Upload
icha-meisyarani-ii
View
222
Download
0
Embed Size (px)
Citation preview
8/2/2019 Ch 15 - Chemical Equilibrium Compatibility Mode
1/16
14/02/2012
1
1
Chemical Equilibrium
Brown, LeMay Ch 15
AP ChemistryMonta Vista High School
2
15.1: Chemical Equilibrium
Occurs when opposing react ions areproceeding at t he same rat e Forward rate = reverse rate of reaction
Ex:
Vapor pressure:rate of vaporization = rate
of condensationSat urat ed solut ion:rate of dissociation =
rate of crystallization
Expressing concentrations: Gases:partial pressures, PX Solut es in li quids:molarity, [X]
8/2/2019 Ch 15 - Chemical Equilibrium Compatibility Mode
2/16
14/02/2012
2
3
Forward reaction: A B Rate = kforward [A]
Reverse reaction: B A Rate = kreverse [B]
or
RT
P
V
nM
RT
PA
A][RT
PB
B][
Forward reaction:
Reverse reaction:
RTPkRate Af
RT
PkRate Br
nRTPV
R = 0.0821 LatmmolK
4
http://www.kentchemistry.com/links/Kinetics/Equilibrium/equilibrium.swf
[B]k[A]k rf
r
f
k
k
[A]
[B]
RT
Pk
RT
Pk Br
Af or
eq
r
f
A
B Kconstantk
k
P
Por
8/2/2019 Ch 15 - Chemical Equilibrium Compatibility Mode
3/16
14/02/2012
3
Kc= kf/kr, at equilibrium, if K> 1, thenmore products at equilib. And if k
8/2/2019 Ch 15 - Chemical Equilibrium Compatibility Mode
4/16
14/02/2012
4
Equilibrium
The equilibrium position is the same at agiven temperature, no matter from whichdirection it is approached.
It is possible to force an equilibrium one wayor the other temporarily by altering thereaction conditions, but once this stress isremoved, the system will return to its
original equilibrium.
7
8
PX
or[X
]
Time
[B] or PB/ RT Equilibrium is established
Figure 1: Reversible reactions
[A] or PA/ RT
[A]0or PA0/ RT
8/2/2019 Ch 15 - Chemical Equilibrium Compatibility Mode
5/16
14/02/2012
5
9
Reversible Reactions and Rate
Reaction
Rate
TimeBackward rate
Forward rate
Equilibrium is established:
Forward rate = Backward rate
When equi l ibr ium is achieved:
[A] [B] and kf/kr = Keq
10
15.2: Law of Mass Action
Derived from rate laws by Guldberg andWaage (1864)
For a balanced chemical reactionin equilibrium:
a A + b B c C + d D
Equilibrium constant expression (Keq):
ba
dc
c[B][A]
[D][C]K
b
B
a
A
d
D
c
Cp
)(P)(P
)(P)(PK
Keq is strictly based on stoichiometry of the reaction (isindependent of the mechanism).
Units: Keq is considered dimensionless (no units)
Cato Guldberg Peter Waage(1836-1902) (1833-1900)
or
8/2/2019 Ch 15 - Chemical Equilibrium Compatibility Mode
6/16
14/02/2012
6
11
Relating Kc and Kp
Convert [A] into PA:
ba
dc
([B]RT)([A]RT)
([D]RT)([C]RT)
b
B
a
A
d
D
c
Cp
)(P)(P
)(P)(PK
baba
dcdc
(RT)[B][A]
(RT)[D][C]
(RT)KKb)(a-d)(c
cp
where n == change in coefficents of products reactants (gases only!)= (c+d) - (a+b)
(RT)Kn
c
RT
P
V
nM RTAPA ][
12
Magnitude of Keq
Since Keq [products]/[reactants], the magnitude ofKeq predicts which reaction direction is favored:
If Keq > 1 then [products] > [reactants]and equilibrium lies to the right
If Keq < 1 then [products] < [reactants]and equilibrium lies to the left
8/2/2019 Ch 15 - Chemical Equilibrium Compatibility Mode
7/16
14/02/2012
7
13
Relationship Between Q and K
Reaction Quotient (Q): The particular ratioof concentration terms that we write for aparticular reaction is called reactionquotient.
For a reaction, A B, Q= [B]/[A]
At equilibrium, Q= K
Reaction Direction: Comparing Q and K
QK, reaction proceeds to left, until Q=K
14
Value of K
For thereference
rxn, A>B,
For thereverse rxn,B>A,
For thereaction,
2A> 2B
For the rxn,
A> C
C> B
K(ref)=[B]/[A]
K= 1/K(ref) K= K(ref)2 K (overall)=K1 X K2
8/2/2019 Ch 15 - Chemical Equilibrium Compatibility Mode
8/16
14/02/2012
8
15
15.3: Types of Equilibria
Homogeneous:all components in samephase (usually gor aq)
N2 (g) + H2 (g) NH3 (g)
3
H
1
N
2NH
P)(P)(P
)(PK
22
3
b
B
a
A
d
D
c
CP
)(P)(P
)(P)(PK
3 21
Fritz Haber(1868 1934)
16
Het erogeneous:different phases
CaCO3 (s) CaO (s) + CO2 (g)
Definit ion: What we use:
][CaCO
)(P[CaO]K
3
CO
eq2
2
COp PK
Even though the concentrations of the solids or liquids donot appear in the equilibrium expression, the substancesmust be present to achieve equilibrium.
Concentrations of pure solids and pure liquids are notincluded in Keq expression because their concentrations donot vary, and are already included in Keq (see p. 548).
8/2/2019 Ch 15 - Chemical Equilibrium Compatibility Mode
9/16
14/02/2012
9
17
15.4: Calculating Equilibrium Constants
Steps to use ICE table:
1. I = Tabulate known initial and equilibriumconcentrations of all species in equilibriumexpression
2. C = Determine the concentration change forthe species where initial and equilibrium areknown
Use stoichiometry to calculateconcentration changes for all otherspecies involved in equilibrium
3. E = Calculate the equilibrium concentrations
18
Ex: Enough ammonia is dissolved in 5.00 L of waterat 25C to produce a solution that is 0.0124 Mammonia. The solution is then allowed to come toequilibrium. Analysis of the equilibrium mixtureshows that [OH1-] is 4.64 x 10-4 M. Calculate Keq at
25C for the reaction:NH3 (aq) + H2O (l) NH4
1+ (aq) + OH1- (aq)
8/2/2019 Ch 15 - Chemical Equilibrium Compatibility Mode
10/16
14/02/2012
10
19
NH3 (aq) + H2O (l) NH41+ (aq) + OH1- (aq)
Initial
Change
Equilibrium
][NH
]][OH[NHK
3
-11
4c
0.0124 M
- x
0.0119 M
0 M 0 M
+ x + x
4.64 x 10-4 M 4.64 x 10-4 M
5-2-4
101.81x0.0119
)10(4.64
NH3 (aq)H2O
(l)NH4
1+ (aq) OH1- (aq)
XXX
x = 4.64 x 10-4
M
20
Ex: A 5.000-L flask is filled with 5.000 x 10-3 mol ofH2 and 1.000 x 10
-2 mol of I2 at 448C. The value ofKeq is 1.33. What are the concentrations of eachsubstance at equilibrium?
H2 (g) + I2 (g) 2 HI (g)
8/2/2019 Ch 15 - Chemical Equilibrium Compatibility Mode
11/16
14/02/2012
11
21
H2 (g) + I2 (g) 2 HI (g)
Initial
Change
Equilibrium
]][I[H
[HI]K
22
2
c
1.000x10-3 M
- x M
(1.000x10-3 x) M
2.000x10-3 M 0 M
- x M + 2x M
(2.000x10-3 x) M 2x M
33.1x)-10x)(2.000-10(1.000
(2x)3-3-
2
4x2 = 1.33[x2 + (-3.000x10-3)x+ 2.000x10-6]0 = -2.67x2 3.99x10-3x+ 2.66x10-6
Using quadratic eqn: x= 5.00x10-4 or 1.99x10-3; x= 5.00x10-4
Then [H2]=5.00x10-4 M; [ I2]=1.50x10
-3 M; [HI]=1.00x10-3 M
H2 (g) I2 (g) HI (g)
22
15.6: Le Chteliers Principle
If a system at equilibrium isdisturbed by a change in: Concentration of one of the
components,
Pressure, or
Temperature
the system will shift its equilibriumposition to counteract the effect ofthe disturbance.
http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/lechv17.swf
Henri Le Chtelier(1850 1936)
8/2/2019 Ch 15 - Chemical Equilibrium Compatibility Mode
12/16
14/02/2012
12
23
4 Changes that do not affect Keq:
1. Concentrat ion
Upon addition of a reactant or product,equilibrium shifts to re-establish equilibriumby consuming part of the added substance.
Upon removal of reactant or product,equilibrium shifts to re-establish equilibriumby producing more of the removed substance.
Ex: Co(H2O)
6
2+ (aq) + 4 Cl1-CoCl4
2- (aq) + 6 H2O (l)
Add HCl, temporarily inc forward rateAdd H2O, temporarily inc reverse rate
24
2. Volume, wi th a gas present (T is constant )
Upon a decrease in V (thereby increasing P),equilibrium shifts to reduce the number of moles of gas.
Upon an increase in V (thereby decreasing P),equilibrium shifts to produce more moles of gas.
Ex: N2
(g) + 3 H2(g) 2 NH
3(g)
If V of container is decreased, equilibrium shifts right. XN2 and XH2
dec
XNH3 inc
3
HN
2
NH
P)(PP
)(PK
22
33
THTN
2
TNH
)P)(XP(X
)P(X
22
34
T
3
HN
2
T
2
NH
P)(XX
P)(X
22
32
T
3
HN
2
NH
P)(XX
)(X
22
3
23
HN
2
NH
P)(
)(K
22
3
Since PT also inc, KP remains constant.
8/2/2019 Ch 15 - Chemical Equilibrium Compatibility Mode
13/16
14/02/2012
13
25
3. Pressure, but not Volume
Usually addit ion of a noble gas, p. 560 Avogadros law: adding more non-reacting particles fills
in the empty space between particles.
In the mixture of red and blue gas particles, below,adding green particles does not stress the system, sothere is no Le Chtelier shift.
26
4. Catalysts
Lower the activation energy of both forward andreverse rxns, therefore increases both forward andreverse rxn rates.
Increase the rate at which equilibrium is achieved,but does not change the ratio of components of
the equilibrium mixture (does not change the Keq)
Energy
Rxn coordinate
Ea, uncatalyzed
Ea, catalyzed
8/2/2019 Ch 15 - Chemical Equilibrium Compatibility Mode
14/16
14/02/2012
14
27
1 Change that does affect Keq:
Temperature: consider heat as a part of the reaction Upon an increase in T, endothermic reaction is favored
(equilibrium shifts to consume the extra heat) Upon a decrease in T, equilibrium shifts to produce more
heat.
Effect on Keq
1. Exothermic equilibria: Reactants Products + heat
Inc T increases reverse reaction rate whichdecreases Keq
2. Endothermic equilibria: Reactants + heat
Products Inc T increases forward reaction rate increases Keq
Ex: Co(H2O)62+ (aq) + 4 Cl1-CoCl4
2- (aq) + 6 H2O (l); H=+?Inc T temporarily inc forward rateDec T temporarily inc reverse rate
28
Vant Hoffs Equation
Vant Hoffs equation shows mathematicallyhow the equilibrium constant is affected bychanges in temp.
ln K2 = -d H0rxn (1 1)
K1 R T2 T1
8/2/2019 Ch 15 - Chemical Equilibrium Compatibility Mode
15/16
14/02/2012
15
29
Effect of Various Changes on Equilibrium
Disturbance Net Direction ofRxn
Effect of Valueof K
Concentration
Increase (reactant) Towards formation ofproduct
None
Decrease(reactant) Towards formation ofreactant
None
Increase (product) Towards formation of
reactant
None
Decrease (product) Towards formation ofproduct
None
30
Effect of Pressure on Equilb.
Pressure
Increase P
(decrease V)
Towardsformation offewer moles ofgas
None
Decrease P
(Increase V)
Towards
formation ofmore moles ofgas
None
Increase P
( Add inertgas, no changein V)
None,concentrationsunchanged
None
8/2/2019 Ch 15 - Chemical Equilibrium Compatibility Mode
16/16
14/02/2012
31
Effect of Temperature on Equilb
Temperature
Increase T
Towardsabsorption ofheat
Increases ifendothermic
Decreases ifexothermic
Decrease T Towards releaseof heat
Increases ifexothermic
Decreases ifendothermic
Catalyst Added None, forwardand reverseequilibriumattained sooner
None