Thermochemistry: Part of Thermodynamics - …chemed.tamu.edu/chem101/pptnotes/Thermochem AF st 3pp 29p.pdf · Dr. Williamson’s Notes on Thermochemistry © vm williamson 6 How many

Dr. Williamsons Notes on Thermochemistry

vm williamson 1

Thermochemistry: Part of

ThermodynamicsDr. Vickie M. Williamson

@vmwilliamsonStudent Version

1

Chemical Thermodynamics! Thermodynamics: study of the

energy changes associated with physical and chemical processes

! Thermochemistry: study of heat changes in chemical reactions and physical changes

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A Chemical Reaction

Copyright 1995 by Saunders College Publishing

C6H12O6(s) + 4KClO3(l) --> 6CO2(g) + 6H2O(g) + 4KCl(s) 3


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Periodic Property: Electronegativity


Na + Cl2

EnergyPotential energy is stored energy Kinetic energy is motion energy

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Heat and Temperature!Heat is energy, not matter!Heat is a type of Kinetic energy-

particles (atoms/molecules) are moving.

!Heat can be gained or lost!Temperature is a measure of heat or

kinetic energy. heat flow

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Heat Units:Joule (J) = kg-m2/s2 =107ergscalorie = 4.184 JBritish thermal unit (BTU) = 1055 J

Conversion: 1 calorie = 4.184 J1 calorie = amount of energy to raise the

temperature of 1 g of water 1oC1 Calorie = 1 food calorie = 1 kcal = 1000

caloriesSo for Snickers: 250.0 Calories

This is 250,000 caloriesHow many J?

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Heat Transfer


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Heat Transfer at the Molecular Level

Copyright 1995 by Saunders College Publishing9


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Heat and Matter


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Calculating Heat! To heat room temperature water for

coffee, the amount of heat needed depends on???

1.

2.

3.

! q =11

Characteristics of the SubstanceHeat capacity: amount of heat required

to raise temperature

Specific heat capacity: amount of heat required to raise temperature of

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Comparative Heat Capacities

200 g of Cu 200 g H2O 1.4 x 106 g H2O 76 J/oC 837 J/oC 5.9 x 106 J/oC

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Some Specific Heat CapacitiesSubstance Specific Heat (J/g.K)

aluminum 0.902copper 0.385gold 0.128water(l) 4.184water(s) 2.06ethanol 2.46O2(g) 0.917N2(g) 1.04

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Calculating Heat

Units on q will depend on units of Specific Heat Specific heat of water =


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How many calories of heat is required to raise the temperature of ______ g of liquid water from 25.0 to ____.0 C?

(A) 1.0 e4 cal (B) 10.5 e3 cal(C) 43.9 e3 cal(D) 10,500. cal

How many joules of heat?

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System vs Surroundings

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! System: Subject(s) involved in the change

! Surroundings: Everything in the systems environment

! Universe:

Thermodynamic Terms

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Thermodynamic Terms

System

Surroundings

Energy Energy

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20

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Exothermic Reactions: Release of Stored Chemical Energy


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Hot and Cold Packs

4Fe(s) + 3O2(g) --> 2Fe2O3(s) + Heat

NH4NO3(s) + Heat --> NH4NO3(aq)

Exothermic Endothermic

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Endothermic and Exothermic Processes

CO2 gas

CO2 solid

Heat absorbed from surroundings (Endothermic)

Heat released to surroundings (Exothermic)E

nerg

y

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A Chemical Reaction


H2(g) + 1/2O2(g) --> H2O(g) + 242 kJ/mol 25

Endothermic and Exothermic Processes

H2(g) + 1/2O2(g)

H2O(g)

H2O(l)

Ene

rgy

Heat absorbed from surroundings(Endothermic)

Heat released to surroundings (Exothermic)

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Energy for the Space Shuttle


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WHY??

WHY??

WHY??

WHY??

Thermodynamics is based upon observations of common experience that have been formulated into Laws

From these few Laws, all of the remaining Laws of Science are deducible by purely logical reasoning

The Laws of Thermodynamics

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The First Law of Thermodynamics

_______________________

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System atState 2 withEnergy E2

Change in Energy = Esys

Esys = Efinal Einitial = E2 E1

If Efinal > Einitial then________ If Efinal < Einitial then________

Heat

Work



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Change of Energy of the System

The First Law!Euniverse = Esys + Esurr = 0

!Esystem = q + w where: Esystem internal energy change

q heat w PV work = PV = (n)RT

!At fixed n(gas) or V: w = 0 and Esystem = qV 33


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Sign Conventions for q and w

Surroundings

System

q ____ q ____

w > 0 w < 0

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P-V Work


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Work in Chemical Systems! Recall: w = (n)RT = PV H2(g) + 1/2O2(g) --> H2O(g)n = nprod nreac = __________= _______Therefore: work is _______

(when work is done __ system)

! Gummy Bear:C6H12O6(s) + 4KClO3(l) --> 6CO2(g) + 6H2O(g) + 4KCl(s)

n = nprod nreac = __________= _______Therefore: work is _____________

(when work is done _____ system) 36


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! What is the internal energy of a system that does _____ kJ of work on the _____________ and releases______kJ of heat to the surrounding?(A) 250 kJ (B) 150 kJ(C) -250 kJ (D) -150 kJ

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Esys, q and wC8H18(l) + 25/2O2(g) --> 8CO2(g) + 9H2O(g)

initial state final state

C8H18(l) + 25/2O2(g)

8CO2(g) + 9H2O(g)

Ene

rgy

Energylost asheat andwork

Energylost asheat

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Sign Conventions for q and w

System does work on surr

Work done on system

Heat absorbed bysystem (endothermic)

Heat released bysystem (exothermic)

Whenever energy (heat or work) is added to a system, the energy of a

system _____________39


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Enthalpy (H)! the total heat content of a system (H)! at constant pressure, the enthalpy change H is: H = qP = Hfinal Hinitial

! At constant P: w = PV and qP = HEsystem = q + w

E = qP PV E = H PV

H = E + PV

! For gases at fixed T: PV = (n)RTH = E + (n)RT 40

! If no work Esystem = _______! If no work and constant pressure then: Esystem = _____ = _____

!H is a state function and 1 factor in thermodynamics

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State Function______________ of the path taken to traverse

from initial state to final state

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State Functions

P, V, T, H, E are state functions

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Exothermic Reaction Energy Profile

H2(g) + 1/2O2(g)

H2O(g)

Heat Evolved = 242 kJ

Hea

t Con

tent

Progress of a Reaction44

Energy Graphs

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System Exothermic = _______ H, IF not work, then also Esystem= _____

System Endothermic = _________ H, IF not work, then also Esystem = _____


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Internal Energy (E)! Defined as the sum of kinetic and

potential energies of all particles in a system

! It is a state function! In a chemical or physical process, where

reactants are converted to products, the change in internal energy (E) is given by:E = Efinal Einitial = Eproducts Ereactants

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Predict the signs for the following:1.H2O(l) H2O(g)2.Chocolate(l) Chocolate (s)3.CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

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1

2

3

Heat Required for Phase Changes

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Calculating Heat Transferred in a Process - Relationships

For temperature changes:Heat(J) = [Mass(g)] [Sp Heat(J/gC)] [Temp change(C)]q =For phase changes:Heat (J) = [Mass(g)] [Hphasechange (J/g)]q =For a specific amount of material or equipment:Heat (J) = [Heat Capacity(J/ C)] [Temp change(C)]q =

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Calculate the heat required to transform 25.00 g of ice at -10.00 C to water at 30.00 C. S solid = 2.04 J/gC S liquid = 4.184 J/gC S gas = 1.40 J/gC Hvap = 2200 J/g Hfus = 335 J/g

!Solid -10.00 to 0.00q =

!Solid 0.00 to Liquid 0.00 phase changeq =

!Liquid 0.00 to 30.00q =

TOTAL = Or ______________J in ___ sig figs 50

Calorimetry! Specific amount of substance in

calorimeter undergoes chemical and/or physical change; resulting energy transfer is monitored by temperature changes

! For an exothermic reaction in solution: |qrxn| = |qcal + qsol| = KT + msT

! If qrxn is losing heat (neg q), then qcal + qsol are gaining heat (pos q).

! Work in absolute, then assign negative OR apply a neg to one side.

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Coffee-Cup Calorimeter

! Pressure is constant! Measures qp (H) 52

Bomb Calorimeter

! Volume is constant, so no work! Measures qV (E) 53

! When 2.00 g of NaOH are dissolved in 500. mL of water in a cup calorimeter, the water changes from 25.00 C to 37.00 C. The heat capacity of the calorimeter is 470. J/ C. What is the heat involved per mole? (molar enthalpy)

! qp =! =! = ____g (4.184J/gC) (12.00C) + 470. J/ C (12.00C)! = ___________________+ ____________! = ______________ J given off for 2.00g NaOH! Mole NaOH = 2.00/40.0g = 0.0500mol! Heat/mole NaOH =! H/mol NaOH = = -6.17 e5 J/mol

or -617 kJ/mol54


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Thermochemical Equations:Constructing Unit Factors

! For the reactionC2H5OH(l) + 3O2(g) --> 2CO2 (g) + 3H2O(l) + 1367 kJ

the following unit factors may be written:

55 Heat is proportional!!

1367 kJ___ mol C2H5OH

1367 kJ___mol O2

1367 kJ1 mol rxn

___ mol O21367 kJ

1 mol C2H5OH 1 mol rxn

1367 kJ____mol CO2

Thermochemical Equations:Reversibility! Reversing the equation changes the sign

but not the magnitude of HH2(g) + 1/2O2(g) --> H2O(g) H = 242 kJ

H2O(g) --> H2 + 1/2O2(g) H = _____kJ

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! If equation is multiplied by a factor n, H must be multiplied by same factor H2(g) + 1/2O2(g) --> H2O(g) H = 242 kJ

2H2(g)+ O2(g) --> 2H2O(g) H = _____kJ

Thermochemical Equations:Proportionality

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! H for individual steps of a reaction sequence may be added to give H for overall reaction

C(s) + O2(g) --> CO2(g) H = 394 kJ/molCO2(g) --> CO(g) + 1/2O2(g) H = +283 kJ/mol

Thermochemical Equations:Combinations of Known H Values

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Hess Law ! Enthalpy change of a process depends

only on the initial and final states; it is independent of the path or the number of steps involved in the process

! If several reactions are added (or subtracted) to give an overall (net) reaction, H terms are also added (or subtracted) accordingly

! If reaction is reversed, the sign of H must be changed accordingly

! If reaction is multiplied, , the value of H must be multiplied accordingly 59

Hess Law: Law of Heat Summation

H2

Reactants

H1 H3

H

H4 H5

Products

H = H1 + H2 + H3 = H4 + H560


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! Problem: Calculate Horxn for: CO(g) + NO(g) --> CO2(g) + 1/2N2(g)given the following data:

CO(g) + 1/2O2(g) --> CO2(g) HoA = 283 kJ O2(g) + N2(g) --> 2NO(g) HoB = 180.6 kJ! Solution:

CO(g) + 1/2O2(g) --> CO2(g) HoA =NO(g) --> 1/2O2(g) + 1/2N2(g) HoC =

Horxn = 61

Find Hrxn from Combinations of Known H Values

! H values __________change significantly with moderate changes in temperature

Thermochemical Equations:Guidelines

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! Hf is the amount of heat released or absorbed when ______of a compound in a specified state is formed from its elements also in their standard states; also referred to as heat of formation

Standard Molar Enthalpy of Formation (Hf)

2C(graphite) + 3H2(g) + 1/2O2(g) --> C2H5OH(l)Hof = 277.7 kJ/mol

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Standard State! Defined as the most stable state of a substance

at __ atm and at some specified temperature, almost always 25C (298 K)

! Liquids = ________________! Gases = ________________________, 8th family! Solids = rest

H2O(l) Br2(l) NO2(g) Hg(l)C(s) 64

Standard States

! Pure substance (liquid or solid): standard state is the pure liquid or solid

! Gases: standard state is the gas at ___ atm pressure

! Solutions: standard state is __ M concentration

Hg(l) NO2(g) HCl(1 M)S(s)

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Hf for Some Elements! The standard molar enthalpy of formation

for any element in its standard state is ____

Substance Hof (kJ/mol)Br2(l) 0Br2(g) 30.91C(diamond) 1.897C(graphite) 0O2(g) 0Na(s) 0

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Hf for Some CompoundsSubstance Hof (kJ/mol)CO2(g) 393.5CaCO3(s) 635.5HBr(g) 36.4H2O(l) 285.8H2O(g) 241.8SO2(g) 296.8SiO2(s) 910.9

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11/1/17 Thermodynamics 68

Enthalpies of Formation

! Physical states of all reactants must be specified; these determine the magnitude of the energy changes H2(g) + 1/2O2(g) --> H2O(g) H = 242 kJ

H2(g) + 1/2O2(g) --> H2O(l) H = 286 kJ

Thermochemical Equations:Guidelines

69


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Stand. Enthalpies of Rxn (Hrxn) and Hof

2C(graphite) + 3H2(g) + 1/2O2(g) --> C2H5OH(l) Horxn = 277.0 kJ/mol

Horxn Hof2C(graphite) + 3H2(g) + 1/2O2(g) --> C2H5OH(g)

Horxn = 235.3 kJ/molHorxn Hof

C2H4(g) + H2O(l) -->C2H5OH(l)Horxn = 510.36 kJ/mol

Horxn Hof 70

For which is the ______= _____?A = ____ B = _____C(s)+ 2H2(g)--> CH4(g)2H2(g)+ O2(g)--> 2H2O(g)1/2 N2(l)+ 3/2 H2(g)--> NH3(g)2Ca(s)+ 2C(s) + 3O2(g)--> 2CaCO3(g)H2(g)+ O2(g)--> H2O(g)

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Hess Law: Schematic Representation

Elements

Reactants

Products

Horxn

n Hof (products)

n Hof (reactants)

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! Hof may be used to calculate Horxn:Horxn = n Hof(products)

n Hof(reactants) n = stoichiometric coefficient! For the general reaction:

aA + bB + ... --> xX + yY + ... Horxn = [xHof(X) + yHof(Y) + ...]

[aHof(A) + bHof(B) + ...]

Calculating Hrxn from Hf

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Calculating Hrxn from Hf CO(g) + NO(g) --> CO2(g) + 1/2N2(g) Ho(reactants) = 1mol Hof(CO) + 1mol Hof(NO)Reactants =Ho(products) = 1 Hof(CO2) + 1/2 Hof(N2)Products =Horxn = products reactants =NOTE: same as before (or very close)

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Hess Law Using Hf Values! Calculate Horxn at 298 K: SiH4(g)+ 2O2(g)--> SiO2(s)+ 2H2O(l) Horeact = [2Hof,O2 + H

of,SiH4

]

Hoprod = [_____________ + ___________]

Horxn = Hof,prod Hof,react Ho = [__________] [_________] = _______ kJ

! Ho = _______________kJ 75


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Bond Energy! Energy necessary to break one mole of

bonds in a gaseous covalent substance at constant temperature and pressure

! Also referred to as bond enthalpy! Always positive in the table. ! Bond-breaking is an endothermic process;

bond-forming is exothermic AB(g) --> A(g) + B(g)

Horxn = Bebreak- BEmake76

Bond Energy

Copyright 1995 by Saunders College Publishing 77

Bond Energy

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Bond Energy and Hrxn

In the gas-phase:Horxn = BE (break) BE (make)

Atoms (g)

Reactants (g)

Products(g)

Horxn

Energy tobreak bonds

Energy released in bond formation

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Bond Energy and Hrxn H2 + Br2 2HBr

2HBr

2H + 2Br

2H + Br2

H2 + Br2

HBrBr = 192 kJ

HHH = 435 kJ - 2HHBr = 736 kJ

Hrxn = 109 kJ

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1

2

3

Carbon-Carbon Bond Energy


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Selected Single Bond Energies*

H C N ClO F P436 414 389 464 569 318 431 H

347 293 351 439 264 330 C159 201 272 209 201 N

138 184 351 205 O159 490 255 F

213 331 P*These are average values in kJ/mol of bonds

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Selected Multiple Bond Energies*

N=N 418NN 946C=N 615CN 891O=O 498

*These are average values in kJ/mol of bonds

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Note trend with single, double, triple bonds: C-C single bond is 347 kJ/mol

C=C 611CC 837C=O 745C=O 799 in CO2CO 1070

Bond Energy and Hrxn

! Horxn = BEb BEm Horxn = [(BEHH +

BEClCl) 2BEHCl] Horxn =

Horxn =! Horxn = __________kJ

H2(g) + Cl2(g) 2HCl(g)


84


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Using a bond energy table, the change in enthalpy is closer to:

CO + H2O CO2 + H2

(A)-40kJ (B) -2000kJ (C) -400kJ (D) H is positiveC2H4 + 3O2 2CO2 + 2H2O

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Break:

Make:

Horxn = _______ ________ = ______ kJ

Summary Calculating H (enthalpy)1.from heating curve:

" q = msT and q = mHfus or vap! m = mass ! s = specific heat of substance !T = Tfinal Toriginal

2. from calorimeter: " |qrxn| = |qwater + qcal |" |qrxn| = | msT + KT |

! K = calorimeter constant3. from Proportionality:

"Given equation + Hrxn , then find H for a certain amount of one of the reactants or products

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4. from Combining Equations (Hess Law):"Given equations + their H values"Use heat is reversible, additive, proportional to H

find for a Target equation

5. from standard Hf values:" Find H for an equation, given Hf values" Hrxn = Sum of product Hf sum of reactant Hf

6. from Bond Energies:" Find H for an equation, given bond energy

values" Hrxn = Sum of bonds broken sum of bonds

made87

Documents

Thermochemistry: Part of Thermodynamics - …chemed.tamu.edu/chem101/pptnotes/Thermochem AF st 3pp 29p.pdf · Dr. Williamson’s Notes on Thermochemistry © vm williamson 6 How many