Post on 23-Oct-2014
transcript
CHEMICAL EQUILIBRIUM
Basic Concepts
Reversible reactions do not go to completion. They can occur in either direction Symbolically, this is represented as:
dD cC bB+aA
Basic Concepts
Chemical equilibrium exists when two opposing reactions occur simultaneously at the same rate. A chemical equilibrium is a reversible
reaction that the forward reaction rate is equal to the reverse reaction rate.
Chemical equilibria are dynamic equilibria. Molecules are continually reacting, even
though the overall composition of the reaction mixture does not change.
Basic Concepts
4
One example of a dynamic equilibrium can be shown using radioactive 131I as a tracer in a saturated PbI2 solution.
solution. into go williodine eradioactiv theof Some
solution. filter the then minutes, few afor Stir 2
I 2 Pb PbI
solution. PbI saturated ain PbI solid Place 1-(aq)
2(aq)
OH
2(s)
2*22
Basic Concepts
5
Graphically, this is a representation of the rates for the forward and reverse reactions for this general reaction.
aA(g) + bB(g) ⇄ cC(g) + dD(g)
Basic Concepts
6
One of the fundamental ideas of chemical equilibrium is that equilibrium can be established from either the forward or reverse direction.
2SO2(g) + O2(g) ⇄ 2SO3(g)
7
The Equilibrium Constant
rate. reverse therepresents which DCkRate
rate. forward therepresents which BAkRate
rr
ff
For a simple one-step mechanism reversible reaction such as:
The rates of the forward and reverse reactions can be represented as:
(g)(g)(g)(g) D C B A
The Equilibrium Constant
10
When system is at equilibrium:Ratef = Rater
BA
DC
k
k
torearrangeswhich
DCkBAk
:give toiprelationsh rate for the Substitute
r
f
rf
The Equilibrium Constant
11
Because the ratio of two constants is a constant we can define a new constant as follows :
kk
K and
KC DA B
f
rc
c
The Equilibrium Constant
Similarly, for the general reaction:
we can define a constant
12
reactions. allfor validis expression This
BA
DCK
products
reactants ba
dc
c
D d C c B b A a (g)(g)(g)(g)
The Equilibrium Constant
13
Kc - Equilibrium constant
Kc - product of the equilibrium concentrations (in M) of the products, each raised to a power equal to its stoichiometric coefficient in the balanced equation, divided by the product of the equilibrium concentrations (in M) of the reactants, each raised to a power equal to its stoichiometric coefficient in the balanced equation.
Kc values are dimensionless because they actually involve a thermodynamic quantity called activity. Activities are directly related to molarity
The Equilibrium Constant
14
Write equilibrium constant expressions for the following reactions at 500oC. All reactants and products are gases at 500oC.
5
23c
235
PCl
ClPClK
ClPClPCl
The Equilibrium Constant
15
HI 2 I + H 22
22
2
c IH
HIK
5
24
3
62
4
c
223
ONH
OHNO=K
OH 6+NO 4O 5 + NH 4
The Equilibrium Constant
Example: One liter of equilibrium mixture from the following system at a high temperature was found to contain 0.172 mole of phosphorus trichloride, 0.086 mole of chlorine, and 0.028 mole of phosphorus pentachloride. Calculate Kc for the reaction.
Equil [ ]’s 0.028 M 0.172 M 0.086 M
16
235 ClPClPCl
The Equilibrium Constant
17
Kc PCl3 Cl2
PCl5 Kc
0.172 0.086 0.028
Kc 0.53
KPCl
PCl Cl
KK
or K K
c' 5
3 2
cc' c
'
c
0 0280172 0 086
19
1 1 10 53 19
.. .
.
. .
235 ClPClPCl
The Equilibrium Constant
18
Example: The decomposition of PCl5 was studied at another temperature. One mole of PCl5 was introduced into an evacuated 1.00 liter container. The system was allowed to reach equilibrium at the new temperature. At equilibrium 0.60 mole of PCl3 was present in the container. Calculate the equilibrium constant at this temperature.
0 0 1.00 Initial
ClPClPCl g2g3g5
M
The Equilibrium Constant
Tanother at 90.0
40.0
60.060.0K
0.60 0.60 0.40 mEquilibriu
0.60+ 0.60+ 0.60- Change
0 0 1.00 Initial
ClPClPCl
'c
g2g3g5
MMM
MMM
M
The Equilibrium Constant
Example : At a given temperature 0.80 mole of N2 and 0.90 mole of H2 were placed in an evacuated 1.00-liter container. At equilibrium 0.20 mole of NH3 was present. Calculate Kc for the reaction.
N 2(g) + 3 H2(g) 2 NH3(g)
Initial 0.80 M 0.90 M 0
Change - 0.10 M - 0.30 M +0.20 M
Equilibrium 0.70 M 0.60 M 0.20 M
Kc NH3 2
N2 H2 3 0.20 2
0.70 0.60 3 0.26
Variation of Kc with the Form of the Balanced Equation
Large equilibrium constants indicate that Large equilibrium constants indicate that most of the reactants are converted to most of the reactants are converted to products.products.
Small equilibrium constants indicate that Small equilibrium constants indicate that only small amounts of products are only small amounts of products are formed.formed.2 main calculations:
Equilibrium Concentrations Kc
Kc equilibrium concentrations
ICE table
Predicting the direction of reactions: The REACTION COEFFICIENT, Qc
Tank of water in Equilibrium
The Reaction Quotient
For this general reaction :
aA +bB cC +dD
Q C c
D d
A aB b
The mass action expression or reaction quotient has the symbol Q. Q has the same form as Kc
The major difference between Q and Kc is that the concentrations used in Q are not necessarily equilibrium values.
The Reaction Quotient
Why do we need another “equilibrium constant” that does not use equilibrium concentrations?
Q will help us predict how the equilibrium will respond to an applied stress - compare Q with Kc.
When
Q = Kc : the system is in equilibrium
Q > Kc : the system goes to the left (), towards reactants
Q < Kc : the system goes to the right (), towards products
The Reaction Quotient
The equilibrium constant for the following reaction is 49 at 450oC. If 0.22 mole of I2, 0.22 mole of H2, and 0.66 mole of HI were put into an evacuated 1.00-liter container, would the system be at equilibrium? If not, what must occur to establish equilibrium?
Calculate Qc, compare with Kc
Use ICE table to calculate equilibrium [ ]s
H2 + I2 ⇄ 2HI
Equilibrium calculations and reaction quotient The equilibrium constant Kc, is 3.00 for the
following reaction at a given temperature. If 1.00 mole of SO2 and 1.00 mole of NO2 are put into an evacuated 2.00 L container and allowed to reach equilibrium, what will be the concentration of each compound at equilibrium?
(g)3(g)2(g)2(g) NO SO NO SO
The equilibrium constant is 49 for the following reaction at 450oC. If 1.00 mole of HI is put into an evacuated 1.00-liter container and allowed to reach equilibrium, what will be the equilibrium concentration of each substance?
H2 + I2 ⇄ 2HI
An equilibrium mixture from the following reaction was found to contain 0.20 mol/L of A, 0.30 mol/L of B, and 0.30 mol/L of C.
(a) What is the value of Kc for this reaction?
(b) If the volume of the reaction vessel were suddenly doubled while the temperature remained constant, what would be the new equilibrium concentrations?
ggg C B A
A 2.00 liter vessel in which the following system is in equilibrium contains 1.20 moles of COCl2, 0.60 moles of CO and 0.20 mole of Cl2.
a) Calculate the equilibrium constant.
b) An additional 0.80 mole of Cl2 is added to the vessel at the same temperature. Calculate the molar concentrations of CO, Cl2, and COCl2 when the new equilibrium is established
g2g2g COCl Cl CO
Comparing Q and Kc
K
Q
K
Q
KQ
Equilibrium Formation of reactants
Formation of products
Q = Kc Q > KcQ < Kc
∴ Equilibrium is aSTABLE condition
Disturbing a System at Equilibrium: Le Chatelier’s Principle
Le Chatelier’s Principle - If a change of conditions (stress) is applied to a system in equilibrium, the system responds in the way that best tends to reduce the stress in reaching a new state of equilibrium.
Some possible stresses to a system at equilibrium are:
1. Changes in concentration of reactants or products.
2. Changes in pressure or volume (for gaseous reactions)
3. Changes in temperature.
Le Chatelier’s Principle
Changes in Concentration of Reactants and/or Products
Look at the following system at equilibrium at 450oC.H2 + I2 ⇄ 2HI Kc = 49
]I][[H
HI][K
22
2
c
- Removing HI: forward – toward product-side
- Removing I2 : backward – toward reactant-side
- Adding H2 : forward – product-side
- Adding HI and removing I2 : backward – reactant-side
- Adding HI and adding H2 : ? Calculate Q
Le Chatelier’s Principle
Changes in Volume • (and pressure for reactions involving gases) Consider the following system
2NO2(g) ⇄
N2O4(g)
Move the piston downward: ⇩ volume, ⇧ pressure
Shift toward products (lower #molecules)
Move the piston upward: ⇧ volume, ⇩ pressureShift toward reactants (greater #molecules)
Le Chatelier’s Principle
Effect of Temperature on Equilibrium Consider the following reaction at equilibrium:
reaction? in thisproduct or reactant aheat Is
kJ/mol 198H SO 2 O SO 2 orxn3(g)2(g)2(g)
2SO2(g) + O2(g) ⇄ 2SO3(g) + 198 kJAdd heat to the system: increase O2 and SO2
Remove heat from the system: increase SO3
Place the system in an ice bath: decrease SO2 and O2, increase SO3
Predictions are only true for EXOTHERMIC reactions
Le Chatelier’s Principle
Effect of Temperature on Equilibrium
Add heat to the system: clear solution
Remove heat from the system: increase precipitate
Place the system in an ice bath: increase precipitate
NH4NO3 (s) + 98.9 kJ ⇄ NH4+
(aq) + NO3-
(aq)
The effect of temperature on reactions in equilibrium depends on the sign of ∆Hrxn
Introduction of a Catalyst Catalysts decrease the activation energy of both the forward
and reverse reaction equally. Catalysts do not affect the position of equilibrium.
The concentrations of the products and reactants will be the same whether a catalyst is introduced or not.
Equilibrium will be established faster with a catalyst.
Le Chatelier’s Principle
Given the reaction below at equilibrium in a closed container at 500oC. How would the equilibrium be influenced by the following?
a) Increasing the temperature
b) Decreasing the external temperature c) Increasing the volume
d) Increasing the pressure
e) Adding He(g) in the system
f) Adding N2(g) in the system
g) Placing solid Pt in the system as catalyst
Backward, decrease productsForward, increase products
Forward, increase products
Backward, decrease products
Forward, increase products
No effect
No effect
A certain indicator was extracted from mayana leaves, and its dissociation in water can be shown in the following thermochemical equation:
At 25oC the predominant color of the neutral extract is light green. Determine the expected color when a solution of the extract is applied with the following stress:
a) Placed in an ice bath
c) Placing sodium carbonate in the solution
b) Placing sodium sulfate in the system
d) Compressing the solution
e) Adding 3 drops of nitric acid f) Adding hot water to the systemg) Adding small amounts of red dye in c)
Yellow
Yellow
Light Green
Light Green
Dark Green
Dark Green
Orange
Equilibrium Constant Expression inGas-Phase Reactions For gas phase reactions the equilibrium
constants can be expressed in partial pressures rather than concentrations.
For gases, the pressure is proportional to the concentration. PV = nRT P = nRT/V n/V = M P= MRT and M = P/RT
For systems involving gases, we can use Kp instead of Kc
Equilibrium Constant Expression inGas-Phase Reactions
Consider this system at equilibrium at 5000C.
Equilibrium Constant Expression inGas-Phase Reactions Using the expression [ ] = P/RT
From the previous slide we can see that the relationship between Kp and Kc is:
reactants) gaseous of moles of (#-products) gaseous of moles of (#=n
RTKKor RTKK npc
ncp
Relationship Between Kp and Kc
Relationship Between Kp and Kc
Nitrosyl bromide, NOBr, is 34% dissociated by the following reaction at 25oC, in a vessel in which the total pressure is 0.25 atmosphere. What is the value of Kp?
g2gg Br + NO 2 NOBr 2
Kc is 49 for the following reaction at 450oC. If 1.0 mole of H2 and 1.0 mole of I2 are allowed to reach equilibrium in a 3.0-liter vessel,
What is the total pressure inside the reaction vessel after equilibrium is reached?
gg2g2 HI 2I H
Heterogeneous Equilibria Heterogeneous equilibria have more
than one phase present. For example, a gas and a solid or a liquid and a
gas.
How does the equilibrium constant differ for heterogeneous equilibria? Pure solids and liquids have activities of unity. Solvents in very dilute solutions have activities that
are essentially unity. The Kc and Kp for the reaction shown above are:
2(g)(s)s3 O32KCl 2KClO
3Op
32c 2
P=K ][O=K
2(g)(l)2(aq)32 COOHCOH
C)25at (F 2CaCaF o-aq
2aqs2
Gibbs’ Free Energy and Equilibrium
G0 is the standard free energy change. G0 is defined for the complete conversion of all
reactants to all products. With superscript 0
G is the free energy change at nonstandard conditions• For example, concentrations other than 1 M or
pressures other than 1 atm.
• Temperature, however, is same as in G0rxn
Gibbs’ Free Energy and Equilibrium
G is related to Go by the following relationship -
quotientreaction =Q
re temperatuabsolute = T
constant gas universal =R
Q log RT 303.2G=G
or lnQ RTG=Go
o
Gibbs’ Free Energy and Equilibrium
At equilibrium, G=0 and Q=Kc. Then we can derive this relationship:
K log RT 2.303 -=G
orK ln RT -=G
: torearrangeswhich
K log RT 303.2G0
orK ln RTG0
0
0
0
0
Gibbs’ Free Energy and Equilibrium
D ofactivity theis C ofactivity theis
B ofactivity theis A ofactivity theis
where=K
dD + cC bB +aA
DC
BA
bB
aA
dD
cC
aa
aa
aa
aa
For the following generalized reaction, the thermodynamic equilibrium constant is defined as follows:
Gibbs’ Free Energy and Equilibrium
General equation; THERMODYNAMIC FORM
All gaseous reactants and products
All solutions of reactants and products
Used when both gaseous and solution forms appear in the chemical equation
Gibbs’ Free Energy and Equilibrium
Calculate the equilibrium constant, Kp, for the following reaction at 25oC.
g2g42 NO 2ON
g2g42 NO 2ON
NON SPONTANEOUSFORWARD
[N2O4] > [NO2]
Kp for the reverse reaction at 25oC can be calculated easily - it is the reciprocal of the above reaction.
2 NO2(g) N2O4(g)
G rxno 4.78 kJ/mol
Kp'
1
Kp
1
0.1456.90
PN2O4 PNO2 2
Gibbs’ Free Energy and Equilibrium
It is difficult to analyze the relationship between free energy and equilibrium constant at non-standard conditions
The basic relationship between ∆G0rxn
and Kc appears only for the standard conditions form
Gorxn K Spontaneity at unit concentration
< 0 > 1Forward reaction spontaneous,
More products than reactants at equilibrium
= 0 = 1 IDEAL system, very RARE
> 0 < 1Reverse reaction spontaneous,
More reactants than products at equilibrium
Gibbs’ Free Energy and Equilibrium
Gibbs’ Free Energy and Equilibrium
Gibbs’ Free Energy and Equilibrium Nitrosyl bromide, NOBr, is 34% dissociated
by the following reaction at 25oC, in a vessel in which the total equilibrium pressure is 0.25 atmosphere.
Calculate the ∆G0rxn.
What is the ∆G when 5.00% of 2.00 atm NOBr has dissociated?
EVALUATION OF EQUILIBRIUM CONSTANTS AT DIFFERENT TEMPERATURES From the value of Ho and K at one
temperature, T1, we can use the van’t Hoff equation to estimate the value of K at another temperature, T2.
OR
The equilibrium constant Kc of the reaction
H2(g) + Br2(g) ⇄ 2HBr(g)
is 1.6 x 105 at 1297 K and 3.5 x 104 at 1495 K.
(a)Is the reaction exothermic or endothermic?(b)Find Kc for the reaction: ½H2(g) + ½Br2(g) ⇄
HBr(g)
(c)Pure HBr is placed in a container of constant volume and heated to 1297 K. What percentage of the HBr is decomposed to Br2 and H2 at equilbrium?
At its normal boiling point of 100oC, the heat of vaporization is 40.66 kJ/mole. What is the equilibrium vapor pressure of water at 25oC?
In the distant future, when hydrogen may be cheaper than coal, steel mills make iron by the reaction
Fe2O3(s) + 3H2(g) ⇄ 2Fe(s) + 3H2O(g)
For this reaction, ∆H0 = 96 kJ/mole and Kc = 8.11 at 1000 K. (a)What percentage of the H2 remains unreacted after the reaction has come to equilibrium at 1000 K?(b) Is this percentage greater or less if the temperature is decreased below 1000 K?