734 CHAPTER 18 Entropy, Free Energy, and Equilibrium The reaction generally is the system.. In part a, the increase in moles of gas gives a large positive value for LlS:'xn' In part b,
Trang 1734 CHAPTER 18 Entropy, Free Energy, and Equilibrium
The reaction generally is the system Therefore,
t.S ~n is t.S ~ys
•
Recall that here per mole means per mole of
reaction as written [ H~ Section 5 6 ]
•
•
•
•
Think About It The results
are consistent with the fact that
production of gases causes an
increase in entropy In part (a), the
increase in moles of gas gives a
large positive value for LlS:'xn' In
part (b), the decrease in moles of
gas gives a large negative value for
LlS:'xn- In part (c), where there is no
change in the number of moles of
gas in the reaction, LlS ~xn is positive
but is also fairly small In general,
in cases where there is no net
change in the number of moles of
gas in a reaction, we cannot predict
whether LlS:'xn will be positive or
negative-but we can predict that
it will be a relatively small number
This makes sense given that gases
invariably have greater entropy than
liquids and solids
For reactions involving only liquids a nd sol id s,
predicting the sign of t S" is more difficult,
but i n many such cas e s an increase in the
total number of molecules and/or ions is
accompan ied by an increase in entropy
The standard entropy values of a large number of substances have been measured in 11K· mol
To · caiCi.iiate · tiie ' s ' t ' lliidarci"entrcipy ' di.ange · for ' li ' reaction (ilS~n)' we look up the standard
entro-pie s of the products and reactant s and use Equation 18.7 Sample Problem 18.2 demonstrates this approach
: Strategy Look up standard entropy values and use Equation 18.7 to calculate LlS ~xn ' Just as we did
: · when we calc ul ated sta ndard enthalpies of ' ' reaction, we consider stoichiometric coefficients to be
dimensionless-giving LlS:'xn units of J / K mol
Setup From Appendix 2, SO[CaC03(s)) = 92.9 J /K mol, SO[CaO(s)] = 39.8 JIK mol,
S O [CO zeg) ] = 213.6 JIK mol, S O [N 2 (g)] = 191.5 J / K mol, S O [H 2(g) ] = 131.0 J/K mol,
S O [NH 3 (g) ] = 193.0 JIK mol, SO[CI2(g)] = 223.0 JIK mol, and SO[HCl(g)] = 187.0 JIK mol
Solution
(a) M:'xn = [SO(CaO) + SO (C0 2)] - [SO(CaC03)]
= [(39.8 JIK mol) + (213.6 J / K mol)] - (92.9 J / K mol)
Entropy Changes in the Surroundings
Next we discuss how ilS ~ urr is calculated When an exothermic process takes place in the
sys-tem, the heat transferred to the surroundings increases the motion of the molecules in the surroundings Consequently, there is an increase in the number of microstates and the entropy
of the surroundings increa ses Conversely, an endothermic process in the system absorbs heat from the surroundings and so decrea ses the entropy of the surroundings by slowing molecular motion Remember that for constant-pres s ure processe s, the heat relea se d or absorbed , q, is equal to the enthalpy change of the system, ilH s y s [ ~~ Sectio n 5 . 3] The change in entropy for
the surroundings, ilS s um is directly proportional to ilH sy s :
The minus sign indicates that a negative enthalpy change in the system (an exothermic process) correspond s to a positive entropy change in the surroundings For an endothermic process, the enthalpy change in the system is a po s itive number and corresponds to a negative entropy change
in the surroundings
Trang 2SECTION 18.3 The Second and Third Laws of Thermodynamics 735
In addition to being directly proportional to b.H s y s , b.S s urr is inversely propOltional to temperature:
Combining the two expressions gives
Equation 18.9
With Equations 18.8 and 18.9 , we can calculate the entropy change in both the syste m and s
ur-roundings for a chemical reaction , and we can determine whether the reaction i s spontaneous
Consider the sy nthesi s of ammonia at 25 ° C:
b.H~xn = -92 6 kJ/mol
From Sample Problem 18.2 ( b) , we have b.S ~ y s
( - 92.6 kJ/mol) into Equation 18.9 , we get
-198.5 J /K mol , and substituting b.H ~ y s
A _ -(-92.6 X 1000) J / mol _
uS surr - 298 K - 311 J/K mol The entropy change for the univer se i s
= -199 J/K· mol + 311 J/K· mol
= 112 J/K mol
Because b.S ~niv is positive, the reaction will be s pontaneous at 25 ° C Keep in mind, though , that
just because a reaction is spontaneous doe s not mean that it will occur at an observable rate The
synthesis of ammonia is, in fact, extremely s low at room temperature Thermodynamics can tell
us whether or not a reaction will occur spo ntaneou sly under specific conditions, but it doe s not
tell u s how fast it will occur
Third Law of Thermodynamics
Finally, we consider the third law of thermodynamics briefly in connection with the
determina-tion of s tandard e ntropy So far we have related entropy to micro sta te s the greater the number of
micro sta te s a system possesses , the larger i s the entropy of the syste m Consider a perfect
crystal-line s ub sta nce at absolute zero (0 K) Under these conditions, there i s essentially no molecular
motion and the number of microstates (W) i s 1 ( there i s only o n e way to arrange the atoms or
molecule s to form a perfect crystal) From Equation 18.2 , we write
S = kin W
=klnl=O
According to the third law of thermodynamics, the entropy of a perfect crystalline substance i s
zero at absolute zero As temperature increa ses, molecular motion increases, causing an increa se
in the number of microstates Thu s, the entropy of any substance at any temperature abo ve 0 K
is greater than zero If the crystalline s ub sta nce is impure or imperfect in any way, then it s
entropy i s greater than zero even at 0 K becau se without perfect order there i s more than one
microstate
The important point about the third law of thermodynamics i s that it enables us to determin e
the absolute entropies of substances Starting with the knowled ge that the entropy of . a pure
crys-., talline s ub sta nce is zero at 0 K , we can measure the increase in entropy of the s ub s tance when it
is heated The change in entropy of a s ub stance, b.S, is the difference between the final and initial
entropy values:
b.S = Sf - Si
where Si i s zero if the substance starts at 0 K Therefore , the measur e d change in entropy is equal
to the absolute entropy at the new temperature
f
A lth o ugh the complete details of these measurements are beyond the scope of thi s book , entropy changes are determined in part
by measur i ng the heat capacity of a substance
[ ~~ Section 5.4] as a function of absolute
temperature
Trang 3736 CHAPTER 18 Entropy, Free Energy, and Equilibrium
Figure 18 4 Entropy increa s es in
a substance a s temperature increa ses
from absolute zero
T em p era ture ( K )
Gas
The entrop y va lue s arri ved at in this way are ca lled absolute entropies becau se they are true
values-unlike s tandard enthalpies of formation, which ar e derived u s ing an arbitrary reference Becau se the tabulated v alue s are determined at I atm, we usually refer to ab so lute entropies as
standard entropie s, So Figure 18.4 shows the incr e a s e in entropy of a substance a s temperature increases from ab so lute zero At 0 K , it ha s a zero entropy value (assuming that it is a perfect crystalline s ub s tance ) As it i s heated , it s entrop y increa s e s gradually at first because of greater molecular motion within the crystal At the melting point , there is a large increase in entropy as the s olid i s transformed int o the liquid Further heating increa ses the entropy of the liquid again due to increa se d molecular motion At the boiling point , there i s a large increa s e in entropy a s a
re s ult of the liquid-to-vapor tran s ition Beyond that temperature , the entropy of the gas continues
to increa s e with increa s ing temperature
18.3 1 U s ing data from Appendix 2, calc ulate t:.s o ( in 1 1K · mol) for
the following reaction:
18.3 2 Using data from Appendix 2, calculate t:.s o ( in 1 1 K· mol ) for
d ) 242 8 1 1 K· mol
e) -2 42.8 1 1 K· mol
Gibbs Free Energy
According to the se cond law of thermodynamic s, D.Suniv > 0 for a s pontaneou s process We are
u s uall y concerned with and usually measure, however, the propertie s of the system rather than tho se of the s urrounding s or tho se of the univer se overall Ther e fore , it i s convenient to have a thermodynamic function that enables u s to determine whether or not a process i s spontaneous by considering the sys tem alone
We begin with Equation 18.5 For a s pontaneou s process,
D.Suniv = D.Ssys + D.Ssurr > 0
Trang 4SECTION 18.4 Gibbs Free Energy 737
Substituting - fUi s y/T for ~SSUIT> we write
~Hsys
~Suniv = ~Ssys + - T > 0
Multiplying both sides of the equation by T gives
Now we have an equation that expresses the second law of thermodynamic s (and predicts whether
or not a proce ss is s pontaneou s) in terms of only the system We no longer need to consider the
surroundings For convenience, we can rearrange the preceding equation, multiply through by -1,
and replace the> sign with a < s ign:
- T~Suni v = ~H sys - T~S s y s < 0
According to this equation, a process carried out at constant pre ss ure and temperature is spo ntaneous
if the changes in enthalpy and entropy of the sys tem are s uch that fUi sys - T~Ssy s is les s than zero
To express the spontaneity of a , proce ss more directly, we introduce another thermodynamic
function called the Gibbs! free energy (G), or simply free energy
Each of the term s in Equation 18.10 pertain s to the system G has units of energy ju s t as Hand
TS do Furthermore, like enthalpy and entropy, free energy is a s tate function The change in free
energy, ~G, of a system for a process that occurs at constant temperature i s
Equation 18.11 enables u s to predict the spontaneity of a process using the change in enthalpy, the
change in entropy, and the ab so lute temperature At constant temperature and pres s ure , for
pro-cesses that are spontaneous as written (in the forward direction ), ~G i s negative For processe s that
are not spontaneous as written but that , are s pontaneous , in the reverse direction , ~G i s po s itive For
systems at equilibrium, ~G is zero
• ~G< 0
• ~G = 0
The reaction is spontaneous in the forward direction (an d nonspontaneou s in
the reverse direction)
The reaction is nonspontaneou s in the forward direction (and spontaneous in
the reverse direction)
The system is at equilibrium
Often we can predict the s ign of ~G for a proces s if we know the signs of ~H and ~S Table 18.3
shows how we can u se Equation 18.11 to make such prediction s
Based on the information in Table 18.3, you may wonder what constitutes a " low" or a
"high" temperature For the example given in the table , O ° C is the temperature that divide s high
from low Water freezes s pontaneously at temperatures below O ° C, and ice melts spo ntaneou sly at
temperature s above O°C At O °C, a system of ice and water is at equilibrium The temperature that
In this context, free energy is the energy
available to do work Thus, i f a particular pr oces s
is accompanied by a release of usable energy ( i.e., if LiG is negative), this fact alone guarantees
that it is spontaneous, and there is no need
to consider what happens to the rest of the
Negative when T~S > fUi
Po s iti ve when T~S < fUi
Always spo nt a neous
1 Jo s iah Willard Gibbs (1839-1903) American physicist One of the founders of thermodynamics Gibbs was a m odes t
and private indi v idual who spent a lmo st a ll hi s professional lif e at Yale Unive r si ty Be cause he published mo st of his work
in obscure journals, Gibb s ne ver ga ined the eminence that hi s contemporary and admirer Jam es Maxwell did Even today,
ve r y few people o ut side of chemistry and physics have eve r heard of Gibbs
Trang 5738 CHAPTER 18 Entropy, Free Energy, and Equilibrium
•
Think About It Spontaneity is
being negative) and by an increase
heat, it should make sense that
the temperature will shift the
making it "more spontaneous."
divides "high" from "low" depends, though , on the individual reaction To determine that
tempera-ture, we must set t1G eq u al to 0 in Equation 18.11 (i.e , the equilibrium condition):
0= t1H - Tt1S
Rearranging to so l ve for T yie ld s
T= t1H
t1S
The temperature that divides high from low for a particular reaction can n ow be calculated if the
val u es of t1H and t1S are known Sample Pr o bl em 18 3 demonstrates the u se of thi s approach
Sample Problem 18.3
According to Table IS.3, a reaction will be spontaneous only at high temperatures if both tlH and
Strategy The temperature that divides high from low is the temperature at which tlH = TtlS
temperature in kelvins; we then convert to degrees Celsius
Practice Problem A reaction will be spontaneous only at low temperatures if both tlH and tlS are
Standard Free-Energy Changes
The introduction of the term ~GO enab l es u s to The"sia'ndardfree~eneiijY ' of'riiaction '(t1G~Xll) is th e free energy change for a reaction when it
write Equation 18.11 as occ ur s under s tandard- s tat e conditions-that i s, when reactant s in th e ir s tandard sta te s are
con-~G O = M ? - T/:,S ve rted to products in their s tandard states The conventions u sed by chemists to define the s tandard
s tates of pure substa n ces and sol uti o n s are
The standard free -energy change for thi s reaction is given by
Equation 18.12 t1G ~xn = [ct1G r (C) + dt1G r ( D ) ) - [at1G r (A) + bt1G r (B))
Equat ion 18.12 can be generalized as follows:
Equation 18.13 t1G~xn = lnt1G r ( pr od uct s) - lmt1G r ( reactants)
•
Trang 6•
where m and n are stoichiometric coefficients The term !1G J is the standard free energy of mation of a compound that is, the free-energy change that occurs when 1 mole of the compound
for-is synthesized from its con s tituent elements , each in its standard state For the combustion of
graphite,
the standard free-energy change (from Equation 18.13) is
!1G~xn = [!1G H C0 2)] - [!1G ~ (C, graphite) + !1G ~ (02)]
As with standard enthalpy of formation, the standard free energy of formation of any element (in
its most stable allotropic form at 1 atm) is defined as zero Thus ,
!1GJ?(C, graphite) = 0 and
Therefore, the standard free-energy change for the reaction in thi s case is equal to the sta ndard free energy of formation of CO2:
!1G ~xn = !1Gf(C02)
Appendix 2 lists the values of !1G J? at 25 ° C for a number of compounds
Sample Problem 18.4 demonstrate s the calculation of s tandard free-energy changes
!:J.Gf[COz(g)] = -394.4 kllmol, !:J.Gf[H20(l)] = -237.2 kJ/mol, and !:J.Gf[MgO(s)] = - 569.6 kl l mo!
All the other substances are elements in their standard states and have, by definition, !:J.G'f = O
Solution
(a) !:J.G ~n = (!:J.Gf[C02(g)] + 2!:J.Gf[H 2 0( I)]) - (!:J.Gf[CHig)] + 2!:J.Gf[02(g)])
= [(-394.4 kllmol) + (2)(-237.2 kJ/mol)] - [(-50.8 kl lmo l ) + (2)(0 kl lmo l )]
that indicates whether or not a process will occur spontaneously under a given set of conditions
What the sign of !1G o tells u s is the same thing that the magnitude of the equilibrium constant (K)
tell s us [ ~~ Section 15.2] A negative !1G o value corresponds to a large K value (products favored
at equilibrium), wherea s a po s itive !1G o value corresponds to a small K value (reactants favored at equilibrium)
Like equilibrium constants, !1G o values change with temperature One of the uses of
Equa-tion 18.11 is to determine the temperature at which a particular equilibrium will begin to favor
a desired product For example, calcium oxide (CaO), also called quicklime, is an extremely valuable inorganic substance with a variety of industrial uses , including water treatment and
Think About It Note that, like standard enthalpies of formation (DJIf), standard free energies of formation (!:J.G7) depend on the state of matter
Using water as an example,
!:J.Gf[H20(l)] = -237.2 kl l mol and
!:J.Gf[H 20(g)] = -228.6 kJ/mo!
Always double-check to make sure you have selected the right value from the table
,
The sign of !:J.GO does indicate whether or not
a process is spontaneous when all reactants a na
products are in their standard states, but this is
very seldom the case
Trang 7740 CHAPTER 18 Entropy, Free Energy, and Equilibrium
An important piece of information for th e chemist re s pon s ible for maximizing CaO
produc-tion i s the temperature at which the decomposition equilibrium of CaC03 begin s to favor products
We can mak e a reliable estimate of that temperature as follows First we calculate b H o and b S o for
the r eact ion at 2S o C, u s ing the data in Appendix 2 To determine b H o , we apply Equation S 1 9:
5" val ue s are · tabulated using joules, whe r eas
Becau se b G o is a large po s itive numb e r, the reaction does not favor product formation at 2S o C
(298 K) And, becau se b H o and b S o are both positive, we know that b G o will be negative (product formation will be favored ) at high temperatures We can determine what constitutes a high tem- perature for this reaction by calculating the temperature at which b G o is zero
At still higher temperatures , b G o becomes increasingly negative, thus favoring product formation
even more Note that in this example we u se d the b H o and b S o values at 2S o C to calculate changes
t o b G o at much higher temperature s Becau s e both b H o and b S o actually change with ture , this approach doe s not give us a truly accurate value for b G o , but it does give us a rea s onably
tempera-good estimate
Equation 18.11 can al so be used to calculate the change in entropy that accompanies a phase
change At the temperature at which a phase change occurs (i.e., the melting point or boiling point
of a s ub sta nc e), the sys tem i s at equilibrium (b G = 0 ) Therefore , Equation 18.11 become s
Trang 8SECTION 18 4 Gibbs Free Energy 741
or
0= tlH - TtlS
tlS = tlH
T
Consider the ice-water equilibrium For the ice-to-water transition, tlH i s the molar heat of fusion
(see Table 12.8) and T is the melting point The entropy change is therefore
6010 Ilmo l
tlSic e + wa t er = 273 K = 22.0 I lK · mol
Thu s, when 1 mole of ice melts at O D C, there i s an incr ease in entropy of 22.0 I l K· mol The
increase in entropy i s con s i ste nt with the increa se in micro s tate s from solid to liquid Co n ve r se l y,
for the water-to-ice transition, the decrease in entropy i s given by
•
tlS = -6010 I l mol = -220 I lK · I
The sa me approach can be applied to the water-to-steam transition In thi s case, tlH i s the heat of
vaporization and T is the boiling point of water Sample Problem 18.5 examines the pha se tran s
i-tion s in benzene
Sample Problem 18.S
The molar heats of fusion and vaporization of benzene are 10.9 and 31.0 kJ / m o l , respectively
Strategy The solid-liquid transition at the melting point and the liquid-vapor transition at the boiling
Setup The melting point of benzene is 5.5 + 273.15 = 278.7 K and the boiling point is 80.1 +
Practice Problem The molar heats of fusion and vaporization of argon are 1.3 and 6.3 kJ / m o l ,
respectively, and argon's melting point and boiling point are -190°C and -186°C, respectively
Calculate the entropy changes for the fusion and vaporization of argon
Checkpoint 18.4 Gibbs Free Energy
a) nonspontaneous at all temperatures
b) spontaneous at all temperatures
d) spontaneous at low temperatures
18 4 2 At what temperature (in 0c) does a
Think About It For the same
transition
•
Trang 9742 CHAPTER 18 Entropy, Free Energy, and Equilibrium
•
Eve n for a react ion that s tarts with all reactan ts
and products in their standard states, as soon
as the reaction begin s , th e conc entra tions of all
species change and standard- st ate conditions
no longer exist
The Q used in Equation 18 1 4 can be eith e r Q ,
( for reactions that take plac e in so lution) o r Qp
( for reactions that take place in the gas ph as e )
Reactant s and products in a chemical reaction are almost always in something other than their
standard sta te s - that is , s olution s u s ually have concentrations other than 1 M and gases usually have pre ss ure s other than 1 atm To determine whether or not a reaction i s s pontaneous, there- fore, we mu s t take into account the actual concentrations and/or pressure s of the species involved
And although we can determine t:: G o from tabulated values, we need to know t:: G to determine
spo ntaneity
The relationship between t:: G and t:: G o, which is derived from thermodynamics, i s
Equation 18.14 t:: G = t:: G o + RTln Q
where R i s the gas constant ( 8.314 11K· mol or 8.314 X 10 - 3 kllK mol), T is the absolute tem
- , pe ' ratiiie' at w hI c h ' the ' reac"ii"on ' takes' pi ace : and ' Q i s the reaction quotient [ ~~ Secti on 15.2] Thus ,
t:: G depend s on two terms: t:: G o and RT In Q For a given reaction at temperature T, the value
the reaction mixtur e
Con s ider the following equilibrium:
Using Equation 18.13 and information from Appendix 2 , we find that t:: G o for thi s reaction at
25 ° C i s 2.60 kl l mol The value of t:: G, however, depends on the pressure s of all three gaseous spe
-cies If we st art with a mixture of gases in which PH 2 = 2 0 atm, PIz = 2.0 atm , and PHI = 3.0 atm, the reaction quotient , Qp, i s
(P HI )2
Qp = ( PHz)(PIz )
= 2.25 Using this value in Equation 18 14 gives
(3.0)2 9.0 (2.0)(2.0) 4 0
Becau se t:: G is positive, we conclude that , s tarting with the se concentrations , the forward reactio n
will not occur s pontaneously a s written Instead, the r eve r se reaction will occur spontaneously and the sys tem will reach equilibrium by consuming part of the HI initially present and producing more H 2 and I2
If, on the other hand, we start with a mixture of gases in which PH = 2.0 atm, PI = 2 0 atm,
1
-
-4
Trang 10SECTION 18.5 Free Energy and Chemical Equilibrium 743
Using this value in Equation 18.14 gives
AG = 2.6 kJ + 8.314 X 10-3 kJ (298 K)(ln 0.25 )
= - 0.8 kJ/mol
With a negative value for AG, the reaction will be spontaneous as written in the forward
direc-tion In this case, the system will achieve equilibrium by consuming so me of the H 2 and 12 to
produce more HI
Sample Problem 18.6 u ses AG o and the reaction quoti e nt to determine in which direction a
reaction is s pontaneou s
•
Sample Problem 18.6
The equilibrium constant, Kp , for the reaction
is 0.1l3 at 298 K, which corresponds to a standard free-energy change of 5.4 kllmo! In a certain
experiment, the initial pressures are PN 0 = 0.453 atm and PNO = 0.122 atm Calculate 6.G for the
Because 6.G is negative, the reaction proceeds spontaneously from left to right to reach equilibrium
Practice Problem A 6.G o for the reaction
HzCg) + 12(g) :;:::, ~ 2HI(g)
is 2.60 kllmol at 25°C Calculate 6.G, and predict the direction in which the reaction is spontaneous
if the starting concentrations are P
H 2 = 3.5 atm, P
I 2 = 1.5 atm, and PHI = 1.75 atm
Practice Problem B What is the minimum partial pressure of 1 2 required for the preceding reaction
to be spontaneous in the forward direction at 25°C if the partial pressures of H 2 and HI are 3.5 and
1.75 atm, respectively?
By definition , AG = 0 and Q = K at equilibrium, where K i s the equilibrium constant Thus,
Think About It Remember, a
can be spontaneous if the starting concentrations of reactants and
products are such that Q < K
In this equation and the o n e that f ollows , K IS _
for reaction s that take place in solution an d Ko
f or reactions that take place in the gas p hase
Trang 11744 CHAPTER 18 Entropy, Free Energy, and Equilibrium
Multimedia
Chemic al Equilibrium equilibrium
., " Multimedia
Ch emical Kinet i cs - interact iv e react i on
coordinate diag ram
•
The sign of 6 GO tells us the same thing that the
magnitude of K tells u s The sign of 6G tells us
the same thing as the comparison of Q and K
values [ ~~ Section 15.4]
Figure 18.5 (a) !:lG o < O At
equilibrium, there is a significant
conversion of reactants to products
(b) !:lG o > O At equilibrium,
reactants are favored over products
In both cases, the net reaction toward
equilibrium is from left to right
(reactants to products) if Q < K and
right to left (products to reactants) if
Q > K At equilibrium, Q = K
18.15 is one of the most important equations in thermodynamics because it enables us to find the
It is s ignificant that Equation 18.15 relates the equilibrium constant to the standard
s how s plots of the free energy of a reacting system versus the extent of the reaction for two tion s Table 18.4 summarizes the relationship between the magnitude of an equilibrium constant
tells us the relative amounts of products and reactants when equilibrium is reached, not the
f - Extent of reaction - - - 1
(b)
GO (products)
Trang 12SECTION 18.5 Free Energy and Chemical Equilibrium 745
Products are favored
Neither products nor reactants are favored
Reactant s are favored
For reaction s with very large or very s mall equilibrium co n sta nt s, it can be ve r y
difficult-so metimes impossible to determine K values by mea s uring the concentrations of the reactant s
and product s Consider, for example, the formation of nitric oxide fro m molecular nitrogen and
molecular oxygen:
•
At 25 ° C, th e equilibrium constant, Kp , is
The very small value of Kp mean s that the concentration of NO at equilibrium will be exceedingly
low and, for all intents and purpo ses, impo ss ible to m eas ure directly In s uch a case, the
equilib-rium constant i s more conveniently d e termined using 6.G o, which can be calcu lated either fro m
tabulated 6.G t' values or from 6.H o and 6.s o
Sample Problems 18 7 and 18.8 s h ow how to u se 6.G o to calculate K and how to u se K to calculate 6.G o , re s pectively
Sample Problem 18.7
Using data from Appendix 2, calculate the equilibrium constant, K p, for the following reaction at
25 ° C :
Strategy Use data from Appendix 2 and Equation 18.13 to calculate t:: G o for the reaction Then use
Equation 1 8 15 to solve for K p
Practice Problem Using data from Appendix 2, calculate the equilibrium constant, K p , for the
following reaction at 25 ° C:
Think About It This is an extremely small equilibrium constant, which is consistent with
We know from everyday experience
spontaneously into its constituent elements at 25 ° C
Trang 13746 CHAPTER 18 Entropy, Free Energy, and Equilibrium
•
Think About It The relatively
large, positive !J.G o , like the very
small K value, corresponds to a
process that lies very far to the left
Note that the K in Equation 18.15
can be any type of K c (K a ' Kb, Ksp'
etc.) or Kp
Sample Problem 18.8
The equilibrium constant, Ksp ' for the dissolution of silver chloride in water at 25°C,
AgCl(s) :;::, = =' Ag +( aq) + Cl - (aq)
is 1.6 X 10-10 Calculate !J.G o for the process
Strategy Use Equation 18.15 to calculate !J.G o
Setup R = 8.314 X 10 - 3 kJlK mol and T = (25 + 273) = 298 K
18 5 1 For the reaction
for which 6.G o = - 1.2 kJ / mol at 25°C
Without knowing the pressures of the reactants and the product, we know that
18.5.3 The !J.G o for the reaction
is -33.3 kJ / mol at 25°C What is the
value of K p?
a) 2 X 10 - 6
b) 7 X 105
c) 3 X 10 - 70 d) 1
e) I X 10 - 1
18.5.4 The K s p for iron(III) hydroxide
[Fe(OH)3J is 1.1 X 10 - 36 at 25°C For the process
determine 6.G o (in kJ/mo l ') at 25°C
Thermodynamics in Living Systems
Many biochemical reactions have a positive I1G o val u e, yet they are essential to the maintenance
of life In living systems, these reactions are coupled to a n energetically favorable process, one