Chapter 5 with Chapter 8'8

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Chapter 5with Chapter 8.8

• Thermochemistry

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Background

• Energy considerations have been key in our
  understanding atomic structure. Energy examples
• Photon energies E hn
• Hydrogen orbital energies E -2.18 x 10-18 J
  (1/n2)
• Kinetic energy operator (h2/8p2m) 2 and the
  potential energy operator ke2/r in the
  Schrödinger Equation
• Ionization energy and Electron Affinity energy
  differences (see usage next slide)
• With the convention that infinite separation has
  a potential energy of zero, lowered (or negative)
  energies implied system stability

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Energetics of NaCl Formation

• NaCl forms a crystal lattice
• Ionize Na takes energy (endothermic)
• Na(g) -gt Na(g) 1e- DH 496 kJ/mol
• Cl has a negative (exothermic) electron
  affinity
• Cl(g) 1e- -gt Cl-(g) DH -349 kJ/mol
• Energy of Na-Cl bond in lattice is -788 kJ/mole
• Overall 496-349-788 -641 kJ/mole so NaCl is
  more stable than Na(g) and Cl(g)

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Kinetic Energy

• Kinetic energy (Ek) is the energy an object
  possesses because of its motion,
• Ek mv2/2
• m mass of the object in kg
• v velocity of object in m/sec2

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Kinetic Energy

• Examples
• Motion of atoms, molecules, ions. This is
  thermal energy. ALL matter has thermal energy.
• Motion of macroscopic objects like a baseball or
  a car (mechanical energy)
• Movement of electrons through a conductor
  (electrical energy)
• Compression and expansion of the spaces between
  molecules in the transmission of sound (acoustic
  energy)

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Potential Energy

• Potential energy results from an objects
  position and when there is a force acting on the
  object.
• Examples
• A ball held above a floor (gravitational energy)
• Energy stored in the separation of two electrical
  charges (electrostatic energy)
• Energy stored in the arrangements of atoms in a
  substance (chemical energy)

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Kinetic and Potential Energy

• Can be interconverted between the two since both
  are forms of energy.
• A force is anything that causes a mass to
  accelerate, as a push or a pull.
• 1 N 1 kgm/sec2 (Unit of Force)
• The energy unit is the Joule, a newton-meter.
• 1 J Nm 1 kgm/sec2 x m or kgm2/s2

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Kinetic and Potential Energy Summary

• Chemical Energy is due to the potential energy
  stored in the arrangements of the atoms in a
  substance.
• If you run a reaction that releases heat, you are
  releasing the energy stored in the bonds.
• Energy because of temperature is associated with
  the kinetic (movement) energy of the atoms and
  molecules.

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The Universe

• Is divided into two parts the system and the
  surroundings.
• The system is the object you are concerned with.
• The surroundings are the rest.
• Closed system Energy, but not matter, may be
  transferred across the boundary separating the
  system from the surroundings (in the form of work
  heat)

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Forms of Transferred Energy

• Energy is the capacity to do work or transfer
  heat.
• Energy can then be transferred as either work or
  heat
• Heat (q) is thermal energy that flows between two
  bodies as the result of a temperature difference
  (always flows from hotter to colder)

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Examples of Energy

• Calculate the energy needed in Joules to lift a
  500 N object 250 cm.
• 1 J 1 Nm
• 500 N250 cm 1 m/100 cm
• 1250 J

Define what a Joule is
Plug and chug (remember the units you are given.)
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Forms of Transferred Energy

• Work Energy used to cause an object to move
  against a force.
• More mathematically defined as
• System Work (w) Force (Fext) x D d i.e. change
  in distance (d)

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Examples of Energy

• Calculate the speed of an Argon atom (in m/sec)
  with a kinetic energy of 1.4x10-20 J.
• 1.4x10-20 J 1/2 39.9 amu x 1.66x10-27 kg v2
• amu
• V Sqrt(21.4x10-20 Nm/(6.62x10-26 kg))
• V sqrt(422960 m2/sec2)
• V 650 m/s

Kinetic Energy (KE) ½ mv2

• Important Information for this problem
• The mass of 1 Ar atom is 39.9 amu
• There are 1.66x10-27 kg/amu

Solve for velocity
Plug and chug and watch units. Remember that 1
J 1 Nm and 1 N 1 Kgm/sec2
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Examples of Energy

• Calculate the energy in Joules of 1 mole of Ar
  atoms from previous slide.
• 1.4 x 10-20 J/atom 6.02x 1023 atoms/mole
• 8.4 x 103 Joules/mol

Do you remember Avagadros Number? Remember that
number refers to a mole of anything, even energy.
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Internal energy

• Not only is energy transferred, but also a system
  has energy because of its temperature
• This is called internal energy, E
• It is the sum of all of the kinetic and potential
  energies of the components of the system.
• Motion through space, movements of atoms relative
  to other atoms, and along an axis are kinetic.
  These are known as the translational, rotational,
  and vibrational energies, respectively.
• Electrostatic energy (between p and e- or
  between cations and anions) is potential energy.

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Internal Energy

• It can be changed by either heat or work flowing
  across the system boundary so that ?EEfinal
  Einitial

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Internal Energy

• First Law of thermodynamics DE q w
• So, internal energy is gained by adding heat or
  work to the system.
• However, Energy cannot be created or destroyed
  (energy conserved)

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The First Law of Thermodynamics
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Exothermic Reaction
Heat
Potential energy
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Endothermic Reaction
Heat
Potential energy
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State Functions

• A property of a system that is determined by
  specifying its condition or its state such as
  temperature, pressure, location, etc.
• State functions depend only on its initial and
  final states and not on the path between them.

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State Functions
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Enthalpy

• As we saw, chemical reactions can absorb or
  release heat.
• However, they also have the ability to do work.
• For example, when a gas is produced, then the gas
  produced can be used to push a piston, thus doing
  work.
• Zn(s) 2H(aq) ? Zn2(aq) H2(g)
• The work performed by the above reaction is
  called pressure-volume work.
• When the pressure is constant,

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P?V Work

• Remember
• System work (Wsys) Force (Fext) x Dh
• For a piston, Fext -PsysA
• and
• Dh DV/A
• so that
• Wsys -PsysA DV/A -PsysDVsys Nm J
• Remember Unit of work and energy is in Joules
  (J)

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Piston as a System
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Example of Internal Energy and P?V Work

• What amount of work is done by a gas when 15 L of
  gas is expanded to 25 L at 2.4 kPa pressure?
• If 2.36 J of heat are absorbed by the gas above,
  what is the change in energy?
• With the work being done in the first part, how
  much heat would it take to keep the internal
  energy of the gas constant?

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Enthalpy
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Enthalpy

• If constant volume w PDV 0, then from the
  first law DE qV
• To characterize processes that occur at
• constant pressure
• Define H E PV
• at constant pressure
• DH DE PDV
• Enthalpy accounts for the internal energy (?E)
  and P?V work at constant pressure.

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Enthalpy
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Enthalpies of Reaction

• For a reaction
• Enthalpy is an extensive property (magnitude DH
  is directly proportional to amount)
• CH4(g) 2O2(g) ? CO2(g) 2H2O(l) DH -890 kJ
• 2CH4(g) 4O2(g) ? 2CO2(g) 4H2O(l) DH -1780
  kJ

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Enthalpies of Reaction

• When we reverse a reaction, we change the sign of
  DH
• CO2(g) 2H2O(l) ? CH4(g) 2O2(g) DH 890 kJ
• Change in enthalpy depends on state of reactants
  and products.
• Since 2H2O(l) ? 2H2O(g) DH 88 kJ
• than
• CH4(g) 2O2(g) ? CO2(g) 2H2O(g) DH -802 kJ

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Enthalpies of Reaction

• Enthalpy Diagram

H2O (g)
Enthalpy
?H - 88 kJ
?H 88 kJ
H2O (l)
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Enthalpies of Reaction

• Example
• Hydrogen peroxide decomposes to water and oxygen
  by the following reaction
• 2H2O2(l) ? 2H2O(l) O2(g) ?H -196 kJ
• Calculate the value of q when 5.00 g of H2O2(l)
  decomposes at constant pressure.
• Hint What is enthalpy of reaction if only 1 mole
  of peroxide is decomposed?

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Enthalpies of Reaction

• Example
• Hydrogen peroxide decomposes to water and oxygen
  by the following reaction
• 2H2O2(l) ? 2H2O(l) O2(g) ?H -196 kJ
• -196 kJ/2 mol peroxide -98 kJ/mol of peroxide
• 5.0 g (1 mol/34 g H2O2) 0.147 mol H2O2
• 0.147 mol H2O2 -98 kJ/mol peroxide
• -14.4 kJ

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Calorimetry

• To measure heat given off in a process such as a
  chemical reaction.
• Two kinds of calorimeters
• Constant pressure (called a coffee cup
  calorimeter)
• Constant volume (called a bomb calorimeter)
• Heat Capacity amount of heat needed to raise an
  objects temperature by 1C or 1 K.

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Calorimetry

• Types of heat capacity
• Molar heat capacity amt of heat to raise 1 mole
  of material 1 oC Ex Waters MHC
  is 75.2 J/moleoC
• Specific heat capacity amt of heat to raise 1
  gram of material 1 oC Ex Waters SH
  is 4.184 J/goC
• Note well use 4.184 J/g-C for aqueous
  solutions too.

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Colorimetry

• Heat capacity Formula
• Specific Heat ____q____
• m x ?T
• C ___J____
• g x C (or K)
• You can calculate heat needed by rearranging
  formula to q mC ?T

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Calorimetry Heat Capacity
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Amount of Heat Example

• The specific heat of graphite is 0.71 J/g-K.
  Calculate the energy needed to raise the
  temperature of 75 kg of graphite from 294 K to
  348 K.
• Given C 0.71 J/g-K m75 kg 75000 g
• Know q mC?T
• Find ?T 348 K 294K 54 K
• Solve q75000 g 0.71J/g-K 54K
• 2,875,000 J or 2900 kJ

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Constant Pressure Calorimetry

m mass of solution C heat capacity of the
calorimeter (J g-1 C-1) (or K-1) ?T the change
in temperature in C or K
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Constant Pressure Calorimetry Example

• A 46.2 g sample of copper is heated to 95.4ºC and
  then placed in a calorimeter containing 75.0 g of
  water at 17.6ºC. The final temperature of both
  the water and the copper is 21.8ºC. What is the
  specific heat of copper?
• Given m46.2 g Cu m75.0g H2O
• Ti 94.4C for Cu Ti 17.6C for
  H2O
• We also know C for water is 4.184 J/mol-C
• and qsol -qcu

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Constant Pressure Calorimetry Example

• Find both ?Ts ?T Tf Ti
• In both cases Tf 21.8C
• For Cu 21.8C 94.4C -73.6C
• For H2O 21.8C 17.6C 4.2C
• Calculate the heat absorbed by water
• qsol mC ?T75.0g4.184 J/mol-C4.2C
• 1318 J (absorbed by solution)

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Constant Pressure Calorimetry Example

• Since qsol -qcu we know the heat given off by
  the copper. And we can substitute into heat
  equation.
• -1318 J 46.2 g C (-73.6C)
• Rearrange and solve
• C -1318 J/(46.2 g (-73.6C))
• C 0.388 J/g-C

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Constant Pressure Calorimetry Example Summary

• Copper and the water are the system, the beaker
  holding the water and everything else is the
  surroundings.
• The copper and water end up at the same temp.
• The energy transferred as heat from the copper to
  the water is negative (temp drops). The qwater
  increases is positive because the temp of water
  increases.
• The values of qwater and qcopper are numerically
  equal but opposite in sign.

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Calorimetry
Bomb Calorimetry (Constant Volume Calorimetry)

• Reaction carried out under constant volume.
• Use a bomb calorimeter.
• Usually study combustion.

qrxn -Ccal x ?T Ccal heat capacity of
calorimeter
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Example of Bomb Calorimetry

• Find the heat of combustion (in kJ/g) of 4.6
  grams of ethanol in a bomb calorimeter with a
  heat capacity of 8 kJ/oC if water temperature
  rises from 25 oC to 35 oC
• q -Ccal DT 8kJ/oC (35 25)oC
• q -80 kJ (this is DE not DH)
• Heat of combustion EtOH(l) -80 kJ/4.6 g
• -17.4
  kJ/g

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Standard Enthalpy

• The enthalpy change for a reaction at standard
  conditions (1 atm , 1 M solutions)
• Symbol DHº
• When using Hesss Law, work by adding the
  equations up to make it look like the desired
  reaction.
• The other parts will cancel out.

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Hesss Law

• If a reaction is carried out in a series of
  steps, ?H for the reaction will equal the sum of
  the enthalpic changes for the individual steps.

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Simple Hesss Law problem

• C(graphite) O2(g) ? CO2(g)     
• ?H 393.51 kJ mol1
• C(diamond) O2(g) ? CO2(g)     
• ?H 395.40 kJ mol1
• Calculate the ?H for the conversion of graphite
  to diamond.

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Simple Hesss Law problem

• C(graphite) O2(g) ? CO2(g)     
• ?H 393.51 kJ
  mol1
• CO2(g) ? C(diamond) O2(g)     
• ?H 395.40 kJ
  mol1
• Add rxns together, cancel out substances that
  appear on both sides and get
• C(graphite) ? C(diamond) ?H 1.9 kJ
  

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Harder Example of Hesss Law

• Given
• C(s) O2(g) ? CO2(g) ?H -393.5kJ/mol
• S(s) O2(g) ? SO2(g) ?H -296.8 kJ/mol
• CS2(l) 3O2(g) ? CO2(g) 2SO2(g)
• ?H -1103.9
  kJ/mol
• Use Hesss Law to calculate enthalpy change for
  the formation of CS2(l) from C(s) and S(s)
• C(s) 2S(s) ? CS2(l)

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• Multiply equations as needed
• C(s) O2(g) ? CO2(g) ?H -393.5kJ/mol
• 2S(s) 2O2(g) ? 2SO2(g) ?H2(-296.8 kJ/mol)
• 
  -593.6 kJ
• CO2(g) 2SO2(g) ? CS2(l) 3O2(g)
• ?H 1103.9
  kJ/mol
• C(s) 2S(s) ? CS2(l) ?H 116.8 kJ
• Reaction 1 was used as is
• Reaction 2 was multiplied by 2 (SO2 cancels)
• Reaction 3 was multiplied by -1 (Get CS2 on
• 
  product side)

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Hesss Law Practice

• Given
  
• calculate DHº for

DHº -1300. kJ
DHº -394 kJ
DHº -286 kJ
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Enthalpies of Formation

• If 1 mol of compound is formed from its
  constituent elements, with each substance in its
  standard state, then the enthalpy change for the
  reaction is called the enthalpy of formation,
  ?Hof .
• Standard conditions (standard state) 1 atm and
  25 oC (298 K).
• If there is more than one state for a substance
  under
• standard conditions, the more stable one is
  used.
• Standard enthalpy of formation of the most stable
  form of an element is zero.

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Enthalpies of Formation
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Given ?Hs 4B 3O2 ? 2B2O3 -2543.8
kJ H2 1/2O2 ? H2O -241.8 kJ B2H6
3O2 ? B2O3 3H2O -2032.9 kJ
Find D Hfo of B2H6
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?Hs 1/2(4B 3O2 ? 2B2O3
-2543.8 kJ) 3(H2 1/2O2 ? H2O -241.8
kJ) -1(B2H6 3O2 ? B2O3 3H2O -2032.9
kJ)
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?Hs 2B 3/2O2 ? B2O3 -1271.9 kJ 3H2
3/2O2 ? 3H2O -725.4 kJ) B2O3 3H2O ?B2H6
3O2 2032.9 kJ)
2B 3H2 ? B2H6
DHfo 35.6 kJ/mol
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Enthalpies of Formation

• Using Enthalpies of Formation of Calculate
  Enthalpies of Reaction
• For a reaction
• Example for C3H8(g) 5O2(g)? 3CO2(g) 4H2O(l)
• ?H 3?Hf(CO2) 4 ?Hf(H2O(l))?Hf(C3H8)
• 
  5?Hf(O2)
• DHorxn 3 (-393.5) 4 (-285.8) -103.8
  5(0)
• DHo rxn -2220 kJ

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Enthalpies of Formation

• Think of reaction like this
• 3C(s) 4H2(g) 5O2 3C 3O2 4H22O2
• C3H8 5O2 ? 3CO2 4H2O

4?Hf
-?Hf
3?Hf
-?Hf
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Practice Problem

• Calculate the standard enthalpy change for the
  reaction below using the standard enthalpies of
  formation.
• 4NH3(g) 5O2(g) ? 4NO(g) 6 H2O(g)
• ?Hfs -46.2 0 90.4
  -241.8
• (kJ/mol)
• Then ?Hrxn 4(90.4) 6(-241.8) (4(-46.2)
  0)
• 361.6 -1450.8 184.8
  -904.4 kJ/mol-rxn

First, get ?Hs from Appendix C (p. 1041 1043)
in book
Then use ?Hrxn of reaction formula
Finally, plug-and-chug. WATCH YOUR SIGNS!!!
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Strengths of Covalent Bonds (Ch. 8.8)

• The energy required to break a covalent bond is
  called the bond dissociation enthalpy, D. That
  is, for the Cl2 molecule, D(Cl-Cl) is given by ?H
  for the reaction
• Cl2(g) ? 2Cl(g).
• When more than one bond is broken
• CH4(g) ? C(g) 4H(g) ?H 1660 kJ
• The bond enthalpy is a fraction of ?H for the
  atomization reaction
• D(C-H) ¼?H ¼(1660 kJ) 415 kJ.
• Bond enthalpies are listed as positive values.

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Enthalpy of Reaction

• Bond Enthalpies and the Enthalpies of Reactions
• We can use bond enthalpies to calculate an
  estimate of the enthalpy for a chemical reaction.
• We recognize that in any chemical reaction bonds
  need to be broken and then new bonds get formed.
• The enthalpy of the reaction is given by the sum
  of bond enthalpies for bonds broken less the sum
  of bond enthalpies for bonds formed.

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Enthalpy of Reaction

• Bond Enthalpies and the Enthalpies of Reactions
• Mathematically, if ?Hrxn is the enthalpy for a
  reaction, then
• We illustrate the concept with the reaction
  between methane, CH4, and chlorine
• CH4(g) Cl2(g) ? CH3Cl(g) HCl(g) ?Hrxn ?

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Strengths of Covalent Bonds
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Strengths of Covalent Bonds

• Bond Enthalpies and the Enthalpies of Reactions
• In this reaction one C-H bond and one Cl-Cl bond
  gets broken while one C-Cl bond and one H-Cl bond
  gets formed.
• The overall reaction is exothermic which means
  than the bonds formed are stronger than the bonds
  broken.
• The above result is consistent with Hesss law.

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Lengths of Covalent Bonds