5 - 1GS105 GS105 Chapter 8 Chemical Reactivity Chemical Equations Moles Reaction Rates Equilibrium Energy in Chemical Reactions Entropy

5 - 2GS105 What is Chemistry? “The study of Matter and its Changes.” Physical Changes Physical Changes = Physical Property Changes in a Physical Property Chemical Changes Chemical Changes = Chemical Property Changes in a Chemical Property Appearance: melting, freezing, evaporation…melting, freezing, evaporation… stretching, molding, cutting…stretching, molding, cutting… Chemical Composition:

5 - 3GS105 Change in the Chemical Composition Burning of Magnesium Chemical Changes Rusting of Iron Decomposing of wood Souring of Milk Examples:

5 - 4GS105 Chemical Reactions Color changeColor change Gas formedGas formed Solid precipitate formedSolid precipitate formed Temperature ChangeTemperature Change Gives heat = exothermicGives heat = exothermic Gets cold = endothermicGets cold = endothermic

5 - 5GS105 Mg + O 2  MgO + Energy Chemical Equations Shows how the Chemical change occurs. Reactants C 3 H 8 + O 2 CO 2 + H 2 O + Energy  Fe + O 2 Fe 2 O 3 ProductsProducts

5 - 6GS105 Chemical equations Chemist’s shorthand to describe a reaction. ReactantsReactants ProductsProducts The state of all substancesThe state of all substances H 2 + O 2 H 2 O + E (g)(g) (g) Any conditions used in the reactionAny conditions used in the reaction heat Same # & type atoms on each sideSame # & type atoms on each side Law of Conservation of Matter Law of Conservation of Matter 2 2 (g) (l) (s) (aq)

5 - 7GS105 Law of Conservation of Matter Question:

5 - 8GS105 Law of Conservation of Matter Solution: Fe + O 2  Fe 2 O 3

5 - 9GS105 Law of Conservation of Matter Question:

5 - 10GS105 Law of Conservation of Matter Solution:

5 - 11GS105 Balancing Equations WB ___W 10 + ___B 8 ___WB ReactantsReactants ProductsProducts Making Hot dogs: How many packages wieners & buns to buy so none is left over. 4540

5 - 12GS105 Ca H Cl Balancing Equations CaCl 2 + H 2 Ca + HCl CaCl 2 + H 2 ReactantsReactants ProductsProducts Step 1 Step 1: Count atoms of each element on both sides of equation

5 - 13GS105 Balancing Equations CaCl 2 + H 2 Ca + HCl CaCl 2 + H 2 Ca H Cl Ca H Cl ReactantsReactants ProductsProducts Step 2 Step 2: Determine which atoms are not balanced. - not balanced

5 - 14GS105 Balancing Equations CaCl 2 + H 2 Ca + HCl CaCl 2 + H 2 ReactantsReactants ProductsProducts not balanced Step 3: Step 3: Balance one element at a time with coefficients in front of formulas until all balanced. (Never change the formula!) Ca H Cl Ca H Cl 2 2

5 - 15GS105 Na P O Mg Cl Balancing Equations Mg 3 (PO 4 ) 2 + NaCl Na 3 PO 4 + MgCl 2 Mg 3 (PO 4 ) 2 + NaCl ReactantsReactants ProductsProducts Step 1 Step 1: Count atoms of each element on both sides of equation

5 - 16GS105 Na P O Mg Cl Balancing Equations Mg 3 (PO 4 ) 2 + NaCl Na 3 PO 4 + MgCl 2 Mg 3 (PO 4 ) 2 + NaCl ReactantsReactants ProductsProducts not balanced Step 2 Step 2: Determine which atoms are not balanced. - not balanced

5 - 17GS105 Na P O Mg Cl Balancing Equations Mg 3 (PO 4 ) 2 + NaCl Na 3 PO 4 + MgCl 2 Mg 3 (PO 4 ) 2 + NaCl ReactantsReactants ProductsProducts not balanced Step 3: Step 3: Balance elements with #’s in front of formulas until all balanced. (Never change the formulas!)

5 - 18GS105 Hints: Start with a metal in a complex compound, or an element that only appears in one formula. (ie Mg here) Na P O Mg Cl Balancing Equations Mg 3 (PO 4 ) 2 + NaCl Na 3 PO 4 + MgCl 2 Mg 3 (PO 4 ) 2 + NaCl ReactantsReactants ProductsProducts not balanced

5 - 19GS105 Hints: Start with an element that only appears in one formula on both sides of the equation. Leave oxygen until last. Balancing Equations CO 2 + H 2 O C 2 H 6 + O 2 CO 2 + H 2 O ReactantsReactants ProductsProducts CHOCHO

5 - 20GS105 Balancing Equations CO 2 + H 2 O C 2 H 6 + O 2 CO 2 + H 2 O CHOCHO CHOCHO ReactantsReactants ProductsProducts Step 1 Step 1: Count atoms of each element on both sides of equation

5 - 21GS105 CO 2 + H 2 O C 2 H 6 + O 2 CO 2 + H 2 O Balancing Equations ReactantsReactants ProductsProducts Step 2 Step 2: Determine which atoms are not balanced. - not balanced CHOCHO CHOCHO

5 - 22GS not balanced CO 2 + H 2 O C 2 H 6 + O 2 CO 2 + H 2 O 2 2 Balancing Equations ReactantsReactants ProductsProducts not balanced Step 3: Step 3: Balance one element at a time with coefficients in front of formulas until all balanced. (Never change the formula!) CHOCHO CHOCHO

5 - 23GS105 CHOCHO CHOCHO CO 2 + H 2 O C 2 H 6 + O 2 CO 2 + H 2 O 2 2 Balancing Equations ReactantsReactants ProductsProducts O 2 Can’t have 3.5 O 2, so multiply equation by 2!

5 - 24GS CO 2 + H 2 O C 2 H 6 + O 2 CO 2 + H 2 O 2 2 Balancing Equations CHOCHO CHOCHO ReactantsReactants ProductsProducts O 2 Can’t have 3.5 O 2, so multiply equation by 2!

5 - 25GS105 ___O 2  ____O 3 ___Cr + ___O 2  ____Cr 2 O 3 ___HNO 2  ___HNO 3 + ___NO + __ H 2 O ___N 2 H 4 O 3  ___N 2 + ___O 2 + ___H 2 O ___N 2 O  ____N 2 + ____O 2 ___NO + ___O 2  ____NO 2 ___CH 4 + ___O 2 --> CO 2 + ____H 2 O Balancing Equations Practice:

5 - 26GS105 ___O 2  ____O 3 ___Cr + ___O 2  ____Cr 2 O 3 ___HNO 2  ___HNO 3 + ___NO + __ H 2 O _ _N 2 H 4 O 3  _ _N 2 + _ _O 2 + _ _H 2 O ___N 2 O  ____N 2 + ____O 2 ___NO + ___O 2  ____NO 2 ___CH 4 + ___O 2 --> CO 2 + ____H 2 O Balancing Equations Solutions:

5 - 27GS105 Types of Chemical Reactions Complete: C 3 H 8 + 5O 2  3CO 2 + 4H 2 O Combustion Incomplete: 2C 3 H 8 + 7O 2  6CO + 8H 2 O C 3 H 8 + 2O 2  3C + 4H 2 O

5 - 28GS105 Combination Reactions Rusting of Iron 4 Fe + 3 O 2  2 Fe 2 O 3 2H 2 + O 2  2H 2 O Explosion of Hydrogen Balloon A + B  C Decomposition Reactions 2 H 2 O 2 2 H 2 O + O 2 Blood with peroxide C  A + B

5 - 29GS105 Single Replacement Reactions Iron Deposits on an Aluminum Pan Al + FeCl 3  Fe + AlCl 3 A + BX  B + AX Double Replacement Reaction BaCl 2(aq) + Na 2 SO 4(aq)  BaSO 4(s) + 2NaCl (aq) AX + BY  BX + AY Insoluble Precipitate Formed

5 - 30GS105 1 pair = 1 dozen = 1 mole = 1 pair = 1 dozen = 1 mole = The Mole 1 mol eggs___ 6.02 x eggs 1 mol Au_______ 6.02 x Au atoms _____1 mole H 2 O_____ 6.02 x H 2 O molecules 6.02 x H 2 O molecules x ,000,000,000,000,000,000,000.

5 - 31GS105 1 car ___ 4 wheels The Mole & Formulas 1 mol cars_ 4 mol wheels 4 mol wheels 1 doz cars 4 doz wheels 1 mole H 2 O 2 mol H 2 mol H 1 mole H 2 O 1 mol O 1 mol O

5 - 32GS105 1 mole = MW in g’s The Mole & Molar Mass 1 mol S_ 32 g S 32 g S 1 mol S_ 32 g S 32 g S 1 mole S = 32 g S 32 g S 1 mol S 32 g S 1 mol S 1 mol C 12 g C 12 g C 1 mol C 12 g C 12 g C 1 mole C = 12 g C 12 g C 1 mol C 12 g C 1 mol C __32 g S _ __32 g S _ 6.02 x atoms S 6.02 x atoms S __32 g S _ __32 g S _ 6.02 x atoms S 6.02 x atoms S __12 g C _ __12 g C _ 6.02 x atoms C 6.02 x atoms C __12 g C _ __12 g C _ 6.02 x atoms C 6.02 x atoms C

5 - 33GS105

5 - 34GS105 Molar Mass Find the MW (or MM) of Glucose; C 6 H 12 O g H = 1 mol H 12 mol H g H 16.0 g O = 1 mol O 6 mol O g O g C 6 H 12 O 6 C 6 H 12 O 6 1 mol C 6 H 12 O g C = 1 mol C 6 mol C g C

5 - 35GS105 Molar Mass Find the MW (MM) of Water; H 2 O 1.0 g H = 1 mol H 2 mol H g H 2.0 g H 16.0 g O = 1 mol O g O 18.0 g H 2 O H 2 O 1 mol H 2 O 1 mol H 2 O_ 18.0 g H 2 O

5 - 36GS105 1 mol H 2 O = 18 g H 2 O Mass to Mole Conversions How many moles of water are in 36 g H 2 O? What should the answer look like? What is Unique to the problem? 36 g H 2 O 36 g H 2 O 1 mol H 2 O mol H 2 O_ 18.0 g H 2 O 18.0 g H 2 O 18.0 g H 2 O 1 mol H 2 O

5 - 37GS105 1 mol H 2 O 18 g H 2 O Mass to Mole Conversions How many molecules of water are in 36 g H 2 O? What should the answer look like? What is Unique to the problem? 36 g H 2 O 36 g H 2 O 1 = molecules H 2 O 1.2 x molecules H 2 O 1 mol H 2 O_ 18.0 g H 2 O 1 mol H 2 O_ 6.02x10 23 molecules H 2 O 6.02x10 23 molecules H 2 O 6.02x10 23 molecules H 2 O _ 1 mol H 2 O

5 - 38GS105 1 mol H 2 O 18 g H 2 O Mass to Mole Conversions How many moles of H are in 36 g H 2 O? What should the answer look like? What is Unique to the problem? 36 g H 2 O 36 g H 2 O 1 mol H mol H = 1 mol H 2 O

5 - 39GS g Gluc = 1 mol Gluc Mole to Mass Conversions How many g’s of Glucose (C 6 H 12 O 6 ) are in 5 mol Glucose? What should the answer look like? What is Unique to the problem? 5 mol Gluc 1 g Glucose g C 6 H 12 O 6 C 6 H 12 O 6 1 mol C 6 H 12 O 6 1 mol C 6 H 12 O g C 6 H 12 O 6

5 - 40GS105 Ratios in Chemical Equations C (s) O 2 (g CO 2(g) C (s) + O 2 (g)  CO 2(g) 1 mol C 1 mol O 2 1 mol O 2 1 mol C CO 2 1 mol CO 2 O 2 1 mol O 2 1 mol CO 2 O 2  1 molecule CO 2 1 atom C + 1 molecule O 2  1 molecule CO 2 12 g C 32 g O 2 32 g O 2 12 g C CO 2 44 g CO 2 O 2 32 g O 2 44 g CO 2 O 2  44 g CO 2 12 gs of C + 32 gs of O 2  44 g CO 2

5 - 41GS105 Ratios in Chemical Equations C (s) O 2 (g CO 2(g) C (s) + O 2 (g)  CO 2(g) 1 mol C 1 mol O 2 1 mol O 2 1 mol C CO 2 1 mol CO 2 O 2 1 mol O 2 1 mol CO 2 O 2 are needed to produce 11 g CO 2 ? How many gs of O 2 are needed to produce 11 g CO 2 ? 12 g C 32 g O 2 32 g O 2 12 g C CO 2 44 g CO 2 O 2 32 g O 2 44 g CO 2

5 - 42GS g CO 2 Ratios in Chemical Equations 32 g O 2 44 g CO 2 O 2 1 mol O 2 1 mol CO 2 So if it takes 32 g O 2 to make 44 g CO 2 then: 11 g CO 2 = g O 2 8 C (s) O 2 (g CO 2(g) C (s) + O 2 (g)  CO 2(g) O 2 are needed to make 11 g CO 2 ? How many gs of O 2 are needed to make 11 g CO 2 ? 32 g O 2 44 g CO 2 X g O 2 11 g CO 2 = 32 g O 2 44 g CO 2 X g O 2 11 g CO 2 = Ratio method: We know:

5 - 43GS105 Ratios in Chemical Equations 32 g O 2 44 g CO 2 O 2 1 mol O 2 1 mol CO 2 Identify any conversion factors. How should the answer look? 11 g CO 2 = g O 2 What is unique to the problem? 8 O 2 32 g O 2 44 g CO 2 C (s) O 2 (g CO 2(g) C (s) + O 2 (g)  CO 2(g) O 2 are needed to make 11 g CO 2 ? How many gs of O 2 are needed to make 11 g CO 2 ? Factor conversion method:

5 - 44GS105 Ratios in Chemical Equations 11 g CO 2 = g O 2 8 O 2 32 g O 2 44 g CO 2 C (s) O 2 (g CO 2(g) C (s) + O 2 (g)  CO 2(g) O 2 are needed to make 11 g CO 2 ? How many gs of O 2 are needed to make 11 g CO 2 ? O 2  11 g CO 2 ? gs C + 8 gs O 2  11 g CO 2 So: 3 O 2  13 g CO 2 ? 5 gs C + 8 gs O 2  13 g CO 2 ? Then could : NO! There is only enough O 2 to make 11 g’s of CO 2. We would still make 11 g’s of CO 2 but have 2 g’s of C left over. O 2 is our limiting reagent.

5 - 45GS105 Moles in Chemical Equations 2 Na (s) Cl 2 (g 2 NaCl 2 Na (s) + Cl 2 (g)  2 NaCl 2 mol Na 1 mol Cl 2 1 mol Cl 2 2 mol Na 2 mol NaCl Cl 2 1 mol Cl 2 2 mol NaCl Cl 2 are needed to produce 2.5 mol NaCl? How many gs of Cl 2 are needed to produce 2.5 mol NaCl? 46 g Na 71 g Cl 2 71 g Cl 2 46 g Na 58.5 g NaCl Cl 2 71 g Cl g NaCl

5 - 46GS105 Moles in Chemical Equations 2 Na (s) Cl 2 (g 2 NaCl 2 Na (s) + Cl 2 (g)  2 NaCl 1 mol Cl g Cl 2 Cl 2 1 mol Cl 2 2 mol NaCl Cl 2 are needed to produce 2.5 mol NaCl? How many gs of Cl 2 are needed to produce 2.5 mol NaCl? = 89 gCl 2 How many moles of Cl 2 are needed? X mol Cl mol NaCl Cl 2 1 mol Cl 2 2 mol NaCl = How many grams of Cl 2 are needed? mol Cl 2 X g Cl 2 = 1 mol Cl 2 71 g Cl

5 - 47GS105 Moles in Chemical Equations 2 Na (s) Cl 2 (g 2 NaCl 2 Na (s) + Cl 2 (g)  2 NaCl 1 mol Cl g Cl 2 Cl 2 1 mol Cl 2 2 mol NaCl Cl 2 are needed to produce 2.5 mol NaCl? How many gs of Cl 2 are needed to produce 2.5 mol NaCl? Identify any conversion factors. How should the answer look? 2.5 mol NaCl = gCl 2 What is unique to the problem? = 89 gCl 2 Cl 2 1 mol Cl 2 2 mol NaCl 70.9 g Cl 2 1 mol Cl 2 Factor conversion method:

5 - 48GS105 H2OH2OH2OH2O H2OH2OH2OH2O more stable Energy in Chemical Reactions Exothermic reaction -  H= heat of reaction Energy Rxn Progress Reactants Products E act = Activation Energy (Gets hot) H 2 + O 2 2H 2 + O 2 2H 2 O + Energy -486 kJ made

5 - 49GS105 Bond Energy 2H 2 + O 2 2H 2 O + Energy 2H-H + O=O  2 H-O-H 2 (436)4984 (464) 1370 kJ used1856 kJ made-=-486 kJ made

5 - 50GS105 Exothermic Reactions Energy Rxn Progress 2Mg + O 2  2MgO + Energy 2H 2 + O 2  2H 2 O + Energy CH 4 + 2O 2  CO 2 + 2H 2 O + Energy -486 kJ made

5 - 51GS105 Bond Energy CH 4 + 2O 2 CO 2 CO 2 4C-H + 2O=O  O=C=O + 2 H-O-H 4 (414)2(498)2 (799) 2652 kJ used3454 kJ made- =- 802 kJ made + 2H 2 O + Energy + 2H 2 O + Energy 4 (464)

5 - 52GS105 more stable Energy in Chemical Reactions Endothermic reaction  H= heat of reaction Energy Rxn Progress Reactants Products E act = Activation Energy (Gets cold)

5 - 53GS105 Energy Rxn Progress Endothermic Reactions Energy is required Products are less stable. Products(Water) Reactants(Ice) +H+H+H+H

5 - 54GS105 Exothermic Reactions Energy Rxn Progress Reactants(Water) Products(Ice) Energy is released Products are more stable. -H-H-H-H

5 - 55GS105 Energy Rxn Progress 2H 2(g) + O 2(g)  2H 2 O + Energy Just because something has the potential to react doesn’t mean it will do so immediately. doesn’t mean it will do so immediately. Just because something has the potential to react doesn’t mean it will do so immediately. doesn’t mean it will do so immediately. H 2(g) O 2(g) may stay together for lifetime without reacting to form water. H 2(g) + O 2(g) may stay together for lifetime without reacting to form water. Reaction Rates

5 - 56GS105 burning Paper burning Paper yellowing Reaction Rates Fast: Slow: Human Aging Explosion Fruit Ripening rusting Nails rusting

5 - 57GS105 They have to have enough E. Reaction Rates For reactants to make products: collideMolecules must collide (If they don’t meet they can’t react: solvents really help)solvents really help) alignedThey have to be aligned correctly. (Parked cars don’t collide) (Car bumpers meeting don’t make a change)

5 - 58GS105 Rates of Reactions Reaction rates can be affected by : polar vs. nonpolar reactant structure( polar vs. nonpolar ) vapor vs liq physical state of reactants ( vapor vs liq. ) medications concentration of reactants ( medications ) sugar cube vs crystals surface area ( sugar cube vs crystals ) hypothermia & metabolism temperature ( hypothermia & metabolism ) H 2 O 2 & blood catalyst ( H 2 O 2 & blood )

5 - 59GS105 Reaction Rates Factors that increase reaction rate: 1. Higher Temperature: Faster cars  More collisions More Energy  More collisions Need Energy of Activation (E act )

5 - 60GS105 Reaction Rates Factors that increase reaction rate: 2. More Reactants (Greater Concentration) : More cars  More collisions 8 blocks: 34 surfaces 8 blocks: 24 surfaces More surface area  More collisions

5 - 61GS105 Reaction Rates Factors that increase reaction rate: 3.Adding a Catalyst: Lower E act  More collisions Uncatalysed reaction

5 - 62GS105 Alter the reaction mechanism but not change the products Uncatalysed reaction Catalysed reaction Lower activation energy A catalyst will: Enzymes Enzymes are biological catalysts. Catalysts

5 - 63GS105 Catalysts

5 - 64GS105 Equilibrium A state where the forward and reverse conditions occur at the same rate. Dynamic Equilibrium I’m in static equilibrium.

5 - 65GS105 Chemical equilibrium Dynamic process Rate of forward Rxn = Rate of reverse Rxn H 2 O (l) H 2 O (g) H 2 O (l) H 2 O (g) (reactant) (product) (reactant) (product) Dynamic Equilibrium Concentration of reactants and products remain constant over time.

5 - 66GS105 Equilibrium and reaction rates A point is ultimately reached where the rates of the forward and reverse reactions are the same. At this point, equilibrium is achieved. Reaction rate Time H 2 O (l) H 2 O (g) H 2 O (l) H 2 O (g) (reactant) (product)

5 - 67GS105 Figure 9.8 2SO 2(g) + O 2(g) 2SO 3(g) At Equilibium SO 2(g) +O 2(g) Initially SO 3(g) Initially

5 - 68GS105 Figure 9.9 2SO 2(g) + O 2(g) 2SO 3(g) At Equilibium SO 2(g) +O 2(g) Initially SO 3(g) Initially

5 - 69GS105 Figure 9.10 N 2(g) + O 2(g) 2NO (g) At Equilibium N 2(g) +O 2(g) Initially NO (g) Initially

5 - 70GS105 Equilibrium Concentration Time KineticEquilibrium Region Concentration of reactants and products remain constant over time.

5 - 71GS105 Le Chatelier’s principle Stress causes shift in equilibrium Adding or removing reagent CaCO 3(s) CaO (s) + CO 2(g) Stress causes shift in equilibrium Adding or removing reagent CaCO 3(s) CaO (s) + CO 2(g) Let CO 2 escape? CO 2 leaves Reaction shifts to the right [ CaCO 3 ] dec, [CaO] inc

5 - 72GS105 Le Chatelier’s principle Stress causes shift in equilibrium Adding or removing reagent CaCO 3(s) CaO (s) + CO 2(g) Stress causes shift in equilibrium Adding or removing reagent CaCO 3(s) CaO (s) + CO 2(g) Close the container so CO 2 doesn’t escape? Add CO 2 Reaction shifts to the left [ CaCO 3 ] Inc, [CaO] dec

5 - 73GS105 Le Chatelier’s principle Stress causes shift in equilibrium Adding or removing reagent N 2(g) + 3 H 2(g) 2 NH 3(g) Stress causes shift in equilibrium Adding or removing reagent N 2(g) + 3 H 2(g) 2 NH 3(g) Add more N 2 ? N2N2N2N2 Reaction shifts to the right [ NH 3 ] inc, [H 2 ] dec

5 - 74GS105 Le Chatelier’s principle Adding or removing reagent N 2(g) + 3 H 2(g) 2 NH 3(g) Adding or removing reagent N 2(g) + 3 H 2(g) 2 NH 3(g) Add more NH 3 ? NH 3 Reaction shifts to the left N 2 [ N 2 ] and [H 2 ] inc

5 - 75GS105 Le Chatelier’s principle Adding Pressure affects an equilibrium with gases N 2(g) + 3 H 2(g) 2 NH 3(g) P Add P? Increasing pressure causes the equilibrium to shift to the side with the least moles of gas. 4 mol of reactants 2 mol of products

5 - 76GS105 Le Chatelier’s principle Temperature can also have an effect. exothermic For exothermic reactions reactants products + heat Raising the temperature shifts it to the left. endothermic For endothermic reactions heat + reactants products Raising the temperature shifts it to the right.

5 - 77GS105 Fe 3+ + SCN - FeSCN 2+ Iron and thiocyanide are in equilibrium with the ferrocyanide ion. 1. What happens when Fe 3+ is added ? 2. What happens when SCN - is added ? 3. What happens when Fe 3+ is removed ? Le Chatelier’s principle

5 - 78GS105 FeCl 3 + 3NH 4 CNS Fe(CNS) 3 + 3NH 4 Cl YellowRed FeCl 3 1. What happens when FeCl 3 is added ? NH 4 CNS 2. What happens when NH 4 CNS is added ? Fe(CNS) 3 3. What happens when Fe(CNS) 3 is removed ? Le Chatelier’s principle

5 - 79GS105 Figure 9.12

5 - 80GS105 Figure 9.13

5 - 81GS105 Without added energy reactions go to disorder Entropy = The amount of disorder: Entropy Randomness factor Nature tends to move spontaneously from a state of lower probably to one of higher probability. If a beaker with marbles is shaken, which beaker (above) will result?