05 Chem Equilibrium
Transcript of 05 Chem Equilibrium
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Chapter 5 Chemical equilibrium
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5.1 The general equilibrium expression
For any chemical reaction, the general reaction
equation can be written as
B
B
0 B Recall the criterion of Gibbs function, at constant T, p
and W=0
m
spontaneity
0 equilibriumrG
mrA G Define: is called the chemical affinity
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if an infinitesimal amount of reaction takes place at constant T
andp, the change in the Gibbs function is given by
B B
B
dG dn
B B
B
dG d
,
B B r m
BT p
GG
is the change of Gibbs function per mole of reaction.r mG
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At constant temperature and pressure,
0r mG
0r mG
0r mG
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We now consider a reaction of ideal gasesA B C Da b c d ln BB B
pRT
p $
$
ln Br m B B B B B
B B B
pG RT
p
$ $
r m B B
B
G $ $ lnB
Br m r m
B
pG G RT
p
$ $
B
Bp
B
pJ
p
$ lnr m r m pG G RT J $
The quantity Jp is called the pressure quotient.
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5.2 Chemical reactions of ideal gases
5.2.1 Standard equilibrium constant
When a chemical reaction attains equilibrium at constanttemperature and pressure
eqln 0r m r m pG G RT J $
We call the standard equilibrium constant and
denote it by K. It is only a function ofT.
eq
pJ
eq
B
B(ideal gases)
B
p
K p
$
$
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2 2 3N (g) + 3H (g) = 2NH (g)
2
3
1 32 2
(NH )
(N ) (H )
eq
eq eq
p
pK
p p
p p
$
$
$ $
31 2 2 32 2N (g)+ H (g)=NH (g)
312 2
3
2
2 2
(NH )
(N ) (H )
eq
eq eq
p
pK
p pp p
$
$
$ $
2
1 2
( )K K$ $ ,1 ,22r m r mG G $ $
r mln /K G RT $ $
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r m, then 0, 0pJ K G A $
r m
, then 0, 0p
J K G A $
r m, then 0, 0pJ K G A $
the forward reaction is
spontaneous
the reverse reaction is
spontaneous.
the reaction is at
equilibrium.
$
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5.2.2 Reactions involving pure condensed phases
and gases
If the pressure of a system does not differ very muchfromp, the chemical potentials of pure condensed
phases approximate the standard state chemical
potentials, i.e.
B(condensed) B(condensed) $
B(condensed) B(condensed) $
The activity of a pure liquid or solid is
1 (pure condensed phases near )B
a p $
Therefore, the condensed substances can be omitted
from the expression of equilibrium constant .B(g)eq
B(g)
B(g)
pK
p
$ $
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For example, for the reaction,12
C(s) + O2(g) = CO(g)
eq
1/2eq
2
(CO,g) /
(O ,g) /
p pK
p p
$
$
$
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5.2.3 Relationship between equilibrium constants of
related reactions
When the equilibrium constant for a particular reaction isnot available, it can be found by combining the
constants for an appropriate set of reactions. For
example
2 2 r m,1 1(1) C(s) + O (g) = CO (g) lnG RT K $ $
12 2 r m,2 22
(2) CO(g) + O (g) = CO (g) lnG RT K $ $
2 r m,3 3(3) CO (g) + C(s) = 2CO(g) lnG RT K $ $
Because (3)=(1)-2(2), we have
r m,3 r m,1 r m,22G G G $ $ $ 2
3 1 2/ ( )K K K$ $ $
5 2 4 M t f ilib i
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5.2.4 Measurement of equilibrium
constants
Physical methodChemical method
5.2.5 Calculation of chemical equilibrium composition
Example: The standard Gibbs function of reaction for
the decomposition
is 11808kJ mol-1 at 2300K. What is the degree of
dissociation of H2O at 2300K and 100kPa?
1
2 2 22
H O(g) H (g)+ O (g)
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Solution The equilibrium constant is 33
r m
118.08 10exp( / ) exp 2.08 10
8.3145 2300K G RT
$ $
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5.2.6 Equilibrium constants in other form
For example
eq
BB
( )B
pK p
1
2 2 22H O(g) H (g)+ O (g)
2 2
2
1/2
H O 1/2
H O
Pap
p pK
p
/ ( ) BpK K p $ $
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B B B B
BB BB y
p py p pK y K
p pp p
B
B B B BBB
B B Bc
nRT
p c RT cc c RT c RT VK Kc p c pp p p
B
B BB
B
B
BBB n
B B
n
p np p pK n K
p n p np p
5 3 T t d d f th
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5.3 Temperature dependence of the
equilibrium constant
2( / )
p
G T HT T
r m r m
2( / )d G T H
dT T
$ $
r m lnG RT K $ $
r m
2
ln Hd K
dT RT
$$
This is called the vant Hoff equation.
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If is temperature independent, carrying out theintegrationr mH
$
If an indefinite integral is taken, the result is
2 2
1 1
r m
2d ln d
K T
K T
HK T
RT
$
$
$
$
2 r m
1 2 1
1 1ln ( )
K H
K R T T
$ $
$
ln K
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A plot of versus 1/Tshould yield a straight line
with slope and intercept C.
r mln HK CRT
$
ln K
ln K
r mH
R
$
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5.4 Effects on the equilibrium
5.4.1 Le Chatelier's principle
equilibrium systems tend to compensate for the effects
of perturbing influences.
If the concentration of a reactant is increased, the
equilibrium position shifts to use up the added reactants.
If the pressure on an equilibrium system is increased,
then the equilibrium position shifts to reduce the pressure.
If the volume of a gaseous equilibrium system is reduced
then the equilibrium position shifts to increase the volume.If the temperature of an endothermic equilibrium system
is increased, the equilibrium position shifts to use up the heat
by producing more products.
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5.4.2 The effect of pressure on the equilibrium
BB B B
B
B
B BB
B B B
B BB
B
B BB
B
B
B
since ( ) ( ) ( )
if 0, then
if 0, then
if 0, no effect
p y p p
K yp p p
p y
p y
$
$ $ $
For example
CaCO3(s) = CaO(s) + CO2(g) pgo backward
N2(g) + 3H2(g) = 2NH3(g) pgo forward
5 4 3 Th ff t f i t t th
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5.4.3 The effect of inert components on the
equilibrium
B
B
B 0 BB
B
B 0 BB
B
if 0, then
if 0, then
n n
n n
BB
B 0 B
( )n p
n n pK
$
$
Suppose nB is componentn0 is inert componentB B
B
B0 B
B
/( )
p pn
n n
$
Therefore, for reactions of , the addition of inert gases, such aswater vapor or nitrogen, will make the reaction produce more products.
B
B
0
5 4 4 Th ff t f t ti ti th
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5.4.4 The effect of concentration ratio on the
equilibrium
For gas reactionA B Y Za b y z
suppose there is no product in the beginning, and let
the mole ratio of reactants B to A be
B
A
nr
n
When, that is, the molar ratio of two reactants is
equal to the ratio of their stoichiometric coefficients, thecontent of products Y and Z have the maximum values of
mole fraction.
br
a
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5.5 Reactions involving real gases
B B ln Bf
RTp
$ $Real gas
B
eq
Br m ln ( ) ln
fG RT RT K
p
$ $$
B B B
eq eqeqB BB
B B B
( ) ( )f p
Kp p
$ $ $
B B Bf p
KK
5 6 Ch i l ilib i i li id i t
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5.6 Chemical equilibriums in liquid mixture
and solution
B B BlnRT a $
5.6.1 Chemical equilibriums in a liquid mixtureunder ordinary pressure
Beq
r m B
Bln ( ) lnG RT a RT K
$ $
Beq
B
B
( )K a $ B B Ba f x
For ideal liquid mixtures
B
B B
B
1 , 1f f B
eq
B
B
( )K x $
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5.6.2 Chemical equilibriums in a solution under
ordinary pressure
A A AlnRT a $
A Beq eq
r m A BBln( ) ( ) lnG RT a a RT K
$ $
A Beq eq
A B
B
( ) ( )K a a $
solute
For ideal dilute solutions
B
eq
B
B
( )b
Kb
$ $
solvent
B B BlnRT a $
equilibrium
A A Aa f x
B B B/a b b$
eq
B
B
bif is mall
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JACOBUS HENRICUS VANT HOFF
JACOBUS HENRICUS VANT HOFF (1852-1911)Dutch physical chemist, received the first Nobel
Prize in chemistry in 1901 for the discovery of the laws
of chemical dynamics and of osmotic pressure. VantHoff was one of the early developers of the laws of
chemical kinetics, developing methods for
determining the order of a reaction; he deduced the
relation between temperature and the equilibrium
constant of a chemical reaction.
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JACOBUS HENRICUS VANT HOFF
In 1874, vant Hoff (and also J.A. Le Bel, independently)
proposed what must be considered one of the most
important ideas in the history of chemistry, namely the
tetrahedral carbon bond. Vant Hoff carried Pasteurs
ideas on asymmetry to the molecular level , and
asymmetry required bonds tetrahedrally distributed
about a central carbon atom. Structural organic
chemistry was born.
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VANT HOFF (1852-1911) Dutch physical chemist
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Home work
5.1 (calculation of equilibrium constant)5.3 (equilibrium and components)
5.10 (calculation of enthalpy, vant Hoff equation)
5.13 (thermodynamics)