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					Thermochemistry


                                                                                           University of Lincoln
                                                                                           presentation

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                  Thermochemistry



•   Enthalpy changes in chemical reactions (+ video)
•   Enthalpy Diagrams
•   Thermochemical Equations
•   Calorimetry and measuring enthalpy changes




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              Energy and Chemistry


• Petrol bombs

• What does this show?

• How to ensure your bonfire burns!

• Why does this happen?

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       Energy and Chemical Reactions

• Formation of new substances
  – Redox reactions
  – Acid-base reactions
  – Precipitation reactions


• Energy released/absorbed
  – Light (chemiluminescence)
  – Electrical energy (electrochemistry)
  – Heat (thermochemistry)

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                                                             Reaction 1
                                                             Energy products>reactants so energy is
               Heat             Heat
                                                             absorbed from the surroundings
                                                             Heat is lost from the surroundings so
SurroundIngs




                                          SurroundIngs
                       System
                                                             Temperature (surroundings) decreases -
                                                             Endothermic
               Endothermic
                                                             e.g. Ba(OH)2.8H2O(s) and NH4Cl(s) (video)


                                                            Reaction 2
                                                            Energy products<reactants so energy flows as
                Heat        Heat                            heat from the system to the surroundings
                       System                               Temperature (surroundings) increases -
                                                            Exothermic

               Exothermic                                   e.g. NaOH(aq) and HCl(aq).


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                    Enthalpy level diagrams
                                                                                               2 mol Na(s) + 2 mol H2O(l)
 The reaction between sodium metal and
 water – metal floats on water –
 effervescent reaction moves metal
 around- yellow flame above the metal-
 no solid residue
2Na(s)  2H 2 O(l)  2NaOH(aq)  H 2 (g)
                                                                                                                  ΔH = -367.5




                                                                                 Enthalpy, H (kJ)
 For You To Do
                                                                                                                  kJ (367.5 kJ
 Draw diagrams for the reactions on the                                                                           of heat is
 previous slide
                                                                                                                  released
 You will need to write balanced chemical
 equations first. For reaction1 assume that
 the products are NH3(g), H2O(l) and
 BaCl2(s) and that rH = +135 kJ mol-1

                                                                                         2 mol NaOH(aq) + 1 mol H2(g)
 Reaction 2 is a straightforward
 neutralisation with a rH = -55 kJ mol-1

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        Thermochemical Equations

 A thermochemical equation is the chemical
 equation for a reaction (including state
 symbols) and the enthalpy of reaction for the
 molar amounts as given by the equation written
 directly after the equation.

2Na(s)  2H 2 O(l)  2NaOH(aq)  H 2 (g)                                                      ΔH 367.5 kJmol 1


   CaO(s)  H 2 O(l)  Ca(OH) 2 (s)                                              Δ H  65.1 kJ mol 1
                                                                                  r




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        Thermochemical Equations

Why do we need state symbols?

In a thermochemical equation it is important to
note state symbols because the enthalpy change
depends on the physical state of the substances.
 2H 2 (g)  O 2 (g)  2H 2 O(g)                                          Δ H  - 483.7 kJ mol -1
                                                                          r



 2H 2 (g)  O 2 (g)  2H 2 O(l)                                        Δ H  - 571.7 kJ mol -1
                                                                        r




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    Thermochemical Equations

              Two important rules

1. When a thermochemical equation is
   multiplied by any factor, the value of H
   for the new equation is obtained by
   multiplying the DH in the original
   equation by that same factor.
2. When a chemical equation is reversed,
   the value of DH is reversed in sign.

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  Using Thermochemical Equations

2H 2 (g)  O 2 (g)  2H 2 O(l)                                            Δ H  - 571.7 kJ mol -1
                                                                           r


• What is the enthalpy change of reaction for the
  formation of 1 mole and 6 moles of water?
• -285.9 kJ mol-1; -1715.1 kJ mol-1
• What is the enthalpy change for the splitting of 1
  mole of water into hydrogen and oxygen gas?
• +285.9 kJ mol-1

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         Using Thermochemical Equations
  Consider the reaction of methane, CH4, burning in the
  presence of oxygen at constant pressure. Given the
  following equation, how much heat could be obtained by
  the combustion of 10.0 grams CH4?

 CH 4(g)  2O 2(g)  CO 2(g)  2H 2 O (l)                                          Δ H o  890.3 kJmol 1
                                                                                    c




 1 g of methane would                                    890.3kJ mol 1
                                                                          55.6 kJ g 1
 give                                                     16.0 g mol 1

Combustion of methane gives 55.6 kJ g-1
10 g of methane would                                           10.0 g  55.6 kJ g 1  556 kJ
give
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                 Measuring enthalpy changes
•Measuring enthalpy changes is called
calorimetry
                                                                                                    Thermometer
•Carry out the reaction in a calorimeter
and measure the temperature change.                                                                                                      HCl(aq)
•Calculate the energy transferred
during the reaction from the
temperature change.                                                                                                          2 polystyrene
                                                                                                                             coffee cups
•Also require the mass of the substance
and the specific heat capacity

Assumptions                                                                                   NaOH (aq)
All the energy change is transferred to
the solution (water)                                                              Coffee-cup calorimeter
No losses of heat to the other
surroundings

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           Need to know
       Specific Heat Capacity
Definition
The amount of energy required to raise the
temperature of a specified mass of an object
(substance) by 1 degree kelvin (K)
units
J g -1K-1 or J kg-1 K-1
Important example
Water 4.184 J g-1 K-1
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                                   An Example:
  25 cm3 of 2.00 mol dm-3 HCl(aq) is mixed with 25
  cm3 of 2.00 mol dm-3 NaOH(aq). The temperature
  rises from 22.5 oC to 34.5 oC. Find the enthalpy
  change for the reaction
  HCl(aq)  NaOH(aq)  NaCl(aq)  H 2 O(l)
                        Q  mcΔc
Q  50g  4.18 Jg K  12.0K      1         1                                                       Q  2508J

             25
         n       2.0mol 0.05mol
            1000
   2508 J of heat is transferred from the reaction of 0.05
   mol HCl with 0.05 mol NaOH
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              An Example continued

           For 1 mol of HCl and NaOH

             2508
         ΔH      50160J 50.2kJ
             0.05

HCl(aq)  NaOH(aq)  NaCl(aq)  H 2 O(l) ΔH  50.2 kJ mol 1




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 Now Try This One
 0.327 g of Zinc powder is added to 55 cm3
 of aqueous copper sulfate solution at 22.8
 oC. The copper sulfate is in excess of that

 needed to react all the zinc. The
 temperature rises to 32.3 oC. Calculate H
 for the following reaction:
Zn(s)  CuSO 4 (aq)  ZnSO 4 (aq)  Cu(s)



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                                                                                                    Thermometer

                                                                                                                                    HCl(aq)



Limitations of this method ??                                                                                                2 polystyrene
                                                                                                                             coffee cups




                                                                                             NaOH (aq)


                                                                                 Coffee-cup calorimeter



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How can we accurately measure
enthalpy changes of combustion
                     Thermometer
reactions?                       Current for
                                                                                                               Stirrer
                                                                                 + ignition
                                                                                 - coil




                                                                                                                                    Needle

A bomb calorimeter                                                                                                                  Gas inlet


                                                                                                                                    Insulated
                                                                                       O2                                           jacket

                                                                                                                                    Steel
                                                                                                                                    bomb

                                                            Ignition coil                          Graphite sample

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        Bomb Calorimetry- measurements
Thermometer
                                Stirrer
               Current
             + for
             -
               ignition
               coil
                                                                                     Some heat from reaction
                                                                                     warms water
                                                      Needle
                                                      Gas inlet
                                                                                     qwater = mc∆T
                                                       Insulated                    Some heat from reaction
                O2                                     jacket                       warms “bomb”
                                                      Steel                         qbomb = heat capacity x ∆T
                                                      bomb

 Ignition             Graphite
 coil                 sample

            Total heat evolved, qtotal = qwater + qbomb
            Total heat from the reaction =qtotal
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Calculating enthalpy changes from calorimetry data


  Calculate enthalpy of combustion of octane.

  C8H18(l) + 25/2 O2(g)                                       8CO2(g) + 9H2O(l)

  • Burn 1.00 g of octane
  • Temp rises from 25.00 to 33.20 oC
  • Calorimeter contains 1200 g water
  • Heat capacity of bomb = 837 J K-1




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Calculating enthalpy changes from calorimetry data
  Step 1: energy transferred from reaction to water.
  q = (4.184 J g-1K-1)(1200 g)(8.20 K) = 41170 J

  Step 2: energy transferred from reaction to bomb.
         q = (bomb heat capacity)(ΔT)
         = (837 J K-1)(8.20 K) = 6860 J

  Step 3:Total energy transferred
         41170 J + 6860 J = 48030 J

  Heat of combustion of 1.00 g of octane = - 48.0 kJ
  For 1 kg = -48 MJ kg-1
 H=-48 kJ x 114 g mol-1=-5472 kJ mol-1

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                                Video




                        Click to link to
                      “Thermochemistry”
                             video




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A case study- Self-heating cans


                                                                                                               Can


                                                                                                              Insert
                                                                                                             Quicklime
                                                                                                            Foil
                                                                                                            separator

                                                                 Button
                                                             Water


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                          The Chemistry
CaO(s) + H2O(l)                                                      Ca(OH)2(s)

quicklime                                                            slaked lime

• Water and quicklime packaged separately
• When mixed, exothermic reaction takes place
  and the temperature of the water increases
• Heat transferred to the drink
• rH = -65.1 kJ mol-1



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How much quicklime is needed to heat up a
              coffee can?
• Think about what information you need
  to know for the calculation before doing
  the calculation – do some research and
  find approximate values

Homework
• Draw an enthalpy level diagram for the
  reaction
• Do the calculation
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FRS1027 Introductory Chemistry

• Hess’s law
• Standard enthalpy of formation
• Calculating enthalpy changes



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                            Hess’s Law
• The enthalpy change on going from
  reactants to products is independent of
  the reaction path taken
• Can be used to calculate enthalpy changes




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                                       Hess’s Law & Energy Level Diagrams
                                       C(s) + O2(g)
                                                                                          Reaction can be
                                                                                          shown as a single

                                             ΔH1 = -110.5 kJ
                                                                                          step or in a two steps.
         ΔH3 = ΔH1 + ΔH2 = -393.5 kJ




                                                                                          ΔHtotal is the same no
                                         CO(g) + ½ O2(g)
                                                                                          matter which path is
                                                                                          followed.
                                             ΔH2 = -283.0 kJ
Energy




                                                                                           ∆H reaction path 1= ∆H
                                                                                          reaction path 2
                                        CO2(g)



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   Standard enthalpy values (Ho)
1. Initial and final species are in their standard
   states
2. The standard state of a substance at a
   specified temperature is its pure form at 1 bar
   (100 kPa). T is usually 298 K (25 oC) but not
   always
3. rHo(298 K) is the standard enthalpy of
   reaction at 298 K

Some Physical States at 298 K
C = graphite; O2 = gas; CH4 = gas; H2O = liquid

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    Standard Enthalpy of Formation
ΔfHo (298 K) = standard molar enthalpy of
formation at 298 K
enthalpy change when 1 mole of compound is
formed from elements in their standard states
at 298 K
(These are available from data books)

H2(g) + 1/2O2(g)        H2O(g)
ΔfHo (H2O, g) = -241.8 kJ mol-1

ΔfHo is zero for elements in their standard
states.
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    Calculating Enthalpy Changes using
     standard enthalpies of formation
      Standard enthalpies of formation and Hess’s
      Law can be used to calculate unknown ∆rHo

      ΔrHo =      ΔHfo (products) -                                             ΔHfo (reactants)

Enthalpy of reaction = sum of the enthalpies of formation of the products (correct
molar amounts) – sum of the enthalpies of formation of the reactants (correct
molar amounts)

           Why is this an application of Hess’s law?



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         Calculating Enthalpy Changes

Calculate ∆cHo for methanol
Standard state of methanol at 298 K is liquid
CH3OH(l) + 3/2O2(g)                                                  CO2(g) + 2H2O(l)
ΔcHo =   ΔHfo (products) -                       ΔHfo (reactants)
= {ΔHfo (CO2) + 2 ΔHfo (H2O)} - {3/2 ΔHfo (O2) + ΔHfo(CH3OH)}
= {(-393.5 kJ) + 2 (-285.8 kJ)} - {0 + (-238.9 kJ)}

ΔcHo(298K)= -726.2 kJ mol-1

Now try the problems on the separate sheet


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                                Video



                    Link to “Hess’s Law”
                           video




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Some Other Problems to do




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Describe the reaction




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 The enthalpy of reaction for black powder

• Black powder is a mixture of potassium nitrate (75%),
  charcoal (13%) and sulfur (12%).
• A simplified equation is:
   2 KNO3( s )  S ( s )  3C( s )  K 2 S ( s )  N 2 ( g )  3CO2 ( g )
   fHo values/kJ mol-1
   KNO3(s)                -494.6                                 K2S(s)                                 -380.70
   CO2(g)                 -393.51
   Calculate the enthalpy of reaction in kJ mol-1 and kJ
   kg-1 of black powder.

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  The reaction of barium hydroxide with ammonium
                       chloride

  Equation
Ba (OH ) 2 .8 H 2O( s )  2 NH 4Cl( s )  2 NH 3( g )  10 H 2O(l )  BaCl 2 ( s )

  fHo values/ kJ mol-1
  Ba(OH)2.8H2O(s)                                           -3345
  NH4Cl(s)                                                  -314
  NH3(g)                                                    -46
  H2O(l)                                                    -286
  BaCl2(s)                                                  -859
  Calculate rHo

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Calculate the standard enthalpy of combustion of
                 octane at 298 K

fHo (octane)= -249.9 kJ mol-1




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FRS1027 Introductory Chemistry


    Bond Dissociation Enthalpies




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                                   Definition
The bond dissociation enthalpy (dissH) for an X-X
diatomic molecule refers to the process:
     X2(g)                   2X(g)

at a given temperature (often 298 K)

dissH = D(X-X) is the bond enthalpy for a specific
process (a particular bond in a molecule)

Breaking bonds is an endothermic process


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                                     D and D
CH4(g)                           CH3(g) + H(g)                                             H=436 kJ mol-1
CH3(g)                           CH2(g) + H(g)                                             H=461 kJ mol-1
CH2(g)                           CH(g) + H(g)                                              H=428 kJ mol-1
CH(g)                            C(g) + H(g)                                               H=339 kJ mol-1

• D depends on the bond
• D is an average value and is obtained from
CH4(g)                           C(g) + H(g)                                               H=1664 kJ mol-1
D (C-H) = 416 kJ mol-1


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   Enthalpy Changes in Chemical Reactions
Enthalpy difference between products and reactants
because different chemical bonds are formed.
Enthalpy change can be estimated from the chemical
bonds that are broken and made.
Breaking bonds is an endothermic process and making
bonds is an exothermic process
Only an estimate because
   •Bond enthalpies are mean values
   •All species are in the gaseous state
A Hess’s law problem


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  Enthalpy changes of combustion for hydrocarbons
  Compound               CH4
                                                                                                                                        Energy
                         Bond                                                      Energy in KJ                 Number                  out KJ
  Bond type              energy                Number broken                       mol-1                        made                    mol-1
  C-C                               347                                      0                            0                       0                0
  C-H                                413                                     4                      1652                          0                0
  C=O                               805                                      0                            0                       2            -1610
  O=O                               498                                      2                       996                          0                0
  O-H                               464                                      0                            0                       4            -1856


                                                                                                        Total
                                                                                                        energy
                                               Total energy                                             out/kJ
                                               in/kJ mol-1                                         2648 mol-1                                  -3466


  Tota enthalpy change of combustion kJ mol-1                                                                                                    -818
Now have a go at the bond enthalpy problems and set up as an Excel spreadsheet. Can you set it up so that
you only need to enter the number of C and H atoms to calculate the enthalpy change ???


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            Acknowledgements
•   JISC
•   HEA
•   Centre for Educational Research and Development
•   School of natural and applied sciences
•   School of Journalism
•   SirenFM
•   http://tango.freedesktop.org




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