Equilibrium

1
Equilibrium

• Chapter 15

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Equilibrium

• Have you ever tried to maintain your balance as
  you walked across a narrow ledge?
• In a chemical reaction balance or equilibrium is
  also maintained.
• You can think of the yields ?? sign as the ledge

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• Equilibrium systems exist in ocean water, blood,
  urine, and many other biological systems
• Chemical reactions for the most part are
  reversible
• You can think of the yield sign ?? as the ledge
  in a chem system

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I. Chemical Equilibrium Concept

• Chemical Equilibrium occurs when opposing
  reactions are proceeding at equal rates
• At equil rate forward rate reverse
• A??B indicates a molar conc.
• Fr A?B rate kf A kf A kr B
• Rr B?A rate kr B B kf cons Kc
• Rearranging formula A kr

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A dynamic equilibrium

• Individual molecules are undergoing change but
  there is no net exchange in the concentration of
  reactants and products
• Does not mean that the concentrations are not
  changing just that the ratio equals a definite
  value
• Look at Habber reaction figure 15.6 text.

• Dihydrogen monoxide
• is also known as hydric acid, and is the major
  component of acid rain.
• contributes to the "greenhouse effect."
• may cause severe burns.
• contributes to the erosion of our natural
  landscape.
• accelerates corrosion and rusting of many metals.
  
• may cause electrical failures and decreased
  effectiveness of automobile brakes.
• has been found in excised tumors of terminal
  cancer patients.

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Law of Mass Action

• Equilibrium can be reached from either direction
• Concentrations of reactants and products are
  expressed as
• aA bB ?? pP qQ
• equil constant Kc Pp Qq products
• Aa
  Qb reactants

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Equilibrium Constant mass action

• Kc depends on the stoichiometry not on mechanics
• Doesnt depend on the initial conc of reactants
  and products
• Doesnt depend on other added sub as long as they
  do not react
• Varies with tempt
• Catalyst do not effect just speeds reaching eq.

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Putting it Together
Question? Main Ideas Details Monitor



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Writing equilibrium expressions

• 2O3 (g) ?? 3O2 (g)
• 2NO (g) Cl2 (g) ?? 2NOCl (aq)
• AgCl (s) ?? Ag (aq) Cl- (aq)
• Kc O23 Kc NOCl2
• O32 NO2 Cl2
• Kc Ag Cl-
• Pure solids and liquids do not effect the
  equilibrium because their conc remain unchanged

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Eq expressed as pressure

• C3H8(g) O2(g) ? CO2(g) H2O(l)
• Kp CO2p3
• C3H8p O2p5
• P partial pressure of the gas
• Kp kc(RT)delta n

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What does kc tell you?

• Ex CO (g) Cl2 (g) ?? COCl2 (g)
• Kc COCl2 4.57 X 109
• CO Cl2
• Kcgtgtgt1 larger numerator reaction goes almost
  totally to products eq lies right favors
  products
• Kcltltlt1 larger denominator

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Le Chateliers Principle

• Henri-Louis Le Chatelier (1858-1936)

• If a system at eq is disturbed by a change in
  temperature, pressure, or the concentration of
  one of the components, the system will shift
  its eq pos so as to counter the effects of the
  distrubance

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Change in Concentration

• Le Châtelier's principle states that if the
  concentration of one of the components of the
  reaction (either product or reactant) is changed,
  the system will respond in such a way as to
  counteract the effect
• If a substance (either reactant or product) is
  removed from a system, the equilibrium will shift
  so as to produce more of that component (and once
  again achieve equilibrium)

14
Change in concentration

• If a substance (either reactant or product) is
  added to a system, the equilibrium will shift so
  as to consume more of that component (and once
  again achieve equilibrium)

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Putting it Together N2(g) 3H2(g) ltgt 2NH3(g)
Question? Main Ideas Details Monitor
How would inceasing H2 change eq
Predict the relative conc of each reactant and product
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• The reaction is driven "to the right" by the
  effects of added H2
• The eq concs will not be identical to the
  original state. However, Kc will be the same. The
  new equilibrium state contains a slightly higher
  concentration of NH3(g), and slightly lower
  concentration of N2(g) (as well as a slightly
  higher concentration of H2(g).

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Change in Volume and Pressure

• A chemical system in equilibrium can respond to
  the effects of pressure also. According to Le
  Châtelier's Rule, if the pressure is increased on
  a system, it will respond by trying to reduce the
  pressure. How does it do this?
• We are primarily concerned with homogeneous
  gaseous reactions
• The stoichiometry of the reaction may lead to a
  greater number of molecules on one side of the
  equation.
• For example, in the Haber reaction, N2(g)
  3H2(g) ltgt 2NH3(g) there are twice as many moles
  of reactants as products

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• If the Haber reaction were in equilibrium, and
  the pressure was increased, the reaction would
  respond to oppose the increase in pressure. It
  could accomplish this by shifting the equilibrium
  to the right (producing NH3(g))
• This would reduce the overall number of moles in
  the reaction, and therefore, lower the pressure
• Systems shift to the side with the fewest number
  of moles if both are the same then no change in
  eq cons occurs

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Changes in Temperature

• The intrinsic value of K does not change when we
  increase concentrations or pressures of
  components in a reaction. However, almost every
  equilibrium constant (K) changes in response to
  changes in temperature.
• We will consider reaction conditions under which
  no work is done, and therefore all energy changes
  associated with reactions will be manifested by
  temperature changes)

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Temperature Changes

• Exothermic reactions are associated with heat
  release when the reaction proceeds in the forward
  direction
• Endothermic reactions are associated with heat
  release when the reaction proceeds in the reverse
  direction (i.e. heat is absorbed in the forward
  direction)

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• These two types of reactions and their associated
  heat changes can be written as
• Exothermic Reactants yield Products Heat
• Endothermic Reactants Heat yield Products
• If temperature is increased, the equilibrium will
  shift so as to minimize the effect of the added
  heat
• The reaction will shift in the appropriate
  direction such that the added heat is absorbed

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• When heat is added to exothermic reactions at
  equilibrium, products will be consumed to produce
  reactants (shift to the LEFT) May also be
  written delta t is negative.
• When heat is added to endothermic reactions at
  equilibrium, reactants will be consumed to
  produce products (shift to the RIGHT) May also be
  written delta t is positive.

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Based on this behavior, what is the effect of T
upon K?

• Assume K 1.0 for an exothermic reaction at
  equilibrium.
• Added heat causes the reaction to shift to the
  left. Reactants lt Products Heat
• Thus, 1.0 must represent a reaction quotient, Q,
  that is too large in comparison to the new value
  of K.
• Thus, the effect of increasing temperature on an
  exothermic reaction is to lower the value of K.
• Conversely, the effect of increasing temperature
  on an endothermic reaction is to increase the
  value of K

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Putting it Together Calc Delta H of formation
for C3H8(g) O2(g) ? CO2(g) H2O(l)
Question? Is the value Exo,or endotherm Main Ideas Details Monitor
How would inc. temp effect eq
How would dec temp eff eq k
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Calculations with eq K

• Example calculating unknown concentrations using
  the eq constant
• CO(g) 3H2(g) lt-gt CH4(g) H20(g)
• At eq 0.3 mol of CO, 0.1 mol H2 and 0.02 mol of
  H20 are in 1.0 liter of a vessel at 1200 k kC is
  3.92 what is the conc of CH4?

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• Kc CH4 H2O
• CO H23
• 3.93 CH4 (.020)
• (0.30) (0.10)3
• CH4 (0.30)(0.10)3 3.93
• (0.020)
• 0.059 mol/l

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Learning CheckPCl5(g)lt-gt PCl3(g) Cl2(g)

• A l.0 liter vessel has a unknown amount of PCl5
  at eq Kc at 250 0C is 0.0415. Calc the unknow
  conc. if 0.02moles of PCl3 and Cl2 are in the
  container. (0.0096)

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Solving linear eq equasions

• CO(g) H20(g) lt-gt C02(g) H2(g)
• Given 1.0 mol of CO2 and H20 in a 50.0 l vessel.
  How many moles are in an eq mix at 1000 oC Ec
  0.58 at 1000oC
• CO(g) H20(g) lt-gt C02(g) H2(g
• I 0.02 0.02 0 0
• C -x -x x x
• E 0.02-x 0.02-x x x

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• 0.58 CO2H2 X2
• CO H2O (0.02-X)2
• ,- 0.76 X2
• (0.02-X)
• (the neg one gives a neg answer x cant be
  neg)
• _ 0.76(0.02-X)X
• 0.0152-0.76XX
• 0.0152 1.76X
• X 0.0086

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H2(g) I2(g) lt-gt 2HI(g)

• What is the eq comp of a reaction mixture
  starting with 0.500 mol each of H2 and I2 in a 1
  l vessel? Kc 49.7 at 458 oC (H2 I2 0.11
  mol/l HI 0.78 mol/l

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Equil with quadratic expressions

• Calc the conc. of the previous problem with 1.00
  molar H2 and 2.00 molar I2 as the starting
  concentrations.
• H2(g) I2(g) lt-gt 2HI(g)
• I 1.00 2.00 0
• C -x -x 2x 49.9 (2x)2
• E 1.00-x 2.00-x 2x (1.00-x)(2.00-x)

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• (1.00-x)(2.00-x) (2x)2
• 49