THERMODYNAMICS. THERMODYNAMICS: Basic Concepts Thermodynamics: (from the Greek therme, meaning...
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Transcript of THERMODYNAMICS. THERMODYNAMICS: Basic Concepts Thermodynamics: (from the Greek therme, meaning...
THERMODYNAMICS
THERMODYNAMICS: Basic Concepts
Thermodynamics: (from the Greek therme, meaning "heat" and, dynamis, meaning "power") is the study of energy conversion between heat and mechanical work, and subsequently the macroscopic variables such as temperature, volume and pressure
Energy: capacity to do work
Work: motion against an opposing force
System: part of the world which we have interest or being investigated.
Surrounding: is where we make our observations
Definition of Work:
• Work is motion against opposing force.• Work is defined as a force acting through a displacement x, the
displacement being in the direction of the force.
xfw
Consider a Gas Expansion Work:
Initial State Final State
P1, V1, T P2, V2, T
Consider a Gas Expansion Work:
:is pressure, opposing external The ,
)( 12
exP
hmgw
hhmgw
VPw
VVPhAPw
A
mgP
ex
exex
ex
)( 12
or
Consider a Gas Expansion Work:
2
1
1
2
2
1
lnln
,2
1
2
1
P
PnRT
V
VnRTw
V
dVnRTw
V
nRTP
PdVPw
PP
dVPw
v
v
in
in
v
v in
ex
v
v ex
and
gas, the of pressure theis
than greater mallyinfinitesi onlyis instant, every at
volume in increase ssimalinfiniteti against done work a For
in
Consider a Gas Expansion Work: the external pressure is infinitetissimally smaller than the internal pressure at all stages of the expansion - reversible process. A reversible change in thermodynamics is a change that can be reversed by an infinitesimal modification of a variable.
in reversible processes, there is maximum amount of work done that could possible extracted from a process
Gas Expansion Work: Sample ProblemA sample of 4.50 g of methane occupies 12.7 L at 310 K.(A)Calculate the work done when the gas expands isothermally against a constant external pressure of 200 Torr until its volume has increased by 3.3 L.(B) Calculate the work that would be done if the same expansion occurred reversibly.
J 87.97
atm L 1
J101.3 atm L 0.87
atm L 0.87
L) torr760
atm 1 torr
process) ble(irreversi formula the use we constantis pressure opposing since A.
:SOLUTION
:
w
w
w
w
VPw ex
3.3)(200(
Consider a Gas Expansion Work:
J -
KmolK J 1-mol g
g
:expansiongas reversible a for B.
:SOLUTION
1-1-
4.167
7.12
)7.123.3(ln310)3145.8()
16
5.4(
ln
w
L
Lw
V
VnRTw
i
f
Definition of HEAT:
Heat is the transfer of energy between two bodies that are at different temperatures.
Heat appears at the boundary of the system.
heat is transferred from the hotter object to the colder one.
Heat is path dependent.
heat specifics
etemperatur in change T mass,m
Tmsq
THE FIRST LAW OF THERMODYNAMICS:
“Energy can neither be created nor destroyed but can be converted from one form to another.” Law of Conservation of Energy.
ENERGY OF THE UNIVERSE:
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surrsysuniv
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EE
EEE
EEE
therefore
system, given a forenergies inchanges the and
0
If a system undergoes an energy change, the surroundings must also undergo a change equal in magnitude but opposite in direction
Total ENERGY of a system; Internal Energy, U
TOTAL ENERGY:
UE
UPEKEE
total
total
0 are PE and KE rest, at system a for
Internal Energy , U: energy associated with the chemical system depends on the thermodynamic parameters such as T, P, V, composition, etc. consists of translational, rotational, vibrational, electronic energies,
as well as intermolecular interactions
Internal Energy, U:
For a given system, we do not know the exact nature of U, we will be only interested in the changes in U for a particular process that the system undergoes.
12 UUU
Mathematical expression of the first law of thermodynamics:
wqU or for an infinitesimal change:
dwdqdU
The change in the internal energy of a system in a given process is the sum of the heat exchange, q, between the system and its surroundings and the work done on or by the system.
HEAT and WORK sign conventions:
Process Sign
Work done by the system on the surroundings
-
Work done on the system by the surroundings
+
Heat absorbed by the system from the surroundings
+
Heat absorbed by the surroundings from the system
-
Constant volume adiabatic bomb calorimeter
Adiabatic means no heat exchange with the surroundings
Sample inside an oxygen filled container is ignited by an electrical discharge.
Heat released is measured by the increase in the temperature of the water.
v
v
v
qU
VPq
wqU
V
0 VP constant at ,
,
ENTHALPHY
In processes carried out under constant P,
PVUHEnthalpy
PVUPVUq
VVPqUU
orVPwqU
p
p
v
or
,
)()(
),(
,
1122
1212
Total enthalpy of a system can not be measured directly, so the change in enthalpy, ΔH, is a more useful value than H itself.
For a constant P process;
VPdUdH
VPUH
change, simalinfiniteti for
Entalphy, H vs. Internal Energy, U
Consider the following reaction:
2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g)
Heat evolved by the reaction is 367.5 kJ, reaction takes place at constant pressure,
qp= H = -367.5KJ.
Internal Energy U:
The volume of the H2(1 mole) generated by the reaction occupies 24.5 L, therefore –PV = -24.5 L at or -2.5 kJ Finally,
U = -367.5 kJ – 2.5 kJ
U = -370 kJ.
- Difference is due to expansion work.
VPHU
Constant Pressure Calorimeter:
A constant-pressure calorimeter made of two plastic cups. The outer cup helps insulate the reacting mixture from the surroundings. Two solutions of known volume containing the reactants at the same temperature are carefully mixed in the calorimeter. The heat produced or absorbed by the reaction can be determined by the temperature change, the quantities and specific heats of the solutions used, and the heat capacity of the calorimeter.
HEAT CAPACITIES:
Calorimetry: measurement of heat changes in chemical and physical processes.
When heat is added to the system, the corresponding temperature rise will depend ona)The amount of heat delivered,b)The amount of the substance,c)The chemical nature and the physical state of the substance,d)The conditions at which the energy is added to the system.
HEAT CAPACITIES:
Thus for a given amount of a substance, change in temperature is related to the heat added:
capacity heat molaris C mol K J
capacity heatis C K J
1-1-
1-
,
,
,
Tn
qC
n
CT
qC
orTCq
HEAT CAPACITY at constant volume
TCnTTCdTCU
dTCdUT
U
T
qC
vv
T
T v
v
vv
)( 12
2
1
or
q U volume, constant at v
HEAT CAPACITY at constant pressure processes
TCnTTCdTCH
dtCdHT
H
T
qC
pp
T
T p
p
pp
)( 12
2
1
or
q H process, pressure constant at p
for gases, Cp > Cv because of the work to be done to the surroundings in constant pressure processes
for condensed phases, Cp and Cv are identical for most purposes
for ideal gases, RCC vp
THERMOCHEMICAL EQUATIONS:
SAMPLE PROBLEM:
How much heat is released when 75 g of octane is burned completely if the enthalpy of combustion is –5,500 kJ/mol C8H18?
C8H18 + 25/2 O2 8CO2 + 9H2O
THERMOCHEMISTRY:
Heat of reaction is the heat change in the transformation of reactants at a given temperature and pressure to products at the same conditions.
For constant pressure processes, the heat of reaction qp is equal to enthalpy change of the reaction, rH.
positive.is H gs.surroundin the from
heatabsorbs system :reactions cEndothermi
negative.is H gs.surroundin
to heat offgives system :reactions rmicExothe
r
r
THERMOCHEMISTRY:
2H2(g) + O2(g) 2H2O (g)
241.8 kJ of heat is given off. The enthalpy change for this process is called standard enthalphy of reaction, rHo
In general, standard enthalpy change of a chemical reaction is the total enthalpy of the products minus the total enthalpy of the reactants
tcoefficien tricstoichiomeis v
enthalpy; molar standard theis H
)reactants(Hv)products(HvHo
ooor
THERMOCHEMISTRY:
)tstanreac(Hv)products(HvH
general; in or
))B(Hb)A(Ha()D(Hd)C(HcH
:is reaction the of enthalpy standard the
dDcCbB Aa
of
of
or
ooooor
formation. of enthalpy molar standard theis H of
- is the enthalpy change when 1 mole of a compound is formed from its constituent elements at 1 bar and 298 K.
THERMOCHEMISTRY:
0))g(O(H ;0))g(H(H
0H :etemperatur particular a at
forms allotropic stable most their inelements for
:H
2o
f2o
f
of
of
H2(g) + 1/2O2(g) H2O(g)
kJ/mole 8.241H of
FORMATION REACTIONS:
formation of 1 mole of a compound from its constituent elements at 1 bar and 298 K.
Example: Formation reaction of NaOH:
Na(s) + ½ O2(g) + ½ H2(g) NaOH(s)
MEASUREMENTS OF of H
DIRECT METHOD: measure fHo of direct synthesis from their elements:
2H2(g) + O2(g) H2O(g)
8.241H
))g(O(H))g(H(H2))g(OH(H2H
)tstanreac(Hv)products(HvH
or
2o
f2o
f2o
fo
r
of
of
or
MEASUREMENTS OF of H
INDIRECT METHOD: 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.
of H for carbon monoxide:
C(graphite) + 1/2O2(g) CO(g)
C(graphite) + O2(g) CO2(g)
CO + 1/2O2(g) CO2(g)
kJ/mole 5.393H or
kJ/mole 0.283H or
MEASUREMENTS OF of H
of H for carbon monoxide:
C(graphite) + O2(g) CO2(g)
CO2(g) 1/2O2(g) + CO(g)
kJ/mole 5.393H or
kJ/mole 0.283H or
C(graphite) + 1/2O2(g) CO(g) kJ/mole 5.110H or
MEASUREMENTS OF of H
C(graphite) + O2(g) CO2(g) + 393.5kJ
kJ/mole 5.393 or HC(graphite) + O2(g) CO2(g)
• Enthalpy changes are additive. • the of the reverse reaction will have the opposite sign• multiplying a reaction by a factor, you also multiply the by the same factor.
or H
or H
Hess’s Law can be rephrased as: ‘ The standard enthalpy of an overall reaction is the sum of the standard enthalpies of the individual reactions into which the reaction might be divided.’
SAMPLE PROBLEM:
Calculate the standard molar enthalpy of formation of acetylene (C2H2) from its elements.
kJ/mole 8.2598H or
kJ/mole 5.393H or
2C(graphite) + H2(g) C2H2(g)
The equations for combustion and the corresponding enthalpy changes are:
(1) C(graphite) + 1/2O2(g) CO2(g)
(2) H2(g) + 1/2O2(g) H2O(l)
(3) 2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
kJ/mole 8.285H or
SAMPLE PROBLEM:o
f H for carbon monoxide:
C(graphite) + O2(g) CO2(g)
CO2(g) 1/2O2(g) + CO(g)
kJ/mole 5.393H or
kJ/mole 0.283H or
C(graphite) + 1/2O2(g) CO(g) kJ/mole 5.110H or
SAMPLE PROBLEM:
TEMPERATURE DEPENDENCE OF ENTHALPY CHANGE
KIRCHHOFFS’s LAW: The difference in enthalpies of a reaction at two different temperatures, T1 and T2, is just the difference in the enthalpies of heating the products and reactants from T1 to T2.
)( 1212 TTCHH pr r
Where Cp is the difference in molar heat capacities between the products and reactants
Bond Dissociation Enthalpy:
Bond Enthalpy: change in enthalpy when bonds form or break in diatomic molecule
Bond Dissociation Enthalpy: average change in enthalpy in polyatomic molecules when individual bonds form or break.
N2(g) 2N(g)
H2O(g) H(g) + OH(g) rHo = 502 kJ/molOH(g) H(g) + O(g) rHo = 427 kJ/mol
SAMPLE PROBLEM:
ENTHALPY CHANGE and BOND ENERGIES:
The Enthalpy of a Reaction can be estimated by the enthalpies of the total number of bonds broken and formed in the reaction.
released energy total- input energy total
(products))(reactantsr
BEBEH o
SAMPLE PROBLEM:
Estimate the enthalpy of combustion of methane:
CH4(g) + O2(g) CO2(g) + H2O(g)
Using bond enthalpies in Table 4.4. Compare your result with that calculated from the enthalpies of formation of products and reactants.
C-H 410 kJ/moleO=O 494 kJ/moleC=O 563 kJ/moleH-O 460 kJ/mole
The Second Law of Thermodynamics
LIMITATION OF THE FIRST LAW:
Does not address whether a particular process is spontaneous or not.
Deals only with changes in energy.
Consider this examples:
Drop a rock from waist-high height, rock will fall spontaneously
Plunger of a spray is presses , gas comes out spontaneously
Metallic sodium is placed in a jar with chlorine gas, reaction occurs
Spontaneous Processes:
Why does the color spread when placing a drop of dye in a glass of clean water?
SPONTANEOUS PROCESSES:
Spontaneous processes are those that can proceed without any outside intervention.
The gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously go back to B.
SPONTANEOUS PROCESSES:
Processes that are spontaneous in one direction are not spontaneous in the reverse direction.
SPONTANEOUS PROCESSES
Processes that are spontaneous at one temperature may be not spontaneous at other temperatures.Above 0C it is spontaneous for ice to melt.Below 0C the reverse process is spontaneous
SPONTANEOUS PROCESSES:
Changes in the extent of disorder.
When we spill a bowl of sugar, why do the grains go everywhere and cause such a mess?
natures’s way to seek disorder. It is easy to create disorder; difficult to create order.
ENTROPY:
Entropy can be thought of as a measure of the randomness or disorder of a system.
It is related to the various modes of motion in molecules.
The Concept of Entropy (S)
Entropy refers to the state of order.
A change in order is a change in the number of ways of arranging the particles and dispersing their energy of motion, and it is a key factor in determining the direction of a spontaneous process.
solid liquid gasmore order less order
crystal + liquid ions in solutionmore order less order
more order less order
crystal + crystal gases + ions in solution
PROCESSES THAT RESULT IN PREDICTABLE ENTROPY CHANGES:
1. Phase Changes
1. Temperature Changes
1. Volume Changes
1. Mixing of substances
1. Increase in number of particles
1. Changes in the number of moles of gaseous components
7. Atomic size or molecular complexity
Spontaneity and the Sign of S
Why does a room with a fragrance bottle at the other end of the room is suddenly filled with the aroma?
Limonene (l) Limonene (g)
S(process) = S (final state) - S(initial state)
A spontaneous process is accompanied by S of positive sign.
Reaction Spontaneity by Inspection
Why do damp clothes become dry when hung outside?
S(g) >> S(l) > S(s)
By inspection alone, decide whether the sublimation of solid carbon dioxide is spontaneous or not. How about the condensation of water?
THERMODYNAMIC DEFINITION OF ENTROPY:
ST
q
V
VlnnR
T
q
V
VnRTln q
process reversible a for
-wq
process, isothermal In
rev
1
2rev
1
2rev
Although for the expansion of gases, they are
applicable to all types of processes at constant
temperature
The Second Law of Thermodynamics:
0SSS
,Thus
0)1V/2Vln(nR
SSS
processes leirreversib for
0T
)V/Vln(nRT
T
)V/Vln(nRTT
q
T
q
SSS
processes; reversible for
:universe the of entropy in Change
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The entropy of an isolated system increases in an irreversible processes and remain unchanged in a reversible process.
ENTROPY CHANGES PHASE CHANGE:
T
HS
T
HS
Hq so,process pressure constant
OHOH
:Consider
vapvap
fusfus
fusrev
)l(2atm1,C0
)s(2
o
SAMPLE PROBLEM
The enthalpy of vaporization of methanol is 35.27 kJ/mol at its normal boiling point of 64.1oC. Calculate (a) the entropy of vaporization of methanol at this temperature
SAMPLE PROBLEM:
Calculate the change in entropy when 1.0 mole of ice at -10oC is heated until it is a liquid at 25oC.
mol-J/K 6.33O(g)H C
mol-J/K 3.75O(l)H C
mol-J/K 7.37O(s)H C
mol-J/K 4.109S
molK/J 0.22S
2p
2p
2p
vap
fus
THIRD LAW OF THERMODYNAMICS:
01lnkS
1W K,0 at WlnkS
dTT
CS
zero; to etemperatur the lower we Suppose
b
b
T
0
p
Every substance has a finite positive entropy, but at absolute zero the entropy maybe come zero, and it does in case of pure perfect crystalline substance.
THIRD LAW OF THERMODYNAMICS:
The entropy of a perfect crystal at 0 K is zero.
It is impossible to reach a temperature of absolute zero
It is impossible to have a (Carnot) efficiency equal to 100% (this would imply Tc = 0).
THIRD LAW OF THERMODYNAMICS:
T = 0, S = 0 T > 0, S > 0
ENTROPY OF CHEMICAL REACTIONS:
)tstanreac(Sv)products(SvS
(B)Sb-(A)Sa-(D)Sd(C)ScS
:by givenis reaction of entropy eth
dDcCbBaA
reaction, alhypothetic a For
ooor
ooooor
SAMPLE PROBLEM:
Calculate the value of the standard molar entropy changes for the following reactions at 298 K.
210.6 :NO 205, :O 191.5, :N
69.9 :OH 205, :O 130.6, :H
213.6 :CO 39.8, :CaO 92.9, :CaCO
mol) (J/K :1bar) K,(298 values S
2NO(g) (g)O (g)N c)
O(l)2H (g)O (g)2H b)
(g)CO CaO(s) (s)CaCO a)
22
222
23
o
22
222
23
The GIBBS Energy:
ST-HG :EnergyGibbs
TS-HG :function a define now we
0TdS-dH
T by multiply
0T
dqdS
0dSdSdS
not? ors spontaneouIs
l)O(gH (g)O(g)H
reaction, the Consider
syssys
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2
22
THE GIBBS ENERGY: Factors affecting
H S G
+ + Positive at low Temp; negative at high Temp; Reaction is spontaneous at forward at high T and spontaneous in reverse direction at low temperature.
+ - Positive at all temperatures, Reaction is spontaneous in the reverse reaction at ll temperatures
- + Negative at all temperatures. Reaction is spontaneous in the forward direction at all T.
- - Negative at low temperatures; positive at high temperatures. Reaction is spontaneous at Low temperatures. Tends to reverse at high temperatures.
The HELMHOLTZ Energy:
0A
:mequilibriu and yspontaneit for criteria
processes. volume and etemperatur constant for
Energy Helmholtz
TS-UA
sys
STANDARD MOLAR GIBBS ENERGY OF FORMATION
)tstanreac(Gv)products(GvG
(B)Gb-(A)Ga-(D)Gd(C)GcG
:by givenis reaction of EnergyGibbs molar standard the
dDcCbBaA
reaction, alhypothetic a For
ooor
ooooor
GIBBS FREE ENERGY, GGIBBS FREE ENERGY, G
∆ ∆GGoo = ∆H = ∆Hoo - T∆S - T∆Soo
Two methods of calculating ∆GTwo methods of calculating ∆Goo
A. Determine ∆HA. Determine ∆Hoorxnrxn and ∆S and ∆Soo
rxnrxn and use GIbbs equation. and use GIbbs equation.
B. Use tabulated values of B. Use tabulated values of free energies of formation, ∆Gfree energies of formation, ∆G ffoo..
∆∆GGoorxnrxn = = ∆G ∆Gff
oo (products) - (products) - ∆G ∆Gffoo (reactants) (reactants)∆∆GGoo
rxnrxn = = ∆G ∆Gffoo (products) - (products) - ∆G ∆Gff
oo (reactants) (reactants)
FREE ENERGIES OF FORMATIONFREE ENERGIES OF FORMATION
Note that ∆G˚Note that ∆G˚ff for an element = 0 for an element = 0
SAMPLE CALCULATION,SAMPLE CALCULATION, ∆G ∆Goorxnrxn
For the combustion of acetyleneFor the combustion of acetylene
CC22HH22(g) + 5/2 O(g) + 5/2 O22(g) --> 2 CO(g) --> 2 CO22(g) + H(g) + H22O(g)O(g)
a)a) by inspection is the reaction spontaneous or not?by inspection is the reaction spontaneous or not?
b)b) Calculate the Calculate the ∆G∆Goorxnrxn using standard molar enthalpies and entropies. using standard molar enthalpies and entropies.
c) Is the reaction spontaneous or not? Is it entropy or enthalpy driven?c) Is the reaction spontaneous or not? Is it entropy or enthalpy driven?
∆∆H (acetylene) 52.26 s = 219.56 g = 68.15H (acetylene) 52.26 s = 219.56 g = 68.15
O g = 205.14O g = 205.14
Co2 h=-393.51 s = 213.74 g = -394.36Co2 h=-393.51 s = 213.74 g = -394.36
H20 h = -241.82 s 188.83 g = -228.57H20 h = -241.82 s 188.83 g = -228.57
FREE ENERGY AND TEMPERATUREFREE ENERGY AND TEMPERATUREIron metal can be produced by reducing its ore (Iron(III) oxide with Iron metal can be produced by reducing its ore (Iron(III) oxide with
graphite:graphite:
2 Fe2 Fe22OO33(s) + 3 C(s) ---> 4 Fe(s) + 3 CO(s) + 3 C(s) ---> 4 Fe(s) + 3 CO22(g)(g)
∆∆HHoorxnrxn = +467.9 kJ = +467.9 kJ ∆S∆Soo
rxnrxn = +560.3 J/K = +560.3 J/K
∆∆GGoorxnrxn = +300.8 kJ = +300.8 kJ
A) Is the reaction spontaneous or not?A) Is the reaction spontaneous or not?
B) At what temperature will the reaction become spontaneous?B) At what temperature will the reaction become spontaneous?At what T does ∆GAt what T does ∆Goo
rxnrxn just change from being (+) to being (-)? just change from being (+) to being (-)?
When ∆GWhen ∆Goorxnrxn = 0 = ∆H = 0 = ∆Hoo
rxnrxn - T∆S - T∆Soorxnrxn
FACT: ∆GFACT: ∆Goorxnrxn is the change in free energy when pure is the change in free energy when pure
reactants convert COMPLETELY to pure products.reactants convert COMPLETELY to pure products. FACT: Product-favored systems have FACT: Product-favored systems have
KKeqeq > 1. > 1.
Therefore, both ∆G˚Therefore, both ∆G˚rxnrxn and K and Keqeq are related to reaction are related to reaction
favorability.favorability.
Thermodynamics and KThermodynamics and Keqeq
KKeqeq is related to reaction favorability and so to ∆G is related to reaction favorability and so to ∆Goorxnrxn..
The larger the value of K the more negative the value The larger the value of K the more negative the value of ∆Gof ∆Goo
rxnrxn
∆ ∆GGoorxnrxn = - RT lnK = - RT lnK
where R = 8.31 J/K•molwhere R = 8.31 J/K•mol
THERMODYNAMICS AND KTHERMODYNAMICS AND Keqeq