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Chem Chapter 14 LEC
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Chapter 14Chemical
Equilibrium
2008, Prentice Hall
Chemistry: A Molecular Approach, 1st Ed.Nivaldo Tro
Roy KennedyMassachusetts Bay Community College
Wellesley Hills, MA
Tro, Chemistry: A Molecular Approach 2
Hemoglobin• protein (Hb) found in red blood
cells that reacts with O2 enhances the amount of O2 that
can be carried through the blood stream
Hb + O2 HbO2
the Hb represents the entire protein – it is not a chemical formula
the represents that the reaction is in dynamic equilibrium
Tro, Chemistry: A Molecular Approach 3
Hemoglobin Equilibrium SystemHb + O2 HbO2
• the concentrations of Hb, O2, and HbO2 are all interdependent
• the relative amounts of Hb, O2, and HbO2 at equilibrium are related to a constant called the equilibrium constant, K the larger the value of K, the more product is found at
equilibrium
• changing the concentration of any one of these necessitates changing the other concentrations to reestablish equilibrium
Tro, Chemistry: A Molecular Approach 4
O2 Transport
Hb + O2
in the lungs, with high concentration of O2, the equilibrium shifts to combine the Hb and O2 together to make more HbO2
in the cells, with low concentration of O2, the equilibrium shifts to break down the HbO2 and increase the amount of free O2
HbO2
O2 in
lungs
Hb HbO2
O2 in
cells
Tro, Chemistry: A Molecular Approach 5
HbF
Hb
Fetal Hemoglobin, HbFHbF + O2 HbFO2
• fetal hemoglobin’s equilibrium constant is larger than adult hemoglobin
• because fetal hemoglobin is more efficient at binding O2, O2 is transferred to the fetal hemoglobin from the mother’s hemoglobin in the placenta
Hb + O2 HbO2O2 HbO2
O2
Hb + O2 HbFO2HbFO2O2
Tro, Chemistry: A Molecular Approach 7
Reaction Dynamics• when a reaction starts, the reactants are consumed and
products are made forward reaction = reactants products therefore the reactant concentrations decrease and the product
concentrations increase as reactant concentration decreases, the forward reaction rate
decreases
• eventually, the products can react to reform some of the reactants reverse reaction = products reactants assuming the products are not allowed to escape as product concentration increases, the reverse reaction rate increases
• processes that proceed in both the forward and reverse direction are said to be reversible reactants products
8
Hypothetical Reaction2 Red BlueTime [Red] [Blue]
0 0.400 0.000
10 0.208 0.096
20 0.190 0.105
30 0.180 0.110
40 0.174 0.113
50 0.170 0.115
60 0.168 0.116
70 0.167 0.117
80 0.166 0.117
90 0.165 0.118
100 0.165 0.118
110 0.164 0.118
120 0.164 0.118
130 0.164 0.118
140 0.164 0.118
150 0.164 0.118
The reaction slows over time,
But the Red molecules never run out!
At some time between 100 and 110 sec,the concentrations of both the Red andthe Blue molecules no longer change –equilibrium has been established.
Notice that equilibrium does not meanthat the concentrations are equal!
Once equilibrium is established, the rateof Red molecules turning into Blue is thesame as the rate of Blue molecules turning into Red
Tro, Chemistry: A Molecular Approach 9
Hypothetical Reaction2 Red Blue
Concentration vs. Time for 2 Red --> Blue
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0 20 40 60 80 100 120 140Time, sec
Con
cent
rati
on
[Red][Blue]
Tro, Chemistry: A Molecular Approach 10
Reaction Dynamics
Time
Rat
e
Rate Forward
Rate Reverse
Initially, only the forwardreaction takes place.
As the forward reaction proceedsit makes products and uses reactants.Because the reactant concentration decreases, the forward reaction slows.As the products accumulate, thereverse reaction speeds up.
Eventually, the reaction proceedsin the reverse direction as fast asit proceeds in the forward direction.At this time equilibrium is established.
Once equilibrium is established,the forward and reverse reactions proceed at the same rate, so theconcentrations of all materialsstay constant.
Tro, Chemistry: A Molecular Approach 11
Dynamic Equilibrium• as the forward reaction slows and the reverse
reaction accelerates, eventually they reach the same rate
• dynamic equilibrium is the condition where the rates of the forward and reverse reactions are equal
• once the reaction reaches equilibrium, the concentrations of all the chemicals remain constantbecause the chemicals are being consumed and
made at the same rate
Tro, Chemistry: A Molecular Approach 12
H2(g) + I2(g) 2 HI(g)at time 0, there are only reactants in the mixture, so only the forward reaction can take place
[H2] = 8, [I2] = 8, [HI] = 0
at time 16, there are both reactants and products in the mixture, so both the forward reaction and reverse reaction can take place
[H2] = 6, [I2] = 6, [HI] = 4
Tro, Chemistry: A Molecular Approach 13
H2(g) + I2(g) 2 HI(g)at time 32, there are now more products than reactants in the mixture − the forward reaction has slowed down as the reactants run out, and the reverse reaction accelerated
[H2] = 4, [I2] = 4, [HI] = 8
at time 48, the amounts of products and reactants in the mixture haven’t changed – the forward and reverse reactions are proceeding at the same rate – it has reached equilibrium
Tro, Chemistry: A Molecular Approach 14
H2(g) + I2(g) 2 HI(g)C
once
ntra
tion
Time Equilibrium Established
Since the [HI] at equilibrium is larger than the [H2] or [I2], we say the position of equilibrium favors products
As the reaction proceeds, the [H2] and [I2] decrease and the [HI] increases
Since the reactant concentrations are decreasing, the forward reaction rate slows down
And since the product concentration is increasing, the reverse reaction rate speeds up
Once equilibrium is established, the concentrations no longer change
At equilibrium, the forward reaction rate is the same as the reverse reaction rate
Tro, Chemistry: A Molecular Approach 15
Equilibrium Equal• the rates of the forward and reverse reactions are equal
at equilibrium• but that does not mean the concentrations of reactants
and products are equal• some reactions reach equilibrium only after almost all
the reactant molecules are consumed – we say the position of equilibrium favors the products
• other reactions reach equilibrium when only a small percentage of the reactant molecules are consumed – we say the position of equilibrium favors the reactants
Tro, Chemistry: A Molecular Approach 16
An Analogy: Population Changes
When Country A citizens feel overcrowded, some will emigrate to Country B. However, as time passes, emigration will occur inboth directions at the same rate, leading topopulations in Country A and Country B that areconstant, though not necessarily equal
Tro, Chemistry: A Molecular Approach 17
Equilibrium Constant• even though the concentrations of reactants and products
are not equal at equilibrium, there is a relationship between them
• the relationship between the chemical equation and the concentrations of reactants and products is called the Law of Mass Action
• for the general equation aA + bB cC + dD, the Law of Mass Action gives the relationship below the lowercase letters represent the coefficients of the balanced
chemical equation always products over reactants
• K is called the equilibrium constant unitless
ba
dc
KBA
DC
Tro, Chemistry: A Molecular Approach 18
Writing Equilibrium Constant Expressions
• for the reaction aA(aq) + bB(aq) cC(aq) + dD(aq) the equilibrium constant expression is
ba
dc
KBA
DC
• so for the reaction 2 N2O5 4 NO2 + O2 the equilibrium constant expression is:
252
24
2
ON
ONO K
Tro, Chemistry: A Molecular Approach 19
What Does the Value of Keq Imply?
• when the value of Keq >> 1, we know that when the reaction reaches equilibrium there will be many more product molecules present than reactant molecules the position of equilibrium favors products
• when the value of Keq << 1, we know that when the reaction reaches equilibrium there will be many more reactant molecules present than product molecules the position of equilibrium favors reactants
Tro, Chemistry: A Molecular Approach 22
Relationships between K and Chemical Equations
• when the reaction is written backwards, the equilibrium constant is inverted
ba
dc
KBA
DCforward
for the reaction aA + bB cC + dDthe equilibrium constant expression is:
for the reaction cC + dD aA + bBthe equilibrium constant expression is:
dc
ba
KDC
BAbackward
forwardbackward
1
KK
Tro, Chemistry: A Molecular Approach 23
Relationships between K and Chemical Equations
• when the coefficients of an equation are multiplied by a factor, the equilibrium constant is raised to that factor
ba
c
KBA
Coriginal
for the reaction aA + bB cC the equilibrium constant expression is:
for the reaction 2aA + 2bB 2cC the equilibrium constant expression is:
nKK originalnew
2
22
2
new
BA
C
BA
C
ba
c
ba
c
K
Tro, Chemistry: A Molecular Approach 24
Relationships between K and Chemical Equations
• when you add equations to get a new equation, the equilibrium constant of the new equation is the product of the equilibrium constants of the old equations
a
b
KA
B1
for the reactions (1) aA bB and (2) bB cC the equilibrium constant expressions are:
for the reaction aA cC the equilibrium constant expression is:
21new KKK
b
c
a
b
a
c
K
B
C
A
B
A
Cnew
b
c
KB
C2
Kbackward = 1/Kforward, Knew = Koldn
Ex 14.2 – Compute the equilibrium constant at 25°C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25°C
K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25°C
Solution:
Concept Plan:
Relationships:
Given:
Find:
N2(g) + 3 H2(g) 2 NH3(g) K1 = 3.7 x 108
K’K
2 NH3(g) N2(g) + 3 H2(g) 81
2107.3
11
KK
NH3(g) 0.5 N2(g) + 1.5 H2(g) 5
21
82
12
102.5'107.3
1'
K
KK
26
Equilibrium Constants for Reactions Involving Gases
• the concentration of a gas in a mixture is proportional to its partial pressure
• therefore, the equilibrium constant can be expressed as the ratio of the partial pressures of the gases
• for aA(g) + bB(g) cC(g) + dD(g) the equilibrium constant expressions are
ba
dc
KBA
DCc
or ba
dc
PP
PPK
BA
DCp
Tro, Chemistry: A Molecular Approach 27
Kc and Kp
• in calculating Kp, the partial pressures are always in atm
• the values of Kp and Kc are not necessarily the samebecause of the difference in unitsKp = Kc when n = 0
• the relationship between them is: nRTKK cp
n is the difference between the number of moles of reactants and moles of products
Tro, Chemistry: A Molecular Approach 28
Deriving the Relationshipbetween Kp and Kc
gas of volume A, of moles ,A AA Vn
V
n
LawGas Ideal thefrom ,AA RTnVP
RT
P
RTRTV
nP
A
AA
A
[A] ngsubstituti
29
Deriving the RelationshipBetween Kp and Kc
RT
PXX
for aA(g) + bB(g) cC(g) + dD(g)
ba
dc
KBA
DCc
substituting
badc
baba
dcdc
ba
dc
RTK
RTPP
RTPP
RTP
RTP
RTP
RTP
K
1
1
1
p
BA
DC
BA
DC
c
ba
dc
PP
PPK
BA
DCp
nbadc RTKRTKK ccp grearrangin
nRTKK cp
Ex 14.3 – Find Kc for the reaction 2 NO(g) + O2(g) 2 NO2(g), given Kp = 2.2 x 1012 @ 25°C
K is a unitless numbersince there are more moles of reactant than product, Kc should be
larger than Kp, and it is
Kp = 2.2 x 1012
Kc
Check:
Solution:
Concept Plan:
Relationships:
Given:
Find:
nRT
KK P
c
Kp Kc
13
1
Kmol
Latm
12
pc
104.5 K298 08206.0
102.2
nRT
KK
2 NO(g) + O2(g) 2 NO2(g)n = 2 3 = -1
Tro, Chemistry: A Molecular Approach 31
Heterogeneous Equilibria• pure solids and pure liquids are materials whose
concentration doesn’t change during the course of a reaction its amount can change, but the amount of it in solution
doesn’t because it isn’t in solution
• because their concentration doesn’t change, solids and liquids are not included in the equilibrium constant expression
• for the reaction aA(s) + bB(aq) cC(l) + dD(aq) the equilibrium constant expression is:
bd
KB
Dc
Tro, Chemistry: A Molecular Approach 32
Heterogeneous Equilibria
2CO
COp
22
c
2
CO
CO
P
PK
K
The amount of C is different, but the amounts of CO and CO2 remains the same. Therefore the amount of C has no effect on the position of equilibrium.
Tro, Chemistry: A Molecular Approach 33
Calculating Equilibrium Constants from Measured Equilibrium Concentrations
• the most direct way of finding the equilibrium constant is to measure the amounts of reactants and products in a mixture at equilibriumactually, you only need to measure one amount – then use
stoichiometry to calculate the other amounts• the equilibrium mixture may have different amounts of
reactants and products, but the value of the equilibrium constant will always be the sameas long as the temperature is kept constant the value of the equilibrium constant is independent of the
initial amounts of reactants and products
34
Initial and Equilibrium Concentrations forH2(g) + I2(g) 2HI(g) @ 445°C
Initial Equilibri
um
Equilibrium
Constant
[H2] [I2] [HI] [H2] [I2] [HI]
0.50 0.50 0.0 0.11 0.11 0.78
0.0 0.0 0.50 0.055 0.055 0.39
0.50 0.50 0.50 0.165 0.165 1.17
1.0 0.5 0.0 0.53 0.033 0.934
]][I[H
[HI]
22
2
eq K
501][0.11][0.1
[0.78]2
50055][0.055][0.
[0.39]2
50165][0.165][0.
[1.17]2
5033][0.53][0.0
[0.934]2
35
Calculating Equilibrium Concentrations• Stoichiometry can be used to determine the equilibrium
concentrations of all reactants and products if you know initial concentrations and one equilibrium concentration
• suppose you have a reaction 2 A(aq) + B(aq) 4 C(aq) with initial concentrations [A] = 1.00 M, [B] = 1.00 M, and [C] = 0. You then measure the equilibrium concentration of C as [C] = 0.50 M.
[A] [B] [C]
initial molarity 1.00 1.00 0
change in concentration
equilibrium molarity 0.50
+0.50-¼(0.50) -½(0.50)
0.75 0.88
Tro, Chemistry: A Molecular Approach 36
Ex 14.6 Find the value of Kc for the reaction2 CH4(g) C2H2(g) + 3 H2(g) at 1700°C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 MConstruct an ICE table for the reaction
for the substance whose equilibrium concentration is known, calculate the change in concentration
+0.035
[CH4] [C2H2] [H2]
initial 0.115 0.000 0.000
change
equilibrium 0.035
37
Ex 14.6 Find the value of Kc for the reaction2 CH4(g) C2H2(g) + 3 H2(g) at 1700°C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 Muse the known change to determine the change in the other materials
add the change to the initial concentration to get the equilibrium concentration in each column
use the equilibrium concentrations to calculate Kc
[CH4] [C2H2] [H2]
initial 0.115 0.000 0.000
change
equilibrium 0.035
+0.035-2(0.035) +3(0.035)
0.045 0.105
020.0
045.0
105.0035.0
CH
HHC
2
3
24
3222
c
K
Tro, Chemistry: A Molecular Approach 38
Expt 1 Expt 2
initial [N2O4] 0 0.0200
initial [NO2] 0.0200 0
change [N2O4]
change [NO2]
equilibrium [N2O4] 0.00452
equilibrium [NO2] 0.0172
The following data were collected for the reaction 2 NO2(g) N2O4(g) at 100°C. Complete the table and
determine values of Kp and Kc for each experiment.
Tro, Chemistry: A Molecular Approach 39
The following data were collected for the reaction 2 NO2(g) N2O4(g) at 100°C. Complete the table and
determine values of Kp and Kc for each experiment.
Expt 1 Expt 2
initial [N2O4] 0 0.0200
initial [NO2] 0.0200 0
change [N2O4] +0.0014 -0.0155
change [NO2] -0.0028 +0.0310
equilibrium [N2O4] 0.0014 0.00452
equilibrium [NO2] 0.0172 0.0310
70.40310.0
00452.0
15.00327.07.4
7.40172.0
0014.0
)73308206.0(
][NO
]O[N
22 Expt C,
1,2 Expt P,
21 Expt C,
2-1CP
22
42C
K
K
K
KK
K
Tro, Chemistry: A Molecular Approach 40
The Reaction Quotient• if a reaction mixture, containing both
reactants and products, is not at equilibrium; how can we determine which direction it will proceed?
• the answer is to compare the current concentration ratios to the equilibrium constant
• the concentration ratio of the products (raised to the power of their coefficients) to the reactants (raised to the power of their coefficients) is called the reaction quotient, Q
ba
dc
QBA
DCc
for the gas phase reactionaA + bB cC + dDthe reaction quotient is:
ba
dc
PP
PPQ
BA
DCp
Tro, Chemistry: A Molecular Approach 41
The Reaction Quotient:Predicting the Direction of Change
• if Q > K, the reaction will proceed fastest in the reverse direction the [products] will decrease and [reactants] will increase
• if Q < K, the reaction will proceed fastest in the forward direction the [products] will increase and [reactants] will decrease
• if Q = K, the reaction is at equilibrium the [products] and [reactants] will not change
• if a reaction mixture contains just reactants, Q = 0, and the reaction will proceed in the forward direction
• if a reaction mixture contains just products, Q = ∞, and the reaction will proceed in the reverse direction
If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Ex 14.7 – For the reaction below, which direction will it proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm & PICl = 0.355 atm
for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
direction reaction will proceed
Solution:
Concept Plan:
Relationships:
Given:
Find:
I2(g) + Cl2(g) 2 ICl(g) Kp = 81.9
QPI2, PCl2, PICl
8.10
102.0114.0
355.0
p
2
ClI
2ICl
p
22
Q
PP
PQ
since Q (10.8) < K (81.9), the reaction will proceed to the right
Ex 14.8 If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and Kc = 2.00 @ 1000°C, find the [CO2]eq for the reaction given.
Units & Magnitude OKCheck:Check: Round to 1 sig fig and substitute back in
Solution:Solve: Solve the equilibrium constant expression for the unknown quantity by substituting in the given amounts
Concept Plan:
Relationships:
Strategize: You can calculate the missing concentration by using the equilibrium constant expression
2 COF2 CO2 + CF4
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
[CO2]eq
Given:
Find:
Sort: You’re given the reaction and Kc. You’re also given the [X]eq of all but one of the chemicals
K, [COF2], [CF4] [CO2]
M10.1
118.0
255.000.2
CF
COFCO
COF
CFCO
2
4
22
c2
22
42c
K
K
22
42c
COF
CFCOK
Tro, Chemistry: A Molecular Approach 45
A sample of PCl5(g) is placed in a 0.500 L container and heated to 160°C. The PCl5 is decomposed into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3 and Cl2 are formed. Determine the equilibrium concentration of
PCl5 if Kc = 0.0635
Tro, Chemistry: A Molecular Approach 46
A sample of PCl5(g) is placed in a 0.500 L container and heated to 160°C. The PCl5 is decomposed into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3 and Cl2 are formed. Determine the equilibrium concentration of
PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
equilibrium
concentration, M? L 500.0
mol 203.0
L 500.0
mol 203.0
M60.2PCl
0635.0
406.0406.0ClPClPCl
PCl
ClPCl
5
c
235
5
23c
K
K
Tro, Chemistry: A Molecular Approach 47
Finding Equilibrium Concentrations When Given the Equilibrium Constant and Initial
Concentrations or Pressures• first decide which direction the reaction will proceed
compare Q to K
• define the changes of all materials in terms of x use the coefficient from the chemical equation for the coefficient of x the x change is + for materials on the side the reaction is proceeding
toward the x change is for materials on the side the reaction is proceeding away
from
• solve for x for 2nd order equations, take square roots of both sides or use the
quadratic formula may be able to simplify and approximate answer for very large or small
equilibrium constants
Tro, Chemistry: A Molecular Approach 48
[I2] [Cl2] [ICl]
initial 0.100 0.100 0.100
change
equilibrium
Ex 14.11 For the reaction I2(g) + Cl2(g) 2 ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures are all
0.100 atm, find the equilibrium concentrations
Construct an ICE table for the reaction
determine the direction the reaction is proceeding
1
100.0100.0
100.0
p
2
ClI
2ICl
p
22
Q
PP
PQ
since Qp(1) < Kp(81.9), the reaction is proceeding forward
Tro, Chemistry: A Molecular Approach 49
[I2] [Cl2] [ICl]
initial 0.100 0.100 0.100
change
equilibrium
Ex 14.11 For the reaction I2(g) + Cl2(g) 2 ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures are all
0.100 atm, find the equilibrium concentrations
represent the change in the partial pressures in terms of x
sum the columns to find the equilibrium concentrations in terms of x
substitute into the equilibrium constant expression and solve for x
2
22
ClI
2ICl
p
100.0
2100.0
100.0100.0
2100.09.81
22
x
x
xx
x
PP
PK
+2xxx0.100x 0.100x 0.100+2x
Tro, Chemistry: A Molecular Approach 50
[I2] [Cl2] [ICl]
initial 0.100 0.100 0.100
change
equilibrium
Ex 14.11 For the reaction I2(g) + Cl2(g) 2 ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures are all
0.100 atm, find the equilibrium concentrations
substitute into the equilibrium constant expression and solve for x
xx
xx
xx
x
x
x
x
9.812100.0100.09.81
2100.09.81100.09.81
2100.0100.09.81
100.0
2100.0
100.0
2100.09.81
2
2
+2xxx0.100x 0.100x 0.100+2x
x
x
xx
0729.0
05.11805.0
9.812100.0100.09.81
Tro, Chemistry: A Molecular Approach 51
[I2] [Cl2] [ICl]
initial 0.100 0.100 0.100
change
equilibrium
Ex 14.11 For the reaction I2(g) + Cl2(g) 2 ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures are all
0.100 atm, find the equilibrium concentrations
substitute x into the equilibrium concentration definition and solve +2xxx
0.100x 0.100x 0.100+2x
x0729.0atm 027.00729.0100.0100.0
2I xP
0.027
atm 027.00729.0100.0100.02Cl xP
0.027
atm 246.00729.02100.02100.0ICl xP
0.246
-0.0729 -0.0729 2(-0.0729)
Tro, Chemistry: A Molecular Approach 52
[I2] [Cl2] [ICl]
initial 0.100 0.100 0.100
change
equilibrium
-0.0729
Ex 14.11 For the reaction I2(g) + Cl2(g) 2 ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures are all
0.100 atm, find the equilibrium concentrations
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K to the given K
0.027 0.027 0.246
-0.0729 2(0.0729)
83
027.0027.0
246.0 2
p
ClI
2ICl
p
22
K
PP
PK
Kp(calculated) = Kp(given) within significant figures
Tro, Chemistry: A Molecular Approach 53
For the reaction I2(g) 2 I(g) the value of Kc = 3.76 x 10-5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to solve for x)
Tro, Chemistry: A Molecular Approach 54
For the reaction I2(g) 2 I(g) the value of Kc = 3.76 x 10-5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?
[I2] [I]
initial 0.500 0
change -x +2x
equilibrium 0.500- x 2x
since [I]initial = 0, Q = 0 and the reaction must proceed forward
25
25
2
2
c
4500.01076.3
500.0
21076.3
I
I
xx
x
x
K
55
For the reaction I2(g) 2 I(g) the value of Kc = 3.76 x 10-5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?
[I2] [I]
initial 0.500 0
change -x +2x
equilibrium 0.500- x 2x
552
255
25
1088.11076.340
41076.31088.1
4500.01076.3
xx
xx
xx
00216.0
42
1088.1441076.31076.3
reasonable be root will oneonly :note 2
4
5255
2
x
x
a
acbbx
Tro, Chemistry: A Molecular Approach 56
For the reaction I2(g) 2 I(g) the value of Kc = 3.76 x 10-5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?
[I2] [I]
initial 0.500 0
change -x +2x
equilibrium 0.498 0.00432
x = 0.00216
0.500 0.00216 = 0.498[I2] = 0.498 M
2(0.00216) = 0.00432[I] = 0.00432 M
52
2
2
c 1075.3498.0
00432.0
I
I K
Tro, Chemistry: A Molecular Approach 57
Approximations to Simplify the Math• when the equilibrium constant is very small, the
position of equilibrium favors the reactants
• for relatively large initial concentrations of reactants, the reactant concentration will not change significantly when it reaches equilibriumthe [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
Tro, Chemistry: A Molecular Approach 58
Checking the Approximation and Refining as Necessary
• we can check our approximation afterwards by comparing the approximate value of x to the initial concentration
• if the approximate value of x is less than 5% of the initial concentration, the approximation is valid
validision approximat the%5%100ionconcentrat initial
eapproximat if
x
Tro, Chemistry: A Molecular Approach 59
[H2S] [H2] [S2]
initial 2.50E-4 0 0
change
equilibrium
Ex 14.13 For the reaction 2 H2S(g) 2 H2(g) + S2(g) @ 800°C, Kc = 1.67 x 10-7. If a 0.500 L flask initially
containing 1.25 x 10-4 mol H2S is heated to 800°C, find the equilibrium concentrations.
Construct an ICE table for the reaction
determine the direction the reaction is proceeding
since no products initially, Qc = 0, and the reaction is proceeding forward
Tro, Chemistry: A Molecular Approach 60
[H2S] [H2] [S2]
initial 2.50E-4 0 0
change
equilibrium
Ex 14.13 For the reaction 2 H2S(g) 2 H2(g) + S2(g) @ 800°C, Kc = 1.67 x 10-7. If a 0.500 L flask initially
containing 1.25 x 10-4 mol H2S is heated to 800°C, find the equilibrium concentrations.
represent the change in the partial pressures in terms of x
sum the columns to find the equilibrium concentrations in terms of x
substitute into the equilibrium constant expression
+x+2x2x2.50E-42x
2x x
24
2
22
22
2c
21050.2
2
SH
SH
x
xxK
61
[H2S] [H2] [S2]
initial 2.50E-4 0 0
change
equilibrium
Ex 14.13 For the reaction 2 H2S(g) 2 H2(g) + S2(g) @ 800°C, Kc = 1.67 x 10-7. If a 0.500 L flask initially
containing 1.25 x 10-4 mol H2S is heated to 800°C, find the equilibrium concentrations.
+x+2x2x2.50E-42x
2x x
24
2
22
22
2c
21050.2
2
SH
SH
x
xxK
since Kc is very small, approximate the [H2S]eq = [H2S]init and solve for x
2.50E-4
24
2
22
22
2c
1050.2
2
SH
SH
xx
K
8
37
1025.6
41067.1
x
5
387
1038.1
4
1025.61067.1
x
x
62
[H2S] [H2] [S2]
initial 2.50E-4 0 0
change
equilibrium
Ex 14.13 For the reaction 2 H2S(g) 2 H2(g) + S2(g) @ 800°C, Kc = 1.67 x 10-7. If a 0.500 L flask initially
containing 1.25 x 10-4 mol H2S is heated to 800°C, find the equilibrium concentrations.
+x+2x2x
2x x
check if the approximation is valid by seeing if x < 5% of [H2S]init
2.50E-4
%5%52.5%1001050.2
1038.14
5
the approximation is not valid!!
63
[H2S] [H2] [S2]
initial 2.50E-4 0 0
change
equilibrium
Ex 14.13 For the reaction 2 H2S(g) 2 H2(g) + S2(g) @ 800°C, Kc = 1.67 x 10-7. If a 0.500 L flask initially
containing 1.25 x 10-4 mol H2S is heated to 800°C, find the equilibrium concentrations.
+x+2x2x2.50E-42x
2x x
254
2
c1038.121050.2
2
xx
K
if approximation is invalid, substitute xcurrent into Kc where it is subtracted and re-solve for xnew
24
37
1022.2
41067.1
x
5
3247
1027.1
4
1022.21067.1
x
x
xcurrent = 1.38 x 10-5
24
2
22
22
2c
21050.2
2
SH
SH
x
xxK
[H2S] [H2] [S2]
initial 2.50E-4 0 0
change
equilibrium
Ex 14.13 For the reaction 2 H2S(g) 2 H2(g) + S2(g) @ 800°C, Kc = 1.67 x 10-7. If a 0.500 L flask initially
containing 1.25 x 10-4 mol H2S is heated to 800°C, find the equilibrium concentrations.
+x+2x2x2.50E-42x
2x x
254
2
c1027.121050.2
2
xx
K
Substitute xcurrent into Kc where it is subtracted and re-solve for xnew. If xnew is the same number, you have arrived at the best approximation.
24
37
1025.2
41067.1
x
5
3247
1028.1
4
1025.21067.1
x
x
xcurrent = 1.27 x 10-5
since xcurrent = xnew, approx. OK
24
2
22
22
2c
21050.2
2
SH
SH
x
xxK
Tro, Chemistry: A Molecular Approach 65
[H2S] [H2] [S2]
initial 2.50E-4 0 0
change
equilibrium
Ex 14.13 For the reaction 2 H2S(g) 2 H2(g) + S2(g) @ 800°C, Kc = 1.67 x 10-7. If a 0.500 L flask initially
containing 1.25 x 10-4 mol H2S is heated to 800°C, find the equilibrium concentrations.
+x+2x2x2.50E-42x
2x x
xcurrent = 1.28 x 10-5
substitute xcurrent into the equilibrium concentration definitions and solve
M 1024.21028.121050.221050.2SH 45442
x
M 1056.21028.122H 552
x
M 1028.1S 52
x
2.24E-4 2.56E-5 1.28E-5
66
[H2S] [H2] [S2]
initial 2.50E-4 0 0
change
equilibrium
Ex 14.13 For the reaction 2 H2S(g) 2 H2(g) + S2(g) @ 800°C, Kc = 1.67 x 10-7. If a 0.500 L flask initially
containing 1.25 x 10-4 mol H2S is heated to 800°C, find the equilibrium concentrations.
+x+2x2x
2.24E-4 2.56E-5 1.28E-5
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K to the given K
724
525
22
22
2c 1067.1
1024.2
1028.11056.2
SH
SH
K
Kc(calculated) = Kc(given) within significant figures
Tro, Chemistry: A Molecular Approach 67
For the reaction I2(g) 2 I(g) the value of Kc = 3.76 x 10-5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?(use the simplifying assumption to solve for x)
Tro, Chemistry: A Molecular Approach 68
For the reaction I2(g) 2 I(g) the value of Kc = 3.76 x 10-5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?
[I2] [I]
initial 0.500 0
change -x +2x
equilibrium 0.500- x 2x
since [I]initial = 0, Q = 0 and the reaction must proceed forward
25
225
2
2
c
4500.01076.3
500.0
2
500.0
21076.3
I
I
x
x
x
x
K
Tro, Chemistry: A Molecular Approach 69
For the reaction I2(g) 2 I(g) the value of Kc = 3.76 x 10-5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?
[I2] [I]
initial 0.500 0
change -x +2x
equilibrium 0.500- x 2x
35
25
25
1017.24
1088.1
41088.1
4500.01076.3
x
x
x%5%434.0%100
500.0
1017.2 3
the approximation is valid!!
Tro, Chemistry: A Molecular Approach 70
For the reaction I2(g) 2 I(g) the value of Kc = 3.76 x 10-5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?
[I2] [I]
initial 0.500 0
change -x +2x
equilibrium 0.500- x 2x
x = 0.00217
0.500 0.00217 = 0.498[I2] = 0.498 M
2(0.00217) = 0.00434[I] = 0.00434 M
52
2
2
c 1078.3498.0
00434.0
I
I K
Tro, Chemistry: A Molecular Approach 71
Disturbing and Re-establishingEquilibrium
• once a reaction is at equilibrium, the concentrations of all the reactants and products remain the same
• however if the conditions are changed, the concentrations of all the chemicals will change until equilibrium is re-established
• the new concentrations will be different, but the equilibrium constant will be the sameunless you change the temperature
Tro, Chemistry: A Molecular Approach 72
Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting the effect various changes in conditions have on the position of equilibrium
• it says that if a system at equilibrium is disturbed, the position of equilibrium will shift to minimize the disturbance disturbances all involve making the system open
Tro, Chemistry: A Molecular Approach 73
An Analogy: Population Changes
When the populations of Country A and Country Bare in equilibrium, the emigration rates between thetwo states are equal so the populations stay constant.
When an influx of population enters Country Bfrom somewhere outside Country A, it disturbs the
equilibrium established between Country A and Country B.
The result will be people moving from Country B into Country A faster than people moving from Country A into Country B. This will continue until a new equilibrium between the populations is established, however the new populations will have different numbers of people than the old ones.
Tro, Chemistry: A Molecular Approach 74
The Effect of Concentration Changes on Equilibrium
• Adding a reactant will decrease the amounts of the other reactants and increase the amount of the products until a new position of equilibrium is found that has the same K
• Removing a product will increase the amounts of the other products and decrease the amounts of the reactants.you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added chemicals or toward side with removed chemicals
• Remember, adding more of a solid or liquid does not change its concentration – and therefore has no effect on the equilibrium
Tro, Chemistry: A Molecular Approach 75
Disturbing Equilibrium:Adding or Removing Reactants
• after equilibrium is established, how will adding a reactant affect the rate of the forward reaction? How will it affect the rate of the reverse reaction? What will this cause? How will it affect the value of K?as long as the added reactant is included in the equilibrium
constant expression i.e., not a solid or liquid
• after equilibrium is established, how will removing a reactant affect the rate of the forward reaction? How will it affect the rate of the reverse reaction? What will this cause? How will it affect the value of K?
Tro, Chemistry: A Molecular Approach 76
Disturbing Equilibrium:Adding Reactants
• adding a reactant initially increases the rate of the forward reaction, but has no initial effect on the rate of the reverse reaction.
• the reaction proceeds to the right until equilibrium is re-established.
• at the new equilibrium position, you will have more of the products than before, less of the non-added reactants than before, and less of the added reactant but not as little of the added reactant as you had before the
addition• at the new equilibrium position, the concentrations of
reactants and products will be such that the value of the equilibrium constant is the same
Tro, Chemistry: A Molecular Approach 77
Disturbing Equilibrium:Removing Reactants
• removing a reactant initially decreases the rate of the forward reaction, but has no initial effect on the rate of the reverse reaction. so the reaction is going faster in reverse
• the reaction proceeds to the left until equilibrium is re-established.
• at the new equilibrium position, you will have less of the products than before, more of the non-removed reactants than before, and more of the removed reactant but not as much of the removed reactant as you had before
the removal• at the new equilibrium position, the concentrations of
reactants and products will be such that the value of the equilibrium constant is the same
Tro, Chemistry: A Molecular Approach 78
The Effect of Adding a Gas to a Gas Phase Reaction at Equilibrium
• adding a gaseous reactant increases its partial pressure, causing the equilibrium to the rightincreasing its partial pressure increases its concentrationdoes not increase the partial pressure of the other gases in
the mixture
• adding an inert gas to the mixture has no effect on the position of equilibriumdoes not effect the partial pressures of the gases in the
reaction
Tro, Chemistry: A Molecular Approach 79
The Effect of Concentration Changes on Equilibrium
When NO2 is added, some of it combines to make more N2O4
Tro, Chemistry: A Molecular Approach 80
The Effect of Concentration Changes on Equilibrium
When N2O4 is added, some of itdecomposes to make more NO2
Tro, Chemistry: A Molecular Approach 81
Effect of Volume Changeon Equilibrium
• decreasing the size of the container increases the concentration of all the gases in the container increases their partial pressures
• if their partial pressures increase, then the total pressure in the container will increase
• according to Le Châtelier’s Principle, the equilibrium should shift to remove that pressure
• the way the system reduces the pressure is to reduce the number of gas molecules in the container
• when the volume decreases, the equilibrium shifts to the side with fewer gas molecules
Tro, Chemistry: A Molecular Approach 82
Disturbing Equilibrium:Changing the Volume
• after equilibrium is established, how will decreasing the container volume affect the total pressure of solids, liquid, and gases? How will it affect the concentration of solids, liquid, solutions, and gases? What will this cause? How will it affect the value of K?
Tro, Chemistry: A Molecular Approach 83
Disturbing Equilibrium: Reducing the Volume• for solids, liquids, or solutions, changing the size of the container has
no effect on the concentration, therefore no effect on the position of equilibrium
• decreasing the container volume will increase the total pressure Boyle’s Law if the total pressure increases, the partial pressures of all the gases will increase
Dalton’s Law of Partial Pressures
• decreasing the container volume increases the concentration of all gases same number of moles, but different number of liters, resulting in a different
molarity
• since the total pressure increases, the position of equilibrium will shift to decrease the pressure by removing gas molecules shift toward the side with fewer gas molecules
• at the new equilibrium position, the partial pressures of gaseous reactants and products will be such that the value of the equilibrium constant is the same
84
Since there are more gas molecules on the reactants
side of the reaction, when the pressure is increased the
position of equilibrium shifts toward the products.
When the pressure is decreased by increasing the volume, the position of equilibrium shifts
toward the side with the greater number of molecules – the
reactant side.
The Effect of Volume Changes on Equilibrium
Tro, Chemistry: A Molecular Approach 85
The Effect of Temperature Changes on Equilibrium Position• exothermic reactions release energy and
endothermic reactions absorb energy• if we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic reaction, it will help us use Le Châtelier’s Principle to predict the effect of temperature changeseven though heat is not matter and not written in a
proper equation
Tro, Chemistry: A Molecular Approach 86
The Effect of Temperature Changes on Equilibrium for Exothermic Reactions
• for an exothermic reaction, heat is a product• increasing the temperature is like adding heat• according to Le Châtelier’s Principle, the equilibrium
will shift away from the added heat• adding heat to an exothermic reaction will decrease
the concentrations of products and increase the concentrations of reactants
• adding heat to an exothermic reaction will decrease the value of K
• how will decreasing the temperature affect the system?
ba
dc
KBA
DCc
aA + bB cC + dD + Heat
Tro, Chemistry: A Molecular Approach 87
The Effect of Temperature Changes on Equilibrium for Endothermic Reactions• for an endothermic reaction, heat is a reactant• increasing the temperature is like adding heat• according to Le Châtelier’s Principle, the equilibrium
will shift away from the added heat• adding heat to an endothermic reaction will
decrease the concentrations of reactants and increase the concentrations of products
• adding heat to an endothermic reaction will increase the value of K
• how will decreasing the temperature affect the system?
ba
dc
KBA
DCc
Heat + aA + bB cC + dD
Tro, Chemistry: A Molecular Approach 88
The Effect of Temperature Changes on Equilibrium
Tro, Chemistry: A Molecular Approach 89
Not Changing the Position of Equilibrium - Catalysts
• catalysts provide an alternative, more efficient mechanism
• works for both forward and reverse reactions• affects the rate of the forward and reverse
reactions by the same factor• therefore catalysts do not affect the position of
equilibrium
Tro, Chemistry: A Molecular Approach 90
Practice - Le Châtelier’s Principle• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with H°
= -198 kJ is at equilibrium. How will each of the following changes affect the equilibrium concentrations of each gas once equilibrium is re-established?adding more O2 to the containercondensing and removing SO3
compressing the gasescooling the containerdoubling the volume of the containerwarming the mixtureadding the inert gas helium to the containeradding a catalyst to the mixture
Tro, Chemistry: A Molecular Approach 91
Practice - Le Châtelier’s Principle• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with H°
= -198 kJ is at equilibrium. How will each of the following changes affect the equilibrium concentrations of each gas once equilibrium is re-established?adding more O2 to the container shift to SO3
condensing and removing SO3 shift to SO3
compressing the gases shift to SO3
cooling the container shift to SO3
doubling the volume of the container shift to SO2
warming the mixture shift to SO2
adding helium to the container no effectadding a catalyst to the mixture no effect