters of acetic anhydride was measured out in this experiment. The density of acetic anhydride is 1.082 g/mL Stoichiometry: Quantitative Analysis • Gravimetric analysis is a common technique used by chemists. • Stoichiometry is the basis of this technique’s results. • Gravimetric Analysis: a sample undergoes a treatment (e.g., heating, forming a solid, decomposition of a solid) that causes a change in the physical state of one of the reactants which then allows it to be isolated from the other reactants. Combustion analysis Reference credit: OpenStax2017 Problem#1: An impure mineral sample, containing Na2SO4 has a mass 0.123 g. The Na2SO4 in the sample was converted to insoluble BaSO4. The mass of BaSO4 produced was 0.177 g. 1. Balance the equation. 2. Write a net ionic equation for this reaction. 3. Determine the mass percent (w/w) of Na2SO4 in the impure mineral sample? Chemical reaction: Na2SO4 (aq) + BaCl2 (aq) → NaCl (aq) + BaSO4(s) Problem#2: • Polyethylene is a polymer used to produce food storage wrap. In a combustion analysis 0.00126 g sample of polyethylene yielded 0.00394 g carbon dioxide and 0.00161 g water. Determine the empirical formula for polyethylene. CxHy + oxygen → 0.00394g CO2 + 0.00161 g H2O Problem#3: • In a combustion analysis, a 6.71 g sample of an unknown fatty acid (CxHyOz) yielded 19.81 g carbon dioxide and 4.51 g of water. • In another experiment, the molecular mass of the unknown fatty acid was determined to be 268.34 g/mol. 6.71 CxHy Oz + oxygen → 19.81 g CO2 + 4.51 g H2O 1. 2. Determine the molecular formula for this fatty acid. Determine the mass of oxygen used to combust the fatty acid. Stoichiometry: Quantitative Analysis • Titration analysis is common techniques used by chemists. • Stoichiometry is the basis of this technique’s results. • Usually liquid samples, where the concentrations (e.g., molarity) of either the titrant (solution in buret) or the solute (in the reaction vessel) is known. Titration set-up Reference credit: imgaracade.com Problem #1: Determine the molarity of an oxalic acid solution, given the following information: • A 20.00 mL sample of oxalic acid, H2C2O4, was titrated with 0.09113 M potassium permanganate, KMnO4. It took 23.24 mL to reach the endpoint of the titration. 2 KMnO4 (aq) + 5 H2C2O4 (aq) + 6 HCl (aq) → 10 CO2 (g) + 2 MnCl2 (aq) + 2 KCl (aq) + 8 H2O (l) Problem#2: 1. Balance the reaction. _ HClO4(aq) + _ Mg(OH)2(aq) → _Mg(ClO4)2 (aq) + _H2O(l) 2. Determine the molarity for HClO4 given the following information. • It takes 31.15 mL of 0.550 M Mg(OH)2 to reach the endpoint of a reaction when 15.0 mL aliquot of HClO4 is titrated. 3. Identify the type of chemical reaction.