2. Chapter 4 Chemical Reactions Chemistry B11

3. Chemical Reactions Chemical change = Chemical reaction Substance(s) is used up (disappear) New substance(s) is formed. Different physical and chemical properties.

4. Chemical Reactions

5. A + B  C + D Reactants Products Chemical Reactions Chemical Equation

6. Chemical Reactions A + B  AB 1. Synthesis reaction (combination) 2H 2 + O 2  2H 2 O AB  A + B 2. Decomposition (analysis) 2NaCl  2Na + Cl 2 A + BC  AC + B 3. Single replacement reaction Fe + CuSO 4  FeSO 4 + Cu AB + CD  AD + CB 4. Double replacement reaction NaCl + AgNO 3  NaNO 3 + AgCl

7. 5. Combustion C 3 H 8 + 5O 2  3CO 2 + 4H 2 O Chemical Reactions Solid (s) Liquid (l) Gas (g) Aqueous (aq) Ca(OH) 2 (s) + 2HCl(g)  CaCl 2 (s) + H 2 O(l) AB + xO 2  yCO 2 + zH 2 O

8. Why balancing? Balance a chemical equation

9. Law of conservation of mass Atoms are neither destroyed nor created. They shift from one substance to another.

10. Balance a chemical equation 1.Begin with atoms that appear in only one compound on the left and right. 2.If an atom occurs as a free element, balance it last. 3.Change only coefficients (not formulas). C 3 H 8 (g) + O 2 (g)  CO 2 (g) + H 2 O(g) last

11. Formula and Molecule Ionic & covalent compounds  Formulaformula of NaCl Covalent compounds  Molecule molecule of H 2 O Formula Weight of NaCl: 23 amu Na + 35.5 amu Cl = 58.5 amu NaCl Molecular Weight of H 2 O: 2 (1 amu H) + 16 amu O = 18 amu H 2 O

12. Mole Mole (mol): formula weight of a substance (in gram). 12g of C = 1 mol C 23g of Na = 1 mol Na 58.5 g of NaCl = 1 mol NaCl 18 g of H 2 O = 1 mol of H 2 O

13. Avogadro’s number (6.02×10 23 ): number of formula units in one mole. 1 mole of apples = 6.02×10 23 apples 1 mole of A atoms = 6.02×10 23 atoms of A 1 mole of A molecules = 6.02×10 23 molecules of A 1 mole of A ions = 6.02×10 23 ions of A Molar mass (g/mol): mass of 1 mole of substance (in gram) (Formula weight) molar mass of Na = 23 g/mol molar mass of H 2 O = 18 g/mol

14. Relationships between amounts of substances in a chemical reaction. Look at the Coefficients! Stoichiometry 2H 2 O (l)  2H 2 (g) + O 2 (g) 2 moles 1 mole 2 2 1 2 liters 1 liter 2 particles 1 particle 2 grams 1 gram

15. A mole B mole mass volume Particle (atom) (molecule) (ion) Particle (atom) (molecule) (ion) mass CH 4 + 2O 2  CO 2 + 2H 2 O 1 step: use coefficient in the balanced equation.

16. CH 4 + 2O 2  CO 2 + 2H 2 O 10 cc O 2 = ? cc CO 2 10 cc O 2 ( 1 cc CO 2 2 cc O 2 ) = 5 cc CO 2 23 mole CH 4 = ? moles H 2 O 23 mole CH 4 ( 2 moles H 2 O 1 mole CH 4 ) = 46 moles H 2 O 40 g CH 4 = ? L CH 4 40 g CH 4 ( 1 mole CH 4 16 g CH 4 ) = 56 L CH 4 22.4 L CH 4 1 mole CH 4 )( STP: 1 mole of substance (gas) = 22.4 L = 22400 cc (cm 3 or mL) 32 g CH 4 = ? moles CO 2 32 g CH 4 ( 1 mole CH 4 16 g CH 4 ) = 2 mole CO 2 1 mole CO 2 1 mole CH 4 )(

17. Limiting Reagents N 2 (g) + O 2 (g)  2NO(g) 1 mole 2 moles 1 mole 4 moles 0 mole 3 moles 2 moles Before reaction: After reaction: Stoichiometry:

18. Limiting Reagents N 2 (g) + O 2 (g)  2NO(g) 1 mole 2 moles 1 mole 4 moles 0 mole 3 moles 2 moles Before reaction: After reaction: Stoichiometry: Used up first Left over

19. Limiting Reagents N 2 (g) + O 2 (g)  2NO(g) 1 mole 2 moles 1 mole 4 moles Limiting reagent 0 mole 3 moles 2 moles Before reaction: After reaction: Stoichiometry: Used up first Left over

20. Limiting Reagents Limiting reagents can control a reaction: N 2 (g) + O 2 (g)  2NO(g) Limiting reagent: is the reactant that is used up first.

21. Limiting Reagents Example: C(s) + O 2 (g)  CO 2 (g) 12g of C 64g of O 2 ? Limiting reagent ? g of CO 2 will be formed Make sure that the chemical equation is balanced. 12g C ( 1 mole C 12g C ) = 1 mole C 64g O 2 ( 1 mole O 2 32g O 2 ) = 2 mole O 2 C is the limiting reagent. C(s) + O 2 (g)  CO 2 (g) 1 mol

22. Limiting Reagents We should use the mass of the limiting reagent. (because it controls our reaction). 12g C ( 1 mole C 12g C )( 1 mole CO 2 1 mole C )( 44g CO 2 1 mole CO 2 ) = 44g CO 2 ? g of CO 2 will be formed: C(s) + O 2 (g)  CO 2 (g) A B 12g 64g ? g

23. Percent Yield Percent yield = actual yield theoretical yield × 100 actual yield: mass of product formed (experimental) theoretical yield: mass of product that should form (according to stoichiometry) N 2 (g) + O 2 (g)  2NO(g) theoretical yield = 40g NO actual yield = 37g NO Percent yield = 37g NO 40g NO × 100 = 92.5% 7.5% error or lost

24. Aqueous Solution (ionic compounds) NaCl (s) Na + (aq) + Cl - (aq) Dissociation (Ionization) aqueous solution: solvent is water H2OH2O AgNO 3 (s) Ag + (aq) + NO 3 - (aq) H2OH2O NaCl AgNO 3 NaCl (aq) + AgNO 3 (aq)  AgCl (s) + NaNO 3 (aq) H2OH2O

25. Molecular equation: 2As 3+ (aq) + 3s 2- (aq)  As 2 S 3 (s) total charge on left side = total charge on right side NaCl (aq) + AgNO 3 (aq)  AgCl (s) + NaNO 3 (aq) Na + (aq) + Ag + (aq) + Cl - (aq) + NO 3 - (aq)  AgCl (s) + Na + (aq) + NO 3 - (aq) Ionic equation: Net ionic equation: Na + (aq) + Ag + (aq) + Cl - (aq) + NO 3 - (aq)  AgCl (s) + Na + (aq) + NO 3 - (aq) Spectator ions Ag + (aq) + Cl - (aq)  AgCl (s)

26. Oxidation and Reduction reactions (redox) Zn(s)  Zn 2+ (aq) + 2e - Zn is oxidized (reducing agent) Cu 2+ (aq) + 2e -  Cu(s)Cu 2+ is reduced (oxidizing agent) oxidation: is the loss of electrons. reduction: is the gain of electrons. Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s) redox reaction

27. Oxidation and Reduction reactions (redox) oxidation: is the gain of oxygen / loss of hydrogen. reduction: is the loss of oxygen / gain of hydrogen. CH 4 (s) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g) redox reaction C gains O and loses H is oxidized (reducing agent) O gains H Is reduced (oxidizing agent) single replacement reactions and combustion reactions  redox reactions double replacement reactions  non redox

28. Heat of reaction C 3 H 8 (s) + 5O 2 (g)  3CO 2 + 4H 2 O + heat (energy) 2HgO(s) + heat (energy)  2Hg(l) + O 2 (g) Endothermic reaction Exothermic reaction All combustion reactions are exothermic.