14-2-1 CHEM 102, Fall 2013, LA TECH Instructor: Dr. Upali Siriwardane e-mail: [email protected]...

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

Instructor: Dr. Upali Siriwardane

e-mail: [email protected]

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


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


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.


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


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


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.


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.


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


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.


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


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


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


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)


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


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 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


K = 1.15 x 102


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


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


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


What is K (Kc) and Kp

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

Kp- equilibrium constant calculated based on partial pressure

Kp =


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 = 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


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


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.


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 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


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)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


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

2X = 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


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


Shifting of Equilibrium

N2O4(g) 2 NO2(g)


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


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


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


Thermodynamics of Equilibrium

a) Enthalpy (DH)

b) Entropy (DS)

c) Free Energy (DG)

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


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


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


Industrial 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


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 17. 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


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

14-2-1 CHEM 102, Fall 2013, LA TECH Instructor: Dr. Upali Siriwardane e-mail: [email protected]...

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Transcript of 14-2-1 CHEM 102, Fall 2013, LA TECH Instructor: Dr. Upali Siriwardane e-mail: [email protected]...