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CHEM 101-GENERAL CHEMISTRY CHAPTER 9

ENERGY & CHEMISTRY

INSTR : FİLİZ ALSHANABLEH

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CHAPTER 9 ENERGY & CHEMISTRY

• Defining Energy • Energy Transformation & Conservation of Energy • Heat Capacity & Calorimetry • Enthalpy • Hess’s Law & Heats of Reaction • Heats of Reactions from Heat of Formation

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Forms of Energy

• Two broad categories of energy: potential energy and kinetic energy. • Potential energy - associated with the relative position of

an object.

• Kinetic energy - associated with motion.

Kinetic energy = 12

mv2

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Forms of Energy

• Internal energy - the combined kinetic and potential energies of atoms and molecules that make up an object or system.

• Chemical energy - energy released or absorbed during a chemical reaction.

• Other forms of energy include radiant, mechanical, thermal, electrical, and nuclear.

• Thermochemistry - the study of the energetic consequences of chemistry involving heat flow.

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Heat and Work

• Heat is the flow of energy between two objects because of a difference in temperature.

• Heat always flows from the warmer object to the cooler object.

• Work is the transfer of energy accomplished by a force moving a mass some distance against resistance. • Pressure-volume work (PV-work) is the most common work type

in chemistry. • Releasing an inflated balloon before it is tied off illustrates an

example of PV-work.

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Energy Units

• Energy = Work done = Force x distance

• The Joule is the SI unit of energy. • 1 Joule = 1 kg m2/s2

• Other energy units include the Btu and the calorie. 1 Btu is the energy required to raise 1 lb of water 1oF.

1 Btu = 1055J 1 calorie is the energy required to raise 1 g water from 14.5 to 15.5 oC. 1 calorie = 4.184 J

W = mass × acceleration × distance = kg × ms2

× m

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Energy Transformation and Conservation of Energy

• During energy transformation, the total energy must be conserved. • The sum of all energy conversions and energy transfers

must equal the total energy present which must remain constant.

• To account for energy transformations and conversions, the system and surroundings must be specified. • System - the part of the universe being considered. • Surroundings - the remainder of the universe. • System + Surroundings = Universe • System and surroundings are separated by a boundary.

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Energy Transformation and Conservation of Energy

• For a system or surroundings, the only possible forms of energy flow are heat, q, and work, w.

• The delta, ∆, means “change in” and is defined as the difference in the final and initial states.

∆E = q + w

∆E = Efinal − Einitial

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Energy Transformation and Conservation of Energy

• The sign resulting from the difference in the final and initial states indicates the direction of the energy flow. • Negative values indicate energy is being released.

• Positive values indicate energy is being absorbed.

q (+) q (-) w (+) w (-) SURROUNDING

SYSTEM

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Example Problem 9.1

• If 515 J of heat is added to a gas that does 218 J of work as a result, what is the change in the energy of the system?

• q = + 515 J (heat added to the system) • w = - 218 J (work done by the system) ∆E = + 515 J + ( - 218 J) = + 297 J

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Problem 9.23

• Which sytem does more work: (a) ∆E = - 436 J, q = 400 J; or (b) ∆E = 317 J, q = 347 J

Answer

(a) ∆E = - 436 J, q = 400 J w = - 836 J (b) ∆E = 317 J, q = 347 J w = - 30 J • System (a) does more work

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Problem 9.24

• In which case is heat added to the sytem: (a) ∆E = - 43 J, w = 40 J; or (b) ∆E = 31 J, w = 34 J

Answer

(a) ∆E = - 43 J, w = 40 J q = - 83 J (b) ∆E = 31 J, w = 34 J q = - 3 J • In non of the cases heat is added to the system

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Energy Transformation and Conservation of Energy

• First law of thermodynamics states that energy can be transformed from one form to another but cannot be created or destroyed.

∆Euniverse = ∆Esurroundings + ∆Esystem = 0

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Waste Energy

• A common way to obtain work from a system is to heat the system. Heat flows in and is converted to work.

• It is impossible to completely convert all heat to work. • Heat not converted to work is considered waste energy,

which may contribute to thermal pollution.

• The efficiency of conversion from heat to work can be expressed as a percentage. • Increases in energy consumption can be offset by

increasing energy efficiencies.

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Waste Energy

• Typical efficiencies of some common energy conversion devices.

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Heat Capacity and Specific Heat

• c : The specific heat capacity, or specific heat, is a physical property of a substance that describes the amount of heat required to raise the temperature of one gram of a substance by 1ºC.

• Cp : The molar heat capacity is a physical property of a substance that describes the amount of heat required to raise the temperature of one mole of a substance by 1ºC. • The amount of heat energy absorbed can be quantified. q = mc∆T

q = nCp∆T

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Heat Capacity and Specific Heat

• Specific heat and molar heat capacities for some common substances.

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Problem 9.31

• A metal radiator is made from 26.0 kg of iron. The specific heat of iron is 0.449 J/gºC. How much heat must be supplied to the radiator to raise its temperature from 25.0 to 55.0ºC.

Answer: ∆T = Tfinal - Tinitial = 55.0 – 25.0 = 30 ºC q = mc∆T = (26.0x103 g)(0.449 J/g ºC)(30 ºC) = 3.50x105 J

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Problem 9.31

• A copper nail and an iron nail of the same mass and initially at the same room temperature are both put into a vessel containing boiling water. Which one would you expect to reach 100 ºC first? Why?

ciron = 0.449 J/gºC and ccopper = 0.384 J/gºC.

Answer: • Copper will reach 100 ºC first. Because its heat capacity

that is specific heat less than specific heat of iron. • Copper required less energy to reach expected temp.

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Example Problem 9.4

• A glass contains 250.0 g of warm water at 78.0ºC. A piece of gold at 2.30ºC is placed in the water. The final temperature reached by this system is 76.9ºC. What was the mass of gold? The specific heat of water is 4.184 J/gºC, and that of gold is 0.129 J/gºC.

Answer: qgold = ‒ qwater mgold x cgold x ∆T = ‒ mwater x cwater x ∆T mgold (0.129 J/gºC)(76.9 ‒ 2.30) = ‒ (250.0g) (4.184 J/gºC)(76.9 ‒ 78.0) mgold = 120 g

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Calorimetry

• Heat flow is measured using a calorimeter.

• A calorimeter measures the heat evolved or absorbed by the system of interest by measuring the temperature change in the surroundings.

• qsystem Reaction System • qsurrounding Calorimetry

qsystem = − qsurroundings

qgained = − qlost

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Calorimetry

• Two types of Calorimeter: • Coffee-Cup Calorimeter • Bomb Calorimeter

• Actual Measurement - temperature change for the calorimeter and

the calorimeter constant are used to determine the amount of heat released by a reaction.

qcalorimeter = Ccalorimeter × ∆Tcalorimeterqreaction = − qcalorimeter

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Coffee-Cup Calorimeter

qreaction = ‒ qalorimeter qreaction = ‒ qwater

qreaction = = ‒ mwater x cwater x ∆T

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Bomb Calorimeter

qreaction = ‒ qalorimeter qreaction = ‒ Ccalorimeter x ∆T

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Example Problem

• 0.450 g of a particular fuel is burned in bomb calorimeter which has calorimeter constant (heat capacity) 118 J/ºC and the temperature increases by 5.28ºC. Calculate the energy density of this fuel, which is the amount of energy liberated per gram of fuel burned.

Answer: qreaction = ‒ qcalorimeter qreaction = ‒ Ccalorimeter x ∆T qcalorimeter = Ccalorimeter x ∆T = (118 J/oC )(5.28oC) = 623.04 J qreaction = ‒ qcalorimeter = ‒ 623.04 J heat evolved when 0.450 g fuel burned Energy density = ‒ 623.04 J / 0.450 g = 1.40x 103 J/g = 1.40 kJ/g

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Enthalpy

• Enthalpy is the heat flow under conditions of constant pressure. • The enthalpy change can be expressed as

ΔH = Hproducts – Hreactants

• State function • ΔH = q at constant pressure

∆H = qp

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Enthalpy

• When a system releases heat, the process is said to be exothermic.

• The value of ∆H is less than zero; the sign on ∆H is negative. CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ΔH < 0

• When a system absorbs heat, the process is said to be endothermic.

• The value of ∆H is greater than zero; the sign on ∆H is positive. H2O(s) → H2O(l) ΔH > 0

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∆H of Phase Changes • Phase changes occur under constant pressure conditions.

• The heat flow during a phase change is an enthalpy change. • During a phase change, temperature does not change with heat

flow due to formation or breaking of intermolecular attractive forces.

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∆H of Phase Changes

• The heat required to convert a liquid to a gas is the heat of vaporization, ∆Hvap.

• ∆Hvap is endothermic with a positive value.

• The heat released to convert a gas to a liquid is the heat of condensation, ∆Hcond.

• ∆Hcond is exothermic with a negative value. ∆Hcond = –∆Hvap

• The value of ∆H for a phase change is compound specific and has

units of kJ/mol. • The heat flow can be calculated using the number of moles of

substance, n, and the value of the enthalpy change. ∆H = n × ∆Hphase change

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∆H of Phase Changes

• Standard molar enthalpies and temperatures for phase changes of water.

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Problem 9.47

• If 14.8 kJ of heat is given off when 1.6 g of HCl condenses from vapor to liquid, what is ∆Hcond for this substances?

(H:1.008, Cl:35.45) Answer: nHCl = 1.6 g / (36.46 g/mol) = 0.044 mol HCl gives off 14.8 kJ of heat ∆Hcond = ‒14.8 kJ / (0.044 mol) ∆Hcond = ‒ 337.2 kJ / mol

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Heat of Reaction • Enthalpy changes can be calculated for chemical reactions, in

addition to temperature changes and phase transitions that is commonly referred to as the heat of reaction. • The enthalpy change for a reaction can be estimated using

bond energies. • During a chemical reaction, reactant bonds are broken and

product bonds are made. • Breaking bonds requires energy. • Making bonds releases energy.

• If the amount of energy released making product bonds is

greater than the amount of energy required to break reactant bonds, the reaction is exothermic. If the energy released is less than the energy required, the reaction is endothermic.

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Bonds and Energy

The combustion of methane breaks 4 C-H bonds and 2 O=O bonds. 2 C=O bonds and 4 O-H bonds are made.

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Thermochemical Equations

• A thermochemical equation summarizes the overall energetics for a chemical reaction. • The sign on the ∆H indicates whether the reaction is

endothermic or exothermic CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ΔHrxn = –891 kJ

• H2O(s) → H2O(l) ΔHrxn = + 6.0 kJ

• H2O(l) → H2O(s) ΔHrxn = – 6.0 kJ

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Thermochemical Equations

• The combustion of methane is an exothermic reaction and releases 891 kJ of heat energy when 1 mole of methane reacts with 2 moles of oxygen.

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ΔHrxn = –891 kJ • For thermochemical equations, if the stoichiometric

coefficients are multiplied by some factor, the heat of reaction must also be multiplied by the same factor.

2CH4(g) + 4O2(g) → 2CO2(g) + 4H2O(l) ΔHrxn = –1,782 kJ

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Hess’s Law and Heats of Reaction

• Hess’s law: the enthalpy change for any process is independent of the particular way the process is carried out.

• In 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. • Enthalpy is a state function.

• A state function is a variable whose value depends only on

the state of the system and not its history.

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Hess’s Law

• Conceptual diagram representing Hess’s law. Enthalpy is a state function, so any convenient path can be used to calculate the enthalpy change.

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An Example on Hess’s Law

N2(g) + 2O2(g) → 2NO2(g) ΔH1 = 68 kJ • This reaction also can be carried out in two distinct steps,

with enthalpy changes designated by ΔH2 and ΔH3.

N2(g) + O2(g) → 2NO(g) ΔH2 = 180 kJ 2NO(g) + O2(g) → 2NO2(g) ΔH3 = – 112 kJ N2(g) + 2O2(g) → 2NO2(g) ΔH2 + ΔH3 = 68 kJ ΔH1 = ΔH2 + ΔH3 = 68 kJ

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Example Problem Consider the following data: Calculate ΔH for the reaction: Answer • Reverse the two reactions:

• Multiply reactions to give the correct numbers of reactants and products:

4x 3x

3 2 2

2 2 2

1 32 2

NH ( ) N ( ) H ( ) H = 46 kJ

2 H ( ) O ( ) 2 H O( ) H = 484 kJ

→ + ∆

+ → ∆ −

g g g

g g g

2 2 2 32 N ( ) 6 H O( ) 3 O ( ) 4 NH ( )+ → +g g g g

2 2 3

2 2 2

1 32 2N ( ) H ( ) NH ( ) H = 46 kJ

2 H O( ) 2 H ( ) O ( ) H = +484 kJ

+ → ∆ −

→ + ∆

g g g

g g g

2 2 3

2 2 2

1 32 2N ( ) H ( ) NH ( ) H = 46 kJ

2 H O( ) 2 H ( ) O ( ) H = +484 kJ

+ → ∆ −

→ + ∆

g g g

g g g

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Example Problem (continued)

• Final reactions:

• Desired reaction: ΔH = +1268 kJ

2 2 3

2 2 2

2 N ( ) 6 H ( ) 4 NH ( ) H = 184 kJ

6 H O( ) 6 H ( ) 3 O ( ) H = +1452 kJ

+ → ∆ −

→ + ∆

g g g

g g g

2 2 2 32 N ( ) 6 H O( ) 3 O ( ) 4 NH ( )+ → +g g g g

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Heats of Reaction for Some Specific Reactions

• Some classes of chemical reactions are given their own labels for heats of reactions. • Heat of combustion, ∆Hcomb • Heat of neutralization, ∆Hneut • Heat of formation, ∆Hf, is the heat of reaction for formation

of substances. • Fractional coefficients are allowed for formation

reactions because only one mole of product can be formed.

C(s) + 12

O2 (g) → CO(g)

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Heats of Reactions from Heat of Formation

• A formation reaction is the chemical reaction by which one mole of a compound is formed from its elements in their standard states.

• The standard state is the most stable form of an element at room temperature, 25oC, and pressure, 1 atm, indicated with a superscript o.

• ∆Hfo = 0 for an element in its standard state. • The enthalpy change for a reaction can be calculated using

heats of formation. ∆H o = vi∆H f

o

i∑ (products)i − vj∆H fo

j∑ (reactants) j

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Example Problem

Calculate ∆H° for the following reaction: 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) Given the following information: ∆Hf° (kJ/mol) H2O(l) –286 NaOH(aq) –470 Answer: ∆H°= [2(∆Hf°)NaOH + 1 (∆Hf°)H2 ] - [2(∆Hf°)Na + 2(∆Hf°)H2O ] ∆H°= [2( -470) + 1 (0.0) ] - [2(0.0) + 2(-286) ] ∆H° = –368 kJ

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Example Problem 9.9

• Ethanol, C2H5OH, has a heat of combustion of 1366.8 kJ/mol. What is the heat of formation of ethanol?

Given the following information: ∆Hf° (kJ/mol) H2O(l) –285.8 CO2(g) –393.5

Answer: C2H5OH(l) + 3O2(g) 2 CO2(g) + 3 H2O(l) ∆H°= ‒1366.8 kJ ∆H°= [2∆Hf°(CO2) + 3 ∆Hf°(H2O)] ‒ [∆Hf°(C2H5OH) + 3 ∆Hf°(O2)] ‒1366.8 kJ = [2x (-393.5) + 3x(-285.5)]- (∆Hf°(C2H5OH) + 3x(0.0)] ∆Hf°(C2H5OH) = ‒ 277.6 kJ/mol

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Energy Density and Fuels

• When deciding the economic merits of a fuel, several factors must be considered. • Technology available to extract the fuel.

• The amount of pollution released by its combustion.

• The fuel’s relative safety.

• The ease of transporting the fuel.

• The fuel’s energy density.

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Energy Density and Fuels

• Energy density is the amount of energy that can be released per gram of fuel burned. • The higher the energy

density of a fuel, the less fuel that must be transported to the customer.

Slide Number 1CHAPTER 9�ENERGY & CHEMISTRYForms of EnergyForms of EnergyHeat and WorkEnergy UnitsEnergy Transformation and Conservation of EnergyEnergy Transformation and Conservation of EnergyEnergy Transformation and Conservation of EnergyExample Problem 9.1Problem 9.23Problem 9.24Energy Transformation and Conservation of EnergyWaste EnergyWaste EnergyHeat Capacity and Specific HeatHeat Capacity and Specific HeatProblem 9.31Problem 9.31Example Problem 9.4CalorimetryCalorimetryCoffee-Cup CalorimeterBomb CalorimeterExample Problem EnthalpyEnthalpyH of Phase Changes∆H of Phase Changes∆H of Phase ChangesProblem 9.47Heat of ReactionBonds and EnergyThermochemical EquationsThermochemical EquationsHess’s Law and Heats of ReactionHess’s LawAn Example on Hess’s LawExample ProblemExample Problem (continued)Heats of Reaction for Some Specific ReactionsHeats of Reactions from Heat of FormationExample Problem Example Problem 9.9Energy Density and FuelsEnergy Density and Fuels