PowerPoint Presentation

CHAPTER 10ENERGY

The nature of ENERGYENERGY The ability to do work or produce heat

LAW OF CONSERVATION OF ENERGYEnergy can neither be created nor destroyedBUTEnergy can be converted from one form to another

POTENTIAL ENERGYEnergy of an object due to its mass and position.KINETIC ENERGYEnergy of an object due to its mass and motion.

Thing to rememberFor Potential or Kinetic energy:When one increases, the other decreasesPATHWAY of the reaction is NOT important [HOW]CHANGE IN LEVEL/ELEVATION is important [WHAT]

ANALOGY:If you have to pay $8 to a friend, you can either:Pay him exactly $8 (one $5 and three $1)Pay him $10 and ask for $2 backIn this scenario: HOW you pay does not matter.WHAT you pay does matter.TEMPERATURE AND HEATTEMPERATUREIt is a measure of the random motions of the molecules/atoms in a substance.HEATFlow of energy due to a difference in temperature

DO NOT confuse Temperature with Heat, they are two different things. Temperature is the measure of heat or lack of heat caused by the fast or slow movement of molecules.

SYSTEM AND SURROUNDINGSSYSTEMThe part of the universe we are concerned withContains the reactants and products of a reaction

SURROUNDINGSEverything else in the universe that is NOT the system

THESE 2 OBJECTS WOULD CONSTITUTE THE SYSTEMEVERYTHING OUTSIDE THESE 2 OBJECTS WOULD BE THE SURROUNDINGSEXOTHERMIC & ENDOTHERMICENDOTHERMICWhen energy flows into the system in the form of heat

EXOTHERMICWhen energy flows out of the system in the form of heat

Exothermic and PE relationshipPotential energy stored in chemical bonds is converted to thermal energy (heat).Reactants have more PE than productsProducts have less PE than reactants

Endothermic and PE relationshipPotential energy stored in chemical bonds of reactants is not enough. Therefore, more energy is necessary (usually provided in the form of heat).Products have more PE than reactantsReactants have less PE than products

THERMODYNAMICSTHERMODYNAMICS is the study of the effects of work, heat, and energy on a system.There are 4 laws of ThermodynamicsLaws 0 3

ZEROTH LAW OF THERMODYNAMICS

FIRST LAW OF THERMODYNAMICSThe energy of the universe is constant means the same thing as the Law of Conservation of Energy

The internal energy of a system (denoted by E) is the sum of all KE and PE of all the particles in the system.

= Change inw = workE = internal energyq = heat

E = q + wTHERMODYNAMIC QUANTITIESNUMBERIndicates the magnitude/size of the energy changeSIGNIndicates the direction of the energy flow

When you add heat into the system, q = (+) , E = (+)

When you lose heat from the system, q = () , E = ()

ENDOTHERMICEXOTHERMIC

MEASURING ENERGY CHANGESCommon units of energy arecalorie (with a lowercase C)Joule (SI unit)1 calorie = 4.184 JoulesThe Calorie that you are familiar with is denoted with a uppercase C because it is actually 1000 calories (1 kilocalorie)

cal J / J cal CONVERSIONSCLASSWORK CONVERSIONSExpress 34.8 cal of energy in Joules.

Express 47.3 J of energy in calories.

Express 0.0348 cal of energy in Joules.

Express 47.3 J of energy in Calories.

Answer:146 JAnswer:11.3 calAnswer:146 JAnswer:0.0113 Cal or kcalSPECIFIC HEAT CAPACITY

SPECIFIC HEAT CAPACITIESTable 10.1 in your textbook (Page 279) You DO NOT need to memorize this table here or the table in the textbook. Use it as a reference to solve problems.

FORMULA involving specific heatCLASSWORK PROBLEMSAnswer:8.1 J -or- 1.9 cal Answer:12 J -or- 2.9 cal CLASSWORK PROBLEMSENTHALPYEnthalpy is the total energy produced () or absorbed (+) by a given reaction.ENTHALPY is denoted by HSo exothermic reaction (+) or () enthalpy?So endothermic reaction (+) or () enthalpy?For a reaction occurring under constant pressure, the CHANGE IN ENTHALPY is the same as the energy that flows as heat.H = Qp

ENTHALPY VALUES?2 mini reactions:Elephants toothpaste (Hydrogen peroxide decomposition)ENDOTHERMIC OR EXOTHERMIC?WHY?POSITIVE OR NEGATIVE ENTHALPY VALUE?

Sulfur dioxide productionENDOTHERMIC OR EXOTHERMIC?WHY?POSITIVE OR NEGATIVE ENTHALPY VALUE?

ENTHALPY PROBLEMSWhen 1 mole of methane (CH4) is burned at constant pressure, 890 kJ of energy is released as heat. Calculate H for a process in which a 5.8-g sample of methane is burned at constant pressure.We know that 1 mole of CH4 releases 890 kJ of heatSo, 1 mole CH4 H = 890 kJ But how many moles in 5.8-g of CH4?And what is the H value for that amount of moles?ENTHALPY PROBLEM # 1 HENTHALPY PROBLEM # 2The reaction that occurs in the heat packs used to treat sports injuries is when Iron metal reacts with Oxygen gas to form solid Iron (III) Oxide. The energy released in this balanced reaction is 1652 kJ. How much heat is released when 1.00g of iron reacts with excess oxygen gas?

4 Fe (s) + 3 O2 (g) 2 Fe2O3 (s)

H = 1652 kJ

ENTHALPY PROBLEM # 2 HExtra Practice Problems WS See Worksheet for problems 3 5Done in class.HESS LAWIn going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

**REMEMBER** the analogy of paying your friend $8. It doesnt matter HOW you pay him as long as you pay it.Hess Law: Case-in-point!One-step reaction:H = 68 kJN2 (g) + 2 O2 (g) 2 NO2 (g)

Two-step reaction:H = 68 kJ N2 (g) + O2 (g) 2 NO (g)H1 = +180 kJ2 NO (g)+ O2 (g) 2 NO2 (g) H2 = 112 kJ

Net Rxn: N2 (g) + 2 O2 (g) 2 NO2 (g)Net H: H1 + H2 = 68 kJ

Characteristics of Enthalpy ChangesIf the order of a reaction is reversed, then the sign of H is also reversed.

Example:

N2 (g) + 2 O2 (g) 2 NO2 (g)H = 68 kJ

2 NO2 (g) N2 (g) + 2 O2 (g) H = 68 kJ

Characteristics of Enthalpy ChangesIf the coefficients in a balanced reaction are multiplied by an integer, the value of H is also multiplied by that same integer.

Example:

N2 (g) + 2 O2 (g) 2 NO2 (g) H = 68 kJ[N2 (g) + 2 O2 (g) 2 NO2 (g)] 4 H = 68kJ 44 N2 (g) + 8 O2 (g) 8 NO2 (g) H = 272 kJ

PROBLEMApply Hess LawApply Characteristics of Enthalpy Changes

Two forms of carbon are graphite and diamond. In each combustion reaction, carbon reacts with oxygen gas to form only carbon dioxide gas. Using the enthalpies of combustion for graphite (-394 kJ/mol) and diamond (-396 kJ/mol), calculate H for the conversion of graphite to diamond. Extra Practice Problems WS See Worksheet for problems 2 4 Done in class.