Transcript Chem 1202

Thermochemistry
Chapter 5
Energy, Heat, Work
Watkins
Chemistry 1422, Chapter 5
1
The Nature of Energy
Energy is what makes things happen.
Matter is what it happens to.
Whenever anything happens to matter,
Energy is involved.
Matter and Energy are the only two
things in the universe, and they are
related:
E = mc2
Watkins
Chemistry 1422, Chapter 5
2
The Nature of Energy (Review)
Energy cannot be created or destroyed during
any process or reaction;
it can be transferred from place to place
it can be converted from one form to another
This fact is called the
First Law of Thermodynamics
Watkins
Chemistry 1422, Chapter 5
3
The Nature of Energy (Review)
A few of the many different forms of energy:
•
•
•
•
•
•
•
•
•
•
Mechanical Work (w = fd = -PDV)
Gravitational Energy (-Gm1m2/d = mgh)
Kinetic Energy (1/2mv2)
1 J = 1 kg•m2•s-2
Coulombic Energy (kz1z2/d)
Internal Energy (DE)
1 cal = 4.184 J
Light Energy (hn) (chapter 6)
Nuclear Energy (mc2)
Electrical Energy (VQ) (chapter 20)
Heat flow (q)
Chemical (Bond) Energy (DHrxn)
Watkins
Chemistry 1422, Chapter 5
4
Thermodynamics (Review)
DEFINITIONS
System (sys)
that part of the universe in
which we are interested.
Surroundings (surr)
the rest of the universe.
Watkins
Chemistry 1422, Chapter 5
5
Thermodynamics (Review)
DEFINITIONS
H2(g) & O2(g)
Internal Energy
Total energy of system
Esys
H2(g) + ½ O2(g) → H2O(g)
Esys
DEsys < 0
H2O(g) → H2(g) + ½ O2(g)
DEsys > 0
H2O(g)
Watkins
We cannot measure Esys
We can measure DEsys
when the system changes
during a process
For example, during a
chemical reaction
DEsys = Efinal - Einitial
Chemistry 1422, Chapter 5
6
First Law of Thermodynamics (Review)
Energy cannot be created or destroyed
during a process; it can only be
converted or transferred.
Therefore ...
• the total energy in the universe is
constant: DEuniv = 0 = DEsys + DEsurr
• any energy transferred to or from
system must be transferred from or to
surroundings
Watkins
Chemistry 1422, Chapter 5
7
First Law of Thermodynamics (Review)
Relating DE to Heat and Work
When a system undergoes a physical or
chemical change, the change in its internal
energy is equal to q, the heat energy flowing
into or out of the system, plus w, the work
(all other forms of energy) coming into or
going out of the system:
DEsys = q + w
Watkins
Chemistry 1422, Chapter 5
8
First Law of Thermodynamics (Review)
q=0
adiabatic
q>0
endothermic
q<0
exothermic
System
Surroundings
q > 0 (+) Heat (–) 0 > q
DEsys = q + w
w = 0 (?)
w>0
w > 0 (+) Work (–) 0 > w
done on system
w<0
done by system
A process occurs in the system
Watkins
Chemistry 1422, Chapter 5
9
State Function Defined
• A State function is an energy term which depends
only on the initial and final states of the system,
not on how the energy is used.
• E and H are state functions, because
DE = Efinal – Einitial
DH = Hfinal – Hintial
• Neither q nor w are state functions (even though
their sum DE is a state function)
Watkins
Chemistry 1422, Chapter 5
10
heating
coil
State Function E
For a given
system, DE
does not
change with
the process
but q and w
do change,
depending
on how the
energy is
used.
Watkins
charged
q<0
q<0
w=0
DE
w<0
discharged
Chemistry 1422, Chapter 5
11
DE = q + w
State Function E
For most chemical systems, the usual
form of work energy is w = -PDV or
expansion/compression work. For
these systems, when the volume is
contant (DV = 0), w = 0
DE = qv
DE is measured as heat flow in a
constant volume process.
Watkins
Chemistry 1422, Chapter 5
12
DE = qV
However, most chemical reactions are
not carried out at constant volume.
They are usually carried out at constant
pressure (in the open atmosphere).
So a new state function was invented for
contant pressure processes.
Watkins
Chemistry 1422, Chapter 5
13
DE = qV
For a contant pressure process (DP = 0);
if the system expands (DV > 0), it does
work on the surroundings:
w = -PDV
DE = q + w = qP - PDV
qP = DE + PDV = DH
DH = qP
State Function H is called Enthalpy
Watkins
Chemistry 1422, Chapter 5
14
State Function H
DE = qV
DH = qP
Think of H as the amount of “heat
energy” contained in the system.
(H can be called the “heat content”)
When a reaction occurs at constant
pressure, the heat that flows (qP)
between system and surroundings is
exactly the enthalpy change DH
Watkins
Chemistry 1422, Chapter 5
15
State Function H
H ("heat content")
• We cannot measure H, only DH = Hfinal - Hinitial = qp
• H is the total heat energy in the system
• qp is heat added to or subtracted from the system at
constant pressure.
Watkins
initial
final
DH < 0
heat out
exo
DH > 0
heat in
endo
final
initial
Chemistry 1422, Chapter 5
H is heat energy
just sitting in the
system ("latent"
heat)
q is heat flow
(heat enrgy being
transferred!)
16
Enthalpies of Reaction
How can you find qp = DHrxn?
Two basically different procedures:
Calculate it
Bond Energies
Hess’s Law
Measure it
Calorimetry
Watkins
Chemistry 1422, Chapter 5
17
Enthalpies of Reaction
(Review: Chapter 8, section 8, page 328)
Bond Dissociation Energy is the energy absorbed (+D)
to break a chemical bond.
Bond breaking is always endothermic
When a chemical bond is formed, “Bond Energy” is the
energy emitted (-D).
Bond making is always exothermic
DHrxn is the sum of all bond breaking and bond
making energies in a reaction.
Watkins
Chemistry 1422, Chapter 5
18
Enthalpies of Reaction
Hess’s Law I
For a chemical reaction at constant pressure:
DHrxn = SH(products) - SH(reactants)
H is an extensive property - the magnitude of H is
directly proportional to amount:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) DH1 = -803 kJ
DH1 = [H(CO2(g)) + H(2H2O(g))] - [H(CH4(g)) + H(2O2(g))]
= [H(CO2(g)) + 2H(H2O(g))] - [H(CH4(g)) + 2H(O2(g))]
[Note that H (kJ) is extensive, but H (kJ/mol) is intensive]
2CH4(g) + 4O2(g)  2CO2(g) + 4H2O(g) DH2 = -1606 kJ
Watkins
Chemistry 1422, Chapter 5
19
Enthalpies of Reaction
Hess’s Law I
For a chemical reaction at constant pressure:
DHrxn = SH(products) - SH (reactants)
When a reaction is reversed, the sign of DH changes:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) DH = -803 kJ
CO2(g) + 2H2O(g)  CH4(g) + 2O2(g) DH = +803 kJ
Watkins
Chemistry 1422, Chapter 5
20
Enthalpies of Reaction
Hess’s Law I
For a chemical reaction at constant pressure:
DHrxn = SH(products) - SH (reactants)
Enthalpy depends on state:
H2O(l)  H2O(g)
DHrxn = +88 kJ/mol
DHrxn = H[H2O(g)] - H[H2O(l)] > 0
H[H2O(g)] > H[H2O(l)]
Watkins
Chemistry 1422, Chapter 5
21
Enthalpies of Reaction
Hess’s Law I
For a chemical reaction at constant pressure:
DHrxn = SH(products) - SH (reactants)
Any reaction can be written as a sum of two or more
other reactions; DH for the overall reaction is the sum
of the DH's of the individual reactions.
Watkins
Chemistry 1422, Chapter 5
22
Enthalpies of Reaction
Hess’s Law I
DHrxn = SH(products) - SH (reactants)
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
2H2O(g)  2H2O(l)
DH1
DH2
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
DH3
[H(CO2(g)) + 2H(H2O(g))] – [H(CH4(g)) + 2H(O2(g))] = -803 kJ
[2H(H2O(l))] – [2H(H2O(g))] = -176 kJ
DH3 = DH1 + DH2 = -979 kJ
To add these two reactions ...
• Cancel common reactant-product pairs
• Write remaining reactants and products
• Add Enthalpies (watch signs!)
Watkins
Chemistry 1422, Chapter 5
23
Enthalpies of Reaction
Hess’s Law II
DHrxn = SH(products) - SH (reactants)
Any reaction can be written as a sum of two or more
other reactions; DH for the overall reaction is the sum
of the DH's of the individual reactions.
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
2H2O(g)  2H2O(l)
DH1
DH2
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
DH3 = DH1 + DH2
DH1 = [1H(CO2(g)) + 2H(H2O(g))] – [1H(CH4(g)) + 2H(O2(g))]
DH2 = [2H(H2O(l))] – [2H(H2O(g))]
DH3 = [1H(CO2(g)) + 2H(H2O(l))] – [1H(CH4(g)) + 2H(O2(g))]
Is there a way to find H?
Watkins
Chemistry 1422, Chapter 5
24
Enthalpies of Reaction
Hess’s Law II
DHrxn = SH(products) - SH (reactants)
We can measure qp = DHrxn.
We cannot measure H(prod), H(react), or H(x),
the molar heat content of any substance x.
But we can measure something that is exactly
equivalent to H(x). We can measure DHof (x),
the Standard Heat of Formation of substance x.
Watkins
Chemistry 1422, Chapter 5
25
Formation Reactions
Hess’s Law II
DHrxn = SH(products) - SH (reactants)
A formation reaction is a chemical reaction, at some
temperature T, which forms substance x from its
elements under the following conditions:
• All reactants are elements
• The product is one mole of substance x
• All reactants and products are at temperature T and
are in their standard states (s, l, g, aq)
• Solids and liquids are pure, gases are at 1 atm,
aqueous solutions are at 1 M.
Watkins
Chemistry 1422, Chapter 5
26
Formation Reactions
Hess’s Law II
An example of a formation reaction is:
2C(graphite) + 2H2(g) + O2(g) → HC2H3O2(l)
At 25 oC, reactant element carbon is in the form
of solid graphite, and elements hydrogen and
oxygen are gases, each at 1 atm; the product,
one mole of pure acetic acid, is a liquid.
The heat of this reaction, qp = DHrxn, is called the
“standard heat of formation” and is given the
special symbol DH°f (kJ/mol). Note that this is
a molar quantity just like H!
Watkins
Chemistry 1422, Chapter 5
27
Formation Reactions
Hess’s Law II
Examples (at 25 oC; listed in Appendix C)
DH°f = - 1670 kJ/mol
H2(g) + S(s) + 2O2(g)  H2SO4(l) DH°f = - 814 kJ/mol
H2(g) + 1/2O2(g)  H2O(l)
DH°f = - 286 kJ/mol
2H2(g) + O2(g)  2H2O(l)
DH°rxn = - 572 kJ
H2O(l) H2(g) + 1/2O2(g)
DH°rxn = + 286 kJ
C(s)  C(diamond)
DH°f = + 1.9 kJ/mol
C(s)  C(graphite)
DH°f = 0 kJ/mol
2Al(s) + 3/2O2(g)  Al2O3(s)
DH°f = 0 for an element in its standard state.
Watkins
Chemistry 1422, Chapter 5
28
Hess’s Law II
Formation Reactions
Why are formation reactions important? Any
chemical reaction can be written as a sum of
reactions derived from formation reactions!
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
DHrxn = ?
(from Appendix C)
• CH4(g) → C(s) + 2H2(g)
DH1 = -DH°f = + 75 kJ
• 2O2(g) → 2O2(g)
DH2 = -2DH°f = -0 kJ
•C(s) + O2(g) → CO2(g)
DH3 = DH°f = - 394 kJ
•2H2(g) + O2(g) → 2H2O(g)
DH4 = 2DH°f = - 484 kJ
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
DHrxn = DH1 + DH2 + DH3 + DH4 = -803 kJ
Watkins
Chemistry 1422, Chapter 5
29
Hess’s Law II
Formation Reactions
Why are formation reactions important? Any
chemical reaction can be written as a sum of
reactions derived from formation reactions!
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
DHrxn = [H(CO2) + 2H(H2O)]
DHrxn = -803 kJ
- [H(CH4) + 2H(O2)]
DHrxn=[DHof(CO2)+2DHof(H2O)]- [DHof(CH4)+2DHof(O2)]
H  DHof
Watkins
Chemistry 1422, Chapter 5
30
Enthalpies of Formation Hess’s Law II
Hess's Law with DH°f will calculate any DHrxn
For any reaction:
aA + bB + ...  cC + dD + ...
DHrxn = [cDHof(C) + dDHof(D)+...]
- [aDHof(A) + bDHof(B)+...]
DHrxn = SpDH°f (products) - SrDH°f (reactants)
p = c, d, ... (stoichiometric coefficients of products)
r = a, b, ... (stoichiometric coefficients of reactants)
Watkins
Chemistry 1422, Chapter 5
31
Enthalpies of Reaction
Hess’s Law II
2 KOH(s) + CO2(g) → K2CO3(s) + H2O(g)
DHfo kJ/mol
-424.7
-393.5
-1150.2
-136.1
(all values from Appendix C)
DHrxn = [1(-1150.2)+1(-136.1)]-[2(-424.7)+1(-393.5)]
DHrxn = -43.4 kJ
Pay close attention to signs!!!
Pay close attention to states!!!
H2O(g) and H2O(l) have different DHof
Watkins
Chemistry 1422, Chapter 5
32
Calorimetry
Calorimetry = measurement of heat flow q.
A calorimeter is an apparatus that measures
heat flow q.
The 1st rule of heat flow: heat always flows
from a hot place (high temperature
source) to a cold place (low temperature
sink). If Tsource = Tsink, q = 0.
Tsource usually decreases and Tsink usually
increases (but not always)
Watkins
Chemistry 1422, Chapter 5
33
Calorimetry
Calorimetry = measurement of heat flow q.
A calorimeter is an apparatus that measures
heat flow q.
Heat capacity = the amount of heat
required to change the temperature of an
object by 1 centigrade degree (°C or K).
The units of heat capacity are J/deg.
The increase in T measures the increased
kinetic energy of the molecules.
Watkins
Chemistry 1422, Chapter 5
34
Heat capacity (J/deg)
The heat capacity of an object depends on
how big the object is. Example: H2O(l )
Calorimetry
(1 g)
(1 mol)
(1 L)
(1 m3 = 1,000 L)
1 cm3
18 cm3
1,000 cm3
1,000,000 cm3
4.18 J/deg
75.3 J/deg
4.18 kJ/deg
4.18 MJ/deg
Watkins
Chemistry 1422, Chapter 5
35
Calorimetry
Heat capacity (J/deg)
The heat capacity of an object also depends
on what the object is.
1 g H2O(l)
1 g Air(g)
1 g NaCl(s)
1 g S(s)
1 g U(s)
4.18 J/deg
1.01 J/deg
0.86 J/deg
0.71 J/deg
0.12 J/deg
Non-metals generally have HIGH heat capacities.
Metals generally have LOW heat capacities.
Watkins
Chemistry 1422, Chapter 5
36
Calorimetry
Heat capacity (J•deg-1) = the amount of heat
required to change the temperature of an
object by 1 degree centigrade (°C or K).
Insulator – an object in which the rate of
heat flow is slow (in part due to a large
heat capacity)
Conductor – an object in which the rate of
heat flow is fast (in part due to a small
heat capacity)
Watkins
Chemistry 1422, Chapter 5
37
Calorimetry
Molar heat capacity
C (J•mol-1•deg-1)
The heat required to change the temperature
of 1 mole of a substance by 1 degree; if n
moles change by DT degrees, then:
q = n•C•DT
Gram heat capacity (“specific heat”)
s (J•g-1•deg-1)
The heat required to change the temperature
of 1 gram of a substance by 1 degree; if m
grams change by DT degrees, then:
q = m•s•DT
Watkins
Chemistry 1422, Chapter 5
38
Calorimetry
q = n•C•DT
q = m•s•DT
DT = Tfinal - Tinitial
To raise the temperature (DT > 0) of a system,
add heat to it (q > 0, endothermic).
To lower the temperature (DT < 0) of a system,
subtract heat from it (q < 0, exothermic).
When does transfering heat to/from a system not
change its temperature? During a phase
change, when the added heat energy breaks
intermolecular bonds instead of speeding up the
molecules.
Watkins
Chemistry 1422, Chapter 5
39
Calorimetry
How much heat must
be added to m grams
of ice at -20 oC to
convert it into m
grams of steam at
120 oC?
(m grams of steam)
Heat added (kJ)
120
vaporization
100
80
60
~
~
oC
40
20
0
-20
fusion
Heating Curve for H2O
(m grams of ice)
Watkins
Chemistry 1422, Chapter 5
40
s1 = 2.09 J/g.deg DHfus = 6.01 kJ/mol
s3 = 4.18 J/g.deg DHvap = 40.67 kJ/mol
s5 = 1.84 J/g.deg
(Water data: Appendix B)
q1=ms1DT: heat added to raise 120
Heat added (kJ)
H2O(s) from -20 to 0 oC
vaporization
Calorimetry
100
q2=(m/18)DHfus (only breaks
intermolecular bonds)
q3=ms3DT: heat added to raise
H2O(l) from 0 to 100 oC
80
60
~
~
oC
40
q4=(m/18)DHvap (only breaks
intermolecular bonds)
q5=ms5DT: heat added to raise
H2O(g) from 100 to 120 oC
Watkins
20
fusion
0
-20
q1
q2
Chemistry 1422, Chapter 5
q3
q5
q4
41
Calorimetry
How much heat must be added to 18.02 grams of ice at -20
oC to convert it into 18.02 grams of steam at 120 oC?
H2O(s, -20oC) → H2O(s, 0oC)
q1 = DH1 = +0.75 kJ
H2O(s, 0oC) → H2O(l, 0oC)
q2 = DH2 = +6.01 kJ
H2O(l, 0oC) → H2O(l, 100oC)
q3 = DH3 = +7.53 kJ
H2O(l, 100oC) → H2O(g, 100oC)
q4 = DH4 = +40.67 kJ
H2O(g, 100oC) → H2O(g, 120oC)
q5 = DH5 = +0.66 kJ
H2O(s, -20oC) → H2O(g, 120oC)
DH = +55.63 kJ
Hess’s Law: Any reaction can be written as the
sum of two or more other reactions.
Watkins
Chemistry 1422, Chapter 5
42
Constant Pressure Calorimetry
Enthalpy of a Solution Reaction
Atmospheric pressure is constant for
an open container!
• Find specific heat of the solution
ssoln
• Weigh solution
msoln
• Measure DTrxn during reaction
• qp = msoln× ssoln× DTrxn = DHrxn
Watkins
Chemistry 1422, Chapter 5
43
Calorimetry
In a constant pressure calorimeter, 25 g of ice at -10 oC is
added to 100 g of water at 25 oC. After thermal equilibrium
is reached, what is the final temperature in the calorimeter?
s(s) = 2.09 J/g.deg
s(l) = 4.18 J/g.deg
DHfus = 6.01 kJ/mol
H2O(25g, s,-10) → H2O(25g, s,0)
q1 = (25)(2.09)(10) J
H2O(25g,s,0) → H2O(25g,l,0)
q2 = (25/18)(6010) J (DT = 0)
H2O(25g,l,0) → H2O(25g,l,T)
q3 = (25)(4.18)(T – 0) J
H2O(100g,l,25) → H2O(100g,l,T)
q4 = (100)(4.18)(T – 25) J
q1 + q2 + q3 + q4 = 0
T=?
Watkins
Chemistry 1422, Chapter 5
44
Constant Volume Calorimetry
Bomb Calorimeter (DV = 0)
• Used for combustion
reactions.
• Measure the heat capacity
of the calorimeter (K) with a
known reaction.
• Measure qV = K•DT = DE for
the unknown reaction.
• Convert DE(comb) to
DH(comb) (not in this class)
DHof is derived from DH(comb)
Watkins
Chemistry 1422, Chapter 5
45