Thermodynamics and Entropy
Transcript of Thermodynamics and Entropy
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Chapters 29 and 35
Thermochemistry
and
Chemical Thermodynamics
Copyright (c) 2011 by Michael A. Janusa, PhD. All rights reserved.
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Thermochemistry
• Thermochemistry is the study of the energy effects that
accompany chemical reactions.
• Why do chemical reactions occur? What is the driving force of
rxn?
• Answer: Stability, wants to get to lower E. For a rxn to take
place spontaneously the products of reaction must be more
stable (lower E) than the starting reactants. Nonspontaneous
means never happen by self.
E
R
P
release E, spon
higher E, less stable, more reactive
E R
P absorb E, nonspon
lower E, more stable, less reactive
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29.1 Reaction Enthalpy
• In chemical reactions, heat is often transferred from
the “system or reaction” to its “surroundings,” or vice
versa.
• system - the substance or mixture of substances
under study in which a change occurs.
• The surroundings are everything in the vicinity of
the thermodynamic system.
system or rxn
surroundings
(
+ into system
- out system
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Heat of Reaction
• Heat flow is defined as the energy that flows
into or out of a system. We follow heat flow
by watching the difference in temperature
between the system and its surroundings.
• Often we follow the surroundings temp
(solvent) and must realize that the opposite is
happening to the system. If system is
absorbing heat from the surroundings than
the temp of the surroundings must be
decreasing. Tsystem (+) Tsurr (-)
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Heat of Reaction • Heat flow or heat of reaction is denoted by the
symbol q and is the amount of heat required to return
a system to the given temperature at the completion
of the reaction.
For an endothermic rxn the sign of q is positive;
heat is absorbed by the system from the
surroundings.
E
P absorb heat, nonspon (endo)
R
q > 0
Surroundings
+q
Tsystem
Tsurr
System
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Heat of Reaction
q < 0
-q
System
Surroundings
For an exothermic rxn, the sign of q is negative;
heat is evolved (released) by the system to the
surroundings.
Tsystem
Tsurr E
R
P
release heat, spon (exo)
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Enthalpy and Enthalpy Change
• The heat absorbed or evolved by a reaction
depends on the conditions under which it
occurs. ex. pressure
• Usually, a reaction takes place in an open vessel,
and therefore under the constant pressure of the
atmosphere.
• heat of this type of reaction is denoted qp; this
heat at constant pressure is named enthalpy and
given symbol H. H is the heat flow at constant
pressure.
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– an extensive property - depends on the quantity of
substance.
– Enthalpy is a state function, a property of a system
that depends only on its present state and is
independent of any previous history of the system.
Enthalpy and Enthalpy Change • Enthalpy, denoted H, is an extensive property of a
substance that can be used to obtain the heat
absorbed or evolved in a chemical reaction at
constant pressure.
o o o
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• The reaction enthalpy for a reaction at a
given temperature and pressure
)reactants((products) HHH
Enthalpy and Enthalpy Change
)i((final) nitialHHH
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• As we already stated the reaction enthalpy is equal to the heat
of reaction at constant pressure. This represents the entire change
in internal energy ( U) minus any expansion “work” done by the
system; therefore we can define enthalpy and internal work by the
• 1st law of thermodynamics:
• In any process, the total change in energy of the system, U, is
equal to the sum of the heat absorbed, q, and the work, w, done by
the system.
• U = qp + w = H + w
Enthalpy and Enthalpy Change
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– Changes in E manifest themselves as exchanges
of energy between the system and surroundings.
– These exchanges of energy are of two kinds; heat
and work - must account for both.
– Heat is energy that moves into or out of a system
because of a temperature difference between
system and surroundings.
– Work is the energy exchange that results when a
force F moves an object through a distance d;
work (w) = F d
In chemical systems, work is defined as a change in
volume at a given pressure, that is:
VPw
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negative sign is to keep sign correct in terms of system. For
expansion, V, will be a positive value but expansion involves
the system doing work on the surroundings and a decrease in
internal energy -- negative keeps it neg. For contraction work,
V, will be a negative value but contraction involves the
surroundings doing work on the system and an increase in
internal energy -- negative keeps it positive (- x - = +).
Giving us the 1st law of thermo is more useful form:
VPHU realize absorb heat (+)
release or evolved heat (-) HW 44
VPw
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29.3 Thermochemical Equations
• A thermochemical equation is the chemical equation for a reaction (including phase labels {important}) in which the equation is given a molar interpretation, and the enthalpy of reaction for these molar amounts is written directly after the equation.
kJ -91.8H );g(NH2)g(H3)g(N 322
If H has a superscript like Ho, means thermo standard
conditions -- 25oC (298K) and 1 atm.
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• The following are two important rules for
manipulating thermochemical equations:
– 1.) When a thermochemical equation is
multiplied by any factor, the value of H for the
new equation is obtained by multiplying the H in
the original equation by that same factor.
– 2.) When a chemical equation is reversed, the
value of H is reversed in sign.
Thermochemical Equations
kJ 967.4 H ; )(4)(2)(4
kJ 483.7- H ; )(2)()(2
o
222
o
222
gOHgOgH
gOHgOgH
kJ 483.7 H ; )()(2)(2
kJ 483.7- H ; )(2)()(2
o
222
o
222
gOgHgOH
gOHgOgH exo
endo
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• Hess’s law of heat summation states that for a chemical equation that can be written as the sum of two or more steps, the enthalpy change for the overall equation is the sum of the enthalpy changes for the individual steps.
• Basically, R & P in individual steps can be added like algebraic quantities in determining overall equation and enthalpy change.
29.5 Hess’s Law
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simple example :
Given: A + D E + C H = X kJ
2A + B 2C H = Y kJ
Question: 2D B + 2E H = ?
2A + 2D 2E + 2C H = 2X kJ
2C 2A + B H = -Y kJ
2D B + 2E H = 2X – Y kJ
_______________________________________
1. Correct side?
2. Correct # moles?
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• For example, suppose you are given the
following GIVEN data:
Hess’s Law
kJ -297H );g(SO)g(O)s(So
22
kJ 198H );g(O)g(SO2)g(SO2o
223
• use these data to obtain the enthalpy change for
the following reaction?
?H );g(SO2)g(O3)s(S2o
32
x2
flip
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• If we multiply the first equation by 2 and
reverse the second equation, they will
sum together to become the third.
(2)kJ) -297(H );g(SO2)g(O2)s(S2o
22
(-1)kJ) 198(H );g(SO2)g(O)g(SO2o
322
kJ -792H );g(SO2)g(O3)s(S2o
32
HW 45
kJ -297H );g(SO)g(O)s(So
22
kJ 198H );g(O)g(SO2)g(SO2o
223
x2
flip
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• The standard enthalpy of formation of a substance,
denoted Hfo, is the enthalpy change for the
formation of one mole of a substance in its
standard state from its component elements in
their standard state (298K & 1 atm).
– Note that the standard enthalpy of formation for a
pure element in its standard state and H+ is zero.
This means elements in their standard state has
Hfo = 0: metals - solids, diatomic gases, H+ ion.
29.6 Standard Enthalpies of
Formation
(molecular scale)
Ag (s) + ½ Cl2 (g) AgCl (s) Hfo AgCl
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• Another way to determine heat of
reaction is the The law of summation
of heats of formation which states that
the enthalpy of a reaction is equal to the
total formation energy of the products
minus that of the reactants.
is the mathematical symbol meaning “the sum
of”, and n is the coefficients of the substances in
the chemical equation.
)reactants()products( o
f
o
f
o HnHnH
Standard Enthalpies of Formation
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Ex. Generic Law of Summation
aA + bB cC + dD
)reactants()products( o
f
o
f
o HnHnH
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A Problem to Consider
– What is the standard reaction enthalpy , Horxn, for
this reaction?
)g(OH6)g(NO4)g(O5)g(NH4 223
molkJ /9.45 0 3.90 8.241:o
fH
)reactants()products( o
f
o
f
o HnHnH
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• Using the summation law:
– Be careful of arithmetic signs as they are a
likely source of mistakes.
)reactants(Hm)products(HnHof
of
o
)]/0(5)/9.45(4[
)]/8.241(6)/3.90(4[
2233
22
molOkJmolOmolNHkJmolNH
OmolHkJOmolHmolNOkJmolNOH o
kJ 906Ho HW 46
)g(OH6)g(NO4)g(O5)g(NH4 223
molkJ /9.45 0 3.90 8.241:o
fH
kJkJkJ
kJkJkJkJ
9066.1836.1089
]0)6.183[()]8.1450(2.361[
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– Entropy, S , is a thermodynamic quantity
that is a measure of the randomness or
disorder of a system.
– The SI unit of entropy is joules per Kelvin
(J/K) and, like enthalpy, is a state function.
35.1.2 The Second Law of
Thermodynamics • The second law of thermodynamics
addresses questions about spontaneity
in terms of a quantity called entropy.
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E
R
P
release E, spon (exo)
E R
P absorb E, nonspon (endo)
Most soluble salts dissolve in water spontaneously; however, most
soluble salts dissolve by an endothermic process.
NH4NO3 (s) NH4+ (aq) + NO3
- (aq) H = 28.1 kJ
There is an increase in molecular disorder or randomness of the system.
Solids: high order/low disorder, high energy
Liquids: middle order/low disorder, medium energy
Gases: low order/high disorder, low energy
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entropy (S) - is a thermodynamic quantity that is a measure of how dispersed the
energy of a system is among the different possible ways that system can contain
energy, typically in J/K units.
One example of entropy is the amount of molecular disorder or randomness in the
system.
S increases as disorder increases and energy decreases
gases have high disorder, low energy
solids have low disorder, high energy
We typically follow the change in entropy in the system so we treat it as a state
property and measure S = Sfinal - Sinitial
+ S = increase in entropy, i.e. disorder increased; - U
- S = decrease in entropy, ie. disorder decreased ; + U
This gets us to the second law of thermo
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Entropy and the Second Law of
Thermodynamics
• The second law of thermodynamics
states that the total entropy of a system
and its surroundings increases for a
spontaneous process.
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The tendency of a system to increase its entropy (+ S) is the second important
factor in determining the spontaneity of a chemical or physical change in addition
to H.
recap:
spontaneous process: (system goes to lower energy state)
favored by - H (exo)
favored by + S (ie. increase disorder)
nonspontaneous process: (system goes to higher energy state)
favored by + H (endo)
favored by - S (ie. decrease in disorder)
Do both need to be true for spon rxn? No, remember soluble salt dissolving
example. The larger term will dictate overall process.
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– As temperature is raised the substance
becomes more disordered as it absorbs
heat and becomes a liquid then a gas,
where entropy > 0; S increases as temp
increase.
– The entropy of a substance is determined
by measuring how much heat is required to
change its temperature per Kelvin degree
(J/K).
35.4 Third Law of Thermodynamics • The third law of thermodynamics states that the
entropy of all perfect crystalline substances
approaches zero as the temperature approaches
absolute zero (Kelvin).
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– Standard state implies 25 oC (298K), 1 atm
pressure, and 1 M for dissolved
substances.(Thermo standard state)
35.5 Standard Reaction Entropy
• The standard entropy of a substance
or ion, also called its absolute entropy,
So, is the entropy value for the standard
state of the species. Similar to heats of
formation, Hfo , except on absolute not
relative scale.
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– This means that elements have nonzero values
for entropy (absolute scale), unlike standard
enthalpies of formation, Hfo , which by
convention, are zero (relative scale).
Standard Entropies and the Third Law
of Thermodynamics
– The symbol So, rather than So, is used for
standard entropies to emphasize that they
originate from the third law and absolute not
relative values.
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)reactants()products( ooo SmnSS
– Even without knowing the values for the entropies
of substances, you can sometimes predict the sign
of So for a reaction.
Entropy Change for a Reaction
• You can calculate the entropy change
for a reaction using a summation law,
similar to the way you obtained Hfo.
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1. A reaction in which a molecule is broken into
two or more smaller molecules.
The entropy usually increases in the
following situations:
Entropy Change for a Reaction
2. A reaction in which there is an increase in the
moles of gases.
3. A process in which a solid changes to liquid
or gas, or a liquid changes to gas.
AB A + B + S
A(g) B(g) + C(g) + S
A(s) B(l) or B(g) + S
B(l) C(g) + S
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Predict S and spon/nonspon based only on entropy for the following rxns:
C2H4 (g) + Br2 (g) BrCH2CH2Br (l)
2 C2H6 (g) + 7 O2 (g) 4 CO2 (g) + 6 H2O (g)
C6H12O6 (s) 2 C2H5OH (l) + 2 CO2 (g)
HW 47
gas to liquid; decrease in disorder; - S;
nonspon based on S only
9 mols gas to 10 mols of gas; increase in
disorder; + S; spon based on S only
solid to liquid/gas (decompose); increase in
disorder; + S; spon based on S only
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– The calculation is similar to that used to obtain
Ho from standard enthalpies of formation.
)l(OH)aq(CONHNH)g(CO)g(NH2 22223
A Problem To Consider
• Calculate the change in entropy, So, at 25oC
for the reaction in which urea is formed from
NH3 and CO2.
Gas to liquid; decrease in disorder; predict - S
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)()()()(2 22223 lOHaqCONHNHgCOgNH
So: 193 J/mol.K 214 174 70
A Problem To Consider
)reactants()products( ooo SmnSS
J/K 356)]/214)(1()/193)(2[(
)]/70)(1()/174)(1[(
23
222
molKJmolCOmolKJmolNH
molKJOmolHmolKJCONHmolNHS o
decrease in disorder as predicted
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– This quantity gives a direct criterion for
spontaneity of reaction.
35.6 Gibbs Free Energy
• The question arises as to how do we decide if enthalpy or
entropy dictates the spontaneity of a reaction. What is the
relationship between H and S?
• The American physicist J. Willard Gibbs introduced the
concept of free energy (sometimes called the Gibbs free
energy), G, which is a thermodynamic quantity defined by
the equation
G= H-T S T – Kelvin scale
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At a given temperature and pressure
G = 0, the reaction gives an equilibrium mixture with
significant amounts of both reactants and products (Temp
transfer point where reaction switches spon/nonspon)
G > 0 , the reaction is nonspontaneous as written, and
reactants do not give significant amounts of product at
equilibrium.
G < 0 , the reaction is spontaneous as written, and the
reactants transform almost entirely to products when
equilibrium is reached.
STHG
Free Energy and Spontaneity • Changes in H an S during a reaction result in a change in free
energy, G , given by the equation
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H S G Description
– (exo)
spon
+disorder
spon
–
spon
Spontaneous at all T
• Lets look at relationship among the signs of H, S and G and
spontaneity. Note that temperature will dictate which will rule. Also realize
T is in K meaning no negative temp.
enthalpy rules at low temp but entropy at very high T
STHG
+ (endo)
non
–disorder
non
+
non
Nonspontaneous at all T
–
– (exo)
Spon
–disorder
non
+ or –
Spontaneous at low T (room); H > T S; - G
Nonspontaneous at high T (1000K); H < T S
+ G
+ (endo)
Non
+disorder
spon
+ or –
Nonspontaneous at low T; H > T S; + G
Spontaneous at high T; H < T S; - G
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– The next example illustrates the calculation
of the standard free energy change, Go,
from Ho and So.
oooSTHG
35.7 Gibbs Energy and Equilibrium
• The standard free energy change, Go,
is the free energy change that occurs
when reactants and products are in their
standard states.
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)g(NH2)g(H3)g(N 322
So: 130.6 191.5 193 J/mol K
Hfo: 0 0 -45.9 kJ/mol
A Problem To Consider
• What is the standard free energy change, Go,
for the following reaction at 25oC?
predict
H, spon
S, nonspon
G, spon
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)reactants(Hm)products(HnHof
of
o
kJmolkJmolNH 8.91 ]0[)]/9.45(2[ 3
)reactants()products( ooo SmnSS
kJ/K -0.197J/K -197 )]/6.130)(3(
)/5.191)(1[()]/193)(2[(
2
23
molKJmolH
molKJmolNmolKJmolNH
– Now substitute into our equation for Go. Note that
So is converted to kJ/K and Kelvin for temp. ooo
STHG
kJ/K) 0.197K)( (298kJ 91.8
kJ 33.1 spon rxn as written
)(2)(3)( 322 gNHgHgN
So: 130.6 191.5 193 J/mol K
Hfo: 0 0 -45.9 kJ/mol
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– By tabulating Gfo for substances, you can
calculate the Go for a reaction by using a
summation law.
)reactants(Gm)products(GnG
of
of
o
Standard Free Energies of Formation
• The standard free energy of formation,
Gfo, of a substance is the free energy
change that occurs when 1 mol of a
substance is formed from its elements in their
stablest states at 1 atm pressure and 25oC.
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44
)(3)(2)(3)( 22252 gOHgCOgOlOHHC
Gfo: -174.8 0 -394.4 -228.6 kJ/mol
A Problem To Consider • Calculate Go for the following reaction at
25oC using std. free energies of formation.
)reactants(Gm)products(GnGof
of
o
]0)/8.174)(1[(
)]/6.228)(3()/4.394)(2[(
52
22
molkJOHHmolC
molkJOmolHmolkJmolCOGo
kJ 8.1299oG spon rxn
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– Here Q is the thermodynamic form of the
reaction quotient ([products]/[reactants] not
necessarily at equil); T in kelvin; R=8.31 J/molK.
QlnRTGGo
Relating Go to the Equilibrium
Constant • The free energy change ( G) when
reactants are in non-standard states
(meaning other than 298K, 1 atm
pressure or 1 M) is related to the
standard free energy change, Go, by
the following equation.
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46
– G represents an instantaneous change in
free energy at some point in the reaction
approaching equilibrium G=0.
Relating Go to the Equilibrium
Constant
– At equilibrium, G=0 and the reaction
quotient Q becomes the equilibrium
constant K.
KlnRTG0o
QRTGG o ln
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– When K > 1 (meaning equil lies to the
right), the ln K is positive and Go is
negative (spon).
– When K < 1 (meaning equil lies to the left),
the ln K is negative and Go is positive
(nonspon).
KlnRTGo
• This result easily rearranges to give the
basic equation relating the standard
free-energy change to the equilibrium
constant.
Relating Go to the Equilibrium
Constant
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48
)l(OH (aq)CONHNH )g(CO)g(NH2 22223
– Rearrange the equation Go= -RTlnK to give
RT
GKln
o
A Problem To Consider • Find the value for the equilibrium constant, K,
at 25oC (298 K) for the following reaction. The
standard free-energy change, Go, at 25oC
equals –13.6 kJ/mol.
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– Substituting numerical values into the equation,
49.5K 298K)J/(mol 31.8
/106.13ln
3 molJK
A Problem To Consider
2401042.2 249.5eK
RT
GKln
o
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– You get the value of GTo at any temperature T by
substituting values of Ho and So at 25 oC into
the following equation.
ooo
T STHG
Calculation of Go at Various
Temperatures
• We typically assume that Ho and So are
essentially constant with respect to
temperature.
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)g(CO)s(CaO)s(CaCO 23
So: 38.2 92.9 213.7 J/mol K
Hfo: -635.1 -1206.9 -393.5 kJ/mol
A Problem To Consider
• Find the Go for the following reaction at
25oC and 1000oC. Relate this to reaction
spontaneity.
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)reactants(Hm)products(HnHof
of
o
kJ 3.178kJ)]9.1206()5.3931.635[(
)reactants()products( ooo SmnSS
kJ/K 0.1590/ 0.159)]9.92()7.2132.38[( KJooo
T STHG
– Now you substitute Ho, So (=0.1590 kJ/K), and
T (=298K) into the equation for Gfo.
)/ 1590.0)( 298(3.17825
KkJKkJGo
Co
kJ 9.13025
o
CoG
So the reaction is
nonspontaneous
at 25oC.
)g(CO)s(CaO)s(CaCO 23
So: 38.2 92.9 213.7 J/mol K
Hfo: -635.1 -1206.9 -393.5 kJ/mol
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53
A Problem To Consider
• Find the Go for the following reaction at
1000oC.
– Now we’ll use 1000oC (1273 K) along with our
previous values for Ho and So because assume
does not change much.
)/ 1590.0)( 1273(3.1781000
KkJKkJGo
Co
kJ 1.241000
o
CoG So the reaction is
spontaneous at
1000oC. You see that this reaction change from nonspon to spon
somewhere between 25oC to 1000oC. How can we determine at
what temp this switch occurred? G=0 is equil, switch point
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– To determine the minimal temperature for
spontaneity, we can set Gº=0 and solve for T.
o
o
oo
ooo
S
HT
STH
STHG 0
)C 848( K 1121K/kJ 1590.0
kJ 3.178T
o
– Thus, CaCO3 should be thermally stable until its
heated to approximately 848 oC.
– This is way you could calculate the normal boiling
point of a liquid. At G=0, the liquid phase and
gas phase will be at equilibrium; temperature at
which switch from liquid to gaseous phase.
HW 48
nonspon < 848oC; CaCO3 stable
spon > 848oC; CaCO3 decomposes easily
)g(CO)s(CaO)s(CaCO 23
l g