Energy
Heat
Work
Heat Capacity
Enthalpy
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© Prof. Zvi C. Koren
20.07.2010
Thermodynamics vs. Kinetics
Thermo
Thermodynamics = Thermo + Dynamics
E
(Note: Absolute E can never be determined by humans!)
Can a reaction occur??? (Is it spontaneous)???
Kinetics
Diamond
vs.
Graphite
Thermodynamically unstable
Kinetically stable
C(di)
2
Thermodynamically stable
?
C(gr)
© Prof. Zvi C. Koren
20.07.2010
Energy
Kinetic Energy
Potential Energy
= K.E. = energy of motion = ½ mv2
= P.E. = energy of position
= m·g·h, g = 9.81 m/s2, h = height
Energy Units
1J
=
=
=
=
1 erg =
3
1 cal ≡
1 BTU =
1 kWhr =
1 Latm ≡
1 erg =
SI unit: Joule (J)
1 kgm2/s2 (“kms”) [Note units of “mv2”]
1 VC
1 Pam3
107 erg,
1 gcm2/s2 (“cgs”)
6.2415 ×1011 eV
R = Gas Constant
4.184 J (exactly)
= energy/molK
1054.35 J
= 0.0821 Latm/molK
3.6 MJ
= 1.99 cal/molK
101.325 J (exactly)
= 8.31 J/molK
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6.2415 ×10 eV
© Prof. Zvi C. Koren 20.07.2010
The First Law of Thermodynamics
Popular Expression: “Law of Conservation of Energy”
Mathematical Expression (and more general):
E =
q + w
heat
work
w: For example, compression or expansion
Note the equation is NOT any of the following:
E = q + w
or
E = q +
w
E is a state function
q and w are path functions
w
(P1,V1,T1)
E1
(P2,V2,T2)
q
E2
E = E2 - E1
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© Prof. Zvi C. Koren
20.07.2010
Work and Heat
Work:
Energy transferred when an object is moved against an opposing force.
Note: Two conditions need to be met: movement and resistance.
Heat:
Energy transferred from a “hot” body (at Thigh) to a “cold” body (at Tlow)
-------------------------------------------------------------------------------------------------------w = – fopposing • d = – fopposing • x
WORK:
needed for direction
popposing
Surroundings fopposing
(piston)
m
System
f = force
d = distance
m
expansion
gas
fopposing = weight = mg
w = – f • d = – mg • h
Also: p = f/A  f = p·A
compression
w = – fopposing • x
= – pA • x = – p • V
w = – P • V
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For piston of constant weight or if open to atmosphere:
popposing = Pinternal
© Prof. Zvi C. Koren
20.07.2010
Qualitative Differences Between q and w
w causes ordered motion
Work that the SYSTEM performs causes the molecules (or
atoms) in an object in the SURROUNDINGS to all move
in the same direction.
q causes random motion
Heat flowing from SYSTEM to SURROUNDINGS
causes the molecules in the surroundings to move in
random directions.
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© Prof. Zvi C. Koren
20.07.2010
Heat Capacity, C
Amount of heat (q) needed to raise a quantity of substance by 1 degree
For: 1 gram
“specific heat capacity”
Units of C  cal (or J)/g•deg
For: 1 mole
“molar heat capacity”
Units of C  cal (or J)/mol•deg
For example, for water:
Cspecific = 1 cal/g•deg, 
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Cmolar = ?
© Prof. Zvi C. Koren
20.07.2010
Gases
Liquids
Solids
Specific Heat Capacities of Some Common Substances
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Substance
Name
C (J/gdeg)
Law of Dulong and Petit:
Al
Aluminum
0.902
Molar Heat Capacities of
Fe
Iron
0.451
Elemental Metals
Cu
Copper
0.385
 3R
Au
Gold
0.128
6.0  0.3 cal/moldeg
H2O(s)
Ice
2.06
Wood
1.76
Concrete
0.88
Note:
Glass
0.84
C(liq) > C(solid)
Granite
0.79
Why?
NH3(l)
Ammonia
4.70
C2H5OH(l) Ethanol
2.46
Water
H2O(l)
4.184 (one of the highest)
(liquid)
H2O(g)
Steam
1.88
O2(g)
Oxygen
0.917
N2(g)
Nitrogen
1.04
© Prof. Zvi C. Koren
20.07.2010
Heating Substances
Let’s heat water or Cu from one temperature to another.
What are the factors that affect how much heat is needed in each case?
qbathtub

qcup
Why?
Start with “C”. Recall its units:
q = C  m  t (Note the overall units)
C  n  t
If: q = + (endothermic), q = – (exothermic)
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© Prof. Zvi C. Koren
20.07.2010
Heat
Transfer
For example:
A hot iron rod is placed in cold water.
Eventually, everything comes to thermal equilibrium
Recall: q = Cmt
Heat Balance ‫מאזן חום בתהליכי העברת חום‬
Σqi = 0 (assuming no heat loss to surroundings)
qhot + qcold = 0
qhot = – qcold
Problems involving 7 parameters in 1 equation
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© Prof. Zvi C. Koren
20.07.2010
Heating Involving Phase Changes (Physical processes)
Heating Curve (Cooling Curve)
g
Vaporization: l  v
Hvap
Tbp
Temp.
(H2O: 2260 J/g)
l
Tmp Fusion: s  l
Hfus
s
(H2O: 333 J/g)
Condensation: v  l, Hcond = ?
For H2O: C(s) = 2.06 J/g·deg
C(l) = 4.184 J/g·deg = 1.0 cal/ g·deg
C(g) = 1.88 J/g·deg
Solidification (or Crystallization): l  s, Hsolid (or cryst) =
time
For example: Calculate the heat (“thermal energy”) required for the
following process: 10.0 g, (ice, -10oC)  (steam, 120oC)
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© Prof. Zvi C. ©
Koren
20.07.2010
Prof. Zvi
C. Koren
Heating/Cooling Curve for Water.
1 mol water is heated from –100C to 1100C.
A constant heating rate of 100 J/min is assumed.
100
Temp. (0C)
vaporization
[Constant Pressure: P = f/A; f = w = mg]
Piston
m
s
0
sl
lv
l
g
fusion
Time (min)
12
100
600
© Prof. Zvi C. Koren
20.07.2010
Calculation of the Heats Involved With Each Step
in the Heating/Cooling Curve
100
Temp. (0C)
vaporization
Name
Value for H2O
C(l)
specific heat capacity
(of liquid)
1.00 cal/g·deg
4.18 J/g ·deg
C( l )
“C-bar”
ΔHfus
molar heat capacity
(of liquid)
18.00 cal/mol·deg
75.2 J/mol ·deg
heat of fusion
333 J/g
ΔHvap
heat of vaporization
2250 J/g
Find the values of C(s) and C(g) of H2O
0
fusion
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Symbol
100
Time (min)
600
© Prof. Zvi C. Koren
20.07.2010
Enthalpy
Enthalpy  H  E + PV
(a convenient definition for H)
open to
atmosphere:
Pinternal = pexternal = constant
pexternal
Pinternal
(Note: Absolute H can never be determined. Why?

H = E
+  (PV)
= q + w +  (PV)
= q - PV + PV, [P]
H = qP
qP
Rxn.
-------------------------------------------------------------------------------------------------V is constant
Note:
(not P)
For constant volume processes, V = 0:
 w = –PV = 0
And
E = qV
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qV
Rxn.
© Prof. Zvi C. Koren
20.07.2010
Calculating Heat of Reaction, Hrxn:
Energy of Reaction, Erxn
Recall,
H
 E
+
PV
 Hrxn = Erxn + (PV)rxn
(PV)rxn
=
(PV)products
(PV)products(s,l,g)

(PV)products(g)
=
-
(PV)reactants
(PV)reactants(s,l,g)
But, for a given quantity,
Vgas >> Vsolid,liquid
 PVgas >> PVsolid,liquid
(PV)reactants(g)
Assume ideal gases: PV = nRT,
 (nRT)products(g) (nRT)reactants(g)
=
RT· ng
 Hrxn

Erxn + RT· ng
H = qP
For example, consider the following rxn.:
2A(s) + B(l) + 4C(g)  2D(g) + E(g)
Hrxn

Erxn + RT· ng
E = qV
Hrxn

Erxn + RT· (3-4)
Hrxn

Erxn - RT
© Prof. Zvi C. Koren 20.07.2010
15
Calculating Heat of Reaction, Hrxn:
Hess’s Law of Heat Summation
If a rxn is made up of other rxns, then the heats are summed.
Why? Because H (like E) is a state function.
For example:
HA
Reactants  Products HA = HB + HC + HD + HE
HB
Problem:
HC
HD
HE
[Recall Born-Haber Cycle]
(1)  (2)  (3)
Find H for rxn (1),
(1) A + 2D  C,
H1 = ?
From the following data: (2) A + 2B  5C,
H2 = 50 kJ
(3)
B  2C + D, H3 = -75 kJ
Solution:
Because

rxn(1) = rxn(2) – 2 • rxn(3),
H1 = H2 – 2 • H3
= 50 – 2 • (-75) = 200 kJ.
By the way, of course it’s the same for E:
E1 = E2 - 2 • E3
But recall:
K1 =
K2 / K3 2
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© Prof. Zvi C. Koren
20.07.2010
Calculating Heat of Reaction, Hrxn:
Heats (or Enthalpies) of Formation
The Standard State
standard state of a substance (at a specific T)
= most stable state of the substance at 1 atm (or 1 bar) at that T.
For example, the standard state for nitrogen:
At 25oC:
N2 (diatomic) and a gas;
At 2,000,000oC: N (monatomic) and a gas, probably even N+;
At -270oC:
crystalline (solid)
At other temps., between Tmp and Tbp, the liquid is most stable
Another example, the standard state for carbon:
At 25oC:
graphite (solid); At 1 atm, graphite is more stable than diamond.
At 2,000,000oC: C (monatomic) and a gas;
(continued)
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© Prof. Zvi C. Koren
20.07.2010
(continued)
“Formation”
Formation is a rxn where:
1 mole of a compound is formed from its elements in their standard
state (most stable form at 1 atm)
For example, formation of CH4:
o
C(s,gr) + 2H2(g)  CH4(g), Hf (CH4) (measured from qP of the rxn)
o
Hf = standard heat (or enthalpy) of formation
Note: (any property) ≡ final – initial
 Hrxn ≡ Hfinal – Hinitial = ΣHproducts – ΣHreactants
So, in effect:
o
Hf (CH4) = H(CH4) – H(C) – 2H(H2)
So, the “enthalpy of formation” of a compound is in effect the “relative
enthalpy” of that compound, that is, its enthalpy relative to the enthalpies of
the elements from which it is composed.
(continued)
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© Prof. Zvi C. Koren
20.07.2010
(continued)
Consider the following combustion reaction:
C6H6(l) + 7½ O2(g)  6 CO2(g) + 3H2O(l)
Recall: (any property) ≡ final – initial
Hrxn ≡ Hfinal – Hinitial = ΣHproducts – ΣHreactants
Hcomb rxn = 6•H(CO2) + 3•H(H2O) – H(C6H6) – 7½ H(O2)
(This last equation is correct but not useful, because we can never know absolute H.
So, we must use relative enthalpies, that is enthalpies of formation, Hf.)
So, build the overall reaction from a series of “formation reactions” (in the Hess-way):
6 • (C + O2  CO2), 6•Hf(CO2)
3 • (H2 + ½ O2  H2O), 3•Hf(H2O)
(6 C + 3 H2  C6H6), Hf(C6H6)
C6H6(l) + 7½ O2(g)  6 CO2(g) + 3H2O(l), Hcomb rxn = ?
o
Hcomb
= 6•Hfo(CO2) + 3•Hfo(H2O) – Hfo(C6H6)
 We can generalize that for any reaction:
o = H o - H o
Hrxn
f
f
P
19
R
Hfo(element) ≡ 0
© Prof. Zvi C. Koren
20.07.2010
20
© Prof. Zvi C. Koren
20.07.2010
Calorimetry
thermometer
Measuring heats of combustion reactions
stirrer
water
O2
Heat Capacity of Calorimeter
(or “Calorimeter Constant”),
Ccalorimeter,
in “energy”/deg:
ignition
wires
“Bomb Calorimeter”
(constant V)
For every experiment use the same overall calorimeter mass.
Calibrate Calorimeter: Use a weighed mass of substance with known ΔHcomb.
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(continued)
© Prof. Zvi C. Koren
20.07.2010
(continued)
Measuring Heats of Reaction (qV) with a “Bomb Calorimeter”:
3 parts: Reaction, Bomb Apparatus, Water
Heat Balance Equation: Σqi = 0  qrxn + qbomb + qwater = 0
– qrxn =
qbomb
+
qwater
= Cbomb· tbomb + (C·m·t)water
[Recall: units of Cbomb = “energy”/deg]
For example, consider the following combustion of octane:
C8H18(l) + 12.5 O2(g)  8 CO2(g) + 9 H2O(l)
1.0 g of octane burns in a constant-volume calorimeter (Cbomb = 837 J/deg) containing
1.20 kg of water. The temperature rises from 25.00oC to 33.20oC. Calculate: (a) the
heat of comb., qV, for this quantity of octane, (b) Hcomb for 1 mole of octane.
Answer to (a):
–qrxn =
qbomb
+
qwater
= Cbomb · tbomb + (C · m · t)water = [Cbomb + (C·m)water]t; t  tf – ti
= [(837 J/deg) + (4.184 J/g·deg) (1200 g)] (8.20oC) = 48.1 kJ
 qV = – 48.1 kJ
Answer to (b): First calculate # of moles: 1.0 g (1 mol/114 g) = 0.0088 mol
 qV = – 48.1 kJ / 0.0088 mol  Ecomb = qV = – 5,500 kJ/mol
qP = Hcomb =
Ecomb +
RT(ng)
= -5.5x106 J/mol + (8.31 J/mol·K)(298 K)(– 4.5) = – 5.5x106 J/mol
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© Prof. Zvi C. Koren
20.07.2010