FITCH Rules
General
“A” students work
(without solutions manual)
~ 10 problems/night.
Office Hours W – F 2-3 pm
Module #11
Thermochemistry
Energy:
Chemistry
Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
G1: Suzuki is Success
G2. Slow me down
G3. Scientific Knowledge is Referential
G4. Watch out for Red Herrings
G5. Chemists are Lazy
C1. It’s all about charge
C2. Everybody wants to “be like Mike”
⎛ qq ⎞
⎛qq
C3. Size Matters
⎟ or = k ⎜
E = k⎜
⎝ d
⎝r +r ⎠
C4. Still Waters Run Deep
C5. Alpha Dogs eat first
1 2
1 2
el
1
2
⎞
⎟
⎠
capacity to do work
Galen, 170
w = ( F )d
Or transfer heat, q
Marie the Jewess, 300
Charles Augustin
James Watt
Coulomb 1735-1806 1736-1819
Jabir ibn
Hawan, 721-815
Luigi Galvani
1737-1798
Count Alessandro
Guiseppe
Antonio Anastasio
Volta, 1747-1827
q
w
Energy
Consumed
Depends on
Both work
And heat
Thomas Graham
Justus von Liebig1805-1869
(1803-1873
J. Willard Gibbs
1839-1903
Richard August
Carl Emil
Erlenmeyer
1825-1909
Ludwig Boltzman
1844-1906
Galileo Galili
1564-1642
An alchemist
Henri Louis
LeChatlier
1850-1936
Amedeo Avogadro
1756-1856
James Joule
(1818-1889)
Evangelista
Torricelli
1608-1647
John Dalton
1766-1844
Rudolph Clausius
1822-1888
Johannes Rydberg
1854-1919
J. J. Thomson
1856-1940
Abbe Jean Picard
1620-1682
William Henry
1775-1836
physician
William Thompson
Lord Kelvin,
1824-1907
Heinrich R. Hertz,
1857-1894
Daniel Fahrenheit
1686-1737
Jacques Charles
1778-1850
1825-1898
Johann Balmer
Max Planck
1858-1947
Blaise Pascal
1623-1662
Robert Boyle,
1627-1691
Georg Simon Ohm
1789-1854
Francois-Marie
Raoult
1830-1901
Svante Arrehenius
1859-1927
Michael Faraday
1791-1867
James Maxwell
1831-1879
Walther Nernst
1864-1941
Isaac Newton
1643-1727
B. P. Emile
Clapeyron
1799-1864
Dmitri Mendeleev
1834-1907
Fritz Haber
1868-1934
Anders Celsius
1701-1744
Germain Henri Hess
1802-1850
Johannes D.
Van der Waals
1837-1923
Thomas Martin
Lowry
1874-1936
Fitch Rule G3: Science is Referential
∆ E = E final − E initial = q + w
Gilbert Newton Lewis
1875-1946
Johannes Nicolaus
Bronsted
1879-1947
Lawrence J. Henderson
1878-1942
Niels Bohr
1885-1962
Erwin Schodinger
1887-1961
Louis de Broglie
(1892-1987)
Friedrich H. Hund
1896-1997
Fritz London
1900-1954
Wolfgang Pauli
1900-1958
Werner Karl
Heisenberg
1901-1976
Linus Pauling
1901-1994
1
Properties and Measurements
Property
Size
Volume
Weight
Unit
m
cm3
gram
Reference State
size of earth
m
mass of 1 cm3 water at specified Temp
(and Pressure)
oC, K
boiling, freezing of water (specified
Pressure)
amu
(mass of 1C-12 atom)/12
atomic mass of an element in grams
atm, mm Hg
earth’s atmosphere at sea level
Temperature
1.66053873x10-24g
quantity
mole
Pressure
Energy, General
Animal hp
heat
BTU
calorie
Kinetic J
Electrostatic
electronic states in atom
Electronegativity F
Heat flow measurements
horse on tread mill
1 lb water 1 oF
1 g water 1 oC
m, kg, s
1 electrical charge against 1 V
Energy of electron in vacuum
Reference state?
To set a “heat flow” scale
1.
Defined conditions: how experiment is performed
open flask, closed flask, pressure
2.
Define the direction of heat flow by giving
a positive or negative number
We would get a different answer if we asked
“Does the “system (sun)” gain energy?”
For this question: the
system (sun) loses heat
+ heat
to the surroundings (earth)
- heat?
To set a “heat flow” scale
1.
Defined conditions: how experiment is performed
open flask, closed flask, pressure
2.
Define the direction of heat flow by giving
a positive or negative number
Does the “system (earth)” gain energy?
+ heat
- heat?
For image above
system (earth) gains heat
from surroundings (sun)
First Law of Thermodynamics: Energy is conserved
No universal change in energy
Just a transfer of energy
2
To set a “heat flow” scale
surroundings
1.
Defined conditions: how experiment is performed open
flask, closed flask, pressure
system
2.
Define the direction of heat flow (q) by giving a positive
or negative number
q is + when heat flows into the system from the
surroundings
3.
q =?
q<0 system
q>0 system
E = constant when System AND surroundings considered!
Properties and Measurements
Property
Size
Volume
Weight
Temperature
1.66053873x10-24g
quantity
mole
Pressure
Energy: Thermal
Unit
m
cm3
gram
Reference State
size of earth
m
mass of 1 cm3 water at specified Temp
(and Pressure)
oC, K
boiling, freezing of water (specified
Pressure)
amu
(mass of 1C-12 atom)/12
atomic mass of an element in grams
atm, mm Hg
earth’s atmosphere at sea level
BTU
calorie
Kinetic J
Energy, of electrons
Electronegativity
Heat Flow
q is - when heat flows out of the system into the
surroundings
Chemical process in the “system” is defined by heat
flow
endothermic q>0
exothermic q<0
Chemical reactions involve Zn( s) + 2 H + (aq ) → Zn 2 + (aq ) + H 2 ( g )
1. heat exchange
As a review:
Heat exchange Constant
At constant
Atm.pressure
Pressure
1 lb water 1 oF
1 g water 1 oC
2kg mass moving at 1m/s
who is oxidized?
who is reduced?
what is the oxidation number on H2?
Who is an oxidizing agent?
H = Greek:
enthalpy
thalpein – to heat
en - in
H for (?) heat
energy of electron in a vacuum
F
into system = +
qP = ∆ H
Subscript
Reminds us that
Pressure is constant
1 atm pressure = constant pressure
This means heat flow, q, is enthalpy change
3
Chemical reactions involve Zn( s) + 2 H + (aq ) → Zn 2 + (aq ) + H 2 ( g )
1. heat exchange
2. work
PressureVolume
work
constant Atm. pressure
∆V
∆ E = E final − E initial = q + w
∆ E P = q P + ( − P∆ V )
∆ E P = ∆ H + ( − P∆ V )
This is a form of Rule G3
Science is referential
Generally final - initial
∆ E P = ∆ H − P∆ V
w = − P∆ V
Change in volume is typically small for
Most reactions
∆ EP ≈ ∆ H
This means we can measure the energy change of
A chemical reaction by measuring the heat exchange
At constant pressure
3 to 8 standard cubic feet of
biogas per pound of manure.
The biogas usually contains 60
to 70% methane.
five Navy Avengers
disappeared in the Bermuda
Triangle on Dec. 5, 194
First example problem will involve methane
We will prove to ourselves that the Pressure-Volume work is a
Small contribution to the total energy change
Methane
Gas
Recovery
At landfills
4
Consider the contribution of volume of gas phase molecules
Consider the contribution of volume change for water in this reaction
CH 4 ( g ) + 2O2 ( g ) → CO2 ( g ) + 2 H 2 O( l )
CH 4 ( g ) + 2O2 ( g ) → CO2 ( g ) + 2 H 2 O( l )
PV = nRT
g ⎤ ⎡ 1cm water ⎤ ⎡ 1L ⎤
*
*
= 0.036 L
[2moleH O ] * ⎡⎢⎣ 18
mol ⎥⎦ ⎢⎣ 1gwater ⎥⎦ ⎢⎣ 10 cm ⎥⎦
3
At constant T:
P( ∆ V ) = ( ∆ n) RT
2
(
)
(l )
3
P( ∆ V ) = n gas final − n gas initial RT
%&$*! Conversions – if
Interested see next slide
PV = [1atm][ 0.036 L]
L ⋅ atm ⎞
⎛
P( ∆ V ) = (1 − 3moles) ⎜ 0.0821
⎟ 298 K
⎝
mol ⋅ K ⎠
P( ∆ V ) = − 48.9316 L ⋅ atm
Most reactions total (q):
PV 1 mole gas
PV 2mole liquid water
%&$*! Conversions – if
Interested see slide after next
.
kJ ⎞
⎛ 01013
P( ∆ V ) = ( − 48.9316 L ⋅ atm) ⎜
⎟ = − 4.9 kJ
⎝ L ⋅ atm ⎠
2mole change
Optional Slide: conversion
⎛ ⎛ kg ⎞ ⎞
⎟⎟
⎜⎜
01013
.
kJ
= 0.0036kJ
L − atm
Energy in kJ
~ 1000
kJ
~
2.5
kJ
~
0.0036 kJ
⎜
⎝
⎟
⎠
By the end of
This module
We will see this
Is “true”
Sig fig tells us that PV energy small compared to q
To set a “heat flow” scale
1.
Defined conditions: how experiment is performed
Constant Pressure
2.
But not on the path taken (state property)
⎛
⎜
⎞
⎟
Heat flow
depends
the conditions
3
3
3
..
105 Pa
Pa ⎞⎞ ⎝ m ⋅ s 2 ⎠
⎛⎛ 101325
J
101325
xx10
⎟ ⎛ kJ ⎞
(( atm
⎟ ( L) ⎛⎜ 10 cm ⎞⎟ ⎛⎜ 1m ⎞⎟ ⎜
atm)) ⎜⎜⎝
kJ
.
⎜
⎟ = 0101325
⎟⎟⎠ ⎜
atm
L ⎠ ⎝ 10 2 cm ⎠ ⎜ ⎛ kg ⋅ m 2 ⎞ ⎟ ⎝ 10 3 J ⎠
⎝
⎝
atm
⎠ ⎜ Pa ⎟
5
3
3
⎜⎜ ⎜
⎝⎝
s2
⎟ ⎟⎟
⎠⎠
⎫
⎧
⎛
⎞
⎛ ⎛ kg ⎞ ⎞
⎜
⎟
⎪
⎪
⎜⎜
3
5
3
3
2⎟ ⎟
⎝
⎠
0101325
.
kJ ⎪ ⎛ 101325
.
x10 Pa ⎞ ⎜ m ⋅ s ⎟ ⎛ 10 cm ⎞ ⎛ 1m ⎞ ⎜
J
⎟ ⎛ kJ ⎞ ⎪
= ⎨⎜
⎜
⎟
⎟
⎜
⎟⎜ 2 ⎟ ⎜
⎬
3
( atm)( L)
L ⎠ ⎝ 10 cm ⎠ ⎛ kg ⋅ m 2 ⎞ ⎟ ⎝ 10 J ⎠ ⎪
atm
⎠ ⎜ Pa ⎟ ⎝
⎪⎝
⎜⎜ ⎜
⎜
⎟
⎟ ⎟⎟
2
⎪
⎪
⎝
⎠
⎝⎝ s ⎠⎠
⎭
⎩
H= enthalpy
q reaction
cons tan tpressure
= ∆ H = H products − H reac tan ts
5
Enthalpy is a state property
(measured under constant pressure, but how measured under that
constant pressure is not important)
Enthalpy is an “extensive” property
q = ∆H > 0
Depends upon the amount present
endothermic
CH 4 ( g ) + 2O2 ( g ) → CO2 ( g ) + 2 H 2 O( l )
H products > H reac tan ts
H 2 O( s ) + heat ⎯⎯→ H 2 O
Think of heat as a reactant
q = ∆H < 0
890kJ of heat is released when 1 mole of methane
Reacts with oxygen
− 890kJ
1moleCH 4 ( g )
exothermic
H products < H reac tan tss
CH 4 ( g ) + 2O2 ( g ) → CO2 ( g ) + H 2 O( l ) + heat
Think of heat as a product
Properties and Measurements
Property
Size
Volume
Weight
Temperature
∆ H = − 890kJ
Unit
m
cm3
gram
Reference State
size of earth
m
mass of 1 cm3 water at specified Temp
(and Pressure)
oC, K
boiling, freezing of water (specified
Pressure)
amu
(mass of 1C-12 atom)/12
atomic mass of an element in grams
atm, mm Hg
earth’s atmosphere at sea level
1.66053873x10-24g
quantity
mole
Pressure
Energy, General
Animal hp
heat
BTU
calorie
Kinetic J
Electrostatic
electronic states in atom
Electronegativity F
Heat flow measurements
horse on tread mill
1 lb water 1 oF
1 g water 1 oC
m, kg, s
1 electrical charge against 1 V
Energy of electron in vacuum
constant pressure, define system vs surroundin
per mole basis (intensive)
⎛ − 890kJ ⎞
⎜⎜
⎟⎟ 2moleCH 4 ( g ) = − 1780kJ
⎝ 1moleCH 4 ( g ) ⎠
(
)
Context for the next example problem
2006 Sept Sci. Am: World Wide Petroleum Usage
Land People Transport
Non29% of total use
transportation
Total transportation
=53%
Land Freight 19%
Air People and Freight 5%
U.S. differs from world in distribution of petroleum use
Transportation = 71.8%
3 Non-transportation
57.8% of Transportation=
Personal land transport
= 41.5% of total U.S. consumption
Data US DOE 2006
6
Example: If you drove an automobile 1.50x102 miles at 17.5 miles/gal
you consume a certain number of gallons of gasoline. If you burn that
number of gallons of gasoline at constant pressure how much heat
would be released? Assume the gasoline is pure octane with a density
of the octane 0.690 g/mL?
∆ H = − 109
. x10 4 kJ
2C8 H18 + 25O2 → 16CO2 ( g ) + 18 H 2 O( g )
Strategy: need moles of octane consumed (Golden Bridge)
miles
mpg
density
gallons
grams
Molar mass
∆H
moles
heat
690gC
gC88H
H88 ⎞⎞ ⎛⎛ 11moleC
moleC88H
H88 ⎞⎞ ⎛ − 109
gal ⎞⎞ ⎛⎛ 33..7852
7852LL⎞⎞ ⎛⎛ 10
1033mL
mL⎞⎞ ⎛⎛ 00..690
. x10 4 kJ ⎞
⎛⎛ 11gal
((150
⎟⎟ ⎜⎜
⎟⎟ ⎜
⎟ =
150..mi
mi))⎜⎝⎜
⎜⎜
⎟⎟ ⎜⎜
⎟⎠⎟ ⎜⎜
⎟
⎟
⎝ 17.5mi ⎠
17..55mi
mi ⎝⎝
17
gal ⎠⎠ ⎝⎝
gal
LL ⎠⎠ ⎝⎝ mLC
mLC88H
H88 ⎠⎠ ⎝⎝ 114
114gC
gC88H
H88 ⎠⎠ ⎝ 2moleC8 H8 ⎠
∆ H = − 1,070,243.955kJ
∆ H = − 107
. x10 kJ 3 sig fig
6
We will use part of this problem
Again:
⎛ 3.7852 L ⎞ ⎛ 10 3 mL ⎞ ⎛ 0.690 gC8 H8 ⎞ ⎛ 1moleC8 H8 ⎞ ⎛ − 109
. x10 4 kJ ⎞ − 124,861kJ
⎟⎜
⎟⎜
⎟ =
⎟⎜
⎜
⎟⎜
gal
⎝ gal ⎠ ⎝ L ⎠ ⎝ mLC8 H8 ⎠ ⎝ 114 gC8 H8 ⎠ ⎝ 2moleC8 H8 ⎠
Rules
1.Enthalpy is an extensive property (depends upon number of moles)
2.Enthalpy change for a reaction is equal in magnitude, but opposite
in sign, to the enthalpy for the reverse reaction
∆ H = − 109
. x10 4 kJ
∆ H = + 109
. x10 4 kJ
ENERGY measurement
a)
change in temperature
b)
some function specific to the material and how
it is organized (bonds)
2C8 H18 + 25O2 → 16CO2 ( g ) + 18 H 2 O( g )
16CO2 ( g ) + 18 H 2 O( g ) → 2C8 H18 + 25O2
3. Enthalpy change depends upon the state of the reactant and
products
∆ H = + 44 kJ
H2 O( l ) → H 2 O( g )
7
ENERGY change is a function of
a) temperature
b) material E = k ⎛⎜ q1q 2 ⎞⎟
ENERGY measurement
a)
change in temperature
b)
some function specific to the material and how
it is organized (bonds)
Pure material
⎝ d ⎠
el
O
H
H
H
O
H
O
H
H
H25
H
O
q = m⋅ c⋅ ∆t
C = heat capacity = heat required to raise the temperature of
the system 1oC
H
ch arg e density ≈
25 oC
q
r
Electron density of water
Shape and charge distribution
On water
C(liquidH2O)=4.18J/g-oC
Liquid water is very strongly organized
due to the polarity of the molecule, so it
has a high specific heat
m = mass
c = specific heat of a pure substance
H
H
H
heat
http://www.lsbu.ac.uk/water/molecule.html
q = C∆ t
)
O
O
H
H
O
H
15 oC
H
H
(
q = C t final − t initial
H
O
O
O
O
H
O
H
H
H
H
H
Electrons on Oxygen sit
“out there” causing large
Electrostatic potential
Oriented on the electrons
units: J/oC
Specific heats, c, of various substances in various physical states
Material
Pb(s)
Pb(l)
Cu(s)
Fe(s)
Cl2(g)
C(s)
CO2(g)
NaCl(s)
Al(s)
C6H6(l)
H2O(g)
C2H5OH(l)
H2O(l)
Specific Heats c, (J/g-K)
0.12803
0.16317
0.382
0.446
0.478
0.71
0.843
0.866
0.89
1.72
1.87
2.43
4.18
Ability to store
heat in a substance
is variable.
8
For the same amount of energy, easier to break
electrostatic attraction of He compared to water with it’s
localized charge
Example: 1.00 cup of water is heated from 25.0 oC to 100.0 oC. How
many joules were used to heat the water?
H
O
O
H
He
He
H
H
O
Mass of water:
He
O
H
H
q = m⋅ c⋅ ∆t
H
H
H
⎤ ⎡ 10 3 mL ⎤ ⎡ 1g ⎤
⎥⎢
⎥ = 236.51g
⎦ ⎣ ⎦ L ⎦ ⎣ 1mL ⎦
⎣ ⎣⎣44cups
⎦ ⎣ ⎦⎦⎣ 1057
. qt
cups
⎡ 1qt ⎤ ⎡
∆ t = Tfinal − Tinitial
He
He
O
1L
⎡ 11qt
qt ⎤⎤⎡ 1L ⎥ ⎢ ⎤
. 100
[100
]⎢ 4]]⎡⎢cups
.100
cup
.cup
[[[100
. cup
cup
⎥⎥.⎢ qt ⎥
]⎢ ⎥ ⎢ 1057
O
H
H
He
∆ t = 100 o C − 25o C
∆ t = 75K
∆ t = 75o C
⎡ 4.18 J ⎤
q = 236.51g ⎢
⎥ 75K
⎣ gK ⎦
H
q = 74148.53
q = 7.41x10 4 J
q = 74.1kJ
Example 2: 1 cup of dry soil (specific heat, c = 0.800 J/gK;
density =1.28 g/cm3). Calculate the Joules required to raise the
temperature of the dry soil from 25oC to 100 oC.
⎡ 1qt ⎤ ⎡
1L
⎤ ⎡ 10 3 mL ⎤ ⎡ 128
. g⎤
. = 302.7 g
⎥
⎢
⎥ = (236.51g )128
⎦ ⎣ L ⎦ ⎣ 1mL ⎦
. cup]⎢
[100
4cups ⎥ ⎢ 1057
. qt ⎥ ⎢
⎣
⎦⎣
⎡ 0.8 J ⎤
q = 302.7 g ⎢
⎥ 75K
⎣ gK ⎦
Lag
Water
Land
q dry soil = 18,164 J = 181
. kJ
q water = 74.1kJ
Because water has a high heat capacity it takes longer than air or soil to warm up and longer to
cool down
9
1. Hot air rises over land
Land
1. Hot air rises
over Lake
2. Cold air from Lake
moves to fill in (Lake
Breeze)
Lake
2. Cold air moves
in from land (Land breeze)
Heat capacity of large bodies
Of water affect human activity
The temperature change in 24 hours in summer for water is not much
leading to big differences between lake and land and thus a lake breeze
Example 3: Heat capacity of the metal block of a car combustion engine.
Assume that a Prius 1.5 L, 4 cylinder 176.6 kg engine block is made of
iron. The specific heat of iron is 440 J/kg-C. If I drive 8 miles twice a
day (to work and back) at an average 42 mpg, what fraction of the total
available enthalpy in 1 gallon of octane is consumed in heating the
engine block from 25oC to 100oC?
⎛ 440 J ⎞
q = (200kg )⎜
⎟ (100 − 25 C ) = 5,828kJ
⎝ kg ⋅ C ⎠
q = m⋅ c⋅ ∆t
⎛ 42miles ⎞ ⎛ trip ⎞ ⎛ 5,828kJ engine heating ⎞ ⎛ 27,730kJ heating ⎞
⎟⎜
⎜
⎟⎜
⎟ =⎜
⎟
1trip
gal
⎝ 1gal ⎠ ⎝ 8miles ⎠ ⎝
⎠ ⎝
⎠
Heating the engine block
Compare to heat available from combustion of 1 gallon of octane (from before
∆ H = − 109
. x10 4 kJ
2C8 H18 + 25O2 → 16CO2 ( g ) + 18 H 2 O( g )
⎛ 3.7852 L ⎞ ⎛ 10 mL ⎞ ⎛ 0.690 gC8 H8 ⎞ ⎛ 1moleC8 H8 ⎞ ⎛ − 109
. x10 4 kJ ⎞ − 124,861kJ
⎟ =
⎟⎜
⎟⎜
⎟⎜
⎟⎜
⎜
gal
⎝ gal ⎠ ⎝ L ⎠ ⎝ mLC8 H8 ⎠ ⎝ 114 gC8 H8 ⎠ ⎝ 2moleC8 H8 ⎠
3
⎛ 5,828kJ ⎞
⎜
⎟ 100 = 22.2%
⎝ 124,861kJ ⎠
Let’s compare to what
I measure
Prius 1
⎛ 39 − 31mpg ⎞
⎜
⎟ 100 = 20.5%
⎝ 39mpg ⎠
Prius 2
⎛ 43 − 34mpg ⎞
⎜
⎟ 100 = 20%
⎝ 43mpg ⎠
10
Properties and Measurements
Property
Size
Volume
Weight
Unit
m
cm3
gram
Reference State
size of earth
m
mass of 1 cm3 water at specified Temp
(and Pressure)
oC, K
boiling, freezing of water (specified
Pressure)
amu
(mass of 1C-12 atom)/12
atomic mass of an element in grams
atm, mm Hg
earth’s atmosphere at sea level
Temperature
1.66053873x10-24g
quantity
mole
Pressure
Energy, General
Animal hp
heat
BTU
calorie
Kinetic J
Electrostatic
electronic states in atom
Electronegativity F
Heat flow measurements
horse on tread mill
1 lb water 1 oF
1 g water 1 oC
m, kg, s
1 electrical charge against 1 V
Energy of electron in vacuum
q reaction = − q calorimeter
If the temperature of the water rises (heat flow into water)
then heat must have been lost from the reaction
constant pressure, define system vs surroundin
per mole basis (intensive) Calorimetry
Example: When 1.00 g of ammonium nitrate, NH4NO3, is
added to 50.0 g of water in a coffee-cup calorimeter, it
dissolves:
NH 4 NO3( s) → NH
+
4 ( aq )
−
3( aq )
+ NO
and the temperature of the water drops from 25.00 to 23.32
oC. Assuming that all the heat absorbed by the reactions
comes from the water, calculate q for the reaction system.
q calorimeter = m ⋅ c ⋅ ∆ t
⎡
J ⎤
q calorimeter = [50.0 g ]⎢ 4.18
[ 23.32 oC − 25.00o C]
gK ⎥⎦
⎣
q calorimeter = − 351J
q reaction = − q calorimeter
q reaction = − ( − 351J )
J⎤
⎡
q calorimeter = ⎢ 209 ⎥ [ − 168
. K]
q reaction = 351J
K⎦
⎣
A 1 degree change in Celsus
q calorimeter = − 35112
. J
Relate reaction heat
To the calorimeter heat
Not all calorimeters can be based on
q reaction = − q calorimeter = − m ⋅ c ⋅ ∆ t
Heat capacity of the calorimeter
{[ ] }
q reaction = − q reaction = − Ccal ∆ t
In some problems
You will
Need to determine
This number
In one step and
Then go on
Is a 1 degree change in Kelvin
11
Example: The reaction between hydrogen and chlorine
H g ( g ) + Cl2 ( g ) → 2 HCl( g )
can be studied in a bomb calorimeter. It is found that when a 1.00 g
sample of H2 reacts completely, the temperature rises from 20.00 to
29.82 oC. Taking the heat capacity of the calorimeter to be 9.33 kJ/oC,
calculate the amount of heat evolved in the reaction.
[ ]
q reaction = − Ccal ∆ t
Heat evolved?
q
Calorimeter heat capacity = 9.33 kJ/oC
Tinitial =20.00 oC
T final = 29.82 oC
kJ ⎤
⎡
q reaction = − ⎢ 9.33 o ⎥ [ 29.82 o C − 20.00 o C ]
C⎦
⎣
q reaction = − 916
. kJ
Example: Salicyclic acid, C7H6O3, is one of the starting materials in the
manufacture of aspirin. When 1.00 g of salicylic acid burns in a bomb
calorimeter, the temperature rises to 32.11oC from 28.91 oC. The
temperature in the bomb calorimeter increases by 2.48oC when the
calorimeter absorbs 9.37 kJ. How much heat is given off when one mole
of salicylic acid is burned?
⎡ 9.37 kJ ⎤
( 32.11 − 29.91)
q reaction = − ⎢
o
1.00 g of C7H6O3
⎣ 2.48 C ⎥⎦
o
Tinitial 32.11 C
Tfinal 29.91oC
q reaction = − 8.3121kJ
9.37kJ required to cause 2.48 oC change
Ccal =
⎛
⎞⎛
⎞
[ ]
gC H O
− 8.3121kJ
⎟⎜
⎟ = − 3158598
kJ
.
(1moleC H O )⎜⎝ 1138
moleC H O ⎠ ⎝ 1gC H O ⎠
???
= − 316kJ
q reaction = − Ccal ∆ t
7
6
7
6
3
7
6
3
3
7
6
3
9.37 kJ
2.48 o C
Example 2 What is the enthalpy change for the reaction
Combine
∆H
NH 4 NO3( s) → NH 4+( aq ) + NO3−( aq )
If exactly 1 g of ammonium nitrate is reacted in a bomb calorimeter
made with 50 g of water and the temperature of the water drops from
25.00 oC to 23.32 oC?
2C8 H18 + 25O2 → 16CO2 ( g ) + 18 H 2 O( g )
calorimetry
Reaction stoichiometry
To get reaction enthalpies
⎡
J ⎤
q reaction = − q calorimeter = − [50.0 g ]⎢ 4.18 o ⎥ [ 23.32 o C − 25.00 o C]
g C⎦
⎣
J ⎤
⎡
q reaction = ⎢ − 209 o ⎥ [ − 168
. o C]
C⎦
⎣
q reaction = 35112
. J
q reaction = 351J
q reaction
cons tan tlabpressure
1 g was reacted
⎡ 351J ⎤ ⎡ 80.05g ⎤ 28100 J 281
. kJ
⎢ 1g ⎥ ⎢ mol ⎥ = mol = mol
⎦
⎦⎣
⎣
2N=28.02
4H=4.04
3O=48.00
80.05
= ∆ H = 351J
NH 4 NO3( s) → NH 4+( aq ) + NO3−( aq )
∆ H = + 281
. kJ
12
NH 4 NO3( s) → NH 4+( aq ) + NO3−( aq )
∆ H = + 281
. kJ
Where did the per/mole go?
1 NH 4 NO3( s )
→ 1 NH 4+( aq ) + 1NO3−( aq )
Thermochemical Equation Rules
The reaction was written as a per/mole
Enthalpy is understood as a per/mole of reactant (or as the reaction
is written)
∆ H = + 281
. kJ
When there are no
Coefficients it is understood that
It is “1”
1. Value of )H applies when products and reactants are at same
temperature, 25oC unless otherwise specified.
2. Sign of )H, indicates whether reaction, when carried out at
constant pressure, is exothermic or endothermic
3. ∆H sign changes when reaction is reversed
NH 4+( aq ) + NO3−( aq ) → NH 4 NO3( s )
∆ H = − 281
. kJ
4. Stoichiometry is important
4. Phases of all species must be specified
5. Values of )H is same regardless of method used to calculate it
(Hess’s Law)
Example illustrating importance of phases
Using a coffee-cup calorimeter, it is found that when an ice cube
weighing 24.6 g melts, it absorbs 8.19 kJ of heat. Calculate
for the phase change represented by the thermochemical equation
H 2 O( s ) → H 2 O( l )
1
H 2 ( g ) + 2 O2 ( g ) → H 2 O( l )
Calculate the enthalpy for the equation above given that:
∆ H = + 5716
. kJ
⎡ 819
. kJ ⎤ ⎡ 18.02 g H2 O ⎤
⎥ 1mole H2O = 6.00kJ
⎥⎢
⎢
⎣ 24.6 gice ⎦ ⎢⎣ mole H2 O ⎥⎦
[
An example of several of the rules using Fuel Cells
Fuel cells use the reaction:
]
2 H 2 O( l ) → 2 H 2 ( g ) + O2 ( g )
Reverse reaction:
∆ H = − 5716
. kJ
2 H 2 ( g ) + O2 ( g ) → 2 H 2 O( l )
scale
∆H =
− 5716
. kJ
= − 286kJ
2
1
H 2 ( g ) + 2 O2 ( g ) → H 2 O( l )
Here we got a number by coming “at it” from an odd direction
13
Hess’s law
Galen, 170
The value of )H for a reaction is the same whether it occurs in one
step or in a series of steps (enthalpy (constant P, T) is a state
function)
Marie the Jewess, 300
Charles Augustin
James Watt
Coulomb 1735-1806 1736-1819
Thomas Graham
Justus von Liebig1805-1869
(1803-1873
Germain Henri Hess
1802-1850
born in Geneva Switzerland
Professor of Chemistry
At St. Petersburg Technological Institute
Example of how Hess’s law is useful
1
2
C( s) + O2 ( g ) → CO( g )
It is difficult to measure the heat evolved for this reaction because it
occurs as the partial burning of carbon in the presence of other reactions
involving the complete burning of carbon
C(s) + O2(g)
This is the
Number we want
CO(g) +1/2O2(g)
But can’t actually measure
Related to 2CO + O2….
Get to the number by an alternative
Path (State function!)
To solve rearrange equations to get CO on right hand side
CO2(g)
∆ H = − 3935
. kJ
C( s) + O2 ( g ) → CO2 ( g )
∆ H = − 566.0kJ
2CO( g ) + O2 ( g ) → 2CO2 ( g )
J. Willard Gibbs
1839-1903
Gilbert Newton Lewis
1875-1946
Jabir ibn
Hawan, 721-815
Luigi Galvani
1737-1798
Count Alessandro
Guiseppe
Antonio Anastasio
Volta, 1747-1827
Richard August
Carl Emil
Erlenmeyer
1825-1909
Ludwig Boltzman
1844-1906
Johannes Nicolaus
Bronsted
1879-1947
Galileo Galili
1564-1642
An alchemist
Henri Louis
LeChatlier
1850-1936
Amedeo Avogadro
1756-1856
James Joule
(1818-1889)
Evangelista
Torricelli
1608-1647
John Dalton
1766-1844
Rudolph Clausius
1822-1888
Johannes Rydberg
1854-1919
J. J. Thomson
1856-1940
Abbe Jean Picard
1620-1682
William Henry
1775-1836
physician
William Thompson
Lord Kelvin,
1824-1907
Heinrich R. Hertz,
1857-1894
Daniel Fahrenheit
1686-1737
Blaise Pascal
1623-1662
Jacques Charles
1778-1850
1825-1898
Johann Balmer
Max Planck
1858-1947
Robert Boyle,
1627-1691
Georg Simon Ohm
1789-1854
Michael Faraday
1791-1867
James Maxwell
1831-1879
Francois-Marie
Raoult
1830-1901
Svante Arrehenius
1859-1927
Walther Nernst
1864-1941
Isaac Newton
1643-1727
B. P. Emile
Clapeyron
1799-1864
Dmitri Mendeleev
1834-1907
Fritz Haber
1868-1934
Anders Celsius
1701-1744
Germain Henri Hess
1802-1850
Johannes D.
Van der Waals
1837-1923
Thomas Martin
Lowry
1874-1936
Fitch Rule G3: Science is Referential
Lawrence J. Henderson
1878-1942
Niels Bohr
1885-1962
Erwin Schodinger
1887-1961
Louis de Broglie
(1892-1987)
Friedrich H. Hund
1896-1997
Fritz London
1900-1954
Wolfgang Pauli
1900-1958
∆ H = − 566.0kJ
2CO( g ) + O2 ( g ) → 2CO2 ( g )
∆ H = + 566.0kJ
+ 566.0kJ
∆H =
2
2CO2 ( g ) → 2CO( g ) + O2 ( g )
∆ H = + 283.0kJ
CO2 ( g ) → CO( g ) + 2 O2 ( g )
∆ H = 0− 3935
. kJ
∆ H = − 3935
. kJ
C( s ) + O2 ( g ) → CO2 ( g )
C( s ) + O2 ( g ) → CO2 ( g )
= + .283
∆ H∆ H
= − 110
5kJ.0kJ
11
CO( g )( g+) O2 ( g )
C( s) CO
+ 22 O
2 ( g22)((gg→
)) → CO
0
∆ H 0 = − 110.5kJ
Werner Karl
Heisenberg
1901-1976
Linus Pauling
1901-1994
1
CO2 ( g ) → CO( g ) + 2 O2 ( g )
1
1
C +O
C( s) + 2 O2(s)( g ) →2(g)CO( g )
CO(g) +1/2O2(g)
CO2(g)
14
Enthalpies of Formation
Rather than getting the enthalpy for each reaction from a bomb
calorimeter use a smaller number of standard reactions from which
Hess’s law can be applied to get all the remainder reactions of
interest
Enthalpy associated with standard reaction is
enthalpy of formation
which is the enthalpy change when one mole of compound is
formed at constant pressure of 1 atm and a fixed temperature,
ordinarily 25oC, from the elements in their stable states at that
pressure and temperature. STP (Standard Temperature and
Pressure)
This allows us to look at enthalpy of compounds
not reactions which reduces total data which must
Be acquired (Chemists are Lazy!!!)
Elements in their stable states at 1atm, 25oC have a standard
molar enthalpy of 0
1atm, 25C
O
− ∆ H Fe
( s)
∆ H Of = − 882 kJ
1
2
N 2 ( g ,1atm,25C ) + O2 ( g ,1atm,25C ) → 1NO2 ( g ,1atm,25C )
Standard molar enthalpy of formation of a compound
From elements in their stable states at 1 atm pressure
25oC
Most )Hfo are negative meaning that formation
of the compound from the elements is ordinarily
exothermic
Elements in their stable states at 1atm, 25oC have a standard
molar enthalpy of 0
Why?
∆ H Of = − 882 kJ
1
2
N 2 ( g ,1atm,25C ) + O2 ( g ,1atm,25C ) → NO2 ( g ,1atm,25C )
Standard molar enthalpy of formation of a compound
Fes → Fes
O
O
∆ H Fe
( s ) = ∆ H Fe ( s )
1 is “understood”
Invoke Rule G5: Chemists are Lazy
1atm, 25C
=0
Products (standard state) - Reactants (standard state) = 0
From elements in their stable states at 1 atm pressure
25oC
Most )Hfo are negative meaning that formation
of the compound from the elements is ordinarily
exothermic
For aqueous ions, the enthalpy is scaled relative to the
proton
∆ H Of H aq+ = 0
15
Do we detect any patterns?
∆H
∆ H Of kJ/mol
O
f
Fe(s)
0
0
Pb(s)
0
0
Li(s)
0
0
Mg(g)
0
0
Mn(s)
0
0
Hg(l)
0
0
Ni(s)
0
0
N2(g)
0
0
O
(s)
0
2
0
P4(s)
0
0
K(s)
0
0
Rb(s)
0
0
0 MOST metals are
0 elemental solids (metal) in
standard state
Al(s)
Ba(s)
Be(s)
Br2(g)
Ca(s)
C(s,graphite)
Cs(s)
Cl2(g)
Cr(s)
Cu(s)
F2(g)
H2(g)
H+(aq)
I2(s)
kJ/mol
∆H
kJ/mol
Sc(s)
Si(s)
Na(s)
S(s,rhombic)
Ti(s)
Zn(s)
0
0
0
0
0
0
Common non-metals
Have specific forms
In which they
Are standard
∑ n∆ H
o
f , products
−
∑ n∆ H
o
f ,reac tan ts
1 atm pressure
25oC
2 Al( s ) + Fe2 O3( s) → 2 Fe( s ) + Al2 O3( s)
} {
∆ H O = ∆ H of , Al2 O3 + 2 ∆ H of , Fe ( s ) − ∆ H of , Fe2O3 s + 2 ∆ H of , Al ( s )
∆H = ∆H
O
o
f , Al2 O3
− ∆H
}
o
f , Fe2 O3 s
2. Elements in standard states can be omitted because heats of
formation are zero
OJO!
Appendix
∆ H O = − 1669.8 − ( − 822.16)
∆ H O = − 1669.8 + 822.16 = − 847.65kJ / mol
Unit
m
cm3
gram
Reference State
size of earth
m
mass of 1 cm3 water at specified Temp
(and Pressure)
oC, K
boiling, freezing of water (specified
Pressure)
amu
(mass of 1C-12 atom)/12
atomic mass of an element in grams
atm, mm Hg
earth’s atmosphere at sea level
Temperature
1.66053873x10-24g
quantity
mole
Pressure
Energy, General
electronic states in atom
Electronegativity
Heat flow measurements
Energy of electron in vacuum
F
constant pressure, define system vs surroundings
per mole basis (intensive)
Standard Molar Enthalpy
25 oC, 1 atm, from stable state
)Hfo Haq+ =0
Calculation of )Ho standard enthalpy change of hot and cold packs
1. The coefficients of products and reactants in the
thermochemical equation must be taken into account
{
Properties and Measurements
Property
Size
Volume
Weight
Group 17 with exception
Of I2 are gases in
Stable, standard state
Calculation of )Ho standard enthalpy change of a reaction
∆HO =
O
f
∆HO =
∑ n∆ H
o
f , products
−
∑ n∆ H
o
f ,reac tan ts
1 atm pressure
25oC
NH 4 NO3,s → NH 4+,aq + NO3−,aq
{
∆ H O = ∆ H of , NH + + ∆ H of , NO −
4 , aq
3 , aq
} − {∆ H
o
f , NH4 NO3 , s
∆ H = { − 132.5 + − 205.0} − { − 365.6}
O
∆ H O = { − 337.5} − { − 365.6} = + 28
Compares well to the calorimetry calc. (28.1kJ/mol)!
kJ
mol
}
Appendix
Compound
NH4NO3,s
NH4+,aq
NO3-aq
MgSO4,s
Mg2+aq
SO42-aq
Fe,s
O2,g
Fe2O3,s
∆Hfo, kJ/mol
-365.6
-132.5
-205.0
-1284.9
-466.8
-909.3
0
0
-1118.4
16
2+
2
MgSO4 ,s ⎯⎯⎯
⎯→ Mg aq
+ SO42,−aq
H O
{
}{
∆ H O = ∆ H of , Mg 2 + + ∆ H of ,SO2 − − ∆ H of , MgSO4 , s
, aq
4 aq
}
∆ H O = { − 466.8 + − 909.3} − { − 1284.9}
kJ
mol
Products are more stable, lower in the energy
well, than reactants
∆ H O = {− 1,3761
. } − { − 1284.9} = − 912
.
http://webmineral.com/data/Hexahydrite.shtml
Example: Calculate the )Ho for the combustion of one mole of
methane CH4 according to the equation
CH 4 ( g ) + 2O2 ( g ) → CO2 ( g ) + 2 H 2 O( g )
Given the standard enthalpies of formation at 25oC, 1 atm from Appendix
kJ/mol
O2(g)
0
∆HO =
n∆ H of , products −
n∆ H of ,reac tan ts
-393.5
CO2(g)
H2O(g)
-241.8
CH4(g)
-74.8
∑
∑
⎡
⎛
⎛
kJ ⎞
kJ ⎞ ⎤
⎟ + (1molCO2 )⎜ − 3935
⎟⎥
∆ H O = ⎢ (2molH 2 O)⎜ − 2418
.
.
molH
O
molCO
⎝
⎠
⎝
2
2⎠⎦
⎣
⎡
⎛
⎛
kJ ⎞ ⎤
kJ ⎞
⎟ + (1molCH 4 )⎜ − 74.8
⎟⎥
− ⎢ (2molO2 )⎜ 0
molCH
molO
⎝
⎠
⎝
2
24 ⎠ ⎦
⎣
http://www.edinformatics.com/math_science/info_water.htm
Liquid water
http://www.lsbu.ac.uk/water/hofmeist.html
http://biochempress.com/Files/IECMD_2003/IECMD_2003_027.pdf
Example: Calculate the )Ho for the combustion of one mole of
methane CH4 according to the equation
CH 4 ( g ) + 2O2 ( g ) → CO2 ( g ) + 2 H 2 O( g )
⎡
⎛
⎛
kJ ⎞ ⎤
kJ ⎞
⎟ + (1molCO2 )⎜ − 3935
⎟⎥
∆ H O = ⎢ (2molH 2 O)⎜ − 2418
.
.
molCO2 ⎠ ⎦
molH 2 O ⎠
⎝
⎝
⎣
⎡
⎛
⎛
kJ ⎞ ⎤
kJ ⎞
⎟ + (1molCH 4 )⎜ − 74.8
⎟⎥
− ⎢ (2molO2 )⎜ 0
molCH 24 ⎠ ⎦
⎝ molO2 ⎠
⎝
⎣
∆ H O = [ − 877.10kJ ] − [ − 74.8kJ ] = − 802.30kJ
Sig figs? ∆ H O = − 802.3kJ
Can also “reverse” the problem (inside out socks)
Example: Calculate the standard enthalpy of formation for octane
∆ H o = − 109
. x10 4 kJ
2C8 H18 + 25O2 → 16CO2 ( g ) + 18 H 2 O( g )
Given the standard enthalpies of formation at 25oC, 1 atm
kJ/mol
∆HO =
∆ H of , products −
∆ H of ,reac tan ts
O2(g)
0
CO2(g)
-393.5
∆ H O = − 109
. x10 4 kJ =
∆ H of , products
-241.8
H2O(g)
CH4(g)
-74.8
−
∆ H of ,reac tan ts
∑
∑
∑
∑
⎡
⎛
⎞
⎛ − 2418
kJ
. kJ ⎞ ⎤
⎟⎟ + 18molH 2 O( g ) ⎜⎜
⎟⎟ ⎥
− 109
. x10 4 kJ = ⎢ 16molCO2 ( g ) ⎜⎜ − 3935
.
molCO
molH
⎝
⎝
⎢⎣
2( g ) ⎠
2 O( g ) ⎠ ⎥
⎦
⎡
⎛ 0kJ ⎞ ⎤
⎛
⎞
kJ
⎟⎥
⎟ + (25molO2 )⎜
− ⎢ (2molC8 H18 )⎜ x
⎝ molO2 ⎠ ⎦
⎝ molC8 H18 ⎠
⎣
(
)
− 109
. x10 4 kJ = [( − 6296kJ ) + ( − 4352.4 kJ ) ] − [ 2 x + 0]
− 109
. x10 4 kJ = − 8704.8kJ − 2 x
(
)
2195.2 kJ = − 2 x
x = − 1097
.
x10 3 kJ
17
For covalent bonds, bond enthalpies depend on?
Bond Enthalpy
The change in enthalpy when 1 mole of bonds is broken in the gaseous
State.
Bond length
Overall structure of the molecule
Rule G3: Science is referential!
Br2 ( g ) → 2 Brg
∆ H = + 193kJ
Cl2 ( g ) → 2Cl g
∆ H = + 243kJ
Which has a stronger bond
Enthalpy?
Bond length is nice, but it doesn’t
Really relate to the Periodic table
AND it isn’t the whole story
Electronegativities?
Can explain some trends
??
Bond
Bond
Length
pm
Pauling's
E.N.
atomic
radii
(pm)
H--H
C--C
Cl-Cl
S-S
Br-Br
N-N
I--I
O-O
74
154
199
205
228
145
267
148
2.2
2.5
3.2
2.6
3
3
2.7
3.5
37
77
99
104
114
70
133
66
Enthalpy
Single Bond
kJ/mol
(Average)
436
348
243
226
193
159
151
145
Bottom line –
atomic radii
∆E.N. seem to “best” determine
bond enthalpies
Bond
Bond
Bond
Length
pm
Pauling's
∆E.N.
atomic
radii
(pm)
Cl-Cl
Br-Br
Br-Br
I--I
199
228
267
0
0
0
99
114
133
Enthalpy
Enthalpy
Single
Single Bond
Bond
kJ/mol
kJ/mol
(Average)
(Average)
243
243
193
193
151
151
H--F
H--Cl
H--Cl
H--Br
H--Br
H--I
92
127
141
161
1.8
1
0.8
0.5
(37) 64
(37) 99
(37) 114
(37) 133
568
568
432
432
366
366
298
298
C--F
C--Cl
C--Cl
C--Br
C--Br
C--I
135
177
194
214
1.5
0.7
0.5
0.2
77 64
77 99
77 114
77 133
488
488
330
330
288
288
216
216
C--F
C--F
C--O
C--O
C--N
C--N
C--C
C--C
135
135
143
143
147
147
154
154
1.5
1.5
1
1
0.5
0.5
0
0
77
77 64
64
77
77 66
66
77
77 70
70
77
77
488
488
360
360
308
308
348
348
H--H
H--H
H--O
H--O
H--N
H--N
H--C
H--C
74
74
96
96
101
101
109
109
0
0
1.3
1.3
0.8
0.8
0.3
0.3
-37
-37
(37)
(37) 66
66
(37)
(37) 70
70
(37)
(37) 77
77
436
436
366
366
391
391
413
413
H--H
H--C
H--N
H--O
74
109
101
96
0
0.3
0.8
1.3
-37
(37) 77
(37) 70
(37) 66
436
413
391
366
Bond
Bond
Bond
Bond
Length
Length
pm
pm
Pauling's
Pauling's
∆E.N.
∆E.N.
atomic
atomic
radii
radii
(pm)
(pm)
Cl-Cl
Cl-Cl
Br-Br
Br-Br
I--I
I--I
199
199
228
228
267
267
00
00
00
99
99
114
114
133
133
Enthalpy
Enthalpy
Single Bond
Single
Bond
kJ/mol
kJ/mol
(Average)
(Average)
243
243
193
193
151
151
H--F
H--F
H--Cl
H--Cl
H--Br
H--Br
H--I
H--I
92
92
127
127
141
141
161
161
1.8
1.8
11
0.8
0.8
0.5
0.5
(37) 64
64
(37)
(37) 99
99
(37)
(37) 114
114
(37)
(37) 133
133
(37)
568
568
432
432
366
366
298
298
C--F
C--F
C--Cl
C--Cl
C--Br
C--Br
C--I
C--I
135
135
177
177
194
194
214
214
1.5
1.5
0.7
0.7
0.5
0.5
0.2
0.2
77 64
64
77
77 99
99
77
77 114
114
77
77 133
133
77
488
488
330
330
288
288
216
216
C--F
C--F
C--O
C--O
C--N
C--N
C--C
C--C
135
135
143
143
147
147
154
154
1.5
1.5
11
0.5
0.5
00
77 64
64
77
77 66
66
77
77 70
70
77
77
77
488
488
360
360
308
308
348
348
H--H
H--H
H--O
H--O
H--N
H--N
H--C
H--C
74
74
96
96
101
101
109
109
00
1.3
1.3
0.8
0.8
0.3
0.3
-37
-37
(37) 66
66
(37)
(37) 70
70
(37)
(37)
77
(37) 77
436
436
366
366
391
391
413
413
H--H
H--C
H--N
H--O
74
109
101
96
0
0.3
0.8
1.3
-37
(37) 77
(37) 70
(37) 66
436
413
391
366
Bond enthalpies increase Single <Double < Triple
But not by multiples of the single bond
C-C
N-N
C-N
C-O
Bond Enthalpy (kJ/mol)
X-X
X=X
347
612
694
159
418
318
293
615
586
351
715
702
X=X
820
1041
941
477
890
879
1075
1053
measured
calc.
measured
calc.
measured
calc.
measured
calc.
http://chemviz.ncsa.uiuc.edu/content/doc-resources-bond.html
18
Examples we have examined about energy so far
Your book’s emphasis on bond enthalpies?
Hydrogen Fuel Cell
Relates to the energy
released
or
taken up
By a reaction
1
∆ H = − 286kJ
H 2 ( g ) + 2 O2 ( g ) → H 2 O( l )
Fossil fuel burning (coal)
If Bonds of Reactants stronger than bonds products endothermic
1
∆ H = − 110.5kJ
C( s) + 2 O2 ( g ) → CO( g )
∆ H = − 3935
. kJ
C( s) + O2 ( g ) → CO2 ( g )
Fossil fuel burning (methane)
We will take up the issue of bond enthalpy when discussing
solids
∆ H = − 890kJ
CH 4 ( g ) + 2O2 ( g ) → CO2 ( g ) + 2 H 2 O( l )
Fossil fuel burning (octane)
∆ H = − 109
. x10 4 kJ
Pure octane
General
FITCH Rules
Chemistry
Energy Density
2C8 H18 + 25O2 → 16CO2 ( g ) + 18 H 2 O( g )
G1: Suzuki is Success
G2. Slow me down
G3. Scientific Knowledge is Referential
G4. Watch out for Red Herrings
G5. Chemists are Lazy
C1. It’s all about charge
C2. Everybody wants to “be like Mike”
⎛ qq ⎞
C3. Size Matters
⎟
E = k⎜
⎝r +r ⎠
C4. Still Waters Run Deep
C5. Alpha Dogs eat first
1 2
el
1
2
19
“A” students work
(without solutions manual)
~ 10 problems/night.
Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
Summary Slides
A + B + heat → C
+q
endothermic
heat
to
system
A + B → heat + C
exothermic
−q
heat
to surroundings
[
]
q = cm t final − t initial = cm∆ t
Office Hours W – F 2-3 pm
c is a measure of the intermolecular interactions; very large for
water = measure of the polarity of the water molecule
q cons tan tpressure = ∆ H = enthalpychange = H products − H reac tan ts
∆ H = + 5716
. kJ
2 H 2 ( g ) + O2 ( g ) → 2 H 2 O( l )
. kJ
− 5716
= − 286kJ
2
1
H 2 ( g ) + 2 O2 ( g ) → H 2 O( l )
An example of a reaction of standard molar enthalpy of formation
1
2
∆ H Of = − 882 kJ
∑ ∆H
∆ H vaporization , H2 O
H 2 O( l ) − H 2 O( g )
o
f , products
−
N 2 ( g ,1atm,25C ) + O2 ( g ,1atm,25C ) → NO2 ( g ,1atm,25C )
∑ ∆H
o
f ,reac tan ts
Unit
m
cm3
gram
Reference State
size of earth
m
mass of 1 cm3 water at specified Temp
(and Pressure)
oC, K
boiling, freezing of water (specified
Pressure)
amu
(mass of 1C-12 atom)/12
atomic mass of an element in grams
atm, mm Hg
earth’s atmosphere at sea level
1.66053873x10-24g
quantity
mole
Pressure
Energy, General
electronic states in atom
Electronegativity
Heat flow measurements
Standard Molar Enthalpy
∆ H = ∆ E + ∆ ( PV )
Says something about
intermolecular
interactions (polarity!)
Properties and Measurements
Property
Size
Volume
Weight
Temperature
An example of Hess’s Law
∆HO =
H 2 O( s ) − H 2 O( l )
2 H 2 O( l ) → 2 H 2 ( g ) + O2 ( g )
∆ H = − 5716
. kJ
∆H =
∆ H fusion , H2 O
Energy of electron in vacuum
F
constant pressure, define system vs surroundings
per mole basis (intensive)
25 oC, 1 atm, from stable state
)Hfo Haq+ =0
∆ H = ∆ E + ( PV ) products − ( PV ) reac tan ts
20
“A” students work
(without solutions manual)
~ 10 problems/night.
Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
Office Hours W – F 2-3 pm
21