Chapter Medley: Chemical Equations and Reactions

Transcription

1 Chapter Medley: Chemical Equations and Reactions To help you with upcoming Chem 101 lab experiments, we will combine the various types of chemical reactions into this mid-quarter chapter medley. 1 I. Writing and Balancing Chemical Equations (Ch 6.6) II. Types of Chemical Reactions A. Combination Reactions (Ch 9.1) B. Decomposition Reactions (Ch 9.1) C. Single-Replacement Reactions (Ch 9.1) D. Double-Replacement Reactions (Ch 9.1) Solubility Rules (Ch 8.4) E. Combustion Reactions (Ch 9.1) F. Oxidation Reduction Reactions (Ch. 9.2 & 9.3) Supplemental Material: Comparing Strength of Agents G. Acid-Base Neutralization Reactions (Ch 10.1, 10.6 & 10.7)

2 2 I. Writing and Balancing Chemical Equations (Ch 6.6) demo: Combustion of ethanol Reactant Products ethanol + oxygen carbon dioxide + water C 2 H 6 O + 3 O 2 2 CO H 2 O Balanced Equation Law of Conservation of Mass An equation coefficient is a multiplier that applies to all atoms and subscripts. It changes the amount, but not the, of the substance. A coefficient of 1 in a balanced equation is not explicitly written.

3 3 Physical states of reactants and products may be indicated: (s) for solid (l) for liquid (g) for gas 4 Fe(s) + 3 O 2 (g) 2 Fe 2 O 3 (s) Reactions of substances in solution are indicated by the symbol (aq), meaning aqueous: Zn(s) + 2 HCl(aq) ZnCl 2 (aq) + H 2 (g)

4 A capital Greek delta ( ) placed above the arrow indicates that the reaction occurs at. The actual temperature required also may be written above or below the arrow o C The symbol for a catalyst may also appear above or below the arrow. A catalyst increases the of the reaction but is not consumed. CH =CH (g) + H (g) Pt 200 o C CH 3 CH 3 (g)

5 5 Guidelines for Balancing Chemical Equations Begin with an element that appears in only a single substance on each side of the equation. If there are several such elements, select the substance that has the largest number of atoms of a single element. To balance an equation, change only coefficients in front. NEVER change. Proceed to balance the atoms of other elements by the same process. Check to see if balancing one element in a substance has caused others to become unbalanced and make necessary corrections. Balance atoms that appear in two or more places on one side of the equation. Check and reduce all coefficients to smallest set of whole numbers (eliminate fractions).

6 6 Practice Problem of Balancing an Equation FeI 2 + Cl 2 FeCl 3 + I 2 (unbalanced) 1. balance Cl atoms FeI Cl 2 2 FeCl 3 + I 2 (unbalanced) 2. balance Fe atoms 2 FeI Cl 2 2 FeCl 3 + I 2 (unbalanced) 3. balance I atoms 2 FeI Cl 2 2 FeCl I 2 balanced Do a final check by counting atoms on each side! 2 Fe 2 Fe 4 I 4 I 6 Cl 6 Cl

7 7 Practice Problem C 3 H 8 O + O 2 CO 2 + H 2 O (unbalanced) 1. balance H (appears once on each side & in largest number) C 3 H 8 O + O 2 CO H 2 O (unbalanced) 2. balance C C 3 H 8 O + O 2 3 CO H 2 O (unbalanced) 3. balance O 6 O atoms in CO O atoms in H 2 O 10 O atoms on product side - 1 O atom in C 3 H 8 O on reactant side 9 O atoms needed from O 2 = 4 ½ molecules of O 2 C 3 H 8 O + 4 ½ O 2 3 CO H 2 O 4. eliminate fractions by multiplying the entire equation by 2; then check the final answer 2 C 3 H 8 O + 9 O 2 6 CO H 2 O balanced 6 C, 16 H, 2+18 = 20 O 6 C, 16 H, 12+8 = 20 O

8 8 Drill Problem Al(s) + NH 4 ClO 4 (s) The white clouds are products of the reaction used to lift the space shuttle. The solid propellants, aluminum powder and ammonium perchlorate, are more convenient to handle as rocket fuel. The white cloud is aluminum oxide. Balance the following reaction involved in the shuttle launch: Al 2 O 3 (s) + AlCl 3 (s) + NO(g) + H 2 O(g)

9 9 Chemical Reactions may have one or more : Formation of a gas Formation of a precipitate (an insoluble solid is formed) Generation of a color or color change Evolution or absorption of (exothermic or endothermic reaction)

10 10 A. Combination Reactions (Ch 9.1) A single product is produced from two or more reactants: X + Y XY 2 H 2 + O 2 2 H 2 O (demo in lecture 1) Figure 9.1 When a hot nail is stuck into a pile of zinc and sulfur, a fiery combination reaction occurs: Zn + S

11 Combination reactions with compounds as reactants: 2 NO 2 + H 2 O 2 2 HNO 3 nitrogen dioxide + hydrogen peroxide 11 Formation of SO 3 + H 2 O H 2 SO 4 acidic oxide + water acid Jābir ibn Hayyān ( AD) was a prominent polymath: a chemist and alchemist, astronomer and astrologer, engineer, geologist, philosopher, physicist, and pharmacist and physician. He is considered by some to be the "Father of Chemistry and is better known under the name of Geber. Geber is credited with the discovery of sulfuric acid.

12 12 B. Decomposition Reactions (Ch 9.1) A single reactant is converted into two or more products. XY X + Y 2 NH 3 3 H 2 + N 2 2 elements PF 5 PF 3 + F 2 compound & element CaCO 3 CaO + CO 2 2

13 13 C. Single-Replacement Reactions (Ch 9.1) An atom or molecule replaces an atom or group of atoms from a compound. Most common = an element replaces another. Some metals replace other metals or hydrogen: Fe + CuSO 4 Cu + FeSO 4 Zn + 2 HCl H 2 + Hydrogen can replace some metals: CuO + H 2 Cu + H 2 O A nonmetal may replace another nonmetal: Cl 2 + NiI 2 I 2 + NiCl 2 F NaCl Cl Order of reactivity: F > Cl > Br > I F 2 will replace all, I 2 will replace

14 14 D. Double-Replacement Reactions (Ch 9.1) Two substances exchange parts: AX + BY AY + BX Complete chemical equation (demo) Co(NO 3 ) 2 (aq) + Na 2 CO 3 (aq) CoCO 3 (s) + 2 NaNO 3 (aq) Ionic equation Soluble ionic compounds (salts) completely into ions in water: Co NO Na + + CO 3 2- CoCO 3 (s) + 2Na + + 2NO 3 - Net ionic equation Co 2+ + CO 3 2- CoCO 3 (s) Na + and NO 3 - are spectator ions. Formation of a drives the reaction. Learn to use the solubility rules to predict the products and to write the ionic and net ionic equation for any double-replacement reaction.

15 15 Solubility Rules to Predict Possible Products (Ch 8.4) Soluble - compound exists as ions in solution Insoluble - compound will form a precipitate No need to memorize these rules, a modified version of this table will be provided on exams or quizzes.

16 16 Practice Problem: Use the solubility rules to predict the product for the following reaction. Then write a balanced chemical equation, the ionic equation, and the net ionic equation for this reaction. Identify the spectator ions. KI (aq) + Pb(NO 3 ) 2 (aq) yellow ppt 2 colorless solutions demo We can write the complete chemical equation based on the fact that this is a double-replacement reaction in which two groups are exchanged: 2 KI (aq) + Pb(NO 3 ) 2 (aq) 2 KNO 3 (?) + PbI 2 (?) Now we check the solubility rules to identify which product is insoluble: 2 KI (aq) + Pb(NO ) (aq) 2 KNO 3 (aq) + PbI 2 (s) 3 2 soluble insoluble

17 2 KI (aq) + Pb(NO 3 ) 2 (aq) 2 KNO 3 (aq) + PbI 2 (s) 17 Compounds that are soluble (aq) exist as ions in solution: 2K + + 2I - + Pb NO 3-2K + + 2NO PbI 2 (s) Ions that exist on both sides of the equation are spectator ions and can be omitted in the net ionic equation: Net Ionic Equation: Spectator Ions: Note that the spectator ions remain present in solution but they DO NOT participate in the reaction.

18 18 Drill problems. Predict the outcomes: KI(aq) + Mg(NO 3 ) 2 (aq) demo CaCl 2 (aq) + AgNO 3 (aq) Na 2 SO 4 (aq) + BaBr 2 (aq)

19 19 E. Combustion Reactions (Ch 9.1) are chemical reactions with oxygen that proceed with the evolution of heat and light: C 2 H 6 O + 3 O 2 2 CO 2 + H 2 O ethanol C 3 H O 2 3 CO 2 + H 2 O propane 2 ZnS + 3 O 2 2 ZnO + SO 2 2 Mg + O 2 combustion reaction combination reaction redox reaction

20 20 F. Oxidation Reduction Reactions (Ch. 9.2 & 9.3) In redox reactions a transfer of electron from one reactant to another occurs. 2 Mg + O 2 oxidation = loss of electrons reduction = gain of electrons

21 21 Assignments of Oxidation Numbers (ox#) 1. In elemental states = Na Cl 2 P 4 O 2 Mg 2. Monatomic ions = charge on the ion Na + +1 S Zn H is usually +1 O is usually -2 F is always -1 in compounds

22 4. a. The sum of all oxidation numbers in a neutral compound = Zero b. For polyatomic ions the sum =. 5. In binary molecules, the more electronegative element is assigned a negative number equal to its charge in binary ionic compounds. CCl 4 Cl has oxidation number (ox#) of -1 ox# C + ox# 4(-1) = 0 ox# C = 22

23 23 Example A: (rules 3 & 4a) P 2 O 5 2(ox# P) + 5(ox# O) = 0 2(ox# P) + 5(-2) = 0 2(ox# P) = ox# P = Oxidation numbers for the elements in P 2 O 5 P = O = -2

24 24 Example B: (rules 2, 3 & 4a) KMnO 4 (ox# K) + (ox# Mn) + 4(ox# O) = 0 (+1) + (ox# Mn) + 4(-2) = 0 (ox# Mn) = Oxidation numbers for the elements in KMnO 4 K = +1 Mn = O = -2

25 25 Example C: (rules 3 & 4b) NO 3 - (ox# N) + 3(ox# O) = -1 (ox# N) + 3(-2) = -1 (ox# N) = Oxidation numbers for the elements in NO 3 - N = O = -2

26 26 Terminology Associated with Redox Processes (Ch 9.3) 2 Mg + O 2 2 MgO oxidation = loss of electrons reduction = gain of electrons Mg is oxidized from 0 to +2 O 2 accepts the electrons from Mg O 2 oxidizes Mg to Mg 2+ O 2 is the agent O 2 is reduced from 0 to -2 Mg transfers electrons to O 2 Mg reduces O 2 to O 2- Mg is the agent

27 27 Practice example: 2 KBr + Cl 2 2 KCl + Br K = no redox Br = oxidized from -1 to 0 Cl = agent Cl = reduced from 0 to -1 Br = agent NOTE: Electron transfer occurs in one direction only! 2 KCl + Br 2 No Reaction Recall order of reactivity: F > Cl > Br > I F 2 is the strongest oxidizing agent, it will replace all. I 2 is the weakest of the halogens, it will replace none.

28 Supplemental Material: Comparing Strength of Agents Recall: Li, Na, K reaction with H 2 O Order of reactivity: Li < Na < K 28 2 Na + 2 H 2 O 2 NaOH + H Na = strong reducing agent Na + = weak oxidizing agent Contrast with Ag, Au, Cu + H 2 O Noble metals do not react, they are weaker reducing agents that alkali metals. Predict the outcomes: Na + AgNO 3 Ag + LiBr

29 29 Demo: Compare the reactivity of Zn and Fe Zn is a stronger reducing agent than Fe. It can be used as a metal to prevent corrosion of iron.

30 30 G. Acid-Base Neutralization Reactions (Ch 10.1, 10.6 & 10.7) Acid + Base Salt + Water HX + BOH BX + HOH reaction

31 31 Arrhenius Acids produce H + in water Fig 10.1 Ionization is the process in which cations and anions are produced from a molecule dissolved in water. HCl hydrochloric acid H + + Cl - ionization in H 2 O HNO 3 nitric acid HClO 4 perchloric acid sulfuric acid H 3 PO 4 phosphoric acid You must know these acids.

32 32 Arrhenius Bases produce OH - in water Fig 10.1 Dissociation is the process in which cations and anions are released from an ionic compound that is dissolved in water. NaOH(s) H 2 O Na + (aq) + OH - (aq) Other common bases: KOH

33 33 Balancing Neutralization Reactions H 3 PO 4 (aq) + NaOH(aq) unbalanced H 3 PO 4 (aq) + NaOH(aq) Na 3 PO 4 (aq) + H 2 O(l) unbalanced H 3 PO 4 (aq) + 3 NaOH(aq) Na 3 PO 4 (aq) + 3 H 2 O(l) 3H + + PO Na + + 3OH - 3Na + + PO H 2 O Net ionic equation:

34 34 Figure 10.6 The acid-base reaction between sulfuric acid and barium hydroxide produces the insoluble salt barium sulfate. Drill Problem: 1. Write the complete, balanced equation for this neutralization, indicating the state of each compound (s), (aq) etc. 2. Write the ionic equation. 3. Identify any spectator ions.