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Introduction to Chemical Introduction to Chemical EquilibriumEquilibrium
Chapter 15Chapter 15
CHEM 160CHEM 160
• Consider colorless frozen N2O4. At room temperature, it decomposes to brown NO2:
N2O4(g) 2NO2(g).• At some time, the color stops changing and we have a
mixture of N2O4 and NO2.• Chemical equilibrium is the point at which the
concentrations of all species are constant.• Using the collision model:
– as the amount of NO2 builds up, there is a chance that two NO2 molecules will collide to form N2O4.
– At the beginning of the reaction, there is no NO2 so the reverse reaction (2NO2(g) N2O4(g)) does not occur.
The Concept of EquilibriumThe Concept of Equilibrium
The Concept of EquilibriumThe Concept of Equilibrium
• The point at which the rate of decomposition: N2O4(g) 2NO2(g)
equals the rate of dimerization: 2NO2(g) N2O4(g).
is dynamic equilibrium.• The equilibrium is dynamic because the reaction has
not stopped: the opposing rates are equal.• Consider frozen N2O4: only white solid is present. On
the microscopic level, only N2O4 molecules are present.
The Concept of EquilibriumThe Concept of Equilibrium
The Concept of EquilibriumThe Concept of Equilibrium
• As the substance warms it begins to decompose: N2O4(g) 2NO2(g)
• A mixture of N2O4 (initially present) and NO2 (initially formed) appears light brown.
• When enough NO2 is formed, it can react to form N2O4: 2NO2(g) N2O4(g).
• At equilibrium, as much N2O4 reacts to form NO2 as NO2 reacts to re-form N2O4:
• The double arrow implies the process is dynamic.N2O4(g) 2NO2(g)
The Concept of EquilibriumThe Concept of Equilibrium
• ConsiderForward reaction: A B Rate = kf[A]Reverse reaction: B A Rate = kr[B]
• At equilibrium kf[A] = kr[B].• For an equilibrium we write:
• As the reaction progresses– [A] decreases to a constant,– [B] increases from zero to a constant.– When [A] and [B] are constant, equilibrium is achieved.
A B
The Concept of EquilibriumThe Concept of Equilibrium
• Alternatively:– kf[A] decreases to a constant,– kr[B] increases from zero to a constant.– When kf[A] = kr[B] equilibrium is achieved.
The Concept of EquilibriumThe Concept of Equilibrium
• Consider
• If we start with a mixture of nitrogen and hydrogen (in any proportions), the reaction will reach equilibrium with a constant concentration of nitrogen, hydrogen and ammonia.
• However, if we start with just ammonia and no nitrogen or hydrogen, the reaction will proceed and N2 and H2 will be produced until equilibrium is achieved.
N2(g) + 3H2(g) 2NH3(g)
The Equilibrium ConstantThe Equilibrium Constant
The Equilibrium ConstantThe Equilibrium Constant
• No matter the starting composition of reactants and products, the same ratio of concentrations is achieved at equilibrium.
• For a general reaction
the equilibrium constant expression is
where Kc is the equilibrium constant.
aA + bB(g) pP + qQ
ba
qpcK
BA
QP
The Equilibrium ConstantThe Equilibrium Constant
• Kc is based on the molarities of reactants and products at equilibrium.
• We generally omit the units of the equilibrium constant.
• Note that the equilibrium constant expression has products over reactants.
The Equilibrium ConstantThe Equilibrium Constant
The Equilibrium Constant in Terms of PressureThe Equilibrium Constant in Terms of Pressure• If KP is the equilibrium constant for reactions
involving gases, we can write:
• KP is based on partial pressures measured in atmospheres.
• We can show thatPA = [A](RT)
ba
qpP
PP
PPK
BA
QP
The Equilibrium ConstantThe Equilibrium Constant
The Equilibrium Constant in Terms of PressureThe Equilibrium Constant in Terms of PressurePA = [A](RT)
• This means that we can relate Kc and KP:
where n is the change in number of moles of gas.• It is important to use:
n = ngas(products) - ngas(reactants)
ncP RTKK
The Equilibrium ConstantThe Equilibrium Constant
The Equilibrium ConstantThe Equilibrium Constant
PhosphorousProblem 1(a)
Phosphorous pentachloride dissociates on heating:
PCl5(g) PCl3(g) + Cl2(g)
If Kc = 3.26 x 10-2 at 191C, what is Kp at this temperature?
Ans.: Kp = 1.24
The Equilibrium ConstantThe Equilibrium Constant
PhosphorousProblem 1(b)
The value of Kc for the following reaction at 900C is 0.28.
S2(g) + 4H2(g) CH4(g) + 2H2S(g)
What is Kp at this temperature?
Ans.: Kp = 3.0 x 10-5
The Equilibrium ConstantThe Equilibrium Constant
PhosphorousProblem 2(a)
Consider the following reaction at 1000C:
CO(g) + 3H2(g) CH4(g) + H2O(g)
At equilibrium, the following concentrations are measured: [CO] = 0.0613 M, [H2] = 0.1839 M, [CH4] = 0.0387,[H2O] = 0.0387 M. Calculate the value of Kc for this reaction.Calculate the value of Kp. ?
Ans.: Kc = 3.93, Kp = 3.60 x 10-4
The Equilibrium ConstantThe Equilibrium Constant
PhosphorousProblem 2(b)
A 5.00 L vessel contained 0.0185 mol of phosphorus trichloride,0.0158 mol of phosphorus pentachloride, and 0.0870 mol chlorine at503 K in an equilibrium mixture. Calculate the value of Kc at thistemperature.
PCl3(g) + Cl2(g) PCl5(g)
Ans.: Kc = 49.1
The Magnitude of Equilibrium ConstantsThe Magnitude of Equilibrium Constants• The equilibrium constant, K, is the ratio of products
to reactants.• Therefore, the larger K the more products are present
at equilibrium.• Conversely, the smaller K the more reactants are
present at equilibrium.• If K >> 1, then products dominate at equilibrium and
equilibrium lies to the right.• If K << 1, then reactants dominate at equilibrium and
the equilibrium lies to the left.
The Equilibrium ConstantThe Equilibrium Constant
The Magnitude of Equilibrium ConstantsThe Magnitude of Equilibrium ConstantsThe Equilibrium ConstantThe Equilibrium Constant
The Magnitude of Equilibrium ConstantsThe Magnitude of Equilibrium Constants• An equilibrium can be approached from any
direction.• Example:
• hasN2O4(g) 2NO2(g)
212.0
ONNO
42
22 cK
The Equilibrium ConstantThe Equilibrium Constant
The Magnitude of Equilibrium ConstantsThe Magnitude of Equilibrium Constants• However,
• has
• The equilibrium constant for a reaction in one direction is the reciprocal of the equilibrium constant of the reaction in the opposite direction.
2NO2(g) N2O4(g)
72.4212.01
NO
ON2
2
42 cK
The Equilibrium ConstantThe Equilibrium Constant
Heterogeneous EquilibriaHeterogeneous Equilibria• When all reactants and products are in one phase, the
equilibrium is homogeneous.• If one or more reactants or products are in a different
phase, the equilibrium is heterogeneous.• Consider:
– experimentally, the amount of CO2 does not seem to depend on the amounts of CaO and CaCO3. Why?
• The concentration of a solid or pure liquid is its density divided by molar mass.
CaCO3(s) CaO(s) + CO2(g)
The Equilibrium ConstantThe Equilibrium Constant
Heterogeneous EquilibriaHeterogeneous EquilibriaThe Equilibrium ConstantThe Equilibrium Constant
Heterogeneous EquilibriaHeterogeneous Equilibria• Neither density nor molar mass is a variable, the
concentrations of solids and pure liquids are constant.• For the decomposition of CaCO3:
• We ignore the concentrations of pure liquids and pure solids in equilibrium constant expressions.
• The amount of CO2 formed will not depend greatly on the amounts of CaO and CaCO3 present.
2
223
COconstant
.COconstantCOCaCO
CaO
cc
c
KK
K
The Equilibrium ConstantThe Equilibrium Constant
• Proceed as follows:– Tabulate initial and equilibrium concentrations (or partial
pressures) given.– If an initial and equilibrium concentration is given for a
species, calculate the change in concentration.– Use stoichiometry on the change in concentration line only
to calculate the changes in concentration of all species.– Deduce the equilibrium concentrations of all species.
• Usually, the initial concentration of products is zero. (This is not always the case.)
• When in doubt, assign the change in concentration a variable.
Calculating Equilibrium ConstantsCalculating Equilibrium Constants
Calculating Equilibrium ConstantsCalculating Equilibrium Constants
PhosphorousProblem 3(a)
Carbon dioxide decomposes at elevated temperatures to carbonmonoxide and oxygen:
2CO2(g) 2CO(g) + O2(g)
At 3000 K, 2.00 mol of CO2 is placed into a 1.00 L container andallowed to come to equilibrium. At equilibrium, 0.90 mol CO2
remains. What is the value for Kc at this temperature?
Ans.: Kc = 0.82
Calculating Equilibrium ConstantsCalculating Equilibrium Constants
PhosphorousProblem 3(b)
Consider the following reaction at 1000C:
CO(g) + 3H2(g) CH4(g) + H2O(g)
The original concentrations of CO and H2 were 0.2000 M and0.3000 M, respectively. At equilibrium, the concentration of CH4 was measured to be 0.0478 M. Calculate the values of Kc and Kp.
Ans.: Kc = 3.91, Kp = 3.60 x 10-4
Calculating Equilibrium ConstantsCalculating Equilibrium Constants
PhosphorousProblem 3(c)
Consider the following reaction at 1000C:
2HI(g) H2(g) + I2(g)
When 4.00 mol of HI was placed into the 5.0 L reaction vessel at458C, the equilibrium mixture was found to contain 0.422 mol I2.Calculate the value of Kc for the decomposition of HI.
Ans.: Kc = 1.79 x 10-2
Calculating Equilibrium ConstantsCalculating Equilibrium Constants
PhosphorousProblem 3(d)
Hydrogen sulfide, a colorless gas with a foul odor, dissociates onheating:
2H2S(g) 2H2(g) + S2(g)
When 0.100 mol H2S was put into a 10.0 L vessel and heated to1132C, it gave an equilibrium mixture containing 0.0285 mol H2.Calculate the value of Kc at this temperature.
Ans.: Kc = 1.35 x 10-6 M
Predicting the Direction of ReactionPredicting the Direction of Reaction• We define Q, the reaction quotient, for a general
reaction
as
where [A], [B], [P], and [Q] are molarities at any time.• Q = K only at equilibrium.
aA + bB(g) pP + qQ
ba
qpQ
BA
QP
Applications of Equilibrium ConstantsApplications of Equilibrium Constants
Predicting the Direction of ReactionPredicting the Direction of Reaction• If Q > K then the reverse reaction must occur to reach
equilibrium (i.e., products are consumed, reactants are formed, the numerator in the equilibrium constant expression decreases and Q decreases until it equals K).
• If Q < K then the forward reaction must occur to reach equilibrium.
Applications of Equilibrium ConstantsApplications of Equilibrium Constants
Calculation of Equilibrium ConcentrationsCalculation of Equilibrium Concentrations• The same steps used to calculate equilibrium
constants are used.• Generally, we do not have a number for the change in
concentration line.• Therefore, we need to assume that x mol/L of a species
is produced (or used).• The equilibrium concentrations are given as algebraic
expressions.
Applications of Equilibrium ConstantsApplications of Equilibrium Constants
Applications of Equilibrium ConstantsApplications of Equilibrium Constants
Phosphorous
Problem 4
The following reaction has an equilibrium constant Kc of 3.07 x 10-4 at 24C:
2NOBr(g) 2NO(g) + Br2(g)
For each of the following compositions, decide whether the reactionis at equilibrium. If not, decide which direction the reaction should go.
(a) [NOBr] = 0.0610 M,[NO] = 0.0151 M, [Br2] = 0.0108 M(b) [NOBr] = 0.115 M,[NO] = 0.0169 M, [Br2] = 0.0142 M(c) NOBr] = 0.181 M,[NO] = 0.0123 M, [Br2] = 0.0201 M
Ans.: a.) goes left; b.) equilibrium; c.) goes right
Applications of Equilibrium ConstantsApplications of Equilibrium Constants
Phosphorous
Problem 5(a)
Nitrogen and oxygen form nitric oxide:
N2(g) + O2(g) 2NO(g) If an equilibrium mixture at 25C contains 0.040 mol/L N2 and0.010 mol/L O2, what is the concentration of NO in this mixture?Kc = 1 x 10-30 at this temperature.
Ans.: 2 x 10-17 M
Applications of Equilibrium ConstantsApplications of Equilibrium Constants
Phosphorous
Problem 5(b)
An equilibrium mixture at 1200 K contains 0.30 mol CO, 0.10 mol H2,and 0.020 mol H2O, plus an unknown amount of CH4 in each liter.What is the concentration of CH4 in this mixture if Kc = 3.92?The reaction is:
CO(g) + 3H2(g) CH4(g) + H2O(g)
Ans.: 0.059 M
Applications of Equilibrium ConstantsApplications of Equilibrium Constants
Phosphorous
Problem 6(a)
The reaction:
CO(g) + H2O(g) CO2(g) + H2(g)
is used to increase the ratio of hydrogen in synthesis gas (mixtures ofCO and H2). Suppose you start with 1.00 mol each of carbonmonoxide and water in a 50.0 L vessel. What is the concentrationof each substance in the equilibrium mixture at 1000C given thatKc = 0.58 at this temperature?
Ans.: [CO] = [H2O] = 0.0114 M, [CO2] = [H2] = 0.0086 M
Applications of Equilibrium ConstantsApplications of Equilibrium Constants
Phosphorous
Problem 6(b)
Hydrogen iodide decomposes to hydrogen gas and iodine gas:
2HI(g) H2(g) + I2(g)
At 800 K, Kc for this reaction is 0.016. If 0.50 mol HI is placed intoa 5.0 L flask, what will be the composition of the mixture atequilibrium?
Ans.: [HI] = 0.080 M, [H2] = [I2] = 0.010 M
Applications of Equilibrium ConstantsApplications of Equilibrium Constants
Phosphorous
Problem 7(a)
N2O4 decomposes to NO2 according to the reaction:
N2O4(g) 2NO2(g)
At 100C, Kc = 0.36. If a 1.00 L flask initially contains 0.100 molN2O4, what will be the concentrations of N2O4 and NO2 at equilibrium?
Ans.: [N2O4] =0.040 M, [NO2] = 0.12 M
Applications of Equilibrium ConstantsApplications of Equilibrium Constants
Phosphorous
Problem 7(b)
Hydrogen and iodine react according to the equation:
H2(g) + I2(g) 2HI(g)
Suppose 1.00 mol H2 and 2.00 mol I2 are placed into a 1.00 L vessel.How many moles of each substance are in the equilibrium mixture at458C if Kc = 49.7 at this temperature?
Ans.: 1.86 mol HI, 0.07 mol H2, 1.07 mol I2
Applications of Equilibrium ConstantsApplications of Equilibrium Constants
Phosphorous
Problem 7(c)
Iodine and bromine react to give iodine monobromide:
I2(g) + Br2(g) 2IBr(g)
What is the equilibrium composition of a mixture at 150C thatinitially contained 0.0015 mol each of iodine and bromine in a 5.0 Lvessel if Kc = 1.2 x 102 at this temperature?
Ans.: [IBr] = 5.1 x 10-4 M, [Br2] = [I2] = 4.7 x 10-5 M
• Consider the production of ammonia
• As the pressure increases, the amount of ammonia present at equilibrium increases.
• As the temperature decreases, the amount of ammonia at equilibrium increases.
• Can this be predicted?• Le Châtelier’s Principle: if a system at equilibrium is
disturbed, the system will move in such a way as to counteract the disturbance.
N2(g) + 3H2(g) 2NH3(g)
Le Châtelier’s PrincipleLe Châtelier’s Principle
Le Châtelier’s PrincipleLe Châtelier’s Principle
Change in Reactant or Product ConcentrationsChange in Reactant or Product Concentrations• Consider the Haber process
• If H2 is added while the system is at equilibrium, the system must respond to counteract the added H2 (by Le Châtelier).
• That is, the system must consume the H2 and produce products until a new equilibrium is established.
• Therefore, [H2] and [N2] will decrease and [NH3] increases.
N2(g) + 3H2(g) 2NH3(g)
Le Châtelier’s PrincipleLe Châtelier’s Principle
Change in Reactant or Product ConcentrationsChange in Reactant or Product ConcentrationsLe Châtelier’s PrincipleLe Châtelier’s Principle
Change in Reactant or Product ConcentrationsChange in Reactant or Product Concentrations• Adding a reactant or product shifts the equilibrium
away from the increase.• Removing a reactant or product shifts the equilibrium
towards the decrease.• To optimize the amount of product at equilibrium, we
need to flood the reaction vessel with reactant and continuously remove product (Le Châtelier).
• We illustrate the concept with the industrial preparation of ammonia
N2(g) + 3H2(g) 2NH3(g)
Le Châtelier’s PrincipleLe Châtelier’s Principle
Change in Reactant or Product ConcentrationsChange in Reactant or Product ConcentrationsLe Châtelier’s PrincipleLe Châtelier’s Principle
Change in Reactant or Product ConcentrationsChange in Reactant or Product Concentrations• N2 and H2 are pumped into a chamber.• The pre-heated gases are passed through a heating
coil to the catalyst bed.• The catalyst bed is kept at 460 - 550 C under high
pressure.• The product gas stream (containing N2, H2 and NH3)
is passed over a cooler to a refrigeration unit.• In the refrigeration unit, ammonia liquefies but not N2
or H2.
Le Châtelier’s PrincipleLe Châtelier’s Principle
Change in Reactant or Product ConcentrationsChange in Reactant or Product Concentrations• The unreacted nitrogen and hydrogen are recycled
with the new N2 and H2 feed gas.• The equilibrium amount of ammonia is optimized
because the product (NH3) is continually removed and the reactants (N2 and H2) are continually being added.
Effects of Volume and PressureEffects of Volume and Pressure• As volume is decreased pressure increases.• Le Châtelier’s Principle: if pressure is increased the
system will shift to counteract the increase.
Le Châtelier’s PrincipleLe Châtelier’s Principle
Le Châtelier’s PrincipleLe Châtelier’s Principle
Phosphorous
Problem 8
Predict the effect of the following concentration changes on thereaction below.
CH4(g) + 2S2(g) CS2(g) + 2H2S(g)
a) (a) Some CH4(g) is removed.(b) Some S2(g) is added.(c) Some CS2(g) is added.(d) Some H2S(g) is removed.(e) Some argon gas is added.
Ans.: a.) goes left; b.) goes right; c.) goes left; d.) goes right ; e.) no effect
Effects of Volume and PressureEffects of Volume and Pressure• That is, the system shifts to remove gases and
decrease pressure.• An increase in pressure favors the direction that has
fewer moles of gas.• In a reaction with the same number of product and
reactant moles of gas, pressure has no effect.• Consider
N2O4(g) 2NO2(g)
Le Châtelier’s PrincipleLe Châtelier’s Principle
Effects of Volume and PressureEffects of Volume and Pressure• An increase in pressure (by decreasing the volume)
favors the formation of colorless N2O4.• The instant the pressure increases, the system is not at
equilibrium and the concentration of both gases has increased.
• The system moves to reduce the number moles of gas (i.e. the forward reaction is favored).
• A new equilibrium is established in which the mixture is lighter because colorless N2O4 is favored.
Le Châtelier’s PrincipleLe Châtelier’s Principle
Le Châtelier’s PrincipleLe Châtelier’s Principle
Phosphorous
Problem 9
Predict the effect of increasing pressure (decreasing volume) oneach of the following reactions.
a) (a) CH4(g) + 2S2(g) CS2(g) + 2H2S(g) (b) H2(g) + Br2(g) 2HBr(g)(c) CO2(g) + C(s) 2CO(g)(d) PCl5(g) PCl3(g) + Cl2(g)(e) N2O4(g) 2NO2(g)
Ans.: a.) no effect; b.) no effect; c.) goes left; d.) goes right ; e.) goes left
Effect of Temperature ChangesEffect of Temperature Changes• The equilibrium constant is temperature dependent.• For an endothermic reaction, H > 0 and heat can be
considered as a reactant.• For an exothermic reaction, H < 0 and heat can be
considered as a product.• Adding heat (i.e. heating the vessel) favors away from
the increase:– if H > 0, adding heat favors the forward reaction,– if H < 0, adding heat favors the reverse reaction.
Le Châtelier’s PrincipleLe Châtelier’s Principle
Effect of Temperature ChangesEffect of Temperature Changes• Removing heat (i.e. cooling the vessel), favors towards
the decrease:– if H > 0, cooling favors the reverse reaction,– if H < 0, cooling favors the forward reaction.
• Consider
for which H > 0.– Co(H2O)6
2+ is pale pink and CoCl42- is blue.
Cr(H2O)6(aq) + 4Cl-(aq) CoCl42-(aq) + 6H2O(l)
Le Châtelier’s PrincipleLe Châtelier’s Principle
Effect of Temperature ChangesEffect of Temperature ChangesLe Châtelier’s PrincipleLe Châtelier’s Principle
Effect of Temperature ChangesEffect of Temperature Changes
– If a light purple room temperature equilibrium mixture is placed in a beaker of warm water, the mixture turns deep blue.
– Since H > 0 (endothermic), adding heat favors the forward reaction, i.e. the formation of blue CoCl4
2-. – If the room temperature equilibrium mixture is placed in a
beaker of ice water, the mixture turns bright pink.– Since H > 0, removing heat favors the reverse reaction
which is the formation of pink Co(H2O)62+.
Cr(H2O)6(aq) + 4Cl-(aq) CoCl42-(aq) + 6H2O(l)
Le Châtelier’s PrincipleLe Châtelier’s Principle
Le Châtelier’s PrincipleLe Châtelier’s Principle
Phosphorous
Problem 10
Predict the effect of increasing temperature on each of the followingreactions. What effect does this change have on Kc?
a) (a) CO(g) + 3H2(g) CH4(g) + H2O(g) H < 0 (b) CO2(g) + C(s) 2CO(g) H > 0(c) 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) H < 0 (d) 2H2O(g) 2H2(g) + O2(g) H > 0
Ans.: a.) goes left, Kc decreases; b.) goes right, Kc increases; c.) goes left, Kc decreases; d.) goes right, Kc increases
The Effect of CatalystsThe Effect of Catalysts• A catalyst lowers the activation energy barrier for the
reaction.• Therefore, a catalyst will decrease the time taken to
reach equilibrium.• A catalyst does not effect the composition of the
equilibrium mixture.
Le Châtelier’s PrincipleLe Châtelier’s Principle