14-2-1CHEM 102, Fall 2013, LA TECH

Instructor: Dr. Upali Siriwardane

e-mail: upali@coes.latech.edu

Office: CTH 311

Phone 257-4941

Office Hours: M,W 8:00-9:30 & 11:00-12:30 am;

Tu,Th, F 8:00 - 10:00 am., or by appointment. 

Test Dates:

Chemistry 102(01) Fall 2013

September 24, 2013 (Test 1): Chapter 13

October 17, 2013 (Test 2): Chapter 14 &15

November 12, 2013 (Test 3) Chapter 16 &17

November 14, 2013 (Make-up test) comprehensive: Chapters 13-17 10:00-11:15 am., CTH 328

14-2-2CHEM 102, Fall 2013, LA TECH

Chapter 14. Chemical Equilibrium 14.1 Fetal Hemoglobin and Equilibrium 61 314.2 The Concept of Dynamic Equilibrium 61 514.3 The Equilibrium Constant (K) 61 814.4 Expressing the Equilibrium Constant in Terms of

Pressure 62214.5 Heterogeneous Equilibria: Reactions Involving Solids

and Liquids 62514.6 Calculating the Equilibrium Constant from Measured

Equilibrium Concentrations 62614.7 The Reaction Quotient: Predicting the Direction of

Change 62914.8 Finding Equilibrium Concentrations 63114.9 Le Châtelier’s Principle: How a System at Equilibrium

Responds to Disturbances 641

14-2-3CHEM 102, Fall 2013, LA TECH

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.

14-2-4CHEM 102, Fall 2013, LA TECH

7) What is the difference between initial [A]i and equilibrium [A]eq concentrations?

Initial @ Equilibrium  

N2O4 NO2 N2O4 NO2 Keq0.00 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

14-2-5CHEM 102, Fall 2013, LA TECH

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

14-2-6CHEM 102, Fall 2013, LA TECH

Equilibrium constant calculations 1) Consider the reaction:

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

At equilibrium, [CO] = 4.14 × 10-6 M;

[Cl2] = 4.14 × 10-6 M; and [COCl2] = 0.0627 M.

Calculate the value of the equilibrium constant.

14-2-7CHEM 102, Fall 2013, LA TECH

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.

14-2-8CHEM 102, Fall 2013, LA TECH

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

14-2-9CHEM 102, Fall 2013, LA TECH

Reaction quotient (Q) calculations2) Consider the reaction system: C2H5OH (aq) + CH3COOH(aq) CH3COOC2H5 +H2O (l)

= ? The concentrations of both ethanol and acetic acid are 0.45

M and the concentration of ethyl acetate is 1.1 M. Use the reaction quotient to determine whether the system is at equilibrium.

14-2-10CHEM 102, Fall 2013, LA TECH

Reaction quotient (Q) calculationsConsider the reaction system: C2H5OH (aq) + CH3COOH(aq) CH3COOC2H5 +H2O (l)

 The concentrations of both ethanol and acetic acid are 0.45 M and the concentration of ethyl acetate is 1.1 M. Use the reaction quotient to determine whether the system is at equilibrium.

= ? and Keq=0.95

14-2-11CHEM 102, Fall 2013, LA TECH

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

14-2-12CHEM 102, Fall 2013, LA TECH

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- 0.263= 0.138 0.525

x-1/2 x

0.40 - 1/2x = 0 + x

14-2-13CHEM 102, Fall 2013, LA TECH

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)

14-2-14CHEM 102, Fall 2013, LA TECH

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?

14-2-15CHEM 102, Fall 2013, LA TECH

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)

14-2-16CHEM 102, Fall 2013, LA TECH

Equilibrium calculations using ICE 3) Consider the reaction A 2B,

where the value of Keq is 1.4 × 10-12.

At equilibrium, the concentration of B is 0.45 M.

What is the concentration of A?

ICE Calculation [A] [B]Initial concentration:    

Change    

Equilibrium concentration:    

 K = 1.4 x 10-12    

        

14-2-17CHEM 102, Fall 2013, LA TECH

Equilibrium calculations using ICE 4) Calculation of unknown concentration of

reactants or products in an equilibrium mixture At 100o C the equilibrium constant (K) for the reaction:

H2(g) + I2(g) 2 HI(g) ; K = 1.15 x 102.

If 0.400 moles of H2 and 0.400 moles. If 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? ICE Calculation [A] [B]Initial concentration:    

Change    

Equilibrium concentration:    

14-2-18CHEM 102, Fall 2013, LA TECH

Equilibrium calculations using ICE 4) H2(g) + I2(g) 2 HI(g) ; K = 1.15 x 102.

If 0.400 moles of H2 and 0.400 moles. If 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?

ICE Calculation [H2] [I2] [HI]Initial concentration:    

Change    

Equilibrium concentration:

   

 K = 1.15 x 102    

14-2-19CHEM 102, Fall 2013, LA TECH

Types of Equilibrium

1) Heterogeneous Equilibrium

2) Heterogeneous Equilibrium

3) Acid Dissociation Constant- Ka

4) Base Dissociation Constant- Kb

5) Autoionization Constant- Kw

6) Solubility Product Constant-Ksp

14-2-20CHEM 102, Fall 2013, LA TECH

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)]

14-2-21CHEM 102, Fall 2013, LA TECH

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

- (aq)

[H3O+

][C2H3O2-]

K =

[H2O][HC2H3O2]

[H3O+

][C2H3O2-]

Ka = K [H2O] =

[HC2H3O2]

14-2-22CHEM 102, Fall 2013, LA TECH

Base Dissociation Constant

NH3 + H2O(l) NH4+ + OH-

[NH4+

][OH-]

K =

[H2O][NH3]

[NH4+

][OH-]

Kb = K [H2O] =

[NH3]

14-2-23CHEM 102, Fall 2013, LA TECH

Autoionization of Water

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

[H3O+

][OH-]

K =

[H2O]2

Kw = K [H2O]2

= [H3O+

][OH-] = 1.0 10

-14

14-2-24CHEM 102, Fall 2013, LA TECH

Solubility Product of Salts in Water

AgCl(s) + H2O (l) Ag+(aq) + Cl-

Ksp = [Ag+

] [Cl-]

Ksp (AgCl) = 1.77 × 10-10

Ksp (BaSO4) = 1.1 x 10-10

14-2-25CHEM 102, Fall 2013, LA TECH

What is K (Kc) and Kp

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

Kp- equilibrium constant calculated based on partial pressure

Kp =

14-2-26CHEM 102, Fall 2013, LA TECH

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

14-2-27CHEM 102, Fall 2013, LA TECH

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

14-2-28CHEM 102, Fall 2013, LA TECH

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 = Kc(RT) Dn

Kc = Kp(RT) -Dn

R = universal gas constant (0.08206 )

T = Kelvin Temperature,

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

atm L

mol K

14-2-29CHEM 102, Fall 2013, LA TECH

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

14-2-30CHEM 102, Fall 2013, LA TECH

Kp = Kc (RT)Dng

= 1.10 x 107

(0.08206 ) (973 K)

= 1.378 x105

atm L

mol K[ ]-1

Partial pressure & Equilibrium Constants

14-2-31CHEM 102, Fall 2013, LA TECH

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.

14-2-32CHEM 102, Fall 2013, LA TECH

Predicting the Direction of a Reaction

14-2-33CHEM 102, Fall 2013, LA TECH

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

14-2-34CHEM 102, Fall 2013, LA TECH

Q Calculation

SO2(g) + NO2(g) NO(g) + SO3(g) (Kc = 85.0 at 460o

C)

   [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

14-2-35CHEM 102, Fall 2013, LA TECH

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?

14-2-36CHEM 102, Fall 2013, LA TECH

Equilibrium Calculation Example

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)

14-2-37CHEM 102, Fall 2013, LA TECH

Equilibrium calculation example

X2

(0.095 - X)Keq = 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

a X2 + b X - c = 0

14-2-38CHEM 102, Fall 2013, LA TECH

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

14-2-39CHEM 102, Fall 2013, LA TECH

Equilibrium Calculation Example

-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

2X = 4.6 x 10

-6 M

X = -4.6 x 10-6

M

14-2-40CHEM 102, Fall 2013, LA TECH

Equilibrium Calculation Example

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.

14-2-41CHEM 102, Fall 2013, LA TECH

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?

14-2-42CHEM 102, Fall 2013, LA TECH

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.

14-2-43CHEM 102, Fall 2013, LA TECH

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

14-2-44CHEM 102, Fall 2013, LA TECH

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

Shifts with TemperatureN2O4(g)

colorless

2NO2(g)

Dark brown

14-2-45CHEM 102, Fall 2013, LA TECH

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

14-2-46CHEM 102, Fall 2013, LA TECH

Shifting of Equilibrium

N2O4(g) 2 NO2(g)

14-2-47CHEM 102, Fall 2013, LA TECH

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

14-2-48CHEM 102, Fall 2013, LA TECH

5) How you would increase the products of following industrially important reactions:a) CO(g) + H2O (g) H2 (g) + CO2 (g);DH= −41.2 kJ/mol

b) N2(g) nitrogen

+ 3H2(g) hydrogen

2NH3(g) ammonia

H = -92.4 kJ mol-1

14-2-49CHEM 102, Fall 2013, LA TECH

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

14-2-50CHEM 102, Fall 2013, LA TECH

Thermodynamics of Equilibrium

a) Enthalpy (DH)

b) Entropy (DS)

c) Free Energy (DG)

(D G is a combined term involving DH, DS and T)

14-2-51CHEM 102, Fall 2013, LA TECH

Probability, Entropy andChemical Equilibrium

14-2-52CHEM 102, Fall 2013, LA TECH

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

14-2-53CHEM 102, Fall 2013, LA TECH

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

14-2-54CHEM 102, Fall 2013, LA TECH

Industrial Production of Ammonia

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

catalysis

high pressure

and temperature

14-2-55CHEM 102, Fall 2013, LA TECH

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

14-2-56CHEM 102, Fall 2013, LA TECH

Haber-Bosch Process

14-2-57CHEM 102, Fall 2013, LA TECH

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

14-2-58CHEM 102, Fall 2013, LA TECH

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.

14-2-59CHEM 102, Fall 2013, LA TECH

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

14-2-60CHEM 102, Fall 2013, LA TECH

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

Shifts with TemperatureN2O4(g)

colorless

2NO2(g)

Dark brown

14-2-61CHEM 102, Fall 2013, LA TECH

Probability, Entropy andChemical Equilibrium

14-2-62CHEM 102, Fall 2013, LA TECH

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

liquid systems, more than solid systems

Chapter 17. Thermodynamics DG = DH -TDS DG = Gibbs Free Energy (- for

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

14-2-63CHEM 102, Fall 2013, LA TECH

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

14-2-64CHEM 102, Fall 2013, LA TECH

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

14-2-65CHEM 102, Fall 2013, LA TECH

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