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Chem 1151: Ch. 5

Chem 1151: Ch. 5. Chemical Reactions. The Chemical Equation. 2H 2 (g) + O 2 (g)  2H 2 O( l ). Reactant(s). Product(s). Identifies reactant(s) and product(s) Identifies states of matter (g = gas, l = liquid, s = solid, aq = aqueous)

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Chem 1151: Ch. 5

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  1. Chem 1151: Ch. 5 Chemical Reactions

  2. The Chemical Equation 2H2(g) + O2 (g)  2H2O(l) Reactant(s) Product(s) • 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 H2 react with 1 molecule of O2) as well as molar relationships (2 moles of H2 react with 1 mole of O2).

  3. Balancing The Chemical Equation N2(g) + H2 (g)  NH3(g) • You cannot change the natural molecular formulas, you can only change the coefficients indicating number of molecules or moles. N2(g) + H2 (g)  2NH3(g) • By multiplying NH3 by coefficient of 2, you now have 2N and 6H. This balances with N2 on reactant side, but not H2. N2(g) + 3H2 (g)  2NH3(g) • Now check your work. Reactant(s) Product(s) 2 N 6 H 2 N 6 H

  4. Balancing The Chemical Equation NO(g) + O2 (g)  NO2(g) nitrogen monoxide • Reactant side has 3O, but product side has 2O. 2NO(g) + O2 (g)  2NO2(g) • Multiply NO2 by coefficient of 2, then multiply NO by 2. • Now check your work. Reactant(s) Product(s) 2 N 4 O 2 N 4 O

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

  6. Redox Reactions • Redox • Combination of Reduction and Oxidation • Oxidation • To combine with oxygen • To lose hydrogen • To lose electrons • To increase in oxidation number 4Fe(s) + 3O2 (g)  2Fe2O3(s) • Reduction • To lose oxygen • To combine with hydrogen • To gain electrons • To decrease in oxidation number 2Fe2O3(s) + 3C(s)  4Fe(s) + 3CO2(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), O2(0), H2(0), Br2(0), Na(0) Rule 2: The O.N. of a simple ion is equal to the charge on the ion. Exs. Na+(+1), Mg2+(+2), Br-(-1) Rule 3: The O.N. of group IA and IIA elements are +1 and +2, respectively. Exs. NaOH (Na = +1), CaCl2 (Ca = +2) Rule 4: The O.N. of H is +1. Exs. HCl (H = +1), H3PO4 (H = +1) Rule 5: The O.N. of oxygen is -2 except in peroxides where it is -1. Exs. H2O (O = -2), H2O2 (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 K2CO3 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 CO2 (O.N of C) + 2(O.N. of O) = 0 2(-2) = -4 (+4) + 2(-2) = 0 CH2O (O.N of C) + 2(O.N. of H) + (O.N. of O) = 0 2(+1) + (-2) = 0 (0) + 2(+1) + (-2) = 0 NO3- (O.N of N) + 3(O.N. of O) = -1 + 3(-2) = -6 (+5) + 3(-2) = -1

  9. Redox e- Transfer S(s) + O2(g)  SO2(g) (O.N of uncombined element = 0) (O.N of uncombined element = 0) (O.N. of S) + 2(O.N. of O) = 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 Mission: Determine O.N. for each atom in the following and identify oxidizing and reducing agents. 4Al(s) + 3O2(g)  2Al2O3(s) (O.N of uncombined element = 0) (O.N of uncombined element = 0) 4(O.N. of Al) + 6(O.N. of O) = 0 + 6(-2) = -12 4(+3) + 6(-2) = 0 • 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

  11. Redox e- Transfer: Examples Mission: Determine O.N. for each atom in the following and identify oxidizing and reducing agents. S2O82-(aq) + 2I-(aq)  I2(aq) + 2SO42-(aq) S2O82-(aq) I2(aq) (O.N of uncombined element = 0) 2(O.N. of S) + 8(O.N. of O) = -2 8(-2) = -16 2(+7) + 8(-2) = -2 2SO42-(aq) 2I-(aq) 2(O.N. of S) + 8(O.N. of O) = -4 2(simple ion charge = -1) = -2 8(-2) = -16 2(+6) + 8(-2) = -4 • 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).  S2O82- Oxidizing agent

  12. Redox e- Transfer: Examples Mission: Determine O.N. for each atom in the following and identify oxidizing and reducing agents. 2KI(aq) + Cl2(aq) 2KCl(aq) + I2(aq) 2KCl(aq) 2KI(aq) 2(O.N. of K) + 2(O.N. of I) = 0 2(O.N. of K) + 2(O.N. of Cl) = 0 2(+1) = +2 2(+1) = +2 2(+1) + 2(-1) = 0 2(+1) + 2(-1) = 0 Cl2(aq) I2(aq) (O.N of uncombined element = 0) (O.N of uncombined element = 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).  Cl2 Oxidizing agent

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

  14. Decomposition Reactions • A single substance is broken down to form 2 or more simpler substances. • These may or may not be redox reactions. A  B + C 2HgO(s)  2Hg(l) + O2(g) Note conservation of mass in balanced equations CaCO3(s)  CaO(s) + CO2(g) Seager SL, Slabaugh MR, Chemistry for Today: General, Organic and Biochemistry, 7th Edition, 2011

  15. Combination Reactions • 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. A + B  C 2Mg(s) + O2(g)  2MgO(s) Redox combustion reaction SO3(g) + H2O(l)  H2SO4(aq) Nonredox reaction, formation of acid rain Seager SL, Slabaugh MR, Chemistry for Today: General, Organic and Biochemistry, 7th Edition, 2011

  16. Single Replacement Reactions • Always redox reactions. • Used to obtain metals from oxide ores. A + BX  B + AX 3C(s) + 2Fe2O3(s)  4Fe(s) + 3CO2(g) H2 (g) Cu(s) (O.N of uncombined element = 0) (O.N of uncombined element = 0) H2O(aq) CuO(s) 1(O.N. of Cu) + 1(O.N. of O) = 0 2(O.N. of H) + 1(O.N. of O) = 0 1(-2) = -2 2(+1) + 1(-2) = 0 1(+2) + 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 • Never redox reactions. • “Partner swapping”. AX + BY  BX + AY HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)

  18. Example 01 • 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. SO2(g) + H2O(l)  H2SO3(aq) Combination (addition) • Nonredox • Combination (addition) HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)

  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) + 2H2O(l)  2KOH(aq) + H2(g) • 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. BaCl2(aq) + Na2CO3(aq)  BaCO3(s) + 2NaCl(aq) • 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. CaCO3(s)  CaO(s) + CO2(g) • Nonredox • Decomposition

  22. Ionic Equations • Ionic compounds and some polar covalent compounds may dissociate in water • H+, Na+, K+, Mg2+, Ca2+, Fe2+, Fe3+, Ag1+, Pb2+, F-, Cl-, Br-, Br-, OH-, NH4+, SO42-, SO32-, PO43-, NO2-, NO3-, CO32- • Reactions can be represented by total ionic equation or net ionic equation • Net Ionic Equation: The actual chemistry that happens Molecular equation HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) Total ionic equation H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq)  Na+(aq) + Cl-(aq)+ H2O(l) Net ionic equation H+(aq) + OH-(aq)  H2O(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 Molecular equation Na2SO4(aq) + BaCl2(aq)  BaSO4(s) + 2NaCl(aq) Total ionic equation 2Na+(aq) + SO42-(aq) + Ba2+(aq) + 2Cl-(aq) BaSO4(s) + 2Na+(aq) + 2Cl-(aq) Net ionic equation SO42-(aq) + Ba2+(aq)  BaSO4(s)

  24. Ionic Equations: Example 02 Molecular equation CaCl2(aq) + Na2CO3(aq)  2NaCl(aq) + CaCO3(s) Total ionic equation Ca2+(aq) + 2Cl-(aq) +2Na+(aq) +CO32-(aq)  CaCO3(s) + 2Cl-(aq) + 2Na+(aq) Net ionic equation Ca2+(aq) +CO32-(aq)  CaCO3(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 2H2(g) +O2(g)  H2O(g) + energy (heat released) • Exothermic: Heat is released to surroundings • Combustion • Acid/base neutralization • Endothermic: Heat is absorbed from surroundings • Water evaporation • Chemical cold packs

  26. The Mole is back! 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. CH4(g) + 2O2(g)  CO2 (g) + 2H2O(l) combustion reaction 1 mol CH4(g) + 2 mol O2(g)  1 mol CO2 (g) + 2 mol H2O(l) • This relationship tells us that we need twice as many moles of O2 compared with CH4 to produce 1 mol CO2 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 CH4 = 16.0 g/mol MW of O2 = 32.0 g/mol MW of CO2 = 44.0 g/mol MW of H2O = 18.0 g/mol 16.0 g CH4(g) + 64.0 g O2(g)  44.0 g CO2 (g) + 36.0 g H2O(l) Mass of CO2 is greater than H2O but only ½ the moles

  28. Applying Stoichiometry to Specific Questions CH4(g) + 2O2(g)  CO2 (g) + 2H2O(l) 1 mol CH4(g) + 2 mol O2(g)  1 mol CO2 (g) + 2 mol H2O(l) • How many moles of O2 are required to react with 1.72 mol CH4? 1.72 mol CH4 = ? mol O2 • How many grams of H2O will be produced when 1.09 mol of CH4 reacts with an excess of O2? 1.09 mol CH4 = ? Grams H2O

  29. Applying Stoichiometry to Specific Questions CH4(g) + 2O2(g)  CO2 (g) + 2H2O(l) 1 mol CH4(g) + 2 mol O2(g)  1 mol CO2 (g) + 2 mol H2O(l) • How many grams of O2 must react with excess CH4 to produce 8.42 g CO2? 8.42 g CO2 = ? Grams O2 Problems 5.38, 5.42, 5.48, 5.50

  30. The Limiting Reactant • Reaction will only continue as long as all necessary reactants are present • Limiting reactant is the one that gives the least amount of product CH4(g) + 2O2(g)  CO2 (g) + 2H2O(l) 1 mol CH4(g) + 2 mol O2(g)  1 mol CO2 (g) + 2 mol H2O(l) • In a combustion reaction of 20.0 g methane and 100.0 g oxygen, how much carbon dioxide (mass) would be produced? Molar ratio CH4(g):CO2 (g) is 1:1 Molar ratio O2(g):CO2 (g) is 2:1 • In this reaction, how much O2 would remain unreacted?  g O2 used in reaction 100.0 g – 80.0 g = 20.0 g O2 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 CaCO3 is heated and produces 235.0 g CaO. What is the percentage yield? MW of CaCO3 = 100.1 g/mol MW of CaO = 56.1 g/mol MW of CO2 = 44.0 g/mol CaCO3 (s)  CaO (s) + CO2 (g)

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