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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
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