Transcript Chapter 15 Chemical Equilibrium

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Equilibrium
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Equilibrium
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Equilibrium
Chapter 15
Chemical Equilibrium
Equilibrium
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Equilibrium
15.1- The Concept of Equilibrium
Define Chemical Equilibrium
The condition in which the concentration of all
reactants and products in a closed system cease
to change with time is called chemical
equilibrium. OR
The rate at which the products are formed front
the reactants equals the rate at which the
reactants are formed by the products
Equilibrium
What are the 3 conditions
necessary for equilibrium?
• Must have a closed system
• Must have a constant temperature
• Ea must be low enough to allow a
reaction.
Equilibrium
15.1- The Concept of Equilibrium
Explain kr [A] = kf [B].
As the concentration of A decreases the
concentration increases no matter what direction
the reaction is going.
Equilibrium
The Concept of Equilibrium
Chemical equilibrium occurs when a reaction and its
reverse reaction proceed at the same rate.
Equilibrium
15.1- The Concept of Equilibrium
Explain the equilibrium mixture.
Once equilibrium is established, the
concentrations of each (reactant/products)
won’t change. They are a mixture.
This does not mean they stopped
reacting, the rate of the reaction is the
same.
Equilibrium
The Concept of Equilibrium
• As a system
approaches equilibrium,
both the forward and
reverse reactions are
occurring.
• At equilibrium, the
forward and reverse
reactions are
proceeding at the same
rate.
Equilibrium
A System at Equilibrium
Once
equilibrium is
achieved, the
amount of each
reactant and
product remains
constant.
Equilibrium
The Equilibrium Constant
• Forward reaction:
N2O4 (g)  2 NO2 (g)
• Rate law:
Rate = kf [N2O4]
Equilibrium
The Equilibrium Constant
• Reverse reaction:
2 NO2 (g)  N2O4 (g)
• Rate law:
Rate = kr [NO2]2
Equilibrium
Depicting Equilibrium
In a system at equilibrium, both the
forward and reverse reactions are being
carried out; as a result, we write its
equation with a double arrow
N2O4 (g)
2 NO2 (g)
Equilibrium
15.2- The Equilibrium Constant
What is the Haber process?
Under high pressure and high temperature,
nitrogen and hydrogen gas creates ammonia
gas.
Equilibrium
15.2- The Equilibrium Constant
Read this section on the Haber process, how
are the concentration of the reactants and the
products change (or not) during the reaction.
At equilibrium, the relative amounts
concentrations of all 3 (H2, N2 and NH3) are all
the same, regardless of whether the starting
mixture was 3:1 molar ratio.
The equilibrium conditions can be reached
from either direction.
Equilibrium
15.2- The Equilibrium Constant
Explain the law of mass action.
This law expresses the relationship between the
concentrations of the reactants and products at
equilibrium in reactions.
aA + bB
 pP + qQ
Equilibrium
15.2- The Equilibrium Constant
Provide and explain the equilibrium-constant
expression.
p
q
[ P] [Q]
Kc 
a
b
[ A] [ B]
Where K is the equilibrium constant, and is
unitless
Equilibrium
15.2- The Equilibrium Constant
Provide and explain the equilibrium-constant
expression.
[ P] p [Q]q
Kc 
[ A]a [ B]b
Kc is an equilibrium constant is a numerical value
obtained when we substitute actual equilibrium
concentration into the expression.
The subscript indicates that concentrations
(expressed in molarity).
Equilibrium
15.2- The Equilibrium Constant
Provide and explain the equilibrium-constant
expression.
p
q
[ P] [Q]  Concentration products
Kc 
a
b
[ A] [ B]  Concentration reactants
Equilibrium
15.2- The Equilibrium Constant
What variables affect the Kc constant value?
Stoichiometry of the reaction and temperature.
Equilibrium
15.2- The Equilibrium Constant
What variables does not affect the Kc
constant value?
The mechanism and initial concentration.
Equilibrium
15.2- The Equilibrium Constant
When writing the equilibrium expression,
what goes in the numerator and denominator?
The products goes in the numerator.
The reactants goes in the denominator
Equilibrium
15.2- The Equilibrium Constant
Go through the Sample Exercise 15.1, then
below SHOW how to do the practice exercise:
Write the Kc expression for H2 + I2  2HI
2
[ HI ]
Kc 
[ H 2][ I 2]
Equilibrium
Write the equilibrium expression for
Keq or Kc this reactions:
1. 2 O3(g) ⇋ 3 O2(g)
3
[O 2]
Kc 
2
[O 3]
Equilibrium
Write the equilibrium expression for
Keq or Kc this reactions:
2. 2 NO(g) + Cl2(g) ⇋ 2 NOCl(g)
2
[ NOCl ]
Kc 
2
[ NO ] [Cl 2]
Equilibrium
Write the equilibrium expression for
Keq or Kc this reactions:
3. Ag+(aq) + 2 NH3(g) ⇋ Ag(NH3)2+(aq)
[ Ag ( NH 3) 2]
Kc 

2
[ Ag ] [ NH 3]
Equilibrium
Write the equilibrium expression for
Keq or Kc this reactions:
4. Cd2+(aq) + 4 Br-(aq) ⇋ CdBr42-(aq)
2
[CdBr 4 ]
Kc 
2
 4
[Cd ] [ Br ]
Equilibrium
Look at the phases, what do they all
have in common?
They are the same phase; all gases or aqueous
solutions
These are termed- homogeneous equilibrium.
Equilibrium
Homogenous equilibrium applies to reactions in
which all reacting species are in the same phase.
2 NO(g) + Cl2(g) ⇋ 2 NOCl(g)
Since we are dealing with gases we need to factor in partial
pressures of the gases.
Provide the equilibrium expression for gases?
(PC)c (PD)d
Kp =
(PA)a (PB)b
Equilibrium
14.2
15.2- The Equilibrium Constant
Provide and explain the Kp expression.
Because pressure is proportional to
concentration for gases in a closed system, the
equilibrium expression can also be written
(PC)c (PD)d
Kp =
(PA)a (PB)b
Equilibrium
Relationship between Kc and Kp
• From the ideal gas law we know that
PV = nRT
• Rearranging it, we get
n
P=
RT
V
Equilibrium
Relationship between Kc and Kp
Plugging this into the expression for Kp for each
substance, the relationship between Kc and Kp
becomes
Kp = Kc (RT)n
Where
n = (moles of gaseous product) − (moles of gaseous reactant)
Equilibrium
Write the equilibrium expression for
Kp for this reactions:
1. 2 O3(g) ⇋ 3 O2(g)
3
[ PO 2]
Kp 
2
[ Po3]
Equilibrium
Write the equilibrium expression for
Kp for this reactions:
2. 2 NO(g) + Cl2(g) ⇋ 2 NOCl(g)
2
[ PNOCl ]
Kp 
2
[ PNO ] [ PCl 2]
Equilibrium
Using the Equilibrium Constant
Exercise 15.2:
In the synthesis of ammonia from nitrogen and
hydrogen (The Haber process),
N2 (g) + 3H2(g) 2NO3 (g)
at temperature of 300 °C, the Kc value is 9.60.
Calculate the Kp.
Provide the Kp for this reaction
2
[ PNO 3]
Kp 
3
[ PN 2] [ PH 2]
Equilibrium
Using the Equilibrium Constant
Exercise 15.2:
In the synthesis of ammonia from nitrogen and
hydrogen (The Haber process),
N2 (g) + 3H2(g) 2NO3 (g)
at temperature of 300 °C, the Kc value is 9.60.
Calculate the Kp.
n
Kp = Kc (RT)
Convert 300 °C to K = 573
Kp = 9.60 (0.0821)(573K) n
Equilibrium
Using the Equilibrium Constant
Provide the Kp for this reaction
2
[ PNO 3]
Kp 
3
[ PN 2] [ PH 2]
∆n = moles of product – moles of reactant
moles of product is 2
moles of reactant is 1+ 3 = 4
∆n = 2– 4 = -2
Equilibrium
Using the Equilibrium Constant
Exercise 15.2:
In the synthesis of ammonia from nitrogen and
hydrogen (The Haber process),
N2 (g) + 3H2(g) 2NO3 (g)
at temperature of 300 °C, the Kc value is 9.60.
Calculate the Kp.
n
Kp = Kc (RT)
Kp = 9.60
((0.0821)(573K))2
Kp = 4.34 x 10-3
Use yx
Equilibrium
Using the Equilibrium Constant
2SO3  2SO2 + O2 at temperature of 1000 K, the
Kc value is 4.08 x10-3. Calculate the Kp
Provide the Kp expression for this reaction.
2
[ PSO 2] [ PO 2]
Kp 
2
[ PSO 3]
Equilibrium
Using the Equilibrium Constant
The ∆n superscript – use the stoichiometry.
2SO3  2SO2 + O2
2+1 (products) - 2 (reactants) = 1
Kp = 4.08 x10-3 (0.0821)(1000K)1
Kp = 0.335
Equilibrium
Calculating BOTH Kc and Kp
The equilibrium concentrations for the reaction between carbon
monoxide and molecular chlorine to form COCl2 (g) at 74°C are
[CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate
the equilibrium constants Kc and Kp.
CO (g) + Cl2 (g)
COCl2 (g)
Provide the Kp expression
[COCl2]
Kc =
[CO][Cl2]
Equilibrium
Calculating BOTH Kc and Kp
The equilibrium concentrations for the reaction between carbon
monoxide and molecular chlorine to form COCl2 (g) at 74°C are
[CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate
the equilibrium constants Kc and Kp.
Substitute the concentrations in the Kc expression.
CO (g) + Cl2 (g)
COCl2 (g)
[COCl2]
0.14
=
= 220
Kc =
[CO][Cl2]
0.012 x 0.054
Kc = 220
Now we can plug n’ chug
Equilibrium
Calculating BOTH Kc and Kp
The equilibrium concentrations for the reaction between carbon
monoxide and molecular chlorine to form COCl2 (g) at 74°C are
[CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate
the equilibrium constants Kc and Kp.
CO (g) + Cl2 (g)
COCl2 (g)
[COCl2]
0.14
=
= 220
Kc =
[CO][Cl2]
0.012 x 0.054
Kp = Kc(RT)n
n = 1 – 2 = -1
R = 0.0821
T = 273 + 74 = 347 K
Kp = 220 x (0.0821 x 347)-1 = 7.7
Equilibrium
Calculating a concentration
The equilibrium constant Kp for the reaction
2NO2 (g)
2NO (g) + O2 (g)
is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO
= 0.400 atm and PNO2 = 0.270 atm?
Provide the Kp expression for this reaction.
Kp =
[PNO]2 [PO2]
[PNO2] 2
Equilibrium
Calculating a concentration
The equilibrium constant Kp for the reaction
2NO2 (g)
2NO (g) + O2 (g)
is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO
= 0.400 atm and PNO2 = 0.270 atm?
Substitute the concentrations in the Kp expression.
Kp =
[0.400] 2 [PO2]
[0.270]2
Rearrange to get PO2 alone and Kp = 158
PO2 = Kp
[0.400]2
[0.270]2
PO2= 158
[0.400]2
[0.270]2
Equilibrium
14.2
Calculating a concentration
The equilibrium constant Kp for the reaction
2NO2 (g)
2NO (g) + O2 (g)
is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO
= 0.400 atm and PNO2 = 0.270 atm?
Solve for the PO2 equilibrium pressure.
PO2= 158
[0.400]2
[0.270]2
PO2 = 158 x (0.400)2/(0.270)2 = 347 atm
Equilibrium
14.2
15.2- The Equilibrium Constant
Explain the relationship between Kc value and
the shift in the equilibrium.
• If K >> 1, the reaction
is product-favored;
• When equilibrium is
achieved, most of the
reactant has been
Equilibrium
converted to product
15.2- The Equilibrium Constant
Explain the relationship between Kc value and
the shift in the equilibrium.
• If K << 1, the reaction is
reactant-favored;
• When equilibrium is
achieved, very little
reactant has been
converted to product
Equilibrium
Comparing Kc values at different
temperatures
Exercise 15.3
H2 + I2  2HI
Kc= 794 at 298K and Kc= 54 at 700K. Is the
formation of HI favored more at the higher
temperature or lower temperature?
At lower temperature because of the higher
Kc value.
Equilibrium
For the following Kc values, which way will
the equilibrium shift?
-4______ Kc= 2.6 x 108____________
product
Kc= 2.15 x 10reactant
reactant
Kc= 2.9 x 10 5 product
_____ Kc= 8.6 x10-6___________
Kc= 1.054__________
Kc= 3.88 x 102
EVEN
____________
product
Equilibrium
15.2- The Equilibrium Constant
Explain the relationship between Kc value and
the direction of the written reaction.
The equilibrium constant of a reaction in the
reverse reaction is the reciprocal of the
equilibrium constant of the forward reaction.
N2O4 (g)
2 NO2 (g)
[NO2]2
2 NO2 (g) Kc = [N O ] = 0.212 at 100C
2 4
[N2O4]
1
K
=
N2O4 (g) c
=
[NO2]2
0.212
= 4.72 at Equilibrium
100C
Manipulating Equilibrium Constants
The equilibrium constant of a reaction that has
been multiplied by a number is the equilibrium
constant raised to a power that is equal to that
number.
N2O4 (g)
[NO2]2
2 NO2 (g) Kc = [N O ]
2 4
2 N2O4 (g)
[NO2]4
4 NO2 (g)Kc = [N O ]2 = (0.212)2 at 100C
2 4
= 0.212 at 100C
Equilibrium
Heterogenous equilibrium applies to reactions in which
reactants and products are in different phases.
CaCO3 (s)
[CaO][CO2]
Kc‘ =
[CaCO3]
[CaCO3]
Kc = [CO2] = Kc‘ x
[CaO]
CaO (s) + CO2 (g)
[CaCO3] = constant
[CaO] = constant
Kp = PCO2
The concentration of solids and pure liquids are not
included in the expression for the equilibrium constant.
Equilibrium
14.2
CaCO3 (s)
CaO (s) + CO2 (g)
PCO 2 = Kp
PCO 2 does not depend on the amount of CaCO3 or CaO
Equilibrium
14.2
Provide the Kc or Kp expression for the following
Cr(s) + 3Ag+ (aq)  Cr+3 (aq) + 3Ag(s)
3
[Cr ]
Kc 
 3
[ Ag ]
3Fe (s) + 4H2O(g)  Fe3O4 (s) + 4 H2(g)
4
[ PH 2]
Kp 
4
[ PH 2 O ]
Equilibrium
Kc
Equilibrium