Energy Energy

The ability to do work or The ability to do work or produce heatproduce heatPotential- Stored energyPotential- Stored energy

Energy stored in chemical Energy stored in chemical bondsbonds

Kinetic- Energy of movementKinetic- Energy of movementEnergy of moving moleculesEnergy of moving molecules

Law of Conservation of Law of Conservation of EnergyEnergy

Energy can not be created Energy can not be created or destroyed, it can only be or destroyed, it can only be converted from one form to converted from one form to another.another.

Chemical Potential Chemical Potential EnergyEnergy

Chemical bonds of a Chemical bonds of a compound contain energycompound contain energy

That energy can be That energy can be converted to do work or is converted to do work or is converted to heat converted to heat

HeatHeat

Heat flows from warm Heat flows from warm objects to cooler objects.objects to cooler objects.

The two objects will The two objects will reach equilibrium and reach equilibrium and record the same record the same temperaturetemperature

Measuring HeatMeasuring Heat Calorie- The amount of Calorie- The amount of

energy required to raise 1 g energy required to raise 1 g of water 1of water 1ooCC1 food Calorie = 1000 calories1 food Calorie = 1000 calories

Joule- The unit of Joule- The unit of measurement for heat energymeasurement for heat energy1 Joule = .2390 calories 1 Joule = .2390 calories

ProblemProblem

How Joules of heat would How Joules of heat would be obtained from a 364 be obtained from a 364 nutritional Calorie nutritional Calorie hamburger?hamburger?

See chart on p. 491 for See chart on p. 491 for conversionsconversions

AnswerAnswer 64 Cal x 64 Cal x 1000 cal1000 cal x x 1 J 1 J = =

1523012 J 1523012 J 1 Cal .2390 cal1 Cal .2390 cal

Specific HeatSpecific HeatThe amount of energy The amount of energy required to raise 1 gram of required to raise 1 gram of any substance 1any substance 1ooCC

Specific heat determines how Specific heat determines how fast or slow that substances fast or slow that substances gain or lose heat. gain or lose heat.

Specific Heat and WaterSpecific Heat and WaterWater has a high specific heatWater has a high specific heat It gains and loses heat energy It gains and loses heat energy very slowlyvery slowly

This is why bodies of water This is why bodies of water stay cool long into spring and stay cool long into spring and warm long into fallwarm long into fall

Calculating Heat ChangeCalculating Heat Changeq = c x m x q = c x m x T T q = Heat gain or loss- Jq = Heat gain or loss- Jc = Specific heat - J/(g . c = Specific heat - J/(g . ooC)C)m = mass gm = mass g T = Change in temperature- T = Change in temperature- ooC C

Calculating Heat ChangeCalculating Heat ChangeHow much heat is required to How much heat is required to raise 2.3 kg of iron 23 raise 2.3 kg of iron 23 ooC?C?

q = c x m x q = c x m x TT q = .449 J/(g.q = .449 J/(g.ooC) x 2300g x 23C) x 2300g x 23ooCCq = 23752.1 J q = 23752.1 J

Calculating Heat ChangeCalculating Heat Change If there is more than one If there is more than one substance you will need to substance you will need to calculate the heat change calculate the heat change separatelyseparately

Ex. Water in an iron pot would Ex. Water in an iron pot would require calculations for both require calculations for both iron and wateriron and water

Determining Specific HeatDetermining Specific Heat 1. Heat the unknown 1. Heat the unknown

substance to a set temperaturesubstance to a set temperature 2. Place the substance in a 2. Place the substance in a

measured mass of watermeasured mass of water 3. Measure the temperature 3. Measure the temperature

change when equilibrium is change when equilibrium is reachedreached

Determining Specific HeatDetermining Specific Heat

4. Calculate the amount of 4. Calculate the amount of energy gained by the water. energy gained by the water. This equals the amount of This equals the amount of energy lost by the substanceenergy lost by the substance

Use the heat value to solve Use the heat value to solve for the specific heat of the for the specific heat of the substancesubstance

Calorimeter LabCalorimeter Lab

Calorimeter- A device Calorimeter- A device used to measure the used to measure the amount of heat gained amount of heat gained or lost by a substanceor lost by a substance

See handout for See handout for instructionsinstructions

ThermochemistryThermochemistryThe study of heat changes The study of heat changes involved in chemical involved in chemical reactionsreactions

Reactions are either Reactions are either exothermic (release energy) exothermic (release energy) or endothermic (absorb or endothermic (absorb energy)energy)

The UniverseThe Universe

A system plus its A system plus its surroundingssurroundings

The System

Surroundings

Enthalpy (H)Enthalpy (H) The heat content of a substance The heat content of a substance

at a constant pressureat a constant pressure There is no way to measure the There is no way to measure the

total enthalpy of a substance but total enthalpy of a substance but you can calculate the change in you can calculate the change in enthalpy for a reactionenthalpy for a reaction

Enthalpy ChangesEnthalpy Changes To calculate enthalpy change To calculate enthalpy change

use the following equationuse the following equation HHrxnrxn = H = Hproductproduct - H - Hreactantsreactants

If If HHrxn rxn is positive the reaction is positive the reaction is endothermic, if negative is endothermic, if negative the reaction is exothermic the reaction is exothermic

Enthalpy changeEnthalpy change

Exothermic reactionExothermic reaction

E

n

t

h

a

l

p

y

4Fe(s) + 3O2(g)

Fe2O3(s)

H= -1625

Product

Reactants

Enthalpy changeEnthalpy change

Endothermic reactionEndothermic reactionE

n

t

h

a

l

p

y

NH4+(aq) + NO3

-(aq)

Product

NH4NO3(s)

Reactant

H= 27 kJ

Heat Gain and EnthalpyHeat Gain and Enthalpy

q = c x m x q = c x m x T T H = HH = Hproductproduct - H - Hreactantsreactants

q = q = H since heat gain or H since heat gain or loss is the same as loss is the same as enthalpy change at enthalpy change at constant pressureconstant pressure

Writing Thermochemical Writing Thermochemical EquationsEquations

1. Write the equation1. Write the equation2. Write the 2. Write the H value to H value to the right of the equationthe right of the equation

Writing Thermochemical Writing Thermochemical EquationsEquations

3. If the reaction is 3. If the reaction is exothermic the exothermic the H value is H value is negative, if endothermic, negative, if endothermic, positivepositive

Enthalpy of CombustionEnthalpy of Combustion

The amount of energy The amount of energy released when one mole released when one mole of a substance is burnedof a substance is burned

HHoocombcomb = Heat of = Heat of

combustion combustion

Changes of StateChanges of State

HHvapvap = Heat of vaporization = Heat of vaporizationHHcondcond = Heat of = Heat of condensationcondensation

HHfus fus = Heat of fusion= Heat of fusionHHsolid solid = Heat of solid= Heat of solid

Changes of stateChanges of state

Changing phase from a Changing phase from a solid to a liquid, liquid to solid to a liquid, liquid to gas or vice versa gas or vice versa requires a change in requires a change in energyenergy

Changes of stateChanges of state

The amount of energy The amount of energy required to convert one required to convert one mole of a substance mole of a substance from a liquid to a gas or from a liquid to a gas or from a liquid to a solidfrom a liquid to a solid

Equations for phase Equations for phase changeschanges

HH22O(l) --> HO(l) --> H22O(g) O(g) HHvapvap=40.7 kJ=40.7 kJ HH22O(g) --> HO(g) --> H22O(l) O(l) HHvapvap=-40.7 kJ=-40.7 kJ HH22O(s) --> HO(s) --> H22O(l) O(l) HHfus fus = 6.01 kJ= 6.01 kJ HH22O(l) --> HO(l) --> H22O(s) O(s) HHvapvap=-6.01 kJ=-6.01 kJ

AssignmentAssignment

P. 500 14, 15, 16, 17P. 500 14, 15, 16, 17p. 504 20-22p. 504 20-22

Lab- Energy ChangeLab- Energy Change

P. 503P. 503Use Microsoft Excel to Use Microsoft Excel to generate graphgenerate graph

Hess’s LawHess’s LawStates that if you can add 2 States that if you can add 2 or more equations to or more equations to produce a final equation for produce a final equation for a reaction, then the sum of a reaction, then the sum of the enthalpy changes for the the enthalpy changes for the individual reactions = the individual reactions = the enthalpy for the final enthalpy for the final reaction.reaction.

Hess’s LawHess’s Law What does this mean? What does this mean? You can predict enthalpy You can predict enthalpy

changes for reactions that can changes for reactions that can not be observed directly.not be observed directly.

Ex. A reaction that takes Ex. A reaction that takes several thousand years. several thousand years. Converting carbon to a Converting carbon to a diamond.diamond.

Applying Hess’s LawApplying Hess’s Law

What is the enthalpy What is the enthalpy change for the following change for the following reaction?reaction?

2S(s) + 3O2S(s) + 3O22(g) (g) 2SO2SO33(g)(g)

Applying Hess’s LawApplying Hess’s LawWe know the following We know the following equations equations

S(s) + OS(s) + O22(g) (g) SOSO22(g) (g) H=-297 kJH=-297 kJ

2SO2SO33(g) (g) 2SO2SO22(g) + O(g) + O22(g) (g) H=198 kJH=198 kJ

Applying Hess’s LawApplying Hess’s Law

1. You need to change 1. You need to change the coefficients in the the coefficients in the two equations to match two equations to match the molar amounts in the molar amounts in the original equationthe original equation

Applying Hess’s LawApplying Hess’s Law 2S(s) + 3O2S(s) + 3O22(g) (g) 2SO 2SO33(g)(g) S(s) + OS(s) + O22(g) (g) SOSO22(g) (g) H=-297 kJH=-297 kJ Since there are 2 mol of S in Since there are 2 mol of S in

the original equation the the original equation the second equation must by second equation must by multiplied by a factor of 2. multiplied by a factor of 2.

Applying Hess’s LawApplying Hess’s Law 22((S(s) + OS(s) + O22(g) (g) SOSO22(g)(g)H=-297 H=-297

kJkJ)) 22S(s) + S(s) + 22OO22(g) (g) 22SOSO22(g) (g) H=H=22(-297 kJ) = -594 kJ(-297 kJ) = -594 kJ Now we have the amount of heat Now we have the amount of heat

required for the reaction of 2 required for the reaction of 2 moles of Smoles of S

Applying Hess’s LawApplying Hess’s Law

Since SO3 is the product in Since SO3 is the product in the original reaction we have the original reaction we have to reverse the second reaction to reverse the second reaction from 2SOfrom 2SO33(g) (g) 2SO2SO22(g) + (g) + OO22(g) (g) H=198 kJ to H=198 kJ to

2SO2SO22(g) + O(g) + O22(g) (g) 2SO 2SO33(g) (g) H= -198 kJH= -198 kJ

Applying Hess’s LawApplying Hess’s Law

2S(s) + 2O2S(s) + 2O22(g) (g) 2SO 2SO22(g) (g) H= -594 kJH= -594 kJ 2SO2SO22(g) + O(g) + O22(g) (g) 2SO2SO33(g) (g) H= -198 kJH= -198 kJ

2S+2O2S+2O22+2SO+2SO22+ O+ O22 2SO 2SO22+2SO+2SO3 3 H= -H= -792kJ792kJ

2S + 3O2S + 3O22 2SO 2SO33 H= -792 kJH= -792 kJ So the original reaction evolves 792 kJ So the original reaction evolves 792 kJ

AssignmentAssignment

P. 508 # 28 and 29P. 508 # 28 and 29

Enthalpy of formationEnthalpy of formation Elements are free atoms. It takes Elements are free atoms. It takes

no heat to form them. They are in no heat to form them. They are in the lowest energy state possible. the lowest energy state possible.

Elements are assigned a value of Elements are assigned a value of 0 kJ for 0 kJ for HHf f since there is no since there is no energy needed to create them energy needed to create them because they are the simplest because they are the simplest substances on Earth substances on Earth

Enthalpy of formationEnthalpy of formation

Compounds are formed by Compounds are formed by combining elements. Energy is combining elements. Energy is required to combine elements, required to combine elements, therefore the therefore the H has a value H has a value greater than 0 if elements form a greater than 0 if elements form a compound. compound.

This is an This is an endothermicendothermic reaction. reaction.

Enthalpy of formationEnthalpy of formation Hrxn = Hrxn = HfHf(product)(product) - - HfHf(reactant)(reactant)

Add up all of the Add up all of the HHff products and subtract the products and subtract the sum of the sum of the HHf f reactantsreactants

This equals the net This equals the net H of the H of the reaction reaction

Reaction SpontaneityReaction Spontaneity Spontaneous Process- a physical Spontaneous Process- a physical

or chemical change that occurs or chemical change that occurs with no outside interventionwith no outside intervention

Most exothermic reactions are Most exothermic reactions are spontaneous and most spontaneous and most endothermic reactions are not endothermic reactions are not spontaneous. spontaneous.

Reaction SpontaneityReaction Spontaneity Not all exothermic reactions are Not all exothermic reactions are

spontaneous and not all spontaneous and not all endothermic reactions are not endothermic reactions are not spontaneousspontaneous

Ice melting is spontaneous but Ice melting is spontaneous but requires energy to occur. This requires energy to occur. This makes in endothermic and makes in endothermic and spontaneousspontaneous

Reaction SpontaneityReaction Spontaneity

What accounts for an What accounts for an endothermic process endothermic process being spontaneous?being spontaneous?

EntropyEntropy

EntropyEntropyThe measure of the The measure of the randomness and disorder of randomness and disorder of a systema system

Substances are more likely Substances are more likely to exist in a high state of to exist in a high state of randomness than in a low randomness than in a low state of randomnessstate of randomness

EntropyEntropySubstances are more likely Substances are more likely to have high entropy than to have high entropy than low entropylow entropy

Spontaneous processes Spontaneous processes always proceed in such a always proceed in such a way that the entropy of the way that the entropy of the universe increasesuniverse increases

EntropyEntropy

Think of a deck of cardsThink of a deck of cardsWhat are the odds that well What are the odds that well shuffled cards will come out shuffled cards will come out in order? in order?

EntropyEntropy

Not very high- 1 in Not very high- 1 in 807,000,000,000,000,000,807,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 000,000,000,000,000,000

Random is favoredRandom is favored

EntropyEntropy

SSsystemsystem = S = Sproductsproducts – S – Sreactantsreactants

If entropy increases in a If entropy increases in a reaction the entropy is reaction the entropy is positivepositive

If entropy decreases then If entropy decreases then entropy is negativeentropy is negative

Predicting entropy Predicting entropy changeschanges

If substances change phase If substances change phase entropy increases as the entropy increases as the energy of the substance energy of the substance increasesincreases

As you go from a solid to a As you go from a solid to a liquid to a gas entropy liquid to a gas entropy increasesincreases

Predicting entropy Predicting entropy changeschanges

If a gas is dissolved in a If a gas is dissolved in a solvent entropy decreasessolvent entropy decreases

Entropy increases when the Entropy increases when the number of gas particles as a number of gas particles as a product is greater than the product is greater than the number of gas particles as a number of gas particles as a reactantreactant

Predicting entropy Predicting entropy changeschanges

Entropy increases as Entropy increases as solids are dissolved in a solids are dissolved in a solventsolvent

Entropy increases as Entropy increases as temperature increasestemperature increases

Entropy and Free EnergyEntropy and Free Energy

SSuniverseuniverse > 0 > 0SSuniverse universe = = SSsystemsystem + + SSsurroundingssurroundings

If a reaction increases the If a reaction increases the entropy of the universe then entropy of the universe then it will be spontaneous it will be spontaneous

Entropy and Free EnergyEntropy and Free Energy

Gibbs free energy determines Gibbs free energy determines if a reaction is spontaneousif a reaction is spontaneous

GGsystemsystem = = HHsystem system - T- T S Ssystemsystem

If If G is negative the reaction G is negative the reaction is spontaneous, if positive the is spontaneous, if positive the reaction is not spontaneousreaction is not spontaneous

P. 517, 518P. 517, 518

Copy chart 16-8, 16-9Copy chart 16-8, 16-9