14-1 CHEM 102, Fall 2011, LA TECH Instructor: Dr. Upali Siriwardane e-mail: [email protected]...

14-1CHEM 102, Fall 2011, LA TECH

Instructor: Dr. Upali Siriwardane

e-mail: [email protected]

Office: CTH 311 Phone 257-4941

Office Hours: M,W 8:00-9:00 & 11:00-12:00 am;

Tu, Th, F 8:00 - 10:00am.

Exams: 11:30-12:45 am, CTH 322.

Sept 22, 2011 (Test 1): Chapter 13

Oct 18, 2011 (Test 2): Chapter 14 &15

Nov 15, 2011 (Test 3): Chapter 17 &18

Comprehensive Final Exam: Nov 17, 2011 :

Chapters 13, 14, 15, 16, 17, and 18

Chemistry 102(01) Fall 2011


Chapter 14. Chemical Equilibrium 14.1Characteristics of Chemical Equilibrium 14.2The Equilibrium Constant 14.3Determining Equilibrium Constants 14.5The Meaning of Equilibrium Constant14.6Using Equilibrium Constants14.7Shifting a Chemical Equilibrium: Le Chatelier's

Principle 14.8Equilibrium at the Nanoscale14.9Controlling Chemical Reactions: The Haber-Bosch

Process


Different types of arrows are used in chemical equations associated with equilibria.

Single arrow

Assumes that the reaction proceeds to completion as written.

Two single-headed arrows

Used to indicate a system in equilibrium.

Two single-headed arrows of different sizes.

May be used to indicate when one side of an equilibrium system is favored.

Chemical equilibrium


Chemical EquilibriumBranch of chemistry dealing with reactions

where reactants and products coexist in a dynamic equilibrium

the rates of forward and backward reactions have comparable rates reaction


Chemical Equilibrium

Equilibrium region.

A point is finally reached where the forward and reverse reactions occur at the same rate.

H2 + I2 2HI

There is no net change in the concentration of any of the species.


Chemical EquilibriumP

art

ial P

ressu

re

Time

HI

I2

H2

Equilibrium

Region

Kinetic

Region


Complete Reaction

Con

cen

trati

on

Time

Kinetic No change Region


Equilibrium

A state where the forward and reverse conditions occur at the same rate.

Dynamic

Equilibrium

I’m in static

equilibrium.


This type of plot

shows the energy

changes during

a reaction.

This type of plot

shows the energy

changes during

a reaction.

Forward and Backward Reactions

H

activation

energy

Pote

nti

al

En

erg

y

Reaction coordinate


Value of K

rate of forward Reaction k+ K = ------------------------------ = --- rate of backward Reaction k-

K = a (infinity) -> Irreversible reactions

K = 0 -> No reaction

K = between 0 and 1 -> Equilibrium reactions


Law of mass ActionDefines an equilibrium constant (K) for the process

j A + k B l C + m D

[C]l[D]m

K = ----------------- ; [A], [B] etc are

[A]j[B]k Equilibrium concentrations

Pure liquid or solid concentrations are not written in the expression.


Equilibrium ExpressionAn equilibrium expression could be written

for any reaction

[HI]2

K = ----------- = 16 L/mol [H2][I2]Keq >> 1 reaction will go mainly to products

Keq ~ 1 reaction will produce roughly equal amounts of product and reactant

Keq << 1 reaction will go mainly to reactants


k is constant at a temperature

Initial @ Equilibrium

N2O4 NO2 N2O4 NO2 Keq

0.00 0.02 0.0014 0.017 0.21

0.00 0.03 0.0028 0.024 0.21

0.00 0.04 0.0045 0.031 0.21

0.02 0.00 0.0045 0.031 0.21

N2O4(g)

colorless

2NO2(g)

Dark brown

K eq [ ]

[ ]NON O

2

2 4

2


Calculating Stepwise EquilibriumAdd two equations with K1 and K2 to get Keq

Keq = K1 x K2

Subtract one equations with K2 from another with K2 to get Keq

Keq = K1 / K2

Doubling K1 to get Keq

Keq = (K1)2 ;tripling Keq = (K1)3 etc.

Reversing a reaction with K1 get Keq

Keq = (K1)½


Stepwise Equilibrium

(1) N2(g) + O2(g) 2NO(g) [NO]

2

Kc1 =

[N2][O2]

(2) 2NO(g) + O2(g) 2NO2(g) [NO2]

2

Kc2 =

[NO]2

[O2]Add to Combine (1.) & (2.)

N2(g) + 2O2(g) 2NO2(g)

[NO]2

[NO2]2

Kc = = Kc1 Kc2

[N2][O2] [NO]2

[O2]



Consider the reactions

2NO + O2 <===> 2 NO2 K = a

2 NO2 <===> N2O4 K = b

The value of the equilibrium constant for the reaction

2NO + O2 <===> N2O4 is

a. a + b

b. ab

c. (a/b)2

d. (ab)2

e. ab/2



Consider the reactions

2NO + O2 <===> 2 NO2 K = a

2 NO2 <===> N2O4 K = b

The value of the equilibrium constant for the reaction

4NO + 2O2 <===> 2 N2O4 is

a. a + b

b. ab

c. (a/b)2

d. (ab)2

e. ab/2


Homogenous equilibrium: Chemical equilibrium where reactants and products are in same phase.

Heterogeneous equilibrium: Chemical Equilibrium where at least one phase of a reactant or product is different from the rest.

Types of Equilibria


Heterogeneous Equilibrium

CaCO3(s) CaO(s) + CO2(g)

[CaO(s)][CO2(g)]

Kc =

[CaCO3(s)]

concentrations of pure solids and liquids

are constant are dropped from expression

Kc = [CO2(g)]


Acid Dissociation ConstantHC2H3O2 (aq) + H2O(l) H3O+ (aq) + C2H3O2

- (aq)

[H3O+

][C2H3O2-]

K =

[H2O][HC2H3O2]

[H3O+

][C2H3O2-]

Ka = K [H2O] =

[HC2H3O2]


Base Dissociation Constant

NH3 + H2O(l) NH4+ + OH-

[NH4+

][OH-]

K =

[H2O][NH3]

[NH4+

][OH-]

Kb = K [H2O] =

[NH3]


Autoionization of Water

H2O (l) + H2O (l) H3O+ + OH-

[H3O+

][OH-]

K =

[H2O]2

Kw = K [H2O]2

= [H3O+

][OH-] = 1.0 10

-14


Pressure Equilibrium Constants Kc & Kp

N2 + 3H2 2NH3

[NH3]2

Kc =

[N2][H2]3

=(PNH3/RT)

2

(PN2/RT) (PH2/RT)3

(PNH3)2 (1/RT)

2

Kc =

(PN2) (1/RT))(PH2)3

(1/RT)3

)

PNH32 (1/RT)2

=

PN2 PH23 (1/RT)(1/RT)3

(1/RT) 2

= Kp

(1/RT)(1/RT)3


Kc vs. Kp

N2 (g) + 3H2 (g) 2NH3 (g)

In General

Kc = Kp (1/RT)Dn

where Dn = #moles gaseous products

- # moles gaseous reactants

(1/RT)2

Kc = Kp = Kp (1/RT)-2

(1/RT)(1/RT)3


What is K (Kc) and Kp

Kc (K) - equilibrium constant calculated based on [A]-Concentrations.

Kp- equilibrium constant calculated based on partial pressure (p)

Kp = K(RT) Dn

R = universal gas constant

T = Kelvin Temperature,

Dn = (sum of stoichiometric coefficients of gaseous products) - (sum of the stoichiometric coefficients of gaseous reactants)


For the following equilibrium, Kc = 1.10 x 107

at 700. o

C. What is the Kp?

2H2 (g) + S2 (g) 2H2S (g)

Kp = Kc (RT)Dng

T = 700 + 273 = 973 K

R = 0.08206

Dng = ( 2 ) - ( 2 + 1) = -1

atm L

mol K

Partial pressure & Equilibrium Constants


Kp = Kc (RT)Dng

= 1.10 x 107

(0.08206 ) (973 K)

= 1.378 x105

atm L

mol K[ ]-1

Partial pressure & Equilibrium Constants


Determining Equilibrium ConstantsICE Method

1. Derive the equilibrium constant expression for the balanced chemical equation

2. Construct a Reaction Table with information (ICE) about reactants and products

3. Include the amounts reacted, x, in the Reaction Table

4. Calculate the equilibrium constant in terms of x


TerminologyInitial concentration:concentration (M) of reactants and products

before the equilibrium is reached.

Equilibrium ConcentrationConcentration (M) of reactants and products

After the equilibrium is reached.


Example: An equilibrium is established by placing 2.00 moles of N2O4(g) in a 5.00 L and heating the flask to 407 K. It was determined that at equilibrium the concentration of the NO2(g) is 0.525 mol/L. What is the value of the equilibrium constant?

N2O4(g) 2 NO2(g)

[NO2]2

Kc =

[N2O4]

N2O4(g) 2 NO2(g)

[Initial] (mol/L) 0.40 0

[Change] -x 2x

[Equilibrium]

0.40- 0.243= 0.138 0.525

x-1/2 x

0.40 - 1/2x = 0 + x


What is the value of the equilibrium constant?

0.525 = 0 + x [NO2]2

Kc =

[N2O4]

0.40 - 1/2x

x = 0.525 [NO2] = 0.40 - 1/2x

= 0.40 - 1/2(0525)

= 0.138

[NO2]2

(0.525)2

Kc = =

[N2O4] 0.138

= 2.00

NO2(g ) N2O4(g)


Equilibrium Calculations

Hydrogen iodide, HI, decomposes according to the equation

2 HI(g) H2(g) + I2(g)

When 4.00 mol of HI placed in a 5.00-L vessel at 458ºC, the equilibrium mixture was found to contain 0.442 mol I2. What is the value of Kc for the reaction?


Initial 4.00/5=.80 0 0

Change -2x x x

Equilibrium 0.80-2x x x=0.442/5

x = 0.0884

Equilibrium concentrations

[HI] = 0.80 - 2x = 0.8 - 2 x 0.0884 = 0.62

[H2] = x = 0.0884

[I2] = x = 0.0884 [H2] [I2] 0.0884 x 0.0884Kc = ---------------- = ------------------------- = 0.0201 [HI]2 (0.62) 2

2 HI(g) H2(g) + I2(g)


Selected Equilibrium Constants


What is the reaction quotient, Q

(Q) is constant in the equilibrium expression when initial concentration of reactants and products are used.

SO2(g)+ NO2(g) NO(g) +SO3(g)

[NO][SO3]

Q = ----------------

[SO2][NO2]

comparing to K and Q provide the net direction to achieve equilibrium.


We can predict the direction of a reaction by calculating the reaction quotient.

Reaction quotient, Q

For the reaction: aA + bB eE + fF

Q has the same form as Kc with one important difference. Q can be for any set of concentrations, not

just at equilibrium.

Q =[E]

e [F]

f

[A]a [B]

b

Equilibrium calculations


Any set of concentrations can be given and a Q calculated. By comparing Q to the Kc

value, we can predict the direction for the reaction.

Q < Kc Net forward reaction will occur.

Q = Kc No change, at equilibrium.

Q > Kc Net reverse reaction will occur.

Reaction quotient


Predicting the Direction of a Reaction


Consider the following reaction:

SO2(g) + NO2(g) NO(g) + SO3(g)

(Kc = 85.0 at 460oC)

Given: 0.040 mole of SO2(g), 0.500 mole of NO2(g), 0.30 mole of NO(g),and 0.020

mole of SO3(g) are mixed in a 5.00 L flask, Determine:

a) The net the reaction quotient, Q.

b) Direction to achieve equilibrium at 460oC.

Q Calculation


Q CalculationSO2(g) + NO2(g) NO(g) + SO3(g) (Kc = 85.0 at 460

oC)

[NO][SO3]

Q = -------------

[SO2][NO2]

0.040 mole 0.500 mole 0.30 mole 0.020 mole [SO2] = -------------; [NO2] = ----------- ; [NO] = ------------;

[SO3] = -----------

5.00 L 5.00L 5.00L 5.00 L

[SO2] = 8 x 10-3

mole/L ; [NO2] =0.1mole/L; [NO] = 0.06 mole/L; [SO3] = 4 x 10-3

mole/L

0.06 (4 x 10-3

)

Q = ------------------ = 0.3

8.0 x 10-3

x 0.1

Therefore the equilibrium shift to right


Equilibrium Calculation Example

A sample of COCl2 is allowed to decompose. The

value of Kc for the equilibrium

COCl2 (g) CO (g) + Cl2 (g)

is 2.2 x 10-10 at 100 oC.

If the initial concentration of COCl2 is 0.095M, what

will be the equilibrium concentrations for each of

the species involved?



COCl2 (g) CO (g) Cl2 (g)

Initial conc., M 0.095 0.000 0.000

Change - X + X + X

in conc. due to reaction

Equilibrium M(0.095 -X) X X

Concentration,

Kc = =[ CO ] [ Cl2 ]

[ COCl2 ]

X2

(0.095 - X)


Equilibrium calculation example

X2

(0.095 - X)Kc = 2.2 x 10-

10 =

Rearrangement gives

X2 + 2.2 x 10-

10 X - 2.09 x 10

-11 = 0

This is a quadratic equation. Fortunately, there is a

straightforward equation for their solution


Quadratic equations

An equation of the form

a X2 + b X + c = 0

Can be solved by using the following

x =

Only the positive root is meaningful in equilibrium problems.

-b + b2 - 4ac

2a



-b + b2 - 4ac

2a

X2 + 2.2 x 10

-10 X - 2.09 x 10

-11 = 0

a b c

X =

X = - 2.2 x 10-10

+ [(2.2 x 10-10

)2 - (4)(1)(- 2.09 x 10

-11)]

1/2

2

X = 4.6 x 10-6

M

X = -4.6 x 10-6

M



Now that we know X, we can solve for the concentration of all of the species.

COCl2 = 0.095 - X = 0.095 M

CO = X = 4.6 x 10-6 M

Cl2 = X = 4.6 x 10-6 M

In this case, the change in the concentration of is COCl2 negligible.


Le Chatelier’s principle

Any stress placed on an equilibrium system will cause the system to shift to minimize the effect of the stress.

You can put stress on a system by adding or removing something from one side of a reaction.

N2(g) + 3H2 (g) 2NH3 (g)

What effect will there be if you added more

ammonia? How about more nitrogen?


Predicting Shifts in Equilibria

Equilibrium concentrations are based on:• The specific equilibrium

• The starting concentrations

• Other factors such as:• Temperature• Pressure• Reaction specific conditions

Altering conditions will stress a system, resulting in an equilibrium shift.


Increase in Concentrationor Partial Pressure

for N2(g) + 3 H2(g) 2 NH3(g)

an increase in N2 and/or H2 concentration or pressure, will cause the equilibrium to shift towards the production of NH3


N2O4(g) 2 NO2(g) ; D H=? (+or -)

Shifts with TemperatureN2O4(g)

colorless

2NO2(g)

Dark brown


Probability, Entropy andChemical Equilibrium


Entropymeasure of the disorder in the systemmore disorder for gaseous systems than

liquid systems, more than solid systems

Chapter 18. Thermodynamics DG = DH -TDS DG = Gibbs Free Energy (- for

spontaneous) DH = Enthalpy DS = Entropy T = Kelvin Temperature


For the following equilibrium reactions:

H2(g) + CO2(g) H2O(g) + CO(g) DH = 40 kJ

Predict the equilibrium shift if:

a) The temperature is increased

b) The pressure is decreased

Predicting Equilibrium Shifts


Changes in pressureIn general, increasing the pressure by decreasing

volume shifts equilibrium towards the side that has the smaller number of moles of gas.

H2 (g) + I2 (g) 2HI (g)

N2O4 (g) 2NO2 (g)

Unaffected by pressureUnaffected by pressure

Increased pressure, shift to leftIncreased pressure, shift to left


Shifting of Equilibrium

N2O4(g) 2 NO2(g)


Equilibrium Systems

product-favored if K > 1

exothermic reactions favor products

increasing entropy in system favors products

at low temperature, product-favored reactions are usually exothermic

at high temperatures, product-favored reactions usually have increase in entropy


Equilibrium Reaction Rates

reactions occur faster in gaseous phase than solids and liquids

reactions rates increase as temperature increases

reactions rates increase as concentration increases

rates increase as particle size decreases

rates increase with a catalyst


Production of Ammonia

N2(g) + 3 H2(g) 2 NH3(g) ; DH = -

catalysis

high pressure

and temperature


Ammonia Synthesis

reaction is slow at room temperature, raising

temperature, increases rate but lowers yield

increasing pressure shifts equilibrium to

products

liquefying ammonia shifts equilibrium to

products

use of catalyst increases rate


Haber-Bosch Process


Decrease in Concentration or Partial Pressure

for N2(g) + 3 H2(g) 2 NH3(g) ; DH = -

likewise, a decrease in NH3 concentration or pressure will cause more NH3 to be produced


Changes in Temperaturefor N2(g) + 3 H2(g) 2 NH3(g) ; DH = -

for an exothermic reaction, an increase in temperature will cause the reaction to shift back towards reactants and vice versa.


Volume Changefor N2(g) + 3 H2(g) 2 NH3(g) ; DH = -

an increase in volume, causes the equilibrium to shift to the left where there are more gaseous molecules

a decrease in volume, causes the equilibrium to shift to the right where there are fewer gaseous molecules


At 100o C the equilibrium constant (K) for the reaction:

H2(g) + I2(g) 2HI(g)

is 1.15 x 102

. If 0.400 moles of H2 and 0.400 moles of I2 are placed

into a 12.0-liter container and allowed to react at this temperature, what is

the HI concentration (moles/liter) at equilibrium?

Calculating Concentrations at Equilibrium


At a certain temperature the value of the equilibrium constant is 3.24 for the

reaction:

H2(g) + CO2(g) H2O(g) + CO(g)

If 0.400 mol H2 and 0.400 mol CO2 are placed in a 1.00 L vessel, what is the

concentration of of CO at equilibrium?

Calculating Concentrations at Equilibrium


Le Chatelier’s principle

Any stress placed on an equilibrium system will cause the system to shift to minimize the effect of the stress.

You can put stress on a system by adding or removing something from one side of a reaction.

N2(g) + 3H2 (g) 2NH3 (g)

What effect will there be if you added more

ammonia? How about more nitrogen?


Predicting Shifts in Equilibria

Equilibrium concentrations are based on:• The specific equilibrium

• The starting concentrations

• Other factors such as:• Temperature• Pressure• Reaction specific conditions

Altering conditions will stress a system, resulting in an equilibrium shift.


Increase in Concentrationor Partial Pressure

for N2(g) + 3 H2(g) 2 NH3(g)

an increase in N2 and/or H2 concentration or pressure, will cause the equilibrium to shift towards the production of NH3


N2O4(g) 2 NO2(g) ; D H=? (+or -)

Shifts with TemperatureN2O4(g)

colorless

2NO2(g)

Dark brown


Probability, Entropy andChemical Equilibrium


Entropymeasure of the disorder in the systemmore disorder for gaseous systems than

liquid systems, more than solid systems

Chapter 18. Thermodynamics DG = DH -TDS DG = Gibbs Free Energy (- for

spontaneous) DH = Enthalpy DS = Entropy T = Kelvin Temperature


For the following equilibrium reactions:

H2(g) + CO2(g) H2O(g) + CO(g); DH = 40 kJ

Predict the equilibrium shift if:

a) The temperature is increased

b) The pressure is decreased

Predicting Equilibrium Shifts


Changes in pressureIn general, increasing the pressure by decreasing

volume shifts equilibria towards the side that has the smaller number of moles of gas.

H2 (g) + I2 (g) 2HI (g)

N2O4 (g) 2NO2 (g)

Unaffected by pressureUnaffected by pressure

Increased pressure, shift to leftIncreased pressure, shift to left


Shifting of Equilibrium

N2O4(g) 2 NO2(g)


Equilibrium Systems

product-favored if K > 1

exothermic reactions favor products

increasing entropy in system favors products

at low temperature, product-favored reactions are usually exothermic

at high temperatures, product-favored reactions usually have increase in entropy


Equilibrium Reaction Rates

reactions occur faster in gaseous phase than solids and liquids

reactions rates increase as temperature increases

reactions rates increase as concentration increases

rates increase as particle size decreases

rates increase with a catalyst


Production of Ammonia

N2(g) + 3 H2(g) 2 NH3(g) ; DH = -catalysis

high pressure

and temperature


Ammonia Synthesis

reaction is slow at room temperature, raising

temperature, increases rate but lowers yield

increasing pressure shifts equilibrium to

products

liquefying ammonia shifts equilibrium to

products

use of catalyst increases rate


Decrease in Concentration or Partial Pressure

for N2(g) + 3 H2(g) 2 NH3(g) ; DH = -

likewise, a decrease in NH3 concentration or pressure will cause more NH3 to be produced


Changes in Temperaturefor N2(g) + 3 H2(g) 2 NH3(g) ; DH = -

for an exothermic reaction, an increase in temperature will cause the reaction to shift back towards reactants and vice versa.


Volume Changefor N2(g) + 3 H2(g) 2 NH3(g) ; DH = -

an increase in volume, causes the equilibrium to shift to the left where there are more gaseous molecules

a decrease in volume, causes the equilibrium to shift to the right where there are fewer gaseous molecules

14-1 CHEM 102, Fall 2011, LA TECH Instructor: Dr. Upali Siriwardane e-mail: [email protected]...

Documents

Transcript of 14-1 CHEM 102, Fall 2011, LA TECH Instructor: Dr. Upali Siriwardane e-mail: [email protected]...