Chemistry deals with matter and its changes CHEMICAL REACTIONS

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1 Chemistry deals with matter and its changes CHEMICAL REACTIONS

2 CHEMICAL EQUATIONS N H 2 2 NH 3 2 N 6 H 2 N 6 H reactants products balanced means equal numbers of atoms of each element on each side When reactants and products are given, only need to change coefficients to achieve balance

3 COMPLETE COMBUSTION Hydrocarbon or other fuel combines with O 2 to produce CO 2 and water. EXAMPLE: Propane C 3 H 8 in a barbeque grill C 3 H O 2 3 CO H 2 O 3C 8H = 3C = 8H 10O = 6O 4O

4 STEPS TO BALANCE A CHEMICAL EQUATION (1) Write an unbalanced equation containing the correct formulas of the reactants and products. (2) Balance atoms of one of the elements. (3) Balance atoms of the remaining elements. (4) Verify that the number of atoms of each element is balanced.

5 EXAMPLE: destruction of marble by acid rain Begin with unbalanced equation HNO 3 (aq) + CaCO 3 (s) CO 2 (g) + H 2 O(l) + Ca(NO 3 ) 2 (aq) focus on N to get 2 HNO 3 (aq) + CaCO 3 (s) CO 2 (g) + H 2 O(l) + 1 Ca(NO 3 ) 2 (aq) next, focus on Ca to get 2 HNO 3 (aq) + 1 CaCO 3 (s) CO 2 (g) + H 2 O(l) + 1 Ca(NO 3 ) 2 (aq) focus on H: 2 on left, so we get 2 HNO 3 (aq) + 1 CaCO 3 (s) CO 2 (g) + 1 H 2 O(l) + 1 Ca(NO 3 ) 2 (aq) left is finished, and it has 1 C so we get 2 HNO 3 (aq) + 1 CaCO 3 (s) 1 CO 2 (g) + 1 H 2 O(l) + 1 Ca(NO 3 ) 2 (aq) Is O balanced? left: = 9 right: = 9 the final result is: 2 HNO 3 (aq) + CaCO 3 (s) CO 2 (g) + H 2 O(l) + Ca(NO 3 ) 2 (aq)

6 EQUATIONS TO BALANCE H 2 S(aq) + Cl 2 (aq) S 8 (s) + HCl(aq) CaO(s) + NH 4 Cl(s) NH 3 (g) + H 2 O(g) + CaCl 2 (s) Al(OH) 3 (s) + HCl(aq) AlCl 3 (aq) + H 2 O( )

7 REACTION TYPES COMBINATION elements react to form compounds S(s) + O 2 (g) SO 2 (g) small compounds combine to make larger ones CaO(aq) + CO 2 (g) CaCO 3 (s) DECOMPOSITION 2 H 2 O 2 H 2 + O 2 H 2 CO 3 H 2 O + CO 2 COMBUSTION organic compound + O 2 C 2 H 5 OH(l) + 3 O 2 2 CO H 2 O

8 REACTION TYPES SINGLE DISPLACEMENT one element replaces another CuSO 4 (aq) + Zn(s) ZnSO 4 (aq) + Cu(s) DOUBLE DISPLACEMENT ions or atoms exchange Precipitation CaCl 2 (aq) + 2 AgNO 3 (aq) 2 AgCl(s) + Ca(NO 3 ) 2 (aq) Acid-Base Reaction HNO 3 + KOH H 2 O + KNO 3 acid base water salt

9 EXCHANGE REACTIONS Double displacement Require a driving force Three driving forces: 1. Precipitate forms 2. Gas forms 3. Weak or non-electrolyte forms (molecular species) AX + BY AB + XY

10 PRECIPITATION REACTIONS Mix Pb(NO 3 ) 2 and KI..what happens? Pb(NO 3 ) 2 (aq) + 2 KI(aq) PbI 2 (s) + 2 KNO 3 (aq) complete ionic equation Pb 2+ (aq) + 2 NO 3 (aq) + 2 K + (aq) + 2 I (aq) PbI 2 (s) + 2 NO 3 (aq) + 2 K + (aq) Eliminate spectator ions net ionic equation Pb 2+ (aq) + 2 I (aq) PbI 2 (s)

11 AQUEOUS SOLUBILITY OF IONIC COMPOUNDS The ions comprising an ionic compound determine its solubility Simplified Rules: (1) Almost all ammonium and alkali metal salts are soluble. (2) Most nitrates, acetates, chlorides, bromides, iodides, and sulfates are soluble. Important exceptions are silver halides, and Ca, Ba, Pb sulfates. (3) Most sulfides, carbonates, phosphates, and hydroxides are insoluble. Important exceptions are the alkali and ammonium salts.

12 IONIC EQUATIONS acid-base neutralization example HNO 3 (aq) + KOH(aq) H 2 O( ) + KNO 3 (aq) H + (aq) + NO 3 (aq) + K + (aq) + OH (aq) H 2 O( ) + NO 3 (aq) + K + (aq) net ionic equation H + (aq) + OH (aq) H 2 O( ) spectator ions precipitation reaction example AgNO 3 (aq) + NaCl(aq) AgCl(s) + NaNO 3 (aq) Ag + (aq) + NO 3 (aq) + Na + (aq) + Cl (aq) AgCl(s) + NO 3 (aq) + Na + (aq) net ionic equation Ag + (aq) + Cl (aq) AgCl(s)

13 GAS FORMATION direct production of gas -or - production of weak acid which decomposes example: H 2 CO 3 HCl(aq) + NaHCO 3 (aq) NaCl(aq) + H 2 CO 3 (aq) sodium bicarbonate H 2 O( ) + CO 2 (g) net ionic equation H + (aq) + HCO 3 (aq) H 2 O( ) + CO 2 (g)

14 ACIDS-BASE NEUTRALIZATION ACID + BASE SALT + WATER HNO 3 + KOH KNO 3 + H 2 O complete ionic equation H + (aq) + NO 3 (aq) + K + (aq) + OH (aq) H 2 O(l) + K + (aq) + NO 3 (aq) eliminate spectator ions net ionic equation H + (aq) + OH (aq) H 2 O(l)

15 ACIDS, BASES, SALTS They are electrolytes: (strong or weak) ACIDS Donate hydrogen ion H + (aq), H 3 O + (aq) HNO 3 (aq) H + (aq) + NO 3 (aq) BASES monoprotic Accept hydrogen ion Raise conc. of OH (aq) ions KOH(aq) K + (aq) + OH (aq) SALTS Ionic compounds Replace H + of acid with positive ion HNO 3 becomes KNO 3

16 Strong Acids and Bases Strong Acids Table 4.2 Hydrochloric, Hydrobromic, Hydroiodic, HCl HBr HI Chloric, HClO 3 Perchloric, HClO 4 Nitric, HNO 3 Sulfuric, HSO 4 Strong Bases Group 1A metal hydroxides (LiOH, NaOH, KOH, RbOH, CsOH) Heavy Group IIA metal hydroxides (Ca(OH) 2, Sr(OH) 2, Ba(OH) 2 )

Reduction reaction: X 2 + 2 e 2 X Oxidation reaction: M M

17 REDOX REACTIONS Reactions where electrons are transferred from one reactant to another. Reduction reaction: X e 2 X Oxidation reaction: M M n+ + n e reduction = accepting e oxidation = losing e The two processes are always linked.

18 OXIDIZING AGENTS O 2, halogens, H 2 O 2, HNO 3, Cr 2 O 7, MnO 4 REDUCING AGENTS H 2, C, metals

19 OXIDATION NUMBERS Oxidation numbers always change in redox reactions Example: acid displacement reaction Mg(s) + 2 HCl(aq) MgCl 2 (aq) + H 2 (g) Mg 0 +2 H 1 0 Cl no change

20 OXIDATION NUMBERS How to determine oxidation numbers (1) The oxid. no. of an atom of a pure element is 0. (2) The oxid. no. of a monatomic ion equals its charge. (3) Some elements have the same oxid. no. in almost all of their compounds and can be used as references for oxid. nos. of other elements in compounds. (4) The sum of the oxid. nos. in a neutral compound is 0; the sum of the oxid. nos. in a polyatomic ion equals the charge on the ion.

21 ACTIVITY SERIES Table 4.5 Relative reactivity of metals lead to the activity series A metal higher in the series will always displace an element below it Metals above H 2 in series will displace it from acid such as HCl to form H 2 (Mg, Zn, Fe) Metals toward bottom are unreactive (Ag, Pt, Au)

22 REACTIONS OF METALS WITH ACIDS FOR MANY METALS metal + acid salt + H 2 Fe(s) + H 2 SO 4 (aq) FeSO 4 (aq) + H 2 (g) SINGLE DISPLACEMENT A + BX AX + B metal acid or metal salt the metal is oxidized

23 USING A BALANCED CHEMICAL EQUATION QUANTITATIVELY Use molar mass to convert moles grams grams moles Use balanced equation to convert moles reactant moles product moles product moles reactant

24 STOICHIOMETRY EXAMPLE (Not Balanced) Pb(NO 3 ) 2 + KI PbI 2 + KNO 3 (Balanced) Pb(NO 3 ) KI PbI KNO 3 2 mol KI reacts with 1 mol Pb(NO 3 ) 2 MW of Pb(NO 3 ) 2 Pb N 2 x 14 = 28. O 6 x 16 = g/mol MW of KI K 39.1 I g/mol 2 mol = g So, g Pb(NO 3 ) 2 reacts completely with g of KI

25 An iron atom weighs roughly twice as much as a sulfur atom. If you want to prepare the compound FeS in a quantitative reaction, approximately what quantities of Fe and S do you combine in the lab? 1. Equal masses of Fe and S? 2. Twice as much Fe as S? 3. Twice as much S as Fe?

26 NITROGLYCERINE DECOMPOSITION 4 C 3 H 5 (NO 3 ) 3 ( ) 12 CO 2 (g) + 10 H 2 O(g) + 6 N 2 (g) + O 2 (g) How much gas is produced per g of nitroglycerine?

27 HOW MUCH CO 2 DOES YOUR CAR PRODUCE? Assume gasoline is octane C 8 H 18 2 C 8 H O 2 16 CO H 2 O 1 gal = 4 qt qt/l = 3.78 L Assume density of 0.80 g/ml 3.78 L x 0.80 g/ml x 10 3 ml/l = 3.02 x 10 3 g octane 3020 g = 26.4 mol octane in 1 gal 114 g/mol 2 mol octane 16 mol CO mol octane 211 mol CO mol CO 2 x 44 g/mol = 9280 g CO 2 1 gal octane (6.7 lb) 9.3 kg CO 2 (20.5 lb) 10,000 mi/year at 25 mpg = 400 gal/year 400 gal octane 8200 lb or 4 tons CO 2

28 LIMITING REACTANT The limiting reactant is used up first The limiting reactant determines the amount of product Need balanced equation to proceed

29 Limiting Reagents 2H 2 + O 2 2H 2 O 10 H 2 and 7 O 2 10 H 2 O and 2 O 2 reaction stops when one of the reactants is depleted What if you had only g Pb(NO 3 ) 2? How much PbI 2 would precipitate?

30 STOICHIOMETRY EXAMPLE (Not Balanced) Pb(NO 3 ) 2 + KI PbI 2 + KNO 3 (Balanced) Pb(NO 3 ) KI PbI KNO 3 2 mol KI reacts with 1 mol Pb(NO 3 ) 2 MW of Pb(NO 3 ) 2 Pb N 2 x 14 = 28. O 6 x 16 = g/mol MW of KI K 39.1 I g/mol 2 mol = g So, g Pb(NO 3 ) 2 reacts completely with g of KI

31 LIMITING REACTANT EXAMPLE If 18.3 g of C 2 H 5 OH reacts with 63.8 g of O 2 (combustion), how many g of CO 2 are produced? Need balanced equation to start Combustion: C 2 H 5 OH + 3 O 2 2 CO H 2 O

32 LIMITING REACTANT cont. C 2 H 5 OH + 3 O 2 2 CO H 2 O 18.3 g 63.8 g? g To find limiting reactant, first need MWs C 2 H 5 OH O 2 2 x 12 = 24 2 x 16 = 32 g/mol 1 x 16 = 16 6 x 1 = 6 46 g/mol Need moles of reactants C 2 H 5 OH O g 46 g/mol = mol 63.8 g = 1.99 mol 32 g/mol

33 LIMITING REACTANT cont. Now, knowing the LR, find the final answer. 1 mol C 2 H 5 OH 2 mol CO mol C 2 H 5 OH mol CO 2 Need MW of CO 2 1 x 12 = 12 2 x 16 = g/mol (0.796 mol) (44 g/mol) = 35.0 g CO 2

34 LIMITING REACTANT Lithium oxide is used on the space shuttle to remove water from the air Li 2 O(s) + H 2 O(g) 2 LiOH(s) Which reactant is limiting if kg of water is to be removed and 82.0 kg of Li 2 O is present? How much product is formed? Li 2 O Need moles of reactants 82.0 kg x 10 3 g/kg x 1 mol = 2744 mol g H 2 O kg x 10 3 g/kg x Li 2 O will be exhausted first 1 mol g = 5552 mol Moles LiOH = 2 x moles Li 2 O Moles LiOH = 2 x 2744 = 5488 moles 5488 moles x g/mol x 10-3 kg/g = kg LiOH

Reactions. Chapter 3 Combustion Decomposition Combination. Chapter 4 Reactions. Exchange reactions (Metathesis) Formation of a precipitate

Reactions. Chapter 3 Combustion Decomposition Combination. Chapter 4 Reactions. Exchange reactions (Metathesis) Formation of a precipitate Reactions Chapter 3 Combustion Decomposition Combination Chapter 4 Reactions Exchange reactions (Metathesis) Formation of a precipitate Formation of a gas Formation of a week or nonelectrolyte Single Displacement