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Energy
Focus on two forms of energy
1. Thermal Energy: the energy of
a system associated with the
kinetic energy of the molecules.
Temperature in Kelvin measures the
average kinetic energy of a system.
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Thermal Energy Heat
Heat: the transfer of thermal energy
between 2 objects at different temps.
Kinetic energy is transferred via
molecular collisions.
hot cold
Chemical Potential Energy
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2. Chemical Potential Energy: the
energy stored within a substance
due to the atoms, bonds, IM forces.
Which has greater potential energy,
one gram of solid water at 0oC or or
one gram of liquid water at 0oC? Why?
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Exothermic Process
Gives off heat (transfer of
thermal energy to surroundings)
Where does the energy come from?
higher chem PE
lower chem PE
2H2 + O2 2H2O + energy
heat to
surroundings
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absorbs heat (transfer of
thermal energy from
surroundings to system)
energy + 2HgO 2Hg + O2
higher energylower energy
Endothermic Process
Where is the energy stored?
http://www.youtube.com/watch?v=_Y1alDuXm6A&feature=youtube_gdata_player
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Potential Energy DiagramA + B C + D
pote
ntial en
erg
y
pote
ntial en
erg
yreaction coordinate reaction coordinate
Ea
EaDH<0 DH>0
exothermic
reaction
endothermic
reaction
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1st Law of
ThermodynamicsTwo ways to change the
Energy (E) of a system.
1. System can exchange heat (q)
with surroundings:
+q if system absorbs heat
(endothermic)
-q if system emits heat
(exothermic)
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1st Law of
Thermodynamics
2. System can exchange work (w)
with surroundings
+w if work done on the system
-w if system does work
(Sign of both q and w is positive
if the system gains energy.)
PV Work
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We will only consider work
associated with a gas changing
volume at constant pressure.
w = -PDV
(note sign)
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DE = q - PDV
Why does the nozzle of a CO2fire extinguisher get cold?
DE = q – PDV = 0 – P(Vf – Vi)
Energy of system drops since the
CO2 gas is expanding against
atmospheric pressure.
Both E & T decrease.
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Chemical Reactions
and the 1st Law
DE = q + w
DE = DH - PDV
Enthalpy (H) is heat flow at constant
pressure (normal lab processes)
Chemical reactions that produce or
use up gases will do PDV work.
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DH and the 1st Law
DE = DH – PDV
= DH - RTDngas
Note: if Dngas = 0, no work &DH = DE
Since PV = nRT
PDV = RTDngas (R = 8.314 J/mol K)
where Dngas =
mol gasproduct – mole gasreactant
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DH and the 1st Law
DE = DH - PDV
What is the change in internal
energy for the following reaction
at 1 atm & 25oC?
2CO(g) + O2(g) 2CO2(g)
DH = -566 kJ
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2CO(g) + O2(g) 2CO2(g)
DH = -566 kJ
2-3 = -1molDE = DH – RTDn
= [-566 x 103J] –
[(8.314J/mol K)(298K)(–1)]
= -564 x 103J = -564 kJ
(per 2 moles of CO)
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Types of Heat Changes
1. Heating/cooling: exchange of
thermal energy
2. Phase change: exchange of
chem PE & thermal energy
3. Chem reactions: exchange of
chem PE & thermal energy
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1. Heating & Cooling
Specific heat: Amt. of heat to
raise 1 g of a substance by 1oC
s =q
m DT
(m = mass in g)
(DT = Tf – Ti) in oC or K
e.g. H2O s = 4.184 J/goC
Fe s = 0.444 J/goC
What are the units of s ?
or 1 K
why
different?
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Specific Heat (s)
s =q
m DT
rearranges to:
q = m s DT = C DT
heat capacity (C) = m s
(C has units of J/oC)
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Specific Heat (s)
466 g of water is heated from
8.50 to 74.60 oC.
Calculate the heat absorbed.
q = m s DT
q = 466g x 4.184J/goC x 66.10oC
= 1.29 x 105 J
Specific Heat: Try It
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How many grams of water
can be heated by 52.3oC
using 978 J of heat?
H2O: s = 4.184 J/goC
2. Phase Change (Change of State)
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For a substance at its melting or
boiling point, T does not change
during the phase change.
What does the added heat do
on the molecular level?
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Heat of Fusion
6.01 kJ + H2O(s) H2O(l)
Ice melts at 0OC
Chem. PE of system increases.
How much water does
6.01 kJ of heat refer to?
What is the “heat of solidification”
for water?
Heat of Vaporization
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DHvap(H2O) = 40.7 kJ/mol
What is the heat change when
218 g water condenses?
Does the chem. PE increase or
decrease?
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3. Chemical Reactions
Methane burns
CH4(g) + 2O2(g)
CO2(g) + 2H2O(l) + 890.4 kJ
DH = -890.4 kJ/molrxn
(exothermic)
Chem PE decreases -
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DH: Rules
Coefficients are always moles.
If reaction reversed, DH sign
changes.
If coefficients in balanced
equation x n, DH x n.
Equation must specify states.
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Give it a Shot !!
Calculate heat evolved when
266 g of white phosphorus burns.
P4(s) + 5O2(g) P4O10(s)
DH = -3013 kJ/molrxn
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Constant Pressure
Calorimetry
Calorimetry: the measurement
of heat changes.
qP = DH (constant pressure,
lab conditions)
In an isolated system, the sum of
the heat changes equals zero.
Sq = 0
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Constant P Calorimetry
Coffee-cup
calorimeter
For a process or reaction in solution,
heat given off by the reaction warms
the water and the cup.
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qsoln is the heat gained by the
water/solution, assumed to have
thermal properties like water.
qcup is the heat gained by
calorimeter cup. If it is a perfect
insulator, qcup = 0
qrxn is the heat given off by the
process/reaction
qsoln + qcup + qrxn = 0
Calorimetry: Try It
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85 g of iron is heated to 143oC and
put in a calorimeter (Ccup = 2.1J/oC)
containing 120 g water at 24.2oC.
The iron + water in the cup reaches
a maximum T= 32.5oC.
What is specific heat of iron?
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Try It !!!
100. mL of 0.500 M HCl is mixed
with 100. mL 0.500 M NaOH, both
at 22.50oC, in a calorimeter with
Ccup = 335 J/oC. Final T = 24.90 oC.
Calculate the reaction heat and the
molar heat of neutralization.
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Different “Heats” (T.6.2)
Heat of Neutralization
Heat of Solution
Heat of Fusion
Heat of Vaporization
Heat of Reaction
Heat of Formation
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Standard
Enthalpies
DH = Hprod - Hreact
Since absolute value of H can’t
be measured (only changes),
need arbitrary reference point.
(sea level analogy)
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Std. H of Formation
DHfo 1 mol of cmpd formed
from its elements
1 atm, 25oC
DHfo of most stable form of an
element is zero.
DHfo are listed in App. 3 in text.
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Std. H of Reaction
DHrxno = [c DHf
o(C) + d DHfo(D)]
– [a DHfo(A) + b DHf
o(B)]
Reaction: aA + bB cC + dD
Thus need to know DHfo
DHrxn = Hprod - Hreact
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Determining DHfo
1. Direct Method: synthesize a
compound from its elements.
Cgraphite + O2(g) CO2(g)
DHrxno = -393.5 kJ
Since DHfo(C) and DHf
o(O2) = 0,
DHfo(CO2) = -393.5kJ
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Determining DHfo
2. Indirect Method:
(Hess’s Law)
DH is same whether
reaction takes place
in one step or many.
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Hess’s Law
(for simplicity, I’ll ignore states !!)
C + 2H2 CH4 DHfo(CH4)= ?
C + O2 CO2 -393.5
2H2 + O2 2H2O -571.6
CH4 + 2O2 CO2 + 2H2O -890.4
DHrxn (kJ)
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Hess’s Law cont.
C + O2 CO2 -393.5
2H2 + O2 2H2O -571.6
CO2 + 2H2O CH4 + 2O2 +890.4
DHrxn (kJ)
C + 2H2 CH4 -74.7
reversed reaction
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Hess’s Law: Try It!
Calculate DHfo of acetylene:
2C(graph) + H2(g) C2H2(g)
Given: DHrxno kJ
C(graph) + O2(g) CO2(g) -393.5
H2(g) + O2(g) H2O(l) -285.8
2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
-2598.8
1
2
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Heat of Solution: DHsoln
heat change upon dissolution
e.g. NaCl(s) Na+(aq) + Cl-(aq)H2O
DHsoln + endothermic
- exothermic
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Solution: 2 Step Process
Solid
crystal
Ions in
gas state
SolutionHeat of
Solution = Hsoln
U + Hhydr = Hsoln
+ -
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Solution: 2 Step Process
NaCl(s) Na+(g) + Cl-(g) 788
Na+(g) + Cl-(g) Na+(aq) + Cl-(aq) -784
NaCl(s) Na+(aq) + Cl- (aq) +4
kJ
Thus DHsoln = 4 kJ