Ch 16 Chemical Equilibrium

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Ch 16 Chemical Equilibrium

• Brady Senese, 4th Ed.

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Chapter 16 Chemical Equilibrium- General Concepts

• When a system is at equilibrium, the forward and
  reverse reaction are proceeding at the same rate
• The concentrations of all species remains
  constant over time, but both the forward and
  reverse reaction never cease
• Equilibrium is signified with double arrows or
  the equal sign

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Dynamic Equilibrium

• Many reactions are reversible this is to say
  that when the reaction appears to be complete,
  amounts of both reactant and product exist.
• We show reversible reactions using a double arrow
  ? or
• The forward reaction is read left to right, while
  the reverse reaction is read right to left.
• HC2H3O2(aq) H2O(l) ?H3O(aq) C2H3O2-(aq)

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Equilibrium- what does it mean?

• The rate of the forward reaction the rate of
  the reverse reaction.
• As product is being made, some is also consumed
  to form reactants
• The process is continually occurring, but the
  amounts present stop changing

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The same equilibrium composition is reached from
either the forward or reverse direction, provided
the overall system composition is the same. Pure
NO2 is brown and pure N2O4 is colorless. The
amber color of the equilibrium mixture indicates
that both species are present at equilibrium.
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• There is a simple relationship among the
  concentrations of the reactants and products for
  any chemical system at equilibrium
• It is called the mass action expression, and is
  derived from thermodynamics (discussed in Chapter
  20)
• Consider the equilibrium

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Four experiments to study the equilibrium among
H2, I2, and HI gases. Different amounts of the
reactants and products are placed in a 10.0 L
reaction vessel at 440oC where the gases
establish equilibrium. When equilibrium is
reached, different amounts of reactants and
products remain.

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• The numerical value of the mass action expression
  is called the reaction quotient, Q

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• The reaction can be evaluated at any
  concentrations
• At equilibrium (and 440oC) for this reaction the
  reaction quotient has the value 49.5 (a unitless
  number)
• This relationship is called the equilibrium law
  for the system

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• The value 49.5 is called the equilibrium
  constant, Kc, and characterizes the system
• For chemical equilibrium to exist, the reaction
  quotient Q must be equal to the equilibrium
  constant Kc
• Consider the general chemical equation

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• The exponents in the mass action expression are
  the same as the stoichiometric coefficients
• At equilibrium
• The form is always products over reactants
  raised to the appropriate powers

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Learning Check

• Write the mass action expressions for the
  following
• 2NO2(g) ? N2O4(g)
• 2CO(g) O2(g) ? 2 CO2(g)
• N2 3H2(g) ?2NH3(g)

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• Various operations can be performed on
  equilibrium expressions
• Changing the direction of equilibrium when the
  direction of an equilibrium is reversed, the new
  equilibrium constant is the reciprocal of the
  original

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• Multiplying the coefficients by a factor when
  the coefficients in an equation are multiplied by
  a factor, the equilibrium constant is raised to a
  power equal to that factor

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• Adding chemical equilibria when chemical
  equilibria are added, their equilibrium constants
  are multiplied

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Learning Check

• For the reaction N2(g) 3H2(g) ?2NH3(g),
  Kc500 for a particular temperature. What would
  be Kc for the following
• 2NH3(g) ? N2(g) 3H2(g)
• ½ N2(g) 3/2 H2(g) ? NH3(g),

0.002
22.4
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The Equilibrium Constant, Kc

• Is a constant value equal to the ratio of product
  concentrations to reactant concentrations raised
  to their respective exponents
• Changes with temperature (Vant Hoff Eqn)

The Equilibrium Constant, Kp

• Based on reactions in which the substances are
  gaseous
• Assumes gas quantities are expressed in
  atmospheres in the mass action expression

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• The size of the equilibrium constant gives a
  measure of how the reaction proceeds
• General statements can be made about the
  equilibrium constant (either Kc or KP)

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Kp vs Kc

• If PVnRT, then P/RTn/V
• Thus, substituting P/RT for molar concentration
  into the Kc results in a pressure-based formula.
• ?n moles of gas in product - moles of gas in the
  reactant.

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Learning Check

• Consider the reaction of 2NO2(g)?N2O4(g)
• If The Kp for the reaction is known to be 0.480
  at 25C,What is the value of Kc at the same
  temperature?

11.7Kc
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The Significance Of K Values

• If K is large, the reaction is efficient (makes
  product effectively)
• If K is small, the reaction is inefficient (makes
  little product)
• If 10-3 ltKlt103, the reaction has much of both
  product and reactant present
• At equilibrium, the QK

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The magnitude of K and the position of
equilibrium. A large amount of product and very
little reactant at equilibrium gives Kgtgt1 (large
K). When , approximately equal amounts
of reactant and product are present at
equilibrium. When Kltlt1, mostly reactant and very
little product are present at equilibrium.
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Learning Check

• Consider the reaction of 2NO2(g)?N2O4(g) if the
  Kp for the reaction is known to be 0.480 at 25C,
  does the reaction favor product or reactant?

Since K is small, this is a reactant favored
reaction, although significant amounts of both
reactant and product are present
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• In a homogeneous reactions, all the reactants and
  products are in the same phase
• Heterogeneous reactions involve more than one
  phase
• For example the thermal decomposition of sodium
  bicarbonate (baking soda)
• Heterogeneous reactions can come to equilibrium
  just like homogeneous systems

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• If NaHCO3 is placed in a sealed container,
  homogeneous equilibrium is established
• The equilibrium law involving pure liquids and
  pure solids can be simplified

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• For a pure liquid or solid, the ratio of amount
  of substance to volume of substance is constant

The concentration of a substance in a solid is
constant. Doubling the number of moles doubles
the volume, but the ratio of moles to volume
remains the same.
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• The equilibrium law for a heterogeneous reaction
  is written without concentrations terms for pure
  solids or pure liquids.
• The equilibrium constants found in tables
  represent all the constants combined

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Learning Check

• Write the mass action expression for the
  following
• Ag(aq) Cl-(aq) ? AgCl(s)
• H3PO4(aq) H2O(l) ? H3O(aq) H2PO4-(aq)
• 3Ca2(aq) 2PO3-4(aq) ? Ca3(PO4)2(s)

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Equilibrium Positions And Shifts

• Equilibrium positions are a combination of
  concentrations that allow QK
• There are an infinite number of possible
  equilibrium positions
• Le Châteliers principle a system at equilibrium
  (QK) when upset by a disturbance (Q?K) will
  shift to offset the stress
• a system is said to shift to the right when the
  forward reaction is dominant (QltK) and shift to
  the left when the reverse direction is dominant
  (QgtK)

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• According to Le Châteliers principle
• If an outside influence upsets an equilibrium,
  the system undergoes a change in the direction
  that counteracts the disturbing influence and, if
  possible, returns the system to equilibrium
• We can consider some common stresses
• Adding or removing a product or reactant
• The equilibrium shifts to remove reactants or
  products that have been added
• The equilibrium shifts to replace reactants or
  products that have been removed

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• Changing the volume
• Reducing the volume of a gaseous reaction causes
  the reaction to decreases the number of molecules
  of gas, if it can
• Moderate pressure changes have a negligible
  effect on reactions involving only liquids or
  solids
• Changing the temperature
• Increasing the temperature shifts a reaction in a
  direction that produces an endothermic
  (heat-absorbing) change
• Decreasing the temperature shifts a reaction in a
  direction that produces an exothermic
  (heat-releasing) change

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How Does This Relate To Thermodynamics?

• ?G0-RT lnKT
• Thus, if you can find ?G0, you can find K.
• Dont forget that the ?G0 that you use must be
  for the temperature that you are using.
• ?H0-T ?S0-RT ln KT
• However,

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• Reactions are frequently run at temperatures
  other that 298 K
• The position of an equilibrium can change because
  the free energy depends on the temperature
• For temperatures near 298 K, we would expect only
  very small changes in the standard enthalpy and
  entropy changes
• For a reaction at temperature T, we can write

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• The free energy change at nonstandard conditions
  is related to the change at standard condition by
  an expression that includes the reaction quotient
  Q
• This important expression allows for any
  concentration or pressure

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• This provides a way to connect standard free
  energy changes for a reaction and the
  equilibrium constant
• Thus, equilibrium constants at various
  temperatures can be quickly and efficiently be
  estimated

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• Catalysts have no effect on the position of
  equilibrium
• Catalysts change how fast a system achieves
  equilibrium, not the relative distribution of
  reactants and products
• Adding an inert gas at constant volume
• If the added gas cannot react with any reactants
  or products it is inert towards the substances in
  the equilibrium
• No concentration changes occur, so Q still equals
  K and no shift in equilibrium occurs

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An Approach to Using Le Châteliers Principle

• Write the mass action expression for the reaction
• Examine the relationship between the affected
  concentration and Q (direct or indirect)
• Compare Q to K
• if the change makes QgtK shifts L
• if the change makes QltK shifts R
• if the change has no effect on Q-- no shift
  expected

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Learning Check

• Cu(H2O)62(aq) 4NH3(aq) ? Cu(NH3)42(aq)
  6H2O(l)
• What is expected if
• NH3(aq) is added

initially, QK NH3 is inverse to Q increasing
NH3 decreases Q QltK shifts right
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Learning Check

• In the reaction N2(g) 3H2(g) ? 2NH3(g)
  What will happen if volume is decreased?

• There are more moles of gas on reactant side,
  hence the equilibrium position should shift
  toward the products
• Increasing pressure temporarily increases Q
• QgtK shifts right then reestablishes equilibrium
  QK

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Learning Check More Shift Predictions

• In the reaction H3PO4(aq) 3OH-(aq) ? 3H2O(l)
  PO4(aq)3- What will happen if PO43- is
  removed?

• decreasing PO43- decreases Q
• QltK shifts right

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Temperature Effects

• The product efficiency of the reaction depends on
  the temperature and whether the reaction is
  endothermic or exothermic
• vant Hoff rule write mass action expression
  with heat term as if it were a concentration
  and predict shifts as before

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Learning Check

• The reaction H3PO4(aq) 3OH-(aq) ? 3H2O(l)
  PO4(aq)3- is exothermic. What will happen if
  the system is cooled?

• since the reaction is exothermic, heat is a
  product
• heat is directly proportional to Q
• decreasing the temperature decreases Q
• QltK shifts to the right

heat
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Catalysts And Equilibrium

• The catalyst lowers the Ea for both the forward
  and reverse reaction.
• The change in Ea affects the kr and kf equally
• Catalysts have no effect on equilibrium

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• Equilibrium calculations can be divided into two
  main categories
• Calculating equilibrium constants from known
  equilibrium concentrations or partial pressures
• Calculating one or more equilibrium
  concentrations or partial pressures using the
  known value of Kc or KP
• Consider the decomposition of N2O4

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• What is Kc?

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Determining Kc From Equilibrium Concentrations

• When all concentrations at equilibrium are known,
  we simply use the mass action expression to
  relate the concentrations to Kc
• Learning Check
• In the reaction H2C2O4 (g) H2O(l) ?H3O (aq)
  HC2O4- (aq) , 1mol H2C2O4 is placed into a liter
  of water. If the equilibrium concentrations of
  H3O (aq) and HC2O4- are both 0.21 M, what is
  value of Kc?

Kc4.4110-2
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Determining Equilibrium Concentrations from Kc

• When all concentrations but one are known, we
  simply use the mass action expression to relate
  the concentrations to Kc
• Learning Check
• In the reaction N2(g) O2(g)?2NO(g), Kc1. If
  the concentrations of N2 and NO are both 2 M,
  what is the concentration of O2 at equilibrium?

O22M
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• More commonly, you will have a set of initial
  conditions and an equilibrium constant
• If a KP describes the system, equilibrium will
  usually be described in terms of partial
  pressures
• If a Kc describes the system, equilibrium will
  usually be describe in terms of concentration
  (molarity, mol/L)
• The Initial, Change, Equilibrium or ICE table
  is a useful way to summarize the problem

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Determining Equilibrium Concentrations from
KcWhen Initial Concentrations Are Given

• Write the mass action expression
• Using concentrations present, compare Q vs K to
  predict shift direction
• Set up ICE table
• allow reaction to proceed as expected, using x
  to represent the quantity
• Substitute equilibrium terms from table into
  mass action expression and solve

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Learning Check

• For the reaction A B ?2D, K 10,000, what are
  the equilibrium concentrations if we start with
  2M A, 2M B, and 5M D?
• A B ? 2D,
• I 2 M 2M 5M
• C -x -x 2x
• E 2-x 2-x 52x

• x1.91
• AB0.09M
• C8.82M

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• Example Ethyl acetate, CH3CO2C2H5, is produced
  from acetic acid and ethanol by the reaction
• At 25oC, Kc4.10 for this reaction. Suppose
  0.100 mol of ethyl acetate and 0.150 mol of water
  are placed in a 1.00 L reaction vessel. What are
  the concentrations of all species at equilibrium?
• ANALYSIS Use an ICE table and the equilibrium
  constant to find the concentrations.

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• This can be solved by putting it in quadratic
  form

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• Negative concentrations are not allowed, so
• A similar procedure can be used to calculate
  partial pressures using KP

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Learning Check

• For the reaction 2A(g)?B(g) given that the
  Kp3.510-16 at 25C, and we place 0.2 atm A
  into the container, what will be the pressure of
  B at equilibrium?
• 2A ? B
• I 0.2 0 atm
• C -2x x
• E 0.2-2x x

2
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Simplifications Dropping x Term in Binomials

• In a mass action expression we sometimes get very
  complicated mathematical problems
• If the equilibrium constant is very small, the
  change to reach equilibrium (x term) is also small

2
2

• x 1.410-17 B 1.410-17
  M
• proof is 0.2- 1.410-17 significant?

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• Example Nitrogen and oxygen react to form
  nitrogen monoxide
• with Kc4.8x10-31. In air at 25oC and 1 atm, the
  N2 concentrations and O2 are initially 0.033 M
  and 0.00810 M. What are the equilibrium
  concentrations?
• ANALYSIS The equilibrium constant is very small,
  very little of the reactants will be converted
  into products

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• Substituting
• N20.033-x0.033 M
• O20.00810-x0.00810 M
• NO2x1.60x10-17 M

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Mass Action Expression (MAE Summary)

• Called Q at any time, and K only when the
  reaction is known to be at equilibrium
• Uses the stoichiometric coefficients as the
  exponent for each reactant
• aA bB ? cC dD

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Equilibrium Constants (Summary)

• When we multiply an equation by a constant, the
  value of the equilibrium constant for the new
  equation is raised to the exponent of the
  multiplier.
• n Rxn ? Kn
• When we reverse an equation, the value of K for
  the new equation is its inverse (1/K)
• When we add reactions, we multiply their K values
  for the net reaction.

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Factors Affecting Quantities Present At
Equilibrium

• Concentration affects the value of Q, and help
  predict the reaction progress
• Pressure and volume of gases both affect the
  value of Q and predict the reaction progress
• Pn/V, hence increasing gas pressure increases
  molarity
• Mn/V, hence increasing volume decreases the
  molarity of a gas
• Temperature affects the value of K
• (endothermic reactions increase K with increasing
  temperature)
• effects may be predicted using heat as a variable
  in the mass action expression (Vant Hoff Rule)