Overview of Thermochemistry

1
THERMOCHEMISTRY
HESS LAW, HEATS OF FORMATION AND PHASE CHANGES
2
CALCULATING HEATS OF RXNS

• There are three ways to calculate the energy of a
  reaction.
• DHmCDT (takes a temperature change, mass, and
  specific heat constant to calculate a ?Hrxn uses
  conservation of energy)
• Enthalpy of Formation (takes data from a
  table and uses it to calculate the energy of a
  reaction)
• Hesss Law (allows us to take two or more chem
  rxns with known ?Hrxn and combine them in such a
  way to calculate the enthalpy of a target
  reaction)

3
HEATS OF FORMATION

• Another method of calculating the enthalpy of a
  reaction is by using heats of formation.
• There are tables of DHform that we can gather
  information from
• Elements are always 0
• DHform is dependent on the number of moles
• We also need to use the equation presented
  earlier

DHrxn ?Hproducts - ?Hreactants
4
HEATS OF FORMATION
Calculate DH for the following reaction 8 Al(s)
3 Fe3O4(s) ? 4 Al2O3(s) 9 Fe(s)
8(0)
3(-1118.4)
4(-1675.7)
9(0)
DHrxn ?Hproducts - ?Hreactants
DHrxn 4(-1675.7)9(0) 8(0)3(-1118.4)
DHrxn (-6,702.8) (-3355.2)
DHrxn -3,347.6 kJ
5
Classwork
Use heats of formations calculations to determine
the combustion of which hydro-carbon will produce
the most energy per mole (CH4 -74.81 kJ/mol
C2H6 -84.68 kJ/mol C3H8 -104.5 C4H10 -126.5
kJ/mol)
CH4 2O2 ? CO2 2H2O 2C2H6 7O2 ? 4CO2
6H2O C3H8 5O2 ? 3CO2 4H2O 2C4H10 13O2 ?
8CO2 10H2O
6
HESS LAW

• The change in energy of a process or reaction is
  a state function, meaning that regardless of the
  path to reach your goal, the energy to get there
  is constant.
• For instance if you want to vaporize a solid, you
  have two pathways.
• You can melt it into a liquid and then vaporize
  it into a gas.
• Or you can sublime the solid directly into a gas.
• Either path gets the desired results and either
  path requires the same amount of heat energy,
  this is Hesss Law.

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The idea that we can calculate ?Hsublimation
by combining the ?Hfus with the ?Hvap
is an illustration of Hess Law.
8

• During any Hesss Law calculation, there are
  two things that we are allowed to do to the given
  reactions in order to manipulate the.
• We can reverse the reaction in order to make the
  products reactants, as long as we change the sign
  of the enthalpy
• We can also increase or decrease the amounts of
  reactants or products by multiplying by a factor,
  as long as we multiply the enthalpy by the same
  factor
• The key is to keep our eye on the prize, the
  goal reaction

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• For example, use Hesss Law to calculate the
  enthalpy of formation for the following reaction
  equation

2 N2(g) 5 O2(g) ? 2 N2O5(g) DH?f ?

• Given the following reaction equations

2NO(g) O2(g)? 2NO2(g) DHrxn -114kJ/mol
2
4NO2(g) O2(g)? 2N2O5(g) DHrxn -110kJ/mol
N2(g) O2(g)? 2NO(g) DHrxn 181kJ/mol
2
2(-114 kJ)(-110 kJ)2(181 kJ) 24 kJ
10

• Example 2
• Given the following information

C2H6?C2H4 H2 137kJ/mol
2H2O?2H2O2 484kJ/mol
2H2O2CO2?C2H43O2 1323kJ/mol
Find the value of ?H for the reaction
2C2H6 7O2 ? 4CO2 6H2O
11

• Example 2
• Rearranging and multiplying

2 C2H6 ? 2 C2H4 2 H2 274kJ/mol
2H2O?2H2O2 484kJ/mol
2C2H46O2 ? 4H2O4CO2 -2646kJ/mol
Find the value of ?H for the reaction
2C2H6 7O2 ? 4CO2 6H2O
12

• Example 2
• Rearranging and multiplying

2 C2H6 ? 2 C2H4 2 H2 274kJ/mol
2H2 O2?2H2O - 484kJ/mol
2C2H46O2 ? 4H2O4CO2 - 2646kJ/mol
Find the value of ?H for the reaction
2C2H6 7O2 ? 4CO2 6H2O
(274kJ)(-484kJ)(-2646kJ) DHrxn
-2856 kJ DHrxn
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Classwork
Before pipelines were built to deliver natural
gas, individual towns and cities contained plants
that produced a fuel known as town gas by passing
steam over red-hot charcoal.
C(s) H2O(g) ? CO(g) H2(g)
Calculate ?H? for this reaction from the
following information..
C(s) ½O2(g) ? CO(g) ?H? -110.53 kJ
CO(g) ½O2(g) ? CO2(g) ?H? -282.98 kJ
C(s) O2(g) ? CO2 (g) ?H? -393.51 kJ
H2(s) ½O2(g) ? H2O(g) ?H? -241.82 kJ
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PHASE CHANGES HEAT

• Energy is required to change the phase of a
  substance
• The amount of heat necessary to melt 1
  mole of substance
• Heat of fusion (?Hfus)
• It takes 6.00 kJ of energy to melt 18 grams of
  ice into liquid water.
• The amount of heat necessary to boil 1
  mole of substance
• Heat of vaporization (?Hvap)
• It takes 40.6 kJ of energy to boil away 18
  grams of water.

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MOLAR HEAT DATA FOR SOME COMMON SUBSTANCES MOLAR HEAT DATA FOR SOME COMMON SUBSTANCES MOLAR HEAT DATA FOR SOME COMMON SUBSTANCES
SUBSTANCE ?Hfus ?Hvap
Mercury, Hg 2.29kJ/mol 59.1kJ/mol
Ethanol, C2H5OH 5.02kJ/mol 38.6kJ/mol
Water, H2O 6.00kJ/mol 40.6kJ/mol
Ammonia, NH3 5.65kJ/mol 23.4kJ/mol
Helium, He 0.02kJ/mol 0.08kJ/mol
Acetone 5.72kJ/mol 29.1kJ/mol
Methanol, CH3OH 3.16kJ/mol 35.3kJ/mol
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• There are 5 distinct sections we can divide the
  curve into
• Ice (solid only)
• Water ice (solid liquid)
• Water only (liquid only)
• Water steam (liquid gas)
• Steam only (gas only)
• We can calculate the amount of energy involved in
  each stage
• There are two types of calculations
• Temperature changes use ?HmC?T
• Phase changes use (mols)?Hfus or (mols)?Hvap

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• If we journey through all of the 5 stages of the
  heating we have 2 phase changes and 3 increases
  in temperatures
• Each stage has its own amnt of energy to absorb
  or release to make the change necessary
• The total energy of the entire process can be
  calculated by combining the energies of each stage

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DHtotal
DHmelting
DHliquid
DHvaporizing
DHgas
DHsolid
DHtotal mCsolidDTn(DHfus)mCliquidDTDHvapDHga
s
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DHtotal
mCsolid DT
n(DHfus)
mCliquidDT
DHmelting
DHliquid
DHvaporizing
DHgas
n(DHvap)
DHsolid
mCgasDT
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SOLID ICE

• Lets say we have 180.0g of ice at 10C, we
  begin heating it on a hot plate with sustained
  continuous heat.
• Heat energy absorbs into the ice increas-ing the
  vibrational or kinetic energy of the ice
  molecules
• The temp will increase will continue to
  increase until just before the ice has enough
  energy to change from solid to liquid (to
  the melting point)

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• We can calc the energy absorbed by the ice
  to this point
• Use Cice2.09J/gC

DHicemCice DT
DHice (180g)(2.09J/gC)(0C-(-10C))
DHice 3762J
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WATER ICE (MELTING)

• Any additional heat absorbed by the ice goes into
  partially breaking the connections between the
  ice molecules.
• There is no change in the KE of the molecules
  (graph flattens out)
• No change in temp
• All of the energy goes into breaking the
  connections

• As long there is solid ice present, the temp
  cannot increase.
• The solid liquid are in equilibrium
  if they are both present

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• The energy required to change from the solid to a
  liquid is called the heat of fusion depends on
  the mols of the substance (DHfus of H2O6000J/mol
  or 6kJ/mol)

• Using the formula DHmelting (mol) DHfus

180g H2O
60,000J
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ALL WATER

• Now all of the particles are free to flow,
• The heat energy gained now goes into the
  vibrational energy of the molecules.
• The temp of the water increases
• The rate of temp increase now depends on the heat
  capacity of liquid water
• Cwater4.18 J/gC

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• The temp continues to increase until
  it just reaches the boiling point
  (for water 100C)
• again, ?HwatermCwater?T

DHwater(180g)(4.18J/gC)(100C-0C)
DHwater 75,240 J
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STEAM WATER (VAPORIZING)

• Any additional heat absorbed by the water goes
  into completely breaking the connections between
  the water molecules.
• Again the heat does not increase the KE of the
  molecules so the temp does not change,
• the energy is used to vaporize the water

• If there are still connections to break or there
  is liquid present, the temp cannot increase.
• The energy required to change from the liquid to
  the vapor phase is called the heat of
  vaporization using ?Hboiling(mol)?Hvap
• ?Hvap of H2O40,600J/mol

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180g H2O
406,000 J
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STEAM ONLY (VAPOR PHASE)

• Again the heat energy goes into the vibrational
  energy of the molecule.
• Rate of temp increase depends on CH2O
  vapor1.84 J/gC
• The temp can increase indefinitely, or until the
  substance decomposes (plasma)
• Well stop at 125C.

32
DHsteam(m)(Csteam)(?T)
DHsteam(180g)(1.84J/gC)(125-100) DHsteam
8280 J
33

• To figure out how much energy we need would need
  all together to heat up the water this much, we
  just need to add up the energy of each step.

DHtotal(3760 J60,000J75,240 J 406,000
J8280 J) DHtotal 553,280 J

• Notice, the majority of the energy is needed for
  the vaporization step.
• The connections between molecules of H2O must be
  broken completely to vaporize

34
Example
How much energy must be lost for 50.0 g of liquid
wax at 85.0C to cool to room temperature at
25.0C? (Csolid wax 2.18 J/gC, m.p. of wax
62.0 C, Cliquid wax2.31 J/gC MM 352.7
g/mol, DHfusion70,500 J/mol)
DHtotal (50g)(2.31J/gC)(62C-85C)
(50g/352.7g/mol)(-70,500J/mol)
(50g)(2.18J/gC)(25C-62C)
mCliquid waxDT
DHliquid wax
DHsolidification
n(DHfusion)
DHtotal (-2656.5 J) (-9994.3 J) (-4033 J)
mCsolid waxDT
DHsolid wax
DHtotal -16,683.8 J
DHtotal DHliquid wax DHsolidification
DHsolid wax
DHtotal mCliquid waxDTn(DHfusion)
mCsolid waxDT
35
Classwork
We have a collection of steam at 173C that
occupies a volume of 30.65 L and a pressure of
2.53 atm. How much energy would it need to lose
to end up as a block of ice at 0.00C?