1. Chem 1151: Ch. 5 Chemical Reactions

2. The Chemical Equation 2H 2 (g) + O 2 (g)  2H 2 O( l )  Identifies reactant(s) and product(s)  Identifies states of matter (g = gas, l = liquid, s = solid, aq = aqueous)  Demonstrates law of conservation of matter in balanced equation  Law of conservation of matter: Atoms are neither created nor destroyed in a chemical reaction, but rearranged to form different molecules.  This equation, as written, describes the reaction between individual molecules (2 molecules of H 2 react with 1 molecule of O 2 ) as well as molar relationships (2 moles of H 2 react with 1 mole of O 2 ). Reactant(s)Product(s)

3. Balancing The Chemical Equation N 2 (g) + H 2 (g)  NH 3 (g)  You cannot change the natural molecular formulas, you can only change the coefficients indicating number of molecules or moles. N 2 (g) + H 2 (g)  2NH 3 (g)  By multiplying NH 3 by coefficient of 2, you now have 2N and 6H. This balances with N 2 on reactant side, but not H 2. N 2 (g) + 3H 2 (g)  2NH 3 (g)  Now check your work. 2 N 6 H 2 N 6 H Reactant(s)Product(s)

4. Balancing The Chemical Equation NO(g) + O 2 (g)  NO 2 (g)  Multiply NO 2 by coefficient of 2, then multiply NO by 2.  Now check your work. 2 N 4 O 2 N 4 O Reactant(s)Product(s) nitrogen monoxide  Reactant side has 3O, but product side has 2O. 2NO(g) + O 2 (g)  2NO 2 (g)

5. Types of Reactions Seager SL, Slabaugh MR, Chemistry for Today: General, Organic and Biochemistry, 7 th Edition, 2011

6. Redox Reactions Oxidation  To combine with oxygen  To lose hydrogen  To lose electrons  To increase in oxidation number Reduction  To lose oxygen  To combine with hydrogen  To gain electrons  To decrease in oxidation number Redox  Combination of Reduction and Oxidation 4Fe(s) + 3O 2 (g)  2Fe 2 O 3 (s) 2Fe 2 O 3 (s) + 3C(s)  4Fe(s) + 3CO 2 (g)

7. Oxidation Numbers (States): Pos. or neg. numbers assigned to elements in chemical formula.  Determined by the following rules: Rule 1: The O.N. of any uncombined element is zero. Exs. Al(0), O 2 (0), H 2 (0), Br 2 (0), Na(0) Rule 2: The O.N. of a simple ion is equal to the charge on the ion. Exs. Na + (+1), Mg 2+ (+2), Br - (-1) Rule 3: The O.N. of group IA and IIA elements are +1 and +2, respectively. Exs. NaOH (Na = +1), CaCl 2 (Ca = +2) Rule 4: The O.N. of H is +1. Exs. HCl (H = +1), H 3 PO 4 (H = +1) Rule 5: The O.N. of oxygen is -2 except in peroxides where it is -1. Exs. H 2 O (O = -2), H 2 O 2 (O = -1) Rule 6: The algebraic sum of all O.N.s of all atoms in complete compound formula equals zero. Rule 7: The algebraic sum of all O.N.s of all atoms in a polyatomic ion is equal to the charge on the ion.

8. Determining Oxidation Numbers K 2 CO 3 CO 2 2(O.N of K) + (O.N. of C) + 3(O.N. of O) = 0 2(+1) + 3(-2) = -4 2(+1) + (+4) + 3(-2) = 0 (O.N of C) + 2(O.N. of O) = 0 2(-2) = -4 (+4) + 2(-2) = 0 CH 2 O(O.N of C) + 2(O.N. of H) + (O.N. of O) = 0 2(+1) + (-2) = 0 (0) + 2(+1) + (-2) = 0 NO 3 - (O.N of N) + 3(O.N. of O) = -1 + 3(-2) = -6 (+5) + 3(-2) = -1

9. Redox e- Transfer (O.N. of S) + 2(O.N. of O) = 0 S(s) + O 2 (g)  SO 2 (g) (O.N of uncombined element = 0) (+4) + 2(-2) = 0  Sulfur is oxidized (combines with O and loses 4 e-, increases O.N. by 4)  Oxygen is reduced (gains 4 e-, decreases O.N. by 4)  Reduction and oxidation processes occur simultaneously  Reducing Agent: Reduces something else and becomes oxidized (loses e-)  Oxidizing Agent: Oxidizes something else and becomes reduced (gains e-) OIL RIG  Oxidation Is Loss (of e-)  Reduction Is Gain (of e-) Note: We can apply these concepts to covalently-bonded elements, but atoms in these compounds do not actually acquire a net charge

10. Redox e- Transfer: Examples 4(O.N. of Al) + 6(O.N. of O) = 0 4Al(s) + 3O 2 (g)  2Al 2 O 3 (s) (O.N of uncombined element = 0) + 6(-2) = -12  Al is oxidized (combines with O and each Al loses 3 e- and increases O.N. by 3)  Al is reducing agent  Oxygen is reduced (each of 6 O gains 2 e- and decreases O.N. by 2)  O is oxidizing agent Mission: Determine O.N. for each atom in the following and identify oxidizing and reducing agents. 4(+3) + 6(-2) = 0

11. Redox e- Transfer: Examples S 2 O 8 2- (aq) + 2I - (aq)  I 2 (aq) + 2SO 4 2- (aq) 2(simple ion charge = -1) = -2  I: O.N. increases from -1 to 0 (loses e-, oxidized). I -  Reducing agent  S: O.N. decreases from +7 to +6 (gains e-, reduced).  S 2 O 8 2- Oxidizing agent Mission: Determine O.N. for each atom in the following and identify oxidizing and reducing agents. 2(O.N. of S) + 8(O.N. of O) = -2 8(-2) = -16 S 2 O 8 2- (aq) 2SO 4 2- (aq) 2(+7) + 8(-2) = -2 2I - (aq) I 2 (aq) (O.N of uncombined element = 0) 2(O.N. of S) + 8(O.N. of O) = -4 8(-2) = -16 2(+6) + 8(-2) = -4

12. Redox e- Transfer: Examples 2KI(aq) + Cl 2 (aq)  2KCl(aq) + I 2 (aq) 2(O.N. of K) + 2(O.N. of I) = 0  I: O.N. increases from -1 to 0 (loses e-, oxidized).  KI reducing agent  Cl: O.N. decreases from 0 to -1 (gains e-, reduced).  Cl 2 Oxidizing agent Mission: Determine O.N. for each atom in the following and identify oxidizing and reducing agents. I 2 (aq) 2KI(aq) 2KCl(aq) 2(O.N. of K) + 2(O.N. of Cl) = 0 2(+1) = +2 2(+1) + 2(-1) = 0 2(+1) = +2 2(+1) + 2(-1) = 0 Cl 2 (aq) (O.N of uncombined element = 0)

13. Types of Reactions A  B + C  Decomposition Reactions  Combination Reactions  Replacement Reactions A + B  C A + BX  B + AX AX + BY  BX + AY Single Double

14. Decomposition Reactions A  B + C  A single substance is broken down to form 2 or more simpler substances.  These may or may not be redox reactions. 2HgO(s)  2Hg( l ) + O 2 (g) CaCO 3 (s)  CaO(s) + CO 2 (g) Note conservation of mass in balanced equations Seager SL, Slabaugh MR, Chemistry for Today: General, Organic and Biochemistry, 7 th Edition, 2011

15. Combination Reactions A + B  C  AKA addition or synthesis reactions.  Two or more substances react to form a single substance.  Reactants may be elements and/or compounds, but product is always a compound. 2Mg(s) + O 2 (g)  2MgO(s) SO 3 (g) + H 2 O( l )  H 2 SO 4 (aq) Redox combustion reaction Nonredox reaction, formation of acid rain Seager SL, Slabaugh MR, Chemistry for Today: General, Organic and Biochemistry, 7 th Edition, 2011

16. Single Replacement Reactions A + BX  B + AX  Always redox reactions.  Used to obtain metals from oxide ores. 3C(s) + 2Fe 2 O 3 (s)  4Fe(s) + 3CO 2 (g) 1(O.N. of Cu) + 1(O.N. of O) = 0 1(-2) = -2 H 2 (g) 1(+2) + 1(-2) = 0 Cu(s) CuO(s) (O.N of uncombined element = 0) H 2 O(aq) (O.N of uncombined element = 0) 2(O.N. of H) + 1(O.N. of O) = 0 2(+1) + 1(-2) = 0  H: O.N. increases from 0 to +1 (loses e-, oxidized). I -  Reducing agent  Cu: O.N. decreases from +2 to 0 (gains e-, reduced).  Cu Oxidizing agent

17. Double Replacement Reactions HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O( l ) AX + BY  BX + AY  Never redox reactions.  “Partner swapping”.

18. Example 01 HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O( l )  Classify each of the reactions represented by the following equations as redox or nonredox. Further reclassify them as decomposition, combination, single- replacement, or double-replacement reactions. SO 2 (g) + H 2 O( l )  H 2 SO 3 (aq) Combination (addition) S+4 O-2 H+1  Nonredox  Combination (addition)

19. Example 02  Classify each of the reactions represented by the following equations as redox or nonredox. Further reclassify them as decomposition, combination, single- replacement, or double-replacement reactions. 2K(s) + 2H 2 O( l )  2KOH(aq) + H 2 (g) K0+1 O-2 H+1 0  K: O.N. increases from 0 to +1 (loses e-, oxidized). I -  Reducing agent  H: O.N. decreases from +1 to 0 (gains e-, reduced).  Oxidizing agent  Single replacement

20. Example 03  Classify each of the reactions represented by the following equations as redox or nonredox. Further reclassify them as decomposition, combination, single- replacement, or double-replacement reactions. BaCl 2 (aq) + Na 2 CO 3 (aq)  BaCO 3 (s) + 2NaCl(aq) Ba+2 Cl Na+1 CO 3 -2  Nonredox  Double-replacement

21. Example 04  Classify each of the reactions represented by the following equations as redox or nonredox. Further reclassify them as decomposition, combination, single- replacement, or double-replacement reactions. Ca+2 C+4 O-2  Nonredox  Decomposition CaCO 3 (s)  CaO(s) + CO 2 (g)

22. Ionic Equations  Ionic compounds and some polar covalent compounds may dissociate in water  H +, Na +, K +, Mg 2+, Ca 2+, Fe 2+, Fe 3+, Ag 1+, Pb 2+, F -, Cl -, Br -, Br -, OH -, NH 4 +, SO 4 2-, SO 3 2-, PO 4 3-, NO 2 -, NO 3 -, CO 3 2-  Reactions can be represented by total ionic equation or net ionic equation  Net Ionic Equation: The actual chemistry that happens HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O( l ) Molecular equation H + (aq) + Cl - (aq) + Na + (aq) + OH - (aq)  Na + (aq) + Cl - (aq) + H 2 O( l ) Total ionic equation Net ionic equation H + (aq) + OH - (aq)  H 2 O( l )  In the total ionic equation, Na+ and Cl- appear as both reactants and products  These are spectator ions  Spectator ions:  Do not participate in reaction  excluded from net ionic equation

23. Ionic Equations: Example 01 Na 2 SO 4 (aq) + BaCl 2 (aq)  BaSO 4 (s) + 2NaCl(aq) Molecular equation 2Na + (aq) + SO 4 2- (aq) + Ba 2+ (aq) + 2Cl - (aq)  BaSO 4 (s) + 2Na + (aq) + 2Cl - (aq) Total ionic equation Net ionic equation SO 4 2- (aq) + Ba 2+ (aq)  BaSO 4 (s)

24. Ionic Equations: Example 02 CaCl 2 (aq) + Na 2 CO 3 (aq)  2NaCl(aq) + CaCO 3 (s) Molecular equation Ca 2+ (aq) + 2Cl - (aq) + 2Na + (aq) + CO 3 2- (aq)  CaCO 3 (s) + 2Cl - (aq) + 2Na + (aq) Total ionic equation Net ionic equation Ca 2+ (aq) + CO 3 2- (aq)  CaCO 3 (s)

25. Energy and Reactions  All chemical reactions have associated energy changes  Energy: Ability to do work or produce change  Energy may be in one of the following forms:  Sound  Light  Electricity  Motion  **Heat  Exothermic: Heat is released to surroundings  Combustion  Acid/base neutralization  Endothermic: Heat is absorbed from surroundings  Water evaporation  Chemical cold packs 2H 2 (g) + O 2 (g)  H 2 O(g) + energy (heat released)

26. http://en.wikipedia.org/wiki/File:Close-up_of_mole.jpg

27. The Mole and Chemical Equations  The Mole can be used to calculate mass relationships in chemical reactions.  Stoichiometry: Study of mass relationships  Coefficients apply to moles or molecules, but not mass. CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O( l ) combustion reaction 1 mol CH 4 (g) + 2 mol O 2 (g)  1 mol CO 2 (g) + 2 mol H 2 O( l )  This relationship tells us that we need twice as many moles of O 2 compared with CH 4 to produce 1 mol CO 2 and 2 moles of water.  It also enables us to figure out how many grams of each substance we would need based on atomic weight/molecular weight (e.g., g/mol) MW of CH 4 = 16.0 g/mol MW of O 2 = 32.0 g/mol MW of CO 2 = 44.0 g/mol MW of H 2 O = 18.0 g/mol 16.0 g CH 4 (g) + 64.0 g O 2 (g)  44.0 g CO 2 (g) + 36.0 g H 2 O( l ) Mass of CO 2 is greater than H 2 O but only ½ the moles

28. Applying Stoichiometry to Specific Questions  How many moles of O 2 are required to react with 1.72 mol CH 4 ? CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O( l ) 1 mol CH 4 (g) + 2 mol O 2 (g)  1 mol CO 2 (g) + 2 mol H 2 O( l ) 1.72 mol CH 4 = ? mol O 2  How many grams of H 2 O will be produced when 1.09 mol of CH 4 reacts with an excess of O 2 ? 1.09 mol CH 4 = ? Grams H 2 O

29. Applying Stoichiometry to Specific Questions CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O( l ) 1 mol CH 4 (g) + 2 mol O 2 (g)  1 mol CO 2 (g) + 2 mol H 2 O( l )  How many grams of O 2 must react with excess CH 4 to produce 8.42 g CO 2 ? 8.42 g CO 2 = ? Grams O 2 Problems 5.38, 5.42, 5.48, 5.50

30. The Limiting Reactant CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O( l ) 1 mol CH 4 (g) + 2 mol O 2 (g)  1 mol CO 2 (g) + 2 mol H 2 O( l )  Reaction will only continue as long as all necessary reactants are present  Limiting reactant is the one that gives the least amount of product  In a combustion reaction of 20.0 g methane and 100.0 g oxygen, how much carbon dioxide (mass) would be produced? Molar ratio CH 4 (g):CO 2 (g) is 1:1 Molar ratio O 2 (g):CO 2 (g) is 2:1  In this reaction, how much O 2 would remain unreacted?  g O 2 used in reaction 100.0 g – 80.0 g = 20.0 g O 2 unused in reaction

31. Reaction Yields Theoretical Yield: How much product you expect based on calculations from molecular equation Actual Yield: How much product is actually recovered.  Usually less than theoretical  Causes:  Experimental/Human error  Side reactions

32. Examples of Reaction Yields In an experiment, 17.0 g of product is obtained from a reaction with a calculated theoretical yield of 34.0 g. What is percentage yield? In the following reaction, 510.0 g of CaCO 3 is heated and produces 235.0 g CaO. What is the percentage yield? CaCO 3 (s)  CaO (s) + CO 2 (g) MW of CaCO 3 = 100.1 g/mol MW of CaO = 56.1 g/mol MW of CO 2 = 44.0 g/mol