Download Chapter 6 Lectures

Thermochemistry
Chapter 6
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1
Overview
Introduce the nature of energy and the
general topics related to energy problems.
Familiarize with the experimental
procedures for measuring heats of reactions.
Hess’s law and its applications based on
enthalpies.
Read the present and new sources of energy.
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2
Law of Conservation
of Energy
Energy can be converted from one form to
another but can neither be created nor
destroyed.
(Euniverse is constant)
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3
The Two Types of Energy
Potential: due to position or composition - can be
converted to work (water before falling from dam,
gasoline, etc.)
Kinetic: due to motion of the object (water falling
and doing work , gasoline burning and driving
engine, etc.)
KE = 1/2 mv2
(m = mass, v = velocity)
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4
Figure 6.1: (a) In the
initial positions, ball A
has a higher potential
energy than ball B. (b)
After A has rolled
down the hill, the
potential energy lost
by A has
been
converted to random
motions
of
the
components of the hill
(frictional heating) and
to the increase in the
potential energy of B.
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5
Total Energy = (PE)A + (KE)B + frictional heat
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6
Energy is a State Function
Depends only on the present state of the
system - not how it arrived there.
It is independent of pathway.
Internal Energy or Total Energy DE,
Enthalpy DH, V, P, T are State Functions
Heat and Work are not state functions
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7
Thermodynamics
State functions are properties that are determined by the state
of the system, regardless of how that condition was achieved.
energy, pressure, volume, temperature
Potential energy of hiker 1 and hiker 2
is the same even though they took
different paths.
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8
Temperature v. Heat
Temperature reflects random motions of
particles, therefore related to kinetic energy
of the system.
Heat involves a transfer of energy between
2 objects due to a temperature difference
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9
Energy Changes in Chemical Reactions
Heat is the transfer of thermal energy between two bodies that
are at different temperatures.
Temperature is a measure of the thermal energy.
Temperature = Thermal Energy
900C
400C
greater thermal energy
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10
Thermochemistry is the study of heat change in chemical
reactions.
The system is the specific part of the universe that is of
interest in the study.
SURROUNDINGS
SYSTEM
open
closed
Exchange: mass & energy
energy
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isolated
nothing 11
System and Surroundings
System: That on which we focus attention
Surroundings: Everything else in the universe
Universe = System + Surroundings
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12
Exo and Endothermic
Heat exchange accompanies chemical
reactions.
Exothermic: Heat flows out of the system
(to the surroundings).
Endothermic: Heat flows into the system
(from the surroundings).
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13
Exothermic process is any process that gives off heat –
transfers thermal energy from the system to the surroundings.
PE stored in chemical bonds are the source of chemical energy.
2H2 (g) + O2 (g)
H2O (g)
2H2O (l) + energy
H2O (l) + energy
Energy lost by system = Energy gained by surrounding
Endothermic process is any process in which heat has to be
supplied to the system from the surroundings.
energy + 2HgO (s)
energy + H2O (s)
2Hg (l) + O2 (g)
H2O (l)
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14
Exothermic versus endothermic.
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15
The study of energy and its conversion is
called thermodynamics.
The law of conservation of energy is called
First Law of Thermodynamics.
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16
First Law
First Law of Thermodynamics:
The energy of the universe is
constant.
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17
Mathematical Definition of the First Law
DE = q + w
DE = change in system’s internal energy
q = heat
w = work
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18
S.Ex. 6.1
• Calc. ΔE, for an endothermic process of
15.6 kJ and work done on the system is 1.4
kJ.
q = +15.6 kJ
w = +1.4 kJ
ΔE = q + w = +15.6 kJ + (+1.4 kJ)
= 17.0 kJ
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19
Figure 6.4: (a) The
piston, moving a
distance ∆h against a
pressure P, does work
on the surroundings.
(b) Since the volume
of a cylinder is the
area of the base times
its height, the change
in volume of the gas is
given by ∆h x A = ΔV.
Expansion
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20
Work
work = force  distance
= F x Δh = PxA x Δh
=
PΔV
ΔV = Vf - Vi
since pressure = force / area,
work = pressure  volume
wsystem = PDV
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21
‘w’ is work done ON the system
•
w = -pΔV
p = external pressure
ΔV = Vf – Vi
= + for expansion → w negative
= -
for compression →
w positive
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22
S. Ex. 6.2
• Calculate the work associated with the
expansion of a gas from 46 L to 64 L at a
constant external pressure of 15 atm.
w = -pΔV
= - 15 atm x (64 L – 46 L)
= - 270 L.atm
-ve, system doing work!!
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23
S. Ex. 6.3
• Hot air balloon- volume change 4.00x106 L to
4.50x106 L, by addition of 1.3x108 J of heat.
q = 1.3x108 J
ΔV = Vf – Vi = (4.50 – 4.00)x106 = 0.50x106 L
w = -p ΔV = - 1.0 atm x 0.50x106 L = - 0.5x106 L.atm
= -5.0x105 L.atm x (101.3 J/1 L.atm) = -5.1x107 J
ΔE = q + w = 1.3x108 J + (- 5.1x107 J)
= 8x107 J
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24
Enthalpy
Enthalpy = H = E + PV
DE = DH  PDV = qP + w
DH = DE + PDV
At constant pressure,
qP = DE + PDV,
where qP = DH at constant pressure
DH = energy flow as heat (at constant pressure)
E, H, P, V are state functions
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25
At constant pressure where PV work is
allowed the change in enthalpy DH is equal
to the energy flow as heat (q or Q).
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26
Signs Convention
Heat Flowing to the system from
surrounding (heating) : DE < 0, q or DH <0
Heat Flowing from the system to the
surrounding (cooling) : DE > 0, q or DH >0
Work done by the system on the
surrounding : W > 0 (-)
Work done on the system by the
surrounding : W < 0 (+)
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27
Thermochemical Equations
Is DH negative or positive?
System gives off heat
Exothermic
DH < 0
890.4 kJ are released for every 1 mole of methane
that is combusted at 250C and 1 atm.
CH4 (g) + 2O2 (g)
CO2 (g) + 2H2O (l) DH = -890.4 kJ
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28
Enthalpy (H) is used to quantify the heat flow into or out of a
system in a process that occurs at constant pressure.
DH = H (products) – H (reactants)
DH = heat given off or absorbed during a reaction at constant pressure
Hproducts < Hreactants
DH < 0
Hproducts > Hreactants
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DH > 0
29
S.Ex. 6.4
• When one mole of CH4 is burnt at const.P,
890 kJ heat was released. Calc. ΔH when
5.8 g CH4 is burnt at const P
• qp = ΔH = -890 kJ/molCH4
5.8 gCH 4 x
1mole
16.0 gCH 4
 0.36mol CH 4
890kJ
0.36mol x 1moleCH


320
kJ
4
DH  320 k J
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30
Heat Capacity
heat absorbed
J
J
C =
=
or
increase in temperature
C
K
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31
The specific heat (s) of a substance is the amount of heat (q)
required to raise the temperature of one gram of the
substance by one degree Celsius.
The heat capacity (C) of a substance is the amount of heat
(q) required to raise the temperature of a given quantity (m)
of the substance by one degree Celsius.
C = ms
Heat (q) absorbed or released:
q = msDt
q = CDt
Dt = tfinal - tinitial
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32
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33
How much heat is given off when an 869 g iron bar cools
from 940C to 50C?
s of Fe = 0.444 J/g • 0C
Dt = tfinal – tinitial = 50C – 940C = -890C
q = msDt = 869 g x 0.444 J/g • 0C x –890C = -34,000 J
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34
Some Heat Exchange Terms
specific heat capacity
heat capacity per gram = J/°C g or J/K g
molar heat capacity
heat capacity per mole = J/°C mol or J/K mol
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35
Calorimetry
The device used experimentally to
determine the heat from chemical reaction
is called a calorimeter.
Calorimetry is the science to measure heat
changes and it is based on the temperature
change.
Heat capacity is a measure of the body or
system ability to absorb heat.
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36
Two Types of Calorimeters
1. Constant Volume Calorimeter
2. Constant Pressure Calorimeter
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37
Constant-Pressure or Coffee Cup Calorimetry
May be used to determine many
Heats such as heat of solution,
neutralization (acid-base), etc.
qlost = qgained
qrxn = - (qwater + qcalorimeter)
qwater = msDt
qcal = Ccal Dt
Reaction at Constant P
DH = qrxn
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38
•
•
•
•
•
•
•
•
•
•
50.0 mL of 1.0M NaOH at 25 oC and
50.0 mL of 1.0M HCl at 25 oC are mixed
Final temp = 31.9 oC; ∆H = ?
q = m.s.∆T
= 100 g x 4.18 J/g.oC x6.9 oC = 2.9x103 J
H+ + OH- →
H2O
0.050 mol 0.050 mol
0.050 mol
∆H = - 2.9x103 J/0.050 mol
= -5.8x104 kJ/mol
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39
S.Ex 6.5
1.00 L of 1.00 M Ba(NO3)2 solution at 25 oC is mixed with
1.00 L of 1.00M Na2SO4 solution at 25 oC the temperature of
the mixture increased to 28.1 oC. Calculate the enthalpy
change per mole of BaSO4 formed.(s = 4.18 J/oC.g)
• Ba2+ (aq) + SO42-(aq) → BaSO4 (s)
• 1 mol
1 mol
1 mol
• q = msΔT = 2000 g x 4.18 J/oC.g x 3.1oC =2.6x104 J
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40
Constant-Volume Calorimetry
DE = q + W
W = -pDV = 0
DE = qV
qsys = qwater + qbomb + qrxn
qsys = 0
qrxn = - (qwater + qbomb)
qwater = msDt
qbomb = CbombDt
Reaction at Constant V
DH = qrxn
No heat enters
or leaves!
DH ~ qrxn
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41
Sample Exercise 6.6
Calculate the energy per gram for burning 1.50g
of CH4 in a constant-volume calorimeter (C=11.3
kJ/0C) where Dt = 7.30C..
Energy released = C x Dt = 11.3x103 x 7.3
= 83 kJ per 1.50g of CH4
83
Energy released per g =
= 55 kJ/g
1.50
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42
The enthalpy of solution (DHsoln) is the heat generated or
absorbed when a certain amount of solute dissolves in a
certain amount of solvent.
DHsoln = Hsoln - Hcomponents
Which substance(s) could be
used for melting ice?
Which substance(s) could be
used for a cold pack?
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43
Thermochemical Equations
•
The stoichiometric coefficients always refer to the number
of moles of a substance
H2O (s)
•
DH = 6.01 kJ
If you reverse a reaction, the sign of DH changes
H2O (l)
•
H2O (l)
H2O (s)
DH = -6.01 kJ
If you multiply both sides of the equation by a factor n,
then DH must change by the same factor n.
2H2O (s)
2H2O (l)
DH = 2 x 6.01 = 12.0 kJ
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44
Thermochemical Equations
•
The physical states of all reactants and products must be
specified in thermochemical equations.
H2O (s)
H2O (l)
DH = 6.01 kJ
H2O (l)
H2O (g)
DH = 44.0 kJ
How much heat is evolved when 266 g of white
phosphorus (P4) burn in air?
P4 (s) + 5O2 (g)
1 mol P4
P4O10 (s)
123.9g
266g
DH = -3013 kJ
-3013 kJ
Q
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Q = -6470 kJ
45
Hess’s Law
Reactants  Products
The change in enthalpy is the same whether
the reaction takes place in one step or a
series of steps.
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46
Calculations via Hess’s Law
1. If a reaction is reversed, DH is also reversed.
N2(g) + O2(g)  2NO(g)
2NO(g)  N2(g) + O2(g)
2.
DH = 180 kJ
DH = 180 kJ
If the coefficients of a reaction are multiplied
by an integer, DH is multiplied by that same
integer.
6NO(g)  3N2(g) + 3O2(g)
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DH = 540 kJ
47
S. Ex. 6.7
C(graphite-s)→ C(diamond-s) ΔH = ?
Given,
Cg(s) + O2(g) → CO2(g) ΔH = -394 kJ
Cd(s) + O2(g) → CO2 (g) ΔH = -396 kJ
Cg(s) + O2(g) → CO2(g) ΔH = -394 kJ
CO2 (g) → Cd(s) + O2(g) ΔH = +396 kJ
_________________________________
Cg(s) → Cd(s)
ΔH = 2 kJ
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48
CO2
+396 kJ
-394 kJ
Cgr
+2 kJ
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Cdia
49
Because there is no way to measure the absolute value of
the enthalpy of a substance, must I measure the enthalpy
change for every reaction of interest?
Establish an arbitrary scale with the standard enthalpy of
0) as a reference point for all enthalpy
formation (DH
f
expressions.
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50
Standard States
Compound
 For
a gas, pressure is exactly 1 atmosphere.
 For a solution, concentration is exactly 1 molar.
 Pure substance (liquid or solid), it is the pure liquid or
solid.
Element
 The
form [N2(g), K(s)] in which it exists at
25°C.
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1 atm and
51
Standard Enthalpies of Formation
defined as
Change in enthalpy that accompanies the
formation of one mole of a compound from its
elements with all substances in their standard
states.
C(s)+ 2H2(g)+ ½ O2(g) → CH3OH (l) ΔHfo= -239 kJ/mol
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52
The standard enthalpy of formation of any
element in its most stable form is zero.
DH0f (O2) = 0
DH0f (C, graphite) = 0
DH0f (O3) = 142 kJ/mol
DH0f (C, diamond) = 1.90 kJ/mol
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53
Standard enthalpy of formation (DH0f) is the heat
change that results when one mole of a compound
is formed from its elements at a pressure of 1 atm.
Example CO2(g) is formed from C(s) and O2(g)
C(s) + O2(g)
CO2(g)
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DH0f
54
Calculate the standard enthalpy of formation of CS2 (l)
given that:
C(graphite) + O2 (g)
CO2 (g) DH0rxn = -393.5 kJ
S(rhombic) + O2 (g)
CS2(l) + 3O2 (g)
SO2 (g)
DH0rxn = -296.1 kJ
CO2 (g) + 2SO2 (g)
0 = -1072 kJ
DHrxn
1. Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic)
CS2 (l)
2. Add the given rxns so that the result is the desired rxn.
C(graphite) + O2 (g)
2S(rhombic) + 2O2 (g)
+ CO2(g) + 2SO2 (g)
CS2 (l) + 3O2 (g)
C(graphite) + 2S(rhombic)
DH0
rxn=
CO2 (g) DH0rxn = -393.5 kJ
2SO2 (g) DH0rxn = -296.1x2 kJ
0 = +1072 kJ
DHrxn
CS2 (l)
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-393.5 + (2x-296.1) + 1072 = 86.3 kJ
55
0 ) is the enthalpy of
The standard enthalpy of reaction (DHrxn
a reaction carried out at 1 atm.
aA + bB
cC + dD
DH0rxn = [ cDH0f (C) + dDH0f (D) ] - [ aDH0f (A) + bDH0f (B) ]
DH0rxn = S nDH0f (products) - S mDHf0 (reactants)
Hess’s Law: When reactants are converted to products, the
change in enthalpy is the same whether the reaction takes
place in one step or in a series of steps.
(Enthalpy is a state function. It doesn’t matter how you get
there, only where you start and end.)
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56
CH4 + 2O2 → CO2 + 2H2O
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57
Change in Enthalpy
Can be calculated from enthalpies of
formation of reactants and products.
DHrxn° = SnpDHf(products)  SnrDHf(reactants)
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59
Benzene (C6H6) burns in air to produce carbon dioxide and
liquid water. How much heat is released per mole of
benzene combusted? The standard enthalpy of formation
of benzene is 49.04 kJ/mol.
2C6H6 (l) + 15O2 (g)
12CO2 (g) + 6H2O (l)
DH0rxn = S nDH0f (products) - S mDHf0 (reactants)
DH0rxn = [ 12DH0f (CO2) + 6DH0f (H2O)] - [ 2DH0f (C6H6)]
DH0rxn = [ 12x–393.5 + 6x–187.6 ] – [ 2x49.04 ] = -5946 kJ
-5946 kJ
= - 2973 kJ/mol C6H6
2 mol
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60
Keep in mind the following
concepts.
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61
S.Ex 6.9
4NH3(g) +7O2(g) → 4NO2(g) +6H2O(l) ΔH=?
ΔHfo(NH3) = -46 kJ/mol
ΔHfo(NO2) = 34 kJ/mol
ΔHfo(H2O) = -286 kJ/mol
ΔH = [6 ΔHfo(H2O) + 4 ΔHfo(NO2)] –
[4 ΔHfo(NH3)+ 7ΔHfo(O2)]
ΔH = [6x-286 + 4x34] – [4x-46 +7x0]
=1396 kJ
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63
S. Ex 6.11
• 2CH3OH + 3O2(g) → 2CO2 + 4 H2O
Calculate heat of combustion per gram of Methanol.
ΔH= 2xΔHf(CO2) + 4x ΔHf(H2O) – 2x ΔHf(CH3OH)
= 2x(-394 kJ) + 4x(-286 kJ) -2x(-239kJ)
= -1454 kJ for 2 mole methanol
-1454 kJ/64.0 g = -22.7 kJ/g
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64
Sample Exercise
Calculate the work done when 50.0 g of
magnesium (Mg) dissolve in excess acid at
1.00 atm and 25.0 0C.
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65
Sample Exercise
Determine the heat of reaction at 298 K for the reaction
which occurs in a welder's acetylene torch:
2 C2H2(g) + 5 O2(g)
4 CO2(g) + 2 H2O(l)
Given the following equations and Ho values
H2(g) + 1/2 O2(g)
2 C(s) + H2(g)
C(s) + O2(g)
H2O(l)
C2H2(g)
CO2(g)
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Ho/kJ = -285.8
Ho/kJ = +226.7
Ho/kJ = -393.5
66
Sample Exercise
A 15.0g nickel is heated to 100.00C and dropped into 55.0g
of water initially at 230C. Assuming an isolated system,
calculate the final temperature of the nickel and water.
Snickel = 0.444 J/0Cg.
Q lost by nickel
=
- Q gained by water
(msDt) nickel
=
-
Dt = tf - ti
(msDt)water
tf is the same.
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ti is different.
67