Chapter 13 Chemical Equilibrium. 13.1 Describing Chemical Equilibrium Reactants Product Reactants ...
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Transcript of Chapter 13 Chemical Equilibrium. 13.1 Describing Chemical Equilibrium Reactants Product Reactants ...
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Chapter 13Chapter 13
Chemical Equilibrium
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13.1 Describing Chemical Equilibrium13.1 Describing Chemical Equilibrium
Reactants Product
Reactants Products
When substances react, they eventually form a mixture of reactants and products in dynamic equilibrium
forward
reverse
For a general reversible reaction:
c C + d Da A + b B
Double arrows show reversible reaction
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The Equilibrium StateThe Equilibrium State
Many reactions do not go to completion instead of reachingChemical Equilibrium: -The state reached when the concentrations of reactants and products remain constant over time.-The rate forward and reverse have become equal
2NO2(g)N2O4(g)
BrownColorless
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The Equilibrium StateThe Equilibrium State2NO2(g)N2O4(g)
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13.2 The Equilibrium Constant 13.2 The Equilibrium Constant KKcc
cC + dDaA + bB
For a homogeneous reaction
Kc = [A]a[B]b
[C]c[D]d
Equilibrium constant expressionwhen concentrations are used
Equilibrium constantReactants
ProductsEquilibrium equation:
Coefficient
=
Kc is temperature dependent
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13.3The Equilibrium Constant 13.3The Equilibrium Constant KKpp
When writing an equilibrium expression for a gaseous reaction in terms of partially pressure, we call it equilibrium constant, Kp
2NO2(g)N2O4(g)
Kp =N2O4
P
NO2 P
2
P is the partial pressure of that component
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ExamplesExamples Write the equilibrium constant, Kc and K’c for the
following reactions:
a. CH4(g) + H2O(g) CO(g) + 3 H2(g)
b. 2 SO2(g) + O2(g) 2 SO3(g)
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ExampleaExampleaWrite Kp and K’p for the following reactions:CO(g)+ 2H2(g) CH3OH(g)
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The Equilibrium Constant The Equilibrium Constant KKcc
Kc =
The equilibrium constant and the equilibrium constant expression are for the chemical equation as written.
N2(g) + 3H2(g)2NH3(g)[NH3]2
[N2][H2]3
=1
Kc
2NH3(g)N2(g) + 3H2(g)[N2][H2]3
[NH3]2
Kc =
´
4NH3(g)2N2(g) + 6H2(g) ´´ K c = ??????
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Relating the Equilibrium Relating the Equilibrium Constant Constant KKp p and Kand Kcc
n
Kp = Kc(RT)n
0.082058K mol
L atmR is the gas constant,
T is the absolute temperature (Kelvin).
is the number of moles of gaseous products minus the number of moles of gaseous reactants.
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ExampleExample Phosphorous pentachloride dissociate on heating
PCl5(g) PCl3(g) + Cl2(g)
If Kc equals 3.26 x 10-2 at 191oC, what is Kp at this temperature?
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ExampleExample Consider the following reaction
2NO (g) + O2 (g) 2NO2 (g)
If Kp = 1.48 x 104 at 184 C, what is Kc?
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13.4Heterogeneous Equilibria13.4Heterogeneous Equilibria
CaO(s) + CO2(g)CaCO3(s)
LimeLimestone
(1)
(1)[CO2]= [CO2]
[CaCO3]
[CaO][CO2]=
Pure solids and pure liquids are not included.
Kc =
Kc = [CO2] Kp = PCO2
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Heterogeneous EquilibriaHeterogeneous Equilibria
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ExampleExample
In the industrial synthesis of hydrogen, mixtures of CO and H2 are enriched in H2 by allowing the CO to react with steam. The chemical equation for this so-called water-gas shift reaction is
CO(g) + H2O(g) CO2(g) + H2(g)
What is the value of Kp at 700K if the partial pressures in an equilibrium mixture at 700K are 1.31 atm of CO, 10.0 atm of H2O, 6.12 atm of CO2, and 20.3 atm of H2?
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ExamplesExamples Consider the following unbalanced reaction
(NH4)2S(s) 2NH3(g) + H2S(g)
An equilibrium mixture of this mixture at a certain temperature was found to have [NH3] = 0.278 M and [H2S] = 0.355 M. What is the value of the equilibrium constant (Kc) at this temperature?
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13.5 Using the Equilibrium 13.5 Using the Equilibrium ConstantConstant
When we know the numerical value of the equilibrium constant, we can make certain judgments about the extent of the chemical reaction
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Predicting the Direction of Predicting the Direction of ReactionReaction
For a chemical system whether in equilibrium or not, we can calculate the value given by the law of mass action
We called it reaction quotient, Qc
cC + dDaA + bB
Qc =[A]t
a[B]tb
[C]tc[D]t
d
Reaction quotient:
The reaction quotient, Qc, is defined in the same way as the equilibrium constant, Kc, except that the concentrations in Qc are not necessarily equilibrium values.
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Using the Equilibrium Using the Equilibrium ConstantConstant
• If Qc = Kc
no net reaction occurs.
• If Qc < Kc
net reaction goes from left to right (reactants to products).• If Qc > Kc
net reaction goes from right to left (products to reactants).
The equilibrium between reactant A and product B: A B
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Finding Equilibrium concentrations Finding Equilibrium concentrations from Initial Concentrationsfrom Initial Concentrations
Steps to follow in calculating equilibrium concentrations from initial concentrationWrite a balance equation for the reactionMake an ICE (Initial, Change, Equilibrium) table,
involvesThe initial concentrationsThe change in concentration on going to equilibrium,
defined as xThe equilibrium concentration
Substitute the equilibrium concentrations into the equilibrium equation for the reaction and solve for x
Calculate the equilibrium concentrations form the calculated value of x
Check your answers
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Calculating Equilibrium Calculating Equilibrium ConcentrationsConcentrations
Sometimes you must use quadratic equation to solve for x, choose the mathematical solution that makes chemical sense
Quadratic equation
ax2 + bx + c = 0
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Finding Equilibrium Finding Equilibrium concentrations from Initial concentrations from Initial ConcentrationsConcentrations
For Homogeneous Mixture
aA(g) bB(g) + cC(g)
Initial concentration (M) [A]initial [B]initial [C]initial
Change (M) - ax + bx + cx
Equilibrium (M) [A]initial – ax [B]initial + bx [C]initial + cx
Kc = [products]
[reactants]
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Finding Equilibrium Finding Equilibrium concentrations from Initial concentrations from Initial ConcentrationsConcentrations
For Heterogenous Mixture
aA(g) bB(g) + cC(s)
Initial concentration (M) [A]initial [B]initial ……..
Change (M) - ax + bx ……..
Equilibrium (M) [A]initial – ax [B]initial + bx ……….
Kc = [products]
[reactants]
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ExamplesExamples The value of Kc for the reaction is 3.0 x 10-2.
Determine the equilibrium concentration if the initial concentration of water is 8.75 M
C(s) + H2O(g) CO(g) + H2(g)
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Using the Equilibrium ConstantUsing the Equilibrium Constant
At 700 K, 0.500 mol of HI is added to a 2.00 L container and allowed to come to equilibrium. Calculate the equilibrium concentrations of H2, I2, and HI . Kc is 57.0 at 700 K.
2HI(g)H2(g) + I2(g)
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ExamplesExamplesConsider the following reaction
I2(g) + Cl2(g) 2ICl(g) Kp = 81.9 (at 25oC)
A reaction mixture at 25oC initially contains PI2 = 0.100 atm, PCl2 = 0.100 atm, and PICl = 0.100 atm. Find the equilibrium pressure of I2, Cl2 and ICl at this temperature.
In which direction does the reaction favored?
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ExampleExampleA flask initially contained hydrogen sulfide at
a pressure of 5.00 atm at 313 K. When the reaction reached equilibrium, the partial pressure of sulfur vapor was found to be 0.15 atm. What is the Kp for the reaction?
2H2S(g) 2H2(g) + S2(g)
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13.6 Le Châtelier’s 13.6 Le Châtelier’s PrinciplePrincipleLe Châtelier’s Principle: If a stress is applied to a reaction mixture at equilibrium, net reaction occurs in the direction that relieves the stress.• The concentration of reactants or
products can be changed.
• The pressure and volume can be changed.
• The temperature can be changed.
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13.7Altering an Equilibrium 13.7Altering an Equilibrium Mixture: ConcentrationMixture: Concentration
2NH3(g)N2(g) + 3H2(g)
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Altering an Equilibrium Mixture: Altering an Equilibrium Mixture: ConcentrationConcentration
Add reactant – denominator in Qc expression becomes larger
◦ Qc < Kc
◦ To return to equilibrium, Qc must be increases◦ More product must be made => reaction shifts to
the right Remove reactant – denominator in Qc expression
becomes smaller
◦ Qc > Kc
◦ To return to equilibrium, Qc must be decreases◦ Less product must be made => reaction shifts to
the left
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Altering an Equilibrium Mixture: Altering an Equilibrium Mixture: ConcentrationConcentration
• the concentration stress of an added reactant or product is relieved by net reaction in the direction that consumes the added substance.
• the concentration stress of a removed reactant or product is relieved by net reaction in the direction that replenishes the removed substance.
In general, when an equilibrium is disturbed by the addition or removal of any reactant or product, Le Châtelier’s principle predicts that
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Altering an Equilibrium Altering an Equilibrium Mixture: ConcentrationMixture: Concentration
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Altering an Equilibrium Altering an Equilibrium Mixture: ConcentrationMixture: Concentration
An equilibrium mixture of 0.50 M N2, 3.00 M H2, and 1.98 M NH3 is disturbed by increasing the N2 concentration to 1.50 M.
Which direction will the net reaction shift to re-establish the equilibrium?
2NH3(g)N2(g) + 3H2(g) at 700 K, Kc = 0.291
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ExampleExample The reaction of iron (III) oxide with carbon monoxide
occurs in a blast furnace when iron ore is reduced to iron metal:
Fe2O3(s) + 3 CO(g) 2 Fe(l) + 3CO2(g)
Use Le Chatellier’s principle to predict the direction of net reaction when an equilibrium mixture is disturbed by:
a. adding Fe2O3
b. Removing CO2
c. Removing CO; also account for the change using the reaction quotient Qc
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ExampleExample Consider the following reaction at equilibrium
CO(g) + Cl2(g) COCl2(g)
Predict whether the reaction will shift left, shift right, or remain unchanged upon each of the following reaction mixture
a. COCl2 is added to the reaction mixture
b. Cl2 is added to the reaction mixture
c. COCl2 is removed from the reaction mixture: also account for the change using the reaction quotient Qc
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13.8 Altering an Equilibrium 13.8 Altering an Equilibrium Mixture: Changes in Pressure Mixture: Changes in Pressure and Volumeand Volume
2NH3(g)N2(g) + 3H2(g) at 700 K, Kc = 0.291
An equilibrium mixture of 0.50 M N2, 3.00 M H2, and 1.98 M NH3 is disturbed by reducing the volume by a factor of 2. Which direction will the net reaction shift?
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Altering an Equilibrium Mixture: Altering an Equilibrium Mixture: Pressure and VolumePressure and Volume
2NH3(g)N2(g) + 3H2(g)
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Altering an Equilibrium Mixture: Altering an Equilibrium Mixture: Pressure and VolumePressure and Volume
• an increase in pressure by reducing the volume will bring about net reaction in the direction that decreases the number of moles of gas.
• a decrease in pressure by enlarging the volume will bring about net reaction in the direction that increases the number of moles of gas.
In general, when an equilibrium is disturbed by a change in volume which results in a corresponding change in pressure, Le Châtelier’s principle predicts that
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Altering an Equilibrium Mixture: Altering an Equilibrium Mixture: Pressure and VolumePressure and Volume
If reactant side has more moles of gas◦ Denominator will be larger
Qc < Kc
To return to equilibrium, Qc must be increased Reaction shifts toward fewer moles of gas (to the
product) If product side has more moles of gas
◦ Numerator will be larger Qc > Kc
To return to equilibrium, Qc must be decreases
Reaction shifts toward fewer moles of gas (to the reactant)
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Altering an Equilibrium Mixture: Altering an Equilibrium Mixture: Pressure and VolumePressure and Volume
Reaction involves no change in the number moles of gas◦ No effect on composition of equilibrium mixture
For heterogenous equilibrium mixture◦ Effect of pressure changes on solids and liquids
can be ignored Volume is nearly independent of pressure
Change in pressure due to addition of inert gas ◦ No change in the molar concentration of
reactants or products◦ No effect on composition
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ExampleExample Consider the following reaction at chemical
equilibrium
2 KClO3(s) 2 KCl(s) + 3O2(g)
a. What is the effect of decreasing the volume of the reaction mixture?
b. Increasing the volume of the reaction mixture?
c. Adding inert gas at constant volume?
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ExamplesExamples Does the number moles of products increases,
decreases or remain the same when each of the following equilibria is subjected to a increase in pressure by decreasing the volume?
◦ PCl5(g) PCl3(g) + Cl2(g)
◦ CaO(s) + CO2(g) CaCO3(s)
◦ 3 Fe(s) + 4H2O(g) Fe3O4(s) + 4 H2(g)
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Altering an Equilibrium Mixture: Altering an Equilibrium Mixture: TemperatureTemperature
• the equilibrium constant for an exothermic reaction (negative H°) decreases as the temperature increases.• Contains more reactant than product• Kc decreases with increasing
temperature• the equilibrium constant for an
endothermic reaction (positive H°) increases as the temperature increases.• Contains more product than reactant• Kc increases with increasing
temperature
In general, when an equilibrium is disturbed by a change in temperature, Le Châtelier’s principle predicts that
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13.9 Altering an Equilibrium 13.9 Altering an Equilibrium Mixture: TemperatureMixture: Temperature
2NH3(g)N2(g) + 3H2(g) H° = -2043 kJ (exothermic )
As the temperature increases, the equilibrium shifts from products to reactants.
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ExampleExample The following reaction is endothermic
CaCO3(s) CaO(s) + CO2(g)
a. What is the effect of increasing the temperature of the reaction mixture?
b. Decreasing the temperature?
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ExamplesExamples In the first step of Ostwald process for the synthesis
of nitric acid, ammonia is oxidized to nitric oxide by the reaction:
2 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(l)
ΔHo = -905 kJ
How does the temperature amount of NO vary with an increases in temperature?
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The Effect of a Catalyst on The Effect of a Catalyst on EquilibriumEquilibrium
Catalyst increases the rate of a chemical reaction◦ Provide a new, lower energy pathway◦ Forward and reverse reactions pass through the
same transition state◦ Rate for forward and reverse reactions increase by
the same factor◦ Does not affect the composition of the equilibrium
mixture◦ Does not appear in the balance chemical equation◦ Can influence choice of optimum condition for a
reaction
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The Effect of a Catalyst on The Effect of a Catalyst on EquilibriumEquilibrium