Chemical Thermodynamics Homework:

1
Chemical ThermodynamicsHomework

• Pg 190
• 30, 32, 34,
• Pg 192
• 58, 60, 62

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Energy

• Energy - the ability to do work.
• Work - when a force is applied to an object.
• There are several types of energy
• Thermal - heat
• Electrical
• Radiant - including light
• Chemical
• Mechanical - like sound
• Nuclear

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Energy

• Energy can be classified as
• Potential energy
• Stored energy - ability to do work.
• Kinetic energy
• Energy of motion - actually doing work.
• Energy can be transferred from one object to
  another. It can also change form.

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Kinetic vs. potential energy
Potential Energy
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Converting potentialto kinetic energy
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Energy and chemical bonds

• During a chemical reaction
• Old bonds break.
• New bonds are formed.
• Energy is either absorbed or released.
• Exothermic Energy is released.
• New bonds are more stable.
• Endothermic Energy is required.
• New bonds are less stable.

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Exothermic
Reactants
Energy
Products
Since excess energy is released, the products are
more stable.
8
Endothermic
Products
Energy
Reactants
Additional energy is required because the
products are less stable.
9
Entropy

• Entropy - a measure of the disorder or
  randomness of a system.
• Disorder is favored over order and may account
  for reaction occurring spontaneously even if it
  is endothermic.

Increased entropy
solid
gas
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Energy and heat

• Thermal energy. Energy of motion of molecules,
  atoms or ions. All materials have this energy if
  at a temperature above 0 K.
• Heat. Thermal energy transfer that results from
  a difference in temperature. Thermal energy
  flows from warm objects to cool ones.

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The laws of thermodynamics

• First
• Energy cannot be created or destroyed but only
  transferred from one body to another or changed
  from one form to another.
• Second
• Every spontaneous change increases the entropy
  of the universe.
• Third
• The entropy of a perfect crystalline substance
  (no disorder) is zero at 0 K.

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Energy units

• Kinetic energy is defined as
• kinetic energy mv2
• m mass and v velocity.
• Joule (J) - the energy required to move a 2 kg
  mass at a speed of 1 m/s. It is a derived SI
  unit.
• J kinetic energy (2 kg) (1 m/s)2
• 1 kg m2 s-2

1 2
1 2
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Energy units

• Calorie (cal)
• Originally defined as the quantity of heat
  required to heat of one gram of water from 15 to
  16 oC.
• It is now defined as 1 cal 4.184 J
• Dietary Calorie
• This is what you see listed on food products.
• It is actually a kilocalorie.

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State functions

• Depend only on the initial and final states of a
  system. They are independent of how the system
  gets from one state to another.
• (i.e. What was illustrated on the previous slide)
• State functions include
• Pressure
• Volume
• Temperature
• Enthalpy

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Enthalpy

• The energy gained or lost (flow of heat) when a
  change takes place under constant pressure.
• DH Hfinal - Hinitial
• Subscripts are used to show the type of change.
• DHvap heat of vaporization
• DHneut heat of neutralization
• DHfusion heat of fusion
• DHsol heat of solution
• DHrxn heat of reaction

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Stoichiometry

• Many reactions are conducted simply for the
  thermal energy that is released.
• Combustion of gasoline, coal, natural gas.
• The thermal energy released can be shown as a
  product in a reaction.
• CH4 (g) 2O2 (g) CO2 (g) 2H2O (l)
  890.32 kJ
• or
• CH4 (g) 2O2 (g) CO2 (g) 2H2O (l)
  DHrxn -890.32 kJ
• When given for a reaction, DH is interpreted in
  terms of moles.

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Stoichiometry

• Determine the thermal energy released when 50.0
  grams of methane is burned in an excess of
  oxygen.
• First, determine the number of moles of methane
  (MM 16.043 u).
• mol CH4 (50.0 g) / (16.043 g/mol)
• 3.12 mol CH4

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Stoichiometry

• Now look at the balanced thermochemical equation.
• CH4 (g) 2O2 (g) CO2 (g) 2H2O (l)
  DHrxn -890.32 kJ
• DHrxn -890.32 kJ / mol CH4 so

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Stoichiometry

• Now look at the balanced thermochemical equation.
• CH4 (g) 2O2 (g) CO2 (g) H2O (l)
  DHrxn -890.32 kJ
• DHrxn -890.32 kJ / mol CH4 so
• Thermal energy released
• (3.12 mol CH4(g)) (-890.32 kJ / mol CH4 )
• - 2.78 x 103 kJ

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Stoichiometry

• The overall reaction in a commercial heat pack
  can be represented as
• 4 Fe(s) 3 O2(g) 2 Fe2O3(s) ?H
  -1652 kJ
• a. How much heat is released when 4.00 mol iron
  is reacted with excess O2?
• How much heat is released when 1.00 mole Fe2O3 is
  produced
• How much heat is released when 10.0 g Fe and 2.00
  g O2 are reacted?

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Hesss law

• The thermal energy given off or absorbed in a
  given change is the same whether it takes place
  in a single step or several steps.
• This is just another way of stating the law of
  conservation of energy.
• If the net change in energy were to differ based
  on the steps taken, then it would be possible to
  create energy -- this cannot happen!

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Hesss law
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Calculating enthalpies

• Thermochemical equations can be combined to
  calculate DHrxn.
• Example.
• 2C(graphite) O2 (g) 2CO (g)
• This cannot be directly determined because CO2
  is always formed.
• However, we can measure the following
• C(graphite) O2 (g) CO2 (g)
  DHrxn -393.51 kJ
• 2CO (g) O2 (g) 2CO2 (g)
  DHrxn -565.98 kJ

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Calculating enthalpies

• By combining the two equations, we can determine
  the DHrxn we want.
• 2 C(graphite) O2 (g) CO2 (g)
  DHrxn -787.02 kJ
• 2CO2 (g) 2CO (g) O2 (g)
  DHrxn 565.98 kJ
• Note.
• Because we need 2 moles of CO2 to be produced in
  the top reaction, the equation and its DHrxn were
  doubled.

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Calculating enthalpies

• Now all we need to do is to add the two equations
  together.
• 2 C(graphite) 2O2 (g) 2CO2 (g)
  DHrxn -787.02 kJ
• 2CO2 (g) 2 CO (g) O2 (g)
  DHrxn 565.98 kJ
• 2 C(graphite) O2 (g) 2 CO (g)
  DHrxn -221.04 kJ
• Note.
• The 2CO2 cancel out, as does one of the O2 on the
  right-hand side.

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Calculating enthalpies

• The real problem with using Hesss law is
  figuring out what equations to combine.
• The most often used equations are those for
  formation reactions.
• Formation reactions
• Reactions in which compounds are formed from
  elements.
• 2 H2 (g) O2 (g) 2 H2O (l) DHrxn
  -571.66 kJ

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Standard enthalpy of formation

• DHfo
• Enthalpy change that results from one mole of a
  substance being formed from its elements.
• All elements are at their standard states.
• The DHfo of an element in its standard state has
  a value of zero.

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Standard enthalpies of formation

• Substance DHfo, kJ/mol
• CaCO3 (s) -1206.92
• CaO (s) -635.09
• CH4 (g) -74.85
• C2H6 (g) -84.67
• CH3OH (l) -238.64
• CH3OH (g) -201.2
• CO (g) -110.52
• CO2 (g) -393.51
• HCl (g) -92.31
• H2O (l) -285.83
• H2O (g) -238.92
• NaCl (s) -411.12
• SO2 (g) -296.83

Standard enthalpy of formation values
are available for a wide range of
substances. Sometimes called the standard molar
enthalpy of formation In addition, separate
values for a substance in different states will
also be given where appropriate.
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Phase change

• We can use DHof values to determine the energy
  required to change from one phase to another.
• Example. Conversion of methanol from a liquid
  to a solid.
• kJ
• C (s) 2 H2 O2 (g) CH3OH (g) DHorxn
  -201.2
• C (s) 2 H2 O2 (g) CH3OH (l) DHorxn
  -238.6

1 2
1 2
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Phase change

• kJ
• C (s) 2H2 O2 (g) CH3OH (g)
  DHorxn -201.2
• CH3OH (l) C (s) 2H2 (g) O2 (g)
  DHorxn 238.6
• CH3OH (l) CH3OH (g) DHorxn
  37.4
• This is not DHovap because the values are at 25
  oC.
• DHovap would be the thermal energy required at
  the boiling point of methanol.

1 2
1 2
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Heat capacity

• Every material will contain thermal energy.
• Identical masses of substances may contain
  different amounts of thermal energy even if at
  the same temperature.
• Heat capacity. The quantity of thermal energy
  required to raise the temperature of an object by
  one degree.
• Specific heat. The amount of thermal energy
  required to raise the temperature of one gram of
  a substance by one degree.

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Specific Heats at 25oC, 1 atm

• Substance SH
• Al(s) 0.90
• Br2 (l) 0.47
• C (diamond) 0.51
• C (graphite) 0.71
• CH2CH2OH (l) 2.42
• CH3(CH2)6CH3 (l) 2.23

• Substance SH
• Fe (s) 0.45
• H2O (s) 2.09
• H2O (l) 4.18
• H2O (g) 1.86
• N2 (g) 1.04
• O2 (g) 0.92

SH specific heat, J g-1 oC-1 or J g-1K -1
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Heat capacity

• Example.
• How many joules must be added to a 50.0 g block
  of aluminum to heat it from 22oC to 85oC?
• Heat required mass x specific heat x DT
• Temperature must be stated in Kelvin!!
• Kelvin oC 273
• (Al specific heat (c) 0.09 J g-1K 1)

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Heat capacity

• Example.
• How many joules must be added to a 50.0 g block
  of aluminum to heat it from 22oC to 85oC?
• Heat required mass x specific heat x DT
• 50.0 g x 0.90 J g-1 K-1 x (358-295)K
• 2.8 kJ
• This is an endothermic change - sign.
• T2 is greater than T1

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Measuring thermal energy changes

• Thermal energy cannot be directly measured.
• We can only measure differences in energy.
• To be able to observe energy changes, we must be
  able to isolate our system from the rest of the
  universe.
• Calorimeter - a device that is used to measure
  thermal energy changes and provide isolation of
  our system.

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Coffee cup calorimeter
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Calorimetry example

• You are given the two solutions listed below.
  Each has an initial temperature of 20.0 oC.
• 50 ml of 0.50 M NaOH
• 50 ml of 0.50 M HCl
• Both are rapidly added to a coffee cup
  calorimeter and stirred. The reaction takes
  place rapidly. The highest temperature is 23.3
  oC. Solution density is 1.0 g/ml.
• Determine the heat of reaction if the specific
  heat of the solution is 4.18 J g-1 oC-1

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Calorimetry example

• First, determine the energy given off.
• 100.0 g (4.18 J g-1 oC-1) (23.3 - 20.0) oC
• - 1.4 x103 J (use - because heat is given
  off)
• Next, determine the moles of HCl or NaOH involved
  in the reaction -- both are the same.
• molHCl (0.5 ) ( 0.05 L)
• 0.025 mol HCl

mol L
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Calorimetry example

• The heat of neutralization for the reaction
• HCl (aq) NaOH (aq) NaCl (aq) H2O
  (l)
• is
• -1.4 x103 J / 0.025 mol
• -5.6 x 104 J/mol
• -56 kJ/mol

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When will a reactionbe spontaneous?

• Spontaneity of a reaction can be determined by a
  study of thermodynamics.
• Thermodynamics can be used to calculate the
  amount of useful work that is produced by some
  chemical reactions.
• The two factors that determine spontaneity are
  enthalpy and entropy.

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Why changes take place
Spontaneous process Takes place naturally with
no apparent cause or stimulus. Nonspontaneou
s process Requires that something be done in
order for it to occur.