Chem Chapter 14 LEC

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


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


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


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


Oxygen Exchange betweenMother and Fetus


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


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]


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.


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


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


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


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


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


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


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


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


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


A Large Equilibrium Constant


A Small Equilibrium Constant


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



• 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



• 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


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


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


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


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.


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


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


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.


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


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


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

42

Q, K, and the Direction of Reaction

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


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


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


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


[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


[I2] [Cl2] [ICl]

initial 0.100 0.100 0.100

change

equilibrium



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


[I2] [Cl2] [ICl]

initial 0.100 0.100 0.100

change

equilibrium



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


[I2] [Cl2] [ICl]

initial 0.100 0.100 0.100

change

equilibrium



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)


[I2] [Cl2] [ICl]

initial 0.100 0.100 0.100

change

equilibrium

-0.0729



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


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)




[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



[I2] [I]

initial 0.500 0

change -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




[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


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


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


[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


[H2S] [H2] [S2]

initial 2.50E-4 0 0

change

equilibrium



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



+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



+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



+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



+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


since xcurrent = xnew, approx. OK

24

2

22

22

2c

21050.2

2

SH

SH

x

xxK


[H2S] [H2] [S2]

initial 2.50E-4 0 0

change

equilibrium



+x+2x2x2.50E-42x

2x x


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



+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



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)




[I2] [I]

initial 0.500 0

change -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




[I2] [I]

initial 0.500 0

change -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!!




[I2] [I]

initial 0.500 0

change -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


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


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


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.


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


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?


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


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


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



When NO2 is added, some of it combines to make more N2O4



When N2O4 is added, some of itdecomposes to make more NO2


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


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?


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

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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

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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




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


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


The Effect of Temperature Changes on Equilibrium

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http://wps.prenhall.com/wps/media/objects/169/173803/NO2andDinitrogenTetraoxide.html


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


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


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

Chem Chapter 14 LEC

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Transcript of Chem Chapter 14 LEC