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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 2Energy products<reactants so energy flows as heat from the system to the surroundings Temperature (surroundings) increases - Exothermice.g. NaOH(aq) and HCl(aq).
Reaction 1 Energy products>reactants so energy is absorbed from the surroundingsHeat is lost from the surroundings soTemperature (surroundings) decreases - Endothermice.g. Ba(OH)2.8H2O(s) and NH4Cl(s) (video)
Endothermic
ExothermicS u r r o u n d I n g s
S u
r r o
u n
d I
n g
s
System
System
Heat Heat
Heat Heat
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Enthalpy level diagrams
For You To DoDraw diagrams for the reactions on the previous slideYou 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
Reaction 2 is a straightforward neutralisation with a rH = -55 kJ mol-1
(g)H2NaOH(aq)O(l)2H2Na(s) 22
The reaction between sodium metal and water – metal floats on water – effervescent reaction moves metal around- yellow flame above the metal- no solid residue
2 mol Na(s) + 2 mol H2O(l)
Enth
alpy
, H
(kJ)
2 mol NaOH(aq) + 1 mol H2(g)
ΔH = -367.5 kJ (367.5 kJ of heat is released
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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.
1r22 molkJ65.1HΔ(s)Ca(OH)O(l)HCaO(s)
122 molkJ367.5ΔH(g)H2NaOH(aq)O(l)2H2Na(s)
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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.
-1r222 mol kJ 483.7- HΔ O(g)2H(g)O(g)2H
-1r222 mol kJ 571.7- HΔ O(l)2H(g)O(g)2H
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Two important rules
Thermochemical Equations
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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• 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
-1r222 mol kJ 571.7- HΔ O(l)2H(g)O(g)2H
Using Thermochemical Equations
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Using Thermochemical EquationsConsider 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?
(l)22(g)2(g)4(g) O2HCO2OCH 1oc kJmol890.3HΔ
kJ556gkJ55.6g10.0 1
Combustion of methane gives 55.6 kJ g-1
1 g of methane would give
11
1gkJ55.6molg16.0
molkJ890.3
10 g of methane would give
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•Measuring enthalpy changes is called calorimetry•Carry out the reaction in a calorimeter and measure the temperature change.•Calculate the energy transferred during the reaction from the temperature change.•Also require the mass of the substance and the specific heat capacity
AssumptionsAll the energy change is transferred to the solution (water) No losses of heat to the other surroundings
Measuring enthalpy changes
HCl(aq)
NaOH (aq)
Thermometer
2 polystyrene coffee cups
Coffee-cup calorimeter
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Need to knowSpecific Heat Capacity
DefinitionThe 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 exampleWater 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
O(l)HNaCl(aq)NaOH(aq)HCl(aq) 2
mcΔcQ
K12.0KgJ4.1850gQ 11 J2508Q
2508 J of heat is transferred from the reaction of 0.05 mol HCl with 0.05 mol NaOH
mol0.05mol2.0100025n
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An Example continued
For 1 mol of HCl and NaOH
kJ50.2J501600.052508ΔH
12 molkJ50.2ΔHO(l)HNaCl(aq)NaOH(aq)HCl(aq)
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Now Try This One0.327 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: Cu(s)(aq)ZnSO(aq)CuSOZn(s) 44
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Limitations of this method ??
HCl(aq)
NaOH (aq)
Thermometer
2 polystyrene coffee cups
Coffee-cup calorimeter
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A bomb calorimeter
How can we accurately measure enthalpy changes of combustion reactions?
NeedleGas inlet
Insulated jacketSteel bomb
+-
O2
ThermometerCurrent for ignition coil
Stirrer
Ignition coil
Graphite sample
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Bomb Calorimetry- Bomb Calorimetry- measurementsmeasurements
Some heat from reaction warms “bomb” qbomb = heat capacity x ∆T
Total heat evolved, qtotal = qwater + qbombTotal heat from the reaction =qtotal
NeedleGas inlet
Steel bomb
+-
O2
ThermometerCurrent for ignition coil
Stirrer
Ignition coil
Graphite sample
Some heat from reaction warms waterqwater = mc∆T
Insulated jacket
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Calculate enthalpy of combustion of Calculate enthalpy of combustion of octane. octane.
CC88HH18(l)18(l) + 25/2 O + 25/2 O2(g)2(g) 8CO8CO2(g)2(g) + + 9H9H22OO(l)(l)
• Burn 1.00 g of octaneBurn 1.00 g of octane• Temp rises from 25.00 to 33.20 Temp rises from 25.00 to 33.20 ooCC• Calorimeter contains 1200 g waterCalorimeter contains 1200 g water• Heat capacity of bomb = 837 J KHeat capacity of bomb = 837 J K-1-1
Calculating enthalpy changes from calorimetry data
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Step 1Step 1: energy transferred from reaction to : energy transferred from reaction to water.water.q = (4.184 J gq = (4.184 J g-1-1KK-1-1)(1200 g)(8.20 K) = 41170 J)(1200 g)(8.20 K) = 41170 J
Step 2Step 2: energy transferred from reaction to : energy transferred from reaction to bomb.bomb.
q = (bomb heat capacity)(∆T)q = (bomb heat capacity)(∆T) = (837 J K= (837 J K-1-1)(8.20 K) = 6860 J)(8.20 K) = 6860 J
Step 3Step 3:Total energy transferred:Total energy transferred 41170 J + 6860 J = 48030 J41170 J + 6860 J = 48030 J
Heat of combustion of 1.00 g of octane = - 48.0 Heat of combustion of 1.00 g of octane = - 48.0 kJkJFor 1 kg = -48 MJ kgFor 1 kg = -48 MJ kg-1-1
H=-48 kJ x 114 g mol-1=-5472 kJ mol-1
Calculating enthalpy changes from calorimetry data
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Video
Click to link to “Thermochemistry
” video
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A case study- Self-heating cans
Water
Can
InsertQuicklimeFoil separator
Button
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The ChemistryCaO(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 DiagramsHess’s Law & Energy Level DiagramsReaction can be shown as a single step or in a two steps. ∆Htotal is the same no matter which path is followed.
∆∆H reaction path H reaction path 1=1=∆H reaction ∆H reaction path 2path 2
C(s) + O2(g)En
ergy
CO2(g)
CO(g) + ½ O2(g)
ΔH3 =
ΔH
1 + Δ
H2 =
-393
.5 k
J
ΔH2 =
-283
.0 k
JΔH
1 = -1
10.5
kJ
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Standard enthalpy values (Ho)
1.Initial and and finalfinal species are in their species are in their standard statesstandard states
2.2.The The standard state of a substancestandard state of a substance at at a specified temperature is its pure form a specified temperature is its pure form at 1 bar (100 kPa)at 1 bar (100 kPa). . T T isis usually 298 K (25 usually 298 K (25 ooC) but not alwaysC) but not always
3.3. rrHHoo(298 K)(298 K) is the is the standard enthalpy standard enthalpy of reactionof reaction at 298 K at 298 K
Some Physical States at 298 KSome Physical States at 298 KC = graphite; OC = graphite; O22 = gas; CH = gas; CH4 4 = gas; H= gas; H22O = O =
liquidliquid
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Standard Enthalpy of FormationStandard Enthalpy of Formation∆∆ffHHoo (298 K) = standard (298 K) = standard molar enthalpy molar enthalpy of formationof formation at 298 K at 298 Kenthalpy change when 1 mole of enthalpy change when 1 mole of compound is formed from elements in compound is formed from elements in their standard states at 298 K their standard states at 298 K (These are available from data books)(These are available from data books)
HH22(g) + 1/2O(g) + 1/2O22(g)(g) HH22O(g)O(g)∆∆ffHHoo (H (H22O, g) = -241.8 kJ molO, g) = -241.8 kJ mol-1-1
∆∆ffHHoo is zero for elements in their is zero for elements in their standard states.standard states.
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Standard enthalpies of formation Standard enthalpies of formation and Hess’s Law can be used to and Hess’s Law can be used to calculate unknown ∆calculate unknown ∆rrHHoo ∆∆rrHHoo = ∆H = ∆Hff
oo (products) - ∆H(products) - ∆Hff
oo
(reactants)(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?
Calculating Enthalpy Changes using standard enthalpies of formation
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Calculate ∆Calculate ∆ccHHoo for methanol for methanolStandard state of methanol at 298 K is liquidCHCH33OH(l) + 3/2OOH(l) + 3/2O22(g)(g) COCO22(g) + 2H(g) + 2H22O(l)O(l) ∆ ∆ccHHoo = ∆H = ∆Hff
oo (products) - ∆H(products) - ∆Hff
oo (reactants)(reactants)
= {∆H= {∆Hffoo (CO(CO22) + 2 ∆H) + 2 ∆Hff
oo (H(H22O)} - {3/2 ∆HO)} - {3/2 ∆Hff
oo (O(O22) + ) +
∆H∆Hffoo(CH(CH33OH)} OH)}
= {(-393.5 kJ) + 2 (-285.8 kJ)} - {0 + (-238.9 kJ)}= {(-393.5 kJ) + 2 (-285.8 kJ)} - {0 + (-238.9 kJ)}
∆∆ccHHoo(298K)= -726.2 kJ mol(298K)= -726.2 kJ mol-1-1
Now try the problems on the separate sheet
Calculating Enthalpy Changes
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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)(2)(2)()()(3 332 ggssss CONSKCSKNO
fHo values/kJ mol-1
KNO3(s) -494.6 K2S(s) -380.70CO2(g) -393.51Calculate 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
fHo values/ kJ mol-1Ba(OH)2.8H2O(s) -3345NH4Cl(s) -314NH3(g) -46H2O(l) -286BaCl2(s) -859Calculate rHo
)(2)(2)(3)(4)(22 10228.)( slgss BaClOHNHClNHOHOHBa
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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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DefinitionThe 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 DCH4(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 fromCH4(g) C(g) + H(g) H=1664 kJ mol-1D (C-H) = 416 kJ mol-1
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Enthalpy Changes in Chemical ReactionsEnthalpy 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 processOnly an estimate because
•Bond enthalpies are mean values•All species are in the gaseous state
A Hess’s law problem
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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 ???
Enthalpy changes of combustion for hydrocarbons Compound CH4
Bond typeBond energy
Number broken
Energy in KJ mol-1
Number made
Energy out KJ mol-1
C-C 347 0 0 0 0C-H 413 4 1652 0 0C=O 805 0 0 2 -1610O=O 498 2 996 0 0O-H 464 0 0 4 -1856
Total energy in/kJ mol-1 2648
Total energy out/kJ mol-1 -3466
Tota enthalpy change of combustion kJ mol-1 -818
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Acknowledgements• JISC• HEA• Centre for Educational Research and Developmen
t• School of natural and applied sciences• School of Journalism• SirenFM• http://tango.freedesktop.org