Chemical Equilibrium Brown, LeMay Ch 15 AP Chemistry Monta Vista High School 1.
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Transcript of Chemical Equilibrium Brown, LeMay Ch 15 AP Chemistry Monta Vista High School 1.
15.1: Chemical Equilibrium
2
Occurs when opposing reactions are proceeding at the same rate Forward rate = reverse rate of
reactionEx:Vapor pressure: rate of vaporization
= rate of condensationSaturated solution: rate of
dissociation = rate of crystallization
Expressing concentrations: Gases: partial pressures, PX Solutes in liquids: molarity, [X]
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:
RT
PkRate A
f
RT
PkRate B
r
nRTPV
R = 0.0821 L•atm
mol•K
4
http://www.kentchemistry.com/links/Kinetics/Equilibrium/equilibrium.swf ( Good link on Equilibrium)
[B]k [A]k rf
r
f
k
k
[A]
[B]
RT
Pk
RT
Pk B
rA
f or
eqr
f
A
B Kconstantk
k
P
Por
5
Kc= kf/kr, at equilibrium, if K> 1, then more products at equilib. And if k<1, then reactants favored at equilb. K=1 (conc. Of reactants and products nearly same at equilibrium)
The magnitude of Kc gives us an indication of how far the reaction has proceeded toward the formation of products, when the equilibrium is achieved.
The larger the value of K, the further the reaction will have proceeded towards completion when equilibrium is reached (more products present at equilibrium).
Equilibrium
6
Equilibrium is dynamic. The forward and reverse rxns occur at the same rate.
There is a spontaneous tendency towards equilibrium. This does not mean that equilibrium will occur quickly, it simply means that there is always a drive TOWARD the equilibrium state. The amount of drive is measured as Free Energy (D G)
The driving force towards equilibrium diminishes as equilibrium is approached. Thus the appearance of products actually decreases the forward impetus of the reaction, making the free energy change less negative.
http://www.youtube.com/watch?v=CMs2WhGY3NE
Equilibrium
7
The equilibrium position is the same at a given temperature, no matter from which direction it is approached.
It is possible to force an equilibrium one way or the other temporarily by altering the reaction conditions, but once this “stress” is removed, the system will return to its original equilibrium.
Figure 1: Reversible reactions
8
PX o
r [X
]
Time →
[B] or PB / RT Equilibrium is established
[A] or PA / RT
[A]0 or PA0 / RT
Reversible Reactions and Rate
9
Reacti
on
R
ate
Time
Backward rate
Forward rate
Equilibrium is established:
Forward rate = Backward rate
When equilibrium is achieved:[A] ≠ [B] and kf/kr = Keq
15.2: Law of Mass Action
10
Derived from rate laws by Guldberg andWaage (1864) For a balanced chemical reaction
in equilibrium:a A + b B ↔ c C + d D
Equilibrium constant expression (Keq):
ba
dc
c [B] [A]
[D] [C] K b
Ba
A
dD
cC
p )(P)(P
)(P)(PK
Keq is strictly based on stoichiometry of the reaction (is independent of the mechanism).
Units: Keq is considered dimensionless (no units)
Cato Guldberg Peter Waage
(1836-1902) (1833-1900)
or
Relating Kc and Kp
11
Convert [A] into PA:
ba
dc
([B]RT)([A]RT)
([D]RT)([C]RT)b
Ba
A
dD
cC
p )(P)(P
)(P )(PK
baba
dcdc
(RT)[B][A]
(RT)[D][C]
(RT) K K b)(a - d)(ccp
where n == change in coefficents of products – reactants (gases only!)= (c+d) - (a+b)
(RT) K nc
RT
P
V
nM RTAPA ][
Magnitude of Keq
12
Since Keq [products]/[reactants], the magnitude of Keq 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”
Relationship Between Q and K
13
Reaction Quotient (Q): The particular ratio of concentration terms that we write for a particular reaction is called reaction quotient.
For a reaction, A B, Q= [B]/[A] At equilibrium, Q= K Reaction Direction: Comparing Q and K Q<K, reaction proceeds to right, until
equilibrium is achieved (or Q=K). Why? Q>K, reaction proceeds to left, until Q=K.
Why?
Value of K
For therxn, A>B,
For the reverse rxn, B >A,
For the reaction,2A > 2B
For the rxn,A > CC > B
K(rxn)= [B]/[A]
K= 1/K(rxn)
K= K(rxn)2 K (overall)= K1 X K2
14
15.3: Types of Equilibria
15
Homogeneous: all components in same phase (usually g or aq)
N2 (g) + H2 (g) ↔ NH3 (g)
3H
1N
2NH
P )(P)(P
)(PK
22
3
bB
aA
dD
cC
P )(P)(P
)(P)(PK
3 21
Fritz Haber(1868 – 1934)
16
Heterogeneous: reactants and products in different phases
CaCO3 (s) ↔ CaO (s) + CO2 (g)
Definition: What we use:
][CaCO
)(P [CaO] K
3
COeq
22COp P K
Even though the concentrations of the solids or liquids do not appear in the equilibrium expression, the substances must be present to achieve equilibrium.
Concentrations of pure solids and pure liquids are not included in Keq expression because their concentrations do not vary, and are “already included” in Keq (see p. 548).
15.4: Calculating Equilibrium Constants
17
ICE tables are used to calculate equilibriumconcentrations of the reactants and productsfrom the initial concentrations.Steps to use “ICE” table:
1. “I” = Tabulate known initial and equilibrium concentrations of all species in equilibrium expression
2. “C” = Determine the concentration change for the species where initial and equilibrium are known
• Use stoichiometry to calculate concentration changes for all other species involved in equilibrium
3. “E” = Calculate the equilibrium concentrations
Sample Problem 1 on Equilibrium:
18
Ex: Enough ammonia is dissolved in 5.00 L of water at 25ºC to produce a solution that is 0.0124 M ammonia. The solution is then allowed to come to equilibrium. Analysis of the equilibrium mixture shows that [OH1-] is 4.64 x 10-4 M. Calculate Keq at 25ºC for the reaction:NH3 (aq) + H2O (l) ↔ NH4
1+ (aq) + OH1- (aq)
NH3 (aq) + H2O (l) ↔ NH41+ (aq) + OH1- (aq)
Initial
Change
Equilibrium
19
][NH
]][OH[NH K
3
-114
c
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
101.81x 0.0119
)10(4.64
NH3 (aq)H2O (l)
NH41+ (aq) OH1- (aq)
XXX
x = 4.64 x 10-4 M
Sample Problem 2 on Equilibrium
20
Ex: A 5.000-L flask is filled with 5.000 x 10-3
mol of H2 and 1.000 x 10-2 mol of I2 at 448ºC. The value of Keq is 1.33. What are the concentrations of each substance at equilibrium?
H2 (g) + I2 (g) ↔ 2 HI (g)
H2 (g) + I2 (g) ↔ 2 HI (g)
Initial
Change
Equilibrium
21
]][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 eq’n: 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)
15.6: Le Châtelier’s Principle
22
If a system at equilibrium is disturbed by a change in: Concentration of one of the components, Pressure, or Temperature
…the system will shift its equilibrium position to counteract the effect of the disturbance.
http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/lechv17.swf
Henri Le Châtelier(1850 – 1936)
4 Changes that do not affect Keq:
23
1. Concentration Upon addition of a reactant or product,
equilibrium shifts to re-establish equilibrium by consuming part of the added substance.
Upon removal of reactant or product, equilibrium shifts to re-establish equilibrium by producing more of the removed substance.Ex: Co(H2O)6
2+ (aq) + 4 Cl1- ↔ CoCl42- (aq) + 6 H2O (l)
•Add HCl, temporarily inc forward rate•Add H2O, temporarily inc reverse rate
2. Volume, with a gas present (T is constant)
24
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 NH3 (g) If V of container is decreased, equilibrium shifts
right. XN2 and XH2
dec XNH3 inc3
HN
2NH
P )(PP
)(PK
22
33
THTN
2TNH
)P)(XP(X
)P(X
22
3 4T
3HN
2T
2NH
P)(XX
P)(X
22
3 2T
3HN
2NH
P)(XX
)(X
22
3
23HN
2NH
P )(
)(K
22
3
Since PT also inc, KP remains constant.
3. Pressure, but not Volume
25
Usually addition of a noble gas, p. 560 Avogadro’s 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, so there is no Le Châtelier shift.
4. Catalysts
26
Lower the activation energy of both forward and reverse rxns, therefore increases both forward and reverse 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
1 Change that does affect Keq:
27
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 which decreases Keq
2. Endothermic equilibria: Reactants + heat ↔ Products
• Inc T increases forward reaction rate increases Keq
Ex: Co(H2O)62+ (aq) + 4 Cl1- ↔ CoCl42- (aq) + 6 H2O (l);
H=+?•Inc T temporarily inc forward rate•Dec T temporarily inc reverse rate
Vant Hoff’s Equation
28
Vant Hoff’s equation shows mathematically how the equilibrium constant is affected by changes in temp.
ln K2 = -d H0 rxn (1 – 1)
K1 R T2 T1
Effect of Various Changes on Equilibrium
Disturbance Net Direction of Rxn
Effect of Value of K
Concentration Increase (reactant)
Towards formation of product
None
Decrease(reactant)
Towards formation of reactant
None
Increase (product)
Towards formation of reactant
None
Decrease (product)
Towards formation of product
None
29
Effect of Pressure on Equilb.
Pressure Increase P (decrease V)
Towards formation of fewer moles of gas
None
Decrease P (Increase V)
Towards formation of more moles of gas
None
Increase P ( Add inert gas, no change in V)
None, concentrations unchanged
None
30