Transcript FRS271 Introductory Chemistry

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
Heat
System
Endothermic
Heat
Heat
System
SurroundIngs
SurroundIngs
Heat
Energy products>reactants so energy is
absorbed from the surroundings
Heat is lost from the surroundings so
Temperature (surroundings) decreases Endothermic
e.g. Ba(OH)2.8H2O(s) and NH4Cl(s) (video)
Reaction 2
Energy products<reactants so energy flows as
heat from the system to the surroundings
Temperature (surroundings) increases Exothermic
Exothermic
e.g. NaOH(aq) and HCl(aq).
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Enthalpy level diagrams
2Na(s)  2H2O(l)  2NaOH(aq)  H2 (g)
For You To Do
Draw diagrams for the reactions on the
previous slide
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
Reaction 2 is a straightforward
neutralisation with a rH = -55 kJ mol-1
2 mol Na(s) + 2 mol H2O(l)
Enthalpy, H (kJ)
The reaction between sodium metal and
water – metal floats on water –
effervescent reaction moves metal
around- yellow flame above the metalno solid residue
ΔH = -367.5
kJ (367.5 kJ
of heat is
released
2 mol NaOH(aq) + 1 mol H2(g)
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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)  2H2O(l)  2NaOH(aq)  H2 (g) ΔH 367.5 kJmol1
CaO(s)  H2O(l)  Ca(OH)2 (s) ΔrH  65.1kJmol1
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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.
2H2 (g)  O2 (g)  2H2O(g) Δr H  - 483.7 kJ mol-1
2H2 (g)  O2 (g)  2H2O(l) Δr H  - 571.7 kJ mol-1
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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
2H2 (g)  O2 (g)  2H2O(l) Δr H  - 571.7 kJ mol-1
• 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?
CH4(g)  2O2(g)  CO2(g)  2H2O(l)
1 g of methane would
give
ΔcHo  890.3 kJmol1
 890.3 kJmol1
1


55.6
kJ
g
16.0 g mol1
Combustion of methane gives 55.6 kJ g-1
10 g of methane would
give
10.0g  55.6 kJg1  556kJ
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Measuring enthalpy changes
•Measuring enthalpy changes is called
calorimetry
•Carry out the reaction in a calorimeter
and measure the temperature change.
Thermometer
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
All the energy change is transferred to
the solution (water)
No losses of heat to the other
surroundings
NaOH (aq)
Coffee-cup calorimeter
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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) H2O(l)
Q  mcΔc
Q  50g 4.18 Jg1 K1  12.0K
Q  2508J
25
n
 2.0 mol 0.05 mol
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
 50160 J  50.2 kJ
0.05
HCl(aq)  NaOH(aq) NaCl(aq) H2O(l) ΔH 50.2kJmol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)  CuSO4 (aq)  ZnSO4 (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
jacket
O2
Steel
bomb
Ignition coil
Graphite sample
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Bomb Calorimetry- measurements
Thermometer
Current
+ for
ignition
coil
Stirrer
Needle
Gas inlet
Insulated
jacket
O2
Steel
bomb
Ignition
coil
Some heat from reaction
warms water
qwater = mc∆T
Some heat from reaction
warms “bomb”
qbomb = heat capacity x ∆T
Graphite
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
ΔH1 = -110.5 kJ
CO(g) + ½ O2(g)
ΔH2 = -283.0 kJ
ΔH3 = ΔH1 + ΔH2 = -393.5 kJ
Energy
C(s) + O2(g)
Reaction can be
shown as a single
step or in a two steps.
ΔHtotal is the same no
matter which path is
followed.
∆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)
ΔcHo =
ΔHfo (products) -
CO2(g) + 2H2O(l)
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:
2KNO3( s )  S( s )  3C( s )  K2 S( s )  N2( g )  3CO2( g )
fHo values/kJ mol-1
KNO3(s)
-494.6
CO2(g)
-393.51
K2S(s)
-380.70
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 .8H2O( s )  2NH 4Cl( s )  2NH3( g ) 10H2O(l )  BaCl2( s )
fHo values/ kJ mol-1
Ba(OH)2.8H2O(s)
NH4Cl(s)
NH3(g)
H2O(l)
BaCl2(s)
Calculate rHo
-3345
-314
-46
-286
-859
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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
Bond type
CH4
Bond
energy
Energy in KJ
mol-1
Number broken
Energy
out KJ
mol-1
Number
made
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
in/kJ mol-1
Total
energy
out/kJ
2648 mol-1
Tota enthalpy change of combustion kJ mol-1
-3466
-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
•
•
•
•
•
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•
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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