Energy Chap. 16. I.Definitions A. Energy Energy is the ability to do work or produce heat I.Definitions

EnergyEnergyChap. 16Chap. 16Chap. 16Chap. 16

I.I. DefinitionsDefinitions

A.A. EnergyEnergy

Energy is the ability to do work or produce heat


A.A. EnergyEnergy

B.B. HeatHeat

Heat is energy moving from one place to another


Heat = EnergyHeat = Energy

For us, these terms will be used synonymously

A.A. EnergyEnergy

B.B. HeatHeat

C.C. TemperatureTemperature

A measure of the kinetic energy of the particles in a substance


A.A. EnergyEnergy

B.B. HeatHeat


D.D. EndothermicEndothermic

A description of a process that absorbs heat


A.A. EnergyEnergy

B.B. HeatHeat


D.D. EndothermicEndothermic

E.E. ExothermicExothermic

A description of a process that gives off heat


Energy of motion

II.II. Types of EnergyTypes of EnergyA.A. KineticKinetic

Moving objects


1.1. MechanicalMechanical

Heat energy (moving particles)


1.1. MechanicalMechanical

2.2. ThermalThermal

Stored energy


B.B. PotentialPotential

Energy that can be released as gravity acts


B.B. PotentialPotential1.1. GravitationalGravitational

Energy stored in chemical bonds


B.B. PotentialPotential1.1. GravitationalGravitational

2.2. ChemicalChemical

Energy in the form of light


B.B. PotentialPotential

C.C. RadiantRadiant

III.III. Measuring Heat (q)Measuring Heat (q)

A.A. UnitsUnitsIII.III. Measuring Heat (q)Measuring Heat (q)

SI unit of energy (work). Work done by applying one Newton force over one meter.

A.A. UnitsUnits1.1. joulejoule


Energy required to heat one gram of water by 1° C.


2.2. caloriecalorie


A nutritional calorie. 1 Cal = 1000 cal


2.2. caloriecalorie

3.3. CalorieCalorie


Equivalent to calories


2.2. caloriecalorie

3.3. CalorieCalorie

4.4. kilocaloriekilocalorie


Equivalent to 1000 calories


2.2. caloriecalorie

3.3. CalorieCalorie



Energy required to heat 1 lb. water by 1º F.


2.2. caloriecalorie

3.3. CalorieCalorie


5.5. BTUBTU


Heat Unit ConversionsHeat Unit Conversions

1 cal = 4.184 J1 cal = 4.184 J

1000 cal = 1 Cal = 1 kcal1000 cal = 1 Cal = 1 kcal

Self Check – Ex. 1Self Check – Ex. 1

A reaction produces 3800 J A reaction produces 3800 J of heat. How many calories of heat. How many calories is this?is this?


A can of soda contains 150 A can of soda contains 150 Calories. How many joules Calories. How many joules of energy is this?of energy is this?

III.III. Measuring Heat (q)Measuring Heat (q)A.A. UnitsUnits

B.B. Heat is related to temperatureHeat is related to temperature

Which has more heat?



Temp. = 35º C

Beaker #1 Beaker #2

Temp. = 65º C



C.C. Heat also depends on. . .Heat also depends on. . .



C.C. Heat also depends on. . .Heat also depends on. . .1.1. Mass of materialMass of material

Which beaker could melt more Which beaker could melt more ice (which has more heat)?ice (which has more heat)?

Beaker #1 Beaker #2

T1 = 85º C

T2 = 85º C



C.C. Heat also depends on. . .Heat also depends on. . .1.1. Mass of materialMass of material

2.2. Type of materialType of material




D.D. Specific HeatSpecific Heat

Amount of heat required to raise the temperature of 1 gram of substance by 1º C




D.D. Specific HeatSpecific Heat1.1. Some material takes a lot of Some material takes a lot of

energy to raise its temperatureenergy to raise its temperature






2.2. Some material takes lessSome material takes less






2.2. Some material takes lessSome material takes less

3.3. For water it’s 1 calorie/g For water it’s 1 calorie/g ºCºC

Specific Heat TableSpecific Heat TableSubstanceSubstance Spec. Heat Spec. Heat (c)(c)

WaterWater 4.184 J/g ºC4.184 J/g ºC

AluminumAluminum 0.89 J/g ºC0.89 J/g ºC

IronIron 0.45 J/g ºC0.45 J/g ºC

CopperCopper 0.387 J/g ºC0.387 J/g ºC

SilverSilver 0.24 J/g ºC0.24 J/g ºC

GoldGold 0.129 J/g ºC0.129 J/g ºC

LeadLead 0.l28 J/g 0.l28 J/g ºCºC




D.D. Specific HeatSpecific Heat

E.E. CalculationCalculation

q = m x c x ∆T


How much heat must be How much heat must be applied to a 25 g chunk of applied to a 25 g chunk of ironiron to raise its temperature to raise its temperature by 100by 100ºC? (cºC? (cironiron = 0.45 J/g ºC) = 0.45 J/g ºC)


How much heat must be How much heat must be applied to a 25 g sample of applied to a 25 g sample of waterwater to raise its temperature to raise its temperature by 100by 100ºC? (cºC? (cironiron = 4.18 J/g ºC) = 4.18 J/g ºC)

IV.IV. Bond EnergyBond Energy

This is endothermic

IV.IV. Bond EnergyBond EnergyA.A. When bonds are broken When bonds are broken

energy is energy is ..

This is endothermic


energy is energy is requiredrequired. (positive). (positive)



B.B. When bonds are formed When bonds are formed energy is energy is ..



B.B. When bonds are formed When bonds are formed energy is energy is releasedreleased. (negative). (negative)

This is exothermic




C.C. The sum of the bond energies The sum of the bond energies gives an estimate of the gives an estimate of the reaction energyreaction energy




C.C. The sum of the bond energies The sum of the bond energies gives an estimate of the gives an estimate of the reaction energyreaction energy1.1. Positive values = endothermicPositive values = endothermic




C.C. The sum of the bond energies The sum of the bond energies gives an estimate of the gives an estimate of the reaction energyreaction energy1.1. Positive values = endothermicPositive values = endothermic

2.2. Negative values = exothermicNegative values = exothermic


Draw Lewis structures for Draw Lewis structures for each substance and calculate each substance and calculate the energy for the reaction the energy for the reaction below.below.

2CO + O2 2CO2

V.V. Enthalpy StoichiometryEnthalpy Stoichiometry

V.V. Enthalpy StoichiometryEnthalpy StoichiometryA.A. Enthalpy represented by Enthalpy represented by HH

and enthalpy change by and enthalpy change by ∆H∆H..



B.B. Enthalpy change for a reaction Enthalpy change for a reaction measured in measured in kJ/molkJ/mol..




C.C. Exothermic reactions have Exothermic reactions have negative ∆H values. (+ negative ∆H values. (+ ∆H∆H for endothermic)





D. Solving problems

Enthalpy StoichiometryEnthalpy Stoichiometry


1.1. Write a balanced equationWrite a balanced equation



2.2. Identify the units of the Identify the units of the unknownunknown




3.3. Write the ‘given’Write the ‘given’




3.3. Write the ‘given’Write the ‘given’

4.4. Fill in conversion factorsFill in conversion factors





D. Solving problems1.1. Finding heatFinding heat


How much heat is produced How much heat is produced when 75g of hydrogen is when 75g of hydrogen is burned in oxygen?burned in oxygen?

2H2 + O2 2H2O ∆H = - 484 kJ





D. Solving problems1.1. Finding heatFinding heat

2.2. Finding massFinding mass


What mass of hydrogen is What mass of hydrogen is required to produce 8500 kJ of required to produce 8500 kJ of energy?energy?

2H2 + O2 2H2O ∆H = - 484 kJ

VI.VI. Enthalpy CalorimetryEnthalpy Calorimetry

VI.VI. Enthalpy CalorimetryEnthalpy CalorimetryA.A. Calorimetry is the science of Calorimetry is the science of

heat measurementheat measurement


1.1. Mass of water in calorimeterMass of water in calorimeter

A.A. Calorimetry is the science of Calorimetry is the science of heat measurementheat measurement

B.B. ΔH can be determined ΔH can be determined experimentally if you measure:experimentally if you measure:



2.2. Initial temperatureInitial temperature






3.3. Final temperatureFinal temperature






3.3. Final temperatureFinal temperature

4.4. Mass of reactant usedMass of reactant used



Calculating Calculating ∆H∆H for rxn for rxn

1.1. Determine the units for your Determine the units for your answer.answer.



2.2. Calculate the heat gained by the Calculate the heat gained by the water.water. QQ = = mc∆Tmc∆T

ccwaterwater = 4.18 J/g = 4.18 J/g ∙ ºC∙ ºC



2.2. Calculate the heat gained by the Calculate the heat gained by the water.water.

3.3. Determine the heat lost by rxn.Determine the heat lost by rxn.

QQgainedgained = - Q = - Qlostlost





4.4. Convert grams reactant into Convert grams reactant into moles reactant.moles reactant.





4.4. Convert grams reactant into Convert grams reactant into moles reactant.moles reactant.

5.5. Divide heat lost (#3) by the Divide heat lost (#3) by the moles reactant (#4).moles reactant (#4).


When a 10.0 g of NaOH is When a 10.0 g of NaOH is added to 200 g of water the added to 200 g of water the temperature goes from 18.2temperature goes from 18.2ºC ºC to 31.6ºC. What is to 31.6ºC. What is ∆H∆H for the for the reaction in kJ/mol?reaction in kJ/mol?

VII.VII. Hess’s LawHess’s Law

VII.VII. Hess’s LawHess’s LawA.A. When a series of equations are When a series of equations are

added together, their enthalpy added together, their enthalpy changes are also addedchanges are also added

2NO (g) + O2NO (g) + O22 (g) 2NO (g) 2NO22 (g) (g) ΔΔHH = -113 kJ= -113 kJ

Add these equationsAdd these equations

NN22 (g) + O (g) + O22 (g) 2NO (g) (g) 2NO (g) ΔΔHH = +181 kJ= +181 kJ

VII.VII. Hess’s LawHess’s LawA.A. When a series of equations are When a series of equations are

added together, their enthalpy added together, their enthalpy changes are also addedchanges are also added

B.B. Equations can be alteredEquations can be altered


1.1. If an equation is reversed . . .If an equation is reversed . . .

A.A. When a series of equations are When a series of equations are added together, their enthalpy added together, their enthalpy changes are also addedchanges are also added


COCO22 (g) CO (g) + ½ O (g) CO (g) + ½ O22 (g) (g) ΔΔHH = +283 kJ= +283 kJ


1.1. If an equation is reversed . . .If an equation is reversed . . .

2.2. If the coefficients are multiplied If the coefficients are multiplied by a factor . . .by a factor . . .

A.A. When a series of equations are When a series of equations are added together, their enthalpy added together, their enthalpy changes are also addedchanges are also added


* Using fractions is perfectly acceptable* Using fractions is perfectly acceptable

2C (g) + O2C (g) + O22 (g) CO (g) (g) CO (g) ΔΔHH = -111 kJ= -111 kJ XX ½ ½


2H2H22 (g) + O (g) + O22 (g) 2H (g) 2H22O (O (ll) ) ΔΔHH = -572 kJ= -572 kJ

HH22 (g) + O (g) + O22 (g) 2H (g) 2H22OO22 ( (ll) ) ΔΔHH = -188 kJ= -188 kJ

2H2H22OO22 ( (ll) 2 H) 2 H22O (O (ll) + O) + O22 (g) (g) ΔΔHH = = ?? kJ kJ


2H2H22 (g) + O (g) + O22 (g) 2H (g) 2H22O (g) O (g) ΔΔHH = -483.6 kJ= -483.6 kJ

3O3O22 (g) 2O (g) 2O33 (g) (g) ΔΔHH = +284 kJ= +284 kJ

3H3H22 (g) + O (g) + O33 (g) 3H (g) 3H22O (g) O (g) ΔΔHH = = ?? kJ kJ

VII.VII. Phase changes and Phase changes and heatheat

The End

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Energy Chap. 16. I.Definitions A. Energy Energy is the ability to do work or produce heat I.Definitions