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Energy Relationships in Chemical Reactions

Chapter 6

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Chapter 6: Energy Relationships in Chemical Reactions

I. Energy, work, heatII. Internal Energy, E

Energy transfer

ΙΙΙ. ΔH (qp) vs ΔE (qv)IV. Thermochemical EquationsV. Heats of Formation; ΔHf

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VI. Hess’ Law• State Functions

Chapter 6: Energy Relationships in Chemical Reactions

VII. Calorimetrya. specific heat/ heat capacityb. Constant Pressure Calorimetry (coffee cup)c. Constant Volume Calorimetry (bomb)

EnergyEnergy is the capacity to do work or transfer heat

•• Thermal energyThermal energy is the energy associated with the random motion of atoms and molecules

•• Chemical energyChemical energy is the energy stored within the bonds of chemical substances

•• Nuclear energyNuclear energy is the energy stored within the collection of neutrons and protons in the atom

•• Electrical energyElectrical energy is the energy associated with the flow of electrons

•• Potential energyPotential energy is the energy available by virtue of an object’s position

•• Kinetic energyKinetic energy is due to an object’s mass and motion

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Energy is the capacity to do work or transfer heat

HeatHeat : Process of energy transfer as a result of temperature difference

WorkWork: Energy transferred due to a change inposition or directed energy transferfrom a process

When energy is transferred from one objectto another it appears as heatheat and/or workwork!

HeatHeat is the transfer of thermal energythermal energy between two bodies that are at different temperatures.

Energy Changes in Chemical Reactions

TemperatureTemperature is a measure of the average thermal energy.

Temperature = Thermal Energy or Heat

900C 400Cgreater thermal energy

Internal Energy, EInternal Energy, E: Total of all possible types of energy within a substance

1.Kinetic Energy; Energy due to mass and motiona) Translationalb) Vibrationalc) Rotational

EEKK = E= ETT + E+ EVV + E+ ERR

2.Potential Energy; Stored energy due to an object’s positiona) Bond – Chemical Energyb) Nucleus – Nuclear Energyc) Atom

EEPP = E= EBB + E+ ENN + E+ EAA

E E TOTALTOTAL = E= EKK + E+ EPP

UnitsUnits : Energy, work, and heat

•• CalorieCalorie – heat needed to increase the temperature of 1.00 gram of water by 1.00°C

• Joule or (Newton *meter)

1 calorie = 4.184 J

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EnergyEnergy is the capacity to do work or transfer heat

HeatHeat : Process of energy transfer as a result of temperature difference

WorkWork: Energy transferred due to a change inposition or directed energy transferfrom a process

When energy is transferred from one objectto another it appears as heat and/or work!

Exothermic processExothermic process :any process that gives off heat –transfers thermal energy from the system to the surroundings.

2H2 (g) + O2 (g) 2H2O (l) + energy; -q

H2O (g) H2O (l) + energy; -q

Endothermic process:Endothermic process: any process in which heat has to be supplied to the system from the surroundings.

energy + 2HgO (s) 2Hg (l) + O2 (g); +q

energy + H2O (s) H2O (l); +q

Change in Enthalpy (ΔH) is the Heat Lost or Gained at Constant Pressure: ΔH = qP

q vs. q vs. ΔΔHH1. q is amount of heat released/absorbed for

a specific amount of material.

2. ΔH is heat released/absorbed by 1 mole of material under constant pressure conditions.

q=nΔH

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WorkWork: Most work done by a chemical reaction involvesexpansion/compression of gases.

2H2 (g) + O2 (g) 2H2O (l) + energy3 moles gas o moles gas

work = -PΔV = -ΔngRT

Expansion: +ΔV so –wGas pushes against atmosphere so system does work or work is lost by system

Compression: -ΔV so + wAtmospheric pressure does work on system orwork is added to system

When energy is transferred from one objectto another it appears as heat and/or work!

ΔE = q + w

Si -28% of Earth’s Crust butwhen in Contact With Air it Forms SiO2.

Si (s) + O2 (g) SiO2 (s) ; ΔE = -908 kJ/mole

InternalEnergy

Si + O2

SiO2

-908 kJ/mole

Energy Reactants = Energy Products + KE Released

When energy is transferred from one objectto another it appears as heat and/or work!

ΔE = q + w

1. Heat of reaction (enthalpy)

Si (s) + O2 (g) -> SiO2 (s) ; ΔE = -908 kJ/mole

Si (s) + O2 (g) -> SiO2 (s) ; ΔΗ = -911 kJ/mole

2. Work

work = -PΔV = -ΔngRT =-(0-1mols)(8.314 X10-3 kJ/moleK)(298 K)-(-1)(2.5kJ) or about +3 KJ

ΔE = -911+ 3 = -908

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Two ways that q can be measured

1. Constant Pressure• qp = ΔE –w• qp =ΔE +PΔV

2. Constant Volume; ΔV =0• qv = ΔE +PΔV• qv= ΔE+ P(0)• qv = ΔE qv ≡ ΔE

qp ≡ ΔΗ H2O (s) H2O (l) ΔH = 6.01 kJ

• The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations

• If you reverse a reaction, the sign of ΔH changes

H2O (l) H2O (s) ΔH = -6.01 kJ

• If you multiply both sides of the equation by a factor n, then ΔH must change by the same factor n.

2H2O (s) 2H2O (l) ΔH = 2 x 6.01 = 12.0 kJ

q = nΔH

(Heat of fusion)

H2O (s) H2O (l) ΔH = 6.01 kJ

• The physical states of all reactants and products must be specified in thermochemical equations.

Thermochemical EquationsThermochemical Equations

H2O (l) H2O (g) ΔH = 44.0 kJ

P4 (s) + 5O2 (g) P4O10 (s) ΔH = -3013 kJ

266 g P4

1 mol P4

123.9 g P4x 3013 kJ

1 mol P4x = 6470 kJ

How much heat is evolved when 266 g of white phosphorus (P4) burn in air? (H = heat of combustion)

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Stoichiometry and Heat of Reaction

Thermite Animation

2 Al (s) + Fe2O3 (s) -> Al2O3(s) + 2 Fe(s)

HessHess’’s Laws Law: When reactants are converted to 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. It doesn’t matter how you get there, only where you start and end.)

NET RXN = RXN 1 + RXN 2 + RXN3 + ......ΔHRXN = ΔHRXN 1 + ΔHRXN 2 + ΔHRXN3

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Thermodynamics

State functionsState functions are properties that are determined by the state of the system, regardless of how that condition was achieved.

Potential energy of hiker 1 and hiker 2is the same even though they took different paths.

energy, pressure, volume, temperature

State Functions

Calculate the standard enthalpy of formation of CS2 (l) given that:C(graphite) + O2 (g) CO2 (g) ΔH0 = -393.5 kJrxn

S(rhombic) + O2 (g) SO2 (g) ΔH0 = -296.1 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) ΔH0 = -1072 kJrxn

1. Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)

2. Add the given rxns so that the result is the desired rxn.

rxnC(graphite) + O2 (g) CO2 (g) ΔH0 = -393.5 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) ΔH0 = -296.1x2 kJrxn

CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) ΔH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)

ΔH0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJrxn

Two Ways Thermochemical Equations Data Can Be Given

1. Thermochemical Equations2. ΔHf

0 Values; Standard Molar Enthalpy of Formation Under Standard State Conditions

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The standard enthalpy of formation of any element in its most stable form is zero.

ΔH0 (O2) = 0f

ΔH0 (O3) = 142 kJ/molf

ΔH0 (C, graphite) = 0fΔH0 (C, diamond) = 1.90 kJ/molf

ΔHf0 (Li2O) = -597.9 kJ/molestands for

2 Li (s) + ½ O2(g) -> Li2O (s); ΔH0 = -597.9 kJ/mole

Standard enthalpy of formationStandard enthalpy of formation (ΔHf0) is the heat

change that results when one mole of a compound is formed from its elements at a pressure of 1 atm (standard state conditions).

Determining ΔH0rxn from ΔH0

f values

aA + bB cC + dD

ΔH0rxn dΔH0 (D)fcΔH0 (C)f= [ + ] - bΔH0 (B)faΔH0 (A)f[ + ]

ΔH0rxn nΔH0 (products)f= Σ mΔH0 (reactants)fΣ-

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol.

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

ΔH0rxn nΔH0 (products)f= Σ mΔH0 (reactants)fΣ-

ΔH0rxn 6ΔH0 (H2O)f12ΔH0 (CO2)f= [ + ] - 2ΔH0 (C6H6)f[ ]

ΔH0rxn = [ 12x–393.5 + 6x–187.6 ] – [ 2x49.04 ] = -5946 kJ

-5946 kJ2 mol

= - 2973 kJ/mol C6H6

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Obtaining ΔHrxn

1. Hess’s Law2. Using ΔHf

0 Values3. Experimentally—Constant Pressure

Calorimetry (Coffee Cup Calorimeter)

Experimentally Determining ΔH from a Temperature Change

• Different quantities of the same substance absorb/release different amounts of heat for same temperature change.

• Different substances can absorb the same amount of heat yet have a different temperature increase

The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius.

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius.

C = ms

Heat (q) absorbed or released:

q = msΔt

q = CΔt

Δt = tfinal - tinitial

s of Fe = 0.444 J/g • 0C

Δt = tfinal – tinitial = 50C – 940C = -890C

q = msΔt = 869 g x 0.444 J/g • 0C x –890C = -34,000 J

How much heat is given off when an 869 g iron bar cools from 940C to 50C?

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Calorimetry;Measurement of Heat Changes

• System; Chemical Reaction• Surroundings; Everything Else (including water

reactants and products are dissolved in)

System

Surroundings

Constant-Pressure Calorimetry

No heat enters or leaves!

qsys = qwater + qcal + qrxn

qsys = 0qrxn = - (qwater + qcal)qwater = msΔtqcal = CcalΔt

6.4

Reaction at Constant PΔH = qrxn/ n

Coffee-Cup Calorimeter

-qSystem = +qSurroundings

-(csolid x masssolid x ΔTsolid) =

+ (cwater x masswater x ΔTwater)

ΔT = Tfinal - Tinitial

Follow-Up Problem 6.4, p.235

Heat Changes at Constant Volume

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Heat Changes at Constant Volume = Energy Changes at Constant Volume

qV = ΔEFor most reactions:

ΔE ~= ΔH

Bomb Calorimeter:

-qsample = +qcalorimeter = heat capacity x ΔT

heat capacity = (heat capacitybomb + heat capacitywater x masswater)

Constant-Volume Calorimetry

No heat enters or leaves!

qsys = qwater + qbomb + qrxn

qsys = 0qrxn = - (qwater + qbomb)qwater = msΔtqbomb = CbombΔt

Reaction at Constant V

ΔH ~ qrxn

ΔH = qrxn

ΔE = qrxn