Gaseous Equilibrium

1
Gaseous Equilibrium

• Equilibrium systems are reaction systems that are
  reversible. Reactants are not completely consumed
  causing the reaction system to try to reach a
  state of equilibrium.
• Reversible means that the reaction may occur in
  the forward direction (thus favoring the
  products) or in the reverse reaction (thus
  favoring the reactants).
• At equilibrium, the rate of the forward reaction
  is equal to rate of the reverse reaction. This
  means that the amounts of all the species at
  equilibrium remain constant.

2
Equilibrium Systems

• 2N2 (g) 3H2 (g) ? 2NH3 (g)
• N2O4(g) ? 2NO2 (g)

3
The Equilibrium Constant

• When partial pressures (in atm) remain constant
  (independent of the original composition, the
  volume of the container, or the total pressure) a
  constant for this system can be calculated and is
  called the equilibrium constant (Kp). Or the
  equilibrium constant can be calculated using the
  concentration of the products and reactants at
  equilibrium (Kc).
• When solving for Kc or Kp you must first decipher
  the equilibrium constant expression from the
  chemical equation
• aA bB ? cC dD

4
Calculating Keq

• Kc CcDd
• AaBb
• or
• Kp (PC)c(PD)d
• (PA)a(PB)b
• Kp Kc(RT)?ng
• R 0.0821 L atm/mol K
• ?ng the change in the of moles of gas in the
  equation (products reactants).

5
The Coefficient Rule

• If the coefficients in a balanced equation are
  multiplied by a factor than the equilibrium
  constant for that equation is raised to that
  power
• K Kn
• For Example N2O4(g) ? 2NO2 (g) K 11
• then for.
• ½ N2O4(g) ? NO2 (g) K ?

6
The Reciprocal Rule

• The equilibrium constants for the forward and
  reverse reactions are reciprocals of each other
• K 1/K
• For Example N2O4(g) ? 2NO2 (g) K 11
• then for.
• 2NO2 (g) ? N2O4 (g) K ?

7
The Rule of Multiple Equilibria

• If a reaction can be expressed as the sum of 2 or
  more reactions, then K for the overall reaction
  is the product of the equilibrium constants for
  those reactions added.
• K1 x K2 K3
• SO2 (g) NO2 (g) ? NO (g) SO3 (g) K3 ?
• SO2 (g) ½ O2 (g) ? SO3 (g) K1 2.2
• NO2 (g) ? NO (g) ½ O2 (g) K2 4.0

8
Example 1

• Consider the reaction by which the air pollutant
  nitrogen monoxide is made from the elements
  nitrogen and oxygen in an automobile engine.
• (a) Write the equilibrium constant expression for
  the reaction.
• (b) At 25 oC, K for this reaction is 4.2 x 10-31.
  Calculate K for the formation of 1mole of this
  product from its foundational elements.
• (c) Find K for the reaction below at 25 oC
• N2 (g) 2O2 (g) ? 2NO2 (g) K
  1.0 x 10-8
• 2NO (g) O2 (g) ? 2NO2 (g) K ?

9
Heterogeneous Equilibrium

• This occurs when there is more than one phase
  present in a chemical equation.
• The position of the equilibrium is independent of
  the amount of solid or pure liquid so they are
  not included in the equilibrium constant
  expression.
• For example, write the equilibrium constant
  expression for the following
• CO2 (g) H2 (g) ? CO (g) H2O (l) K
  
• I2 (s) ? I2 (g) K

10
Example 2

• Write the expression for K for
• (a) the reduction of black solid copper (II)
  oxide with hydrogen to form copper metal and
  steam.
• (b) the reaction of steam with red hot coke to
  form a mixture of hydrogen and carbon monoxide,
  called water gas.

11
Example 3 - Determining K

• When given the partial pressures, fairly simple,
  apply the equation previously discussed.
• Solid ammonium chloride is sometimes used as a
  flux in soldering because it decomposes on
  heating into ammonia gas and hydrogen chloride
  gas.
• The HCl formed removes oxide films from metals
  to be soldered. In a certain equilibrium system
  at 400 oC, 22.6 g of ammonium chloride is
  present the partial pressures of ammonia and
  hydrogen chloride are 2.5 atm and 4.8 atm,
  respectively. Calculate the K.

12
Example 4 - An Equilibrium Table

• However, it is a bit more difficult when your
  given only one partial pressure _at_ equilibrium.
  This is when an equilibrium table will be useful
• Consider the equilibrium system
• 2HI (g) ? H2 (g) I2 (g)
• Originally, a system contains only HI at a
  pressure of 1.00 atm at 520 oC. The equilibrium
  partial pressure of H2 is found to be 0.10 atm.
  Calculate
• (a) PI2 at equilibrium
• (b) PHI at equilibrium
• (c) K

13
Understanding K

• When K gt 1 the reaction favors the products, the
  forward reaction (?Go lt 0).
• When K lt 1 the reaction favors the reactants, the
  reverse reaction (?Go gt 0).
• When K 1 neither the forward nor the reverse is
  favored (?Go 0).

14
The reaction quotient, Q

• This is basically calculated the same way as K.
  Except the values used for Q are not values at
  equilibrium. This value will allow you to predict
  which direction the reaction will move towards
• If Q lt K then the reaction will move in the
  forward direction
• If Q gt K then the reaction will move in the
  reverse reaction
• If Q K then the reaction is already at
  equilibrium

15
Solving for equilibrium values from initial
values ICE Tables

• More commonly K is used to determine the
  equilibrium partial pressures (concentrations) of
  reactants and products from the original partial
  pressures (or concentrations). To do this
• Using the balanced chemical equation for the
  reaction, write the expression for K.
• Prepare a table that will allow you to write the
  Initial amounts, the Change that occurs, and the
  Equilibrium amounts. This will help organize your
  information ICE tables.
• Express the equilibrium partial pressures of all
  species in terms of a single unknown, x. Remember
  that the changes in partial pressures are related
  through the coefficients of the balanced
  equation.
• Substitute the equilibrium terms into the
  expression for K. This will give an algebraic
  equation that must be solved for x.
• Once x is found, refer back to the table and
  calculate the equilibrium partial pressures.

16
Example 5

• Assume that the reaction for the formation of
  gaseous hydrogen fluoride from hydrogen and
  fluorine gases has an equilibrium constant of
  1.15 x 102 at a certain temperature. In a
  particular experiment, 3.000 mol of each
  component was added to a 1.500 L flask.
• a. Write the equilibrium constant expression for
  the reaction.
• b. What direction will the reaction shift to
  reach equilibrium? Explain.
• c. Calculate the equilibrium concentrations of
  all species.

17
Example 6

• Gaseous NOCl decomposes to form nitrogen monoxide
  and chlorine gases. At 35 oC the equilibrium
  constant is 1.6 x 10-5. In an experiment, 1.0 mol
  of NOCl is placed in a 2.0 L flask.
• a. Write the equilibrium constant expression for
  this reaction.
• b. What are the equilibrium concentrations of all
  species?

18
Le Chateliers PrincipleEffects of
Concentration, Pressure, Temperature

• Concentration adding or removing reactants or
  products
• If a species is added the reaction will shift
  away from that species, in order to restore
  equilibrium.
• If a species is removed the reactions will shift
  towards that species in order to restore
  equilibrium.

19

• Pressure compressing or expanding the system
• When a system is compressed (increase in
  pressure) the reaction will shift towards the
  side with the least amount of moles.
• When the system is expanded (decrease in
  pressure) the reaction will shift towards the
  side with more moles.

20

• Temperature increasing or decreasing the
  temperature
• When temperature is increased the reaction will
  shift away from the heat.
• So for the following endothermic reaction which
  way will the reaction shift? Will the K value
  decreases or increase? Why?
• N2O4 (g) ? 2NO2 (g) H 57.2 kJ
• When the temperature is decreased the reaction
  will shift towards the heat.
• So for the following endothermic reaction which
  way will the reaction shift? Will the K value
  decrease or increase? Why?
• N2O4 (g) ? 2NO2 (g) H 57.2 kJ

21
vant Hoff Equation

• The equilibrium constant changes with temperature
  (this is the only one of the three stresses
  that changes the value of K), to determine this
  change the vant Hoff equation maybe used

22
Equilibrium Gibbs Free Energy

• Relationship between standard Gibbs free energy
  and the equilibrium constant (or reaction
  quotient)
• ?Go -RT lnK
• Gibbs free energy and standard Gibbs free energy
• ?G ?Go RT lnQ

23
Example 7

• Consider the equilibrium reaction for the
  formation of gaseous ammonia from gaseous
  nitrogen and hydrogen, where ?Go -33.3 kJ/mol.
  For each of the following mixtures of reactants
  and products below at 25 oC, predict the
  direction in which the system will shift to reach
  equilibrium. If the reaction is at equilibrium
  then solve for the equilibrium constant.
• a. Pammonia 1.00 atm, Pnitrogen 1.47 atm,
  Phydrogen 1.00 x 10-2 atm
• b. Pammonia Pnitrogen Phydrogen 1.00 atm

24
MC 1

• CuO(s) H2(g) ltgt Cu(s) H2O(g) ?H -
  2.0 kJ/mol
• When the substances in the equation above are at
  equilibrium at pressure P and temperature T, the
  equilibrium can be shifted to favor the products
  by
• (A) increasing the pressure by means of a moving
  piston at constant T(B) increasing the pressure
  by adding an inert gas such as nitrogen(C)
  decreasing the temperature(D) allowing some
  gases to escape at constant P and T(E) adding a
  catalyst

25
MC 2

• HgO(s) 4 I H2O ?? HgI42 2 OH ?H lt 0
  Consider the equilibrium above. Which of the
  following changes will increase the concentration
  of HgI42
• (A) Increasing the concentration of OH(B)
  Adding 6 M HNO3(C) Increasing the mass of HgO
  present(D) Increasing the temperature(E) Adding
  a catalyst

26
MC 3

• In which of the following systems would the
  number of moles of the substances present at
  equilibrium NOT be shifted by a change in the
  volume of the system at constant temperature?
• (A) CO(g) NO(g) ltgt CO2(g) 1/2 N2(g)(B)
  N2 (g) 3 H2 (g) ltgt 2 NH3(g)(C) N2 (g) 2
  O2 (g) ltgt 2 NO2(g)(D) N2O4 (g) ltgt 2
  NO2(g)(E) NO(g) O3 (g) ltgt NO2(g) O2(g)

27
MC 4

• PCl3(g) Cl2(g) ? PCl5(g) energy
• Some PCl3 and Cl2 are mixed in a container at
  200 C and the system reaches equilibrium
  according to the equation above. Which of the
  following causes an increase in the number of
  moles of PCl5 present at equilibrium?
• I. Decreasing the volume of the container II.
  Raising the temperature III. Adding a mole of He
  gas at constant volume
• (A) I only(B) II only(C) I and III only(D) II
  and III only(E) I, II, and III

28
MC 5

• 4 HCl(g) O2(g) ? 2 Cl2(g) 2 H2O(g)
• Equal numbers of moles of HCl and O2 in a closed
  system are allowed to reach equilibrium as
  represented by the equation above. Which of the
  following must be true at equilibrium?
• I. HCl must be less than Cl2. II. O2
  must be greater than HCl. III. Cl2 must
  equal H2O.
• (A) I only(B) II only(C) I and III only(D) II
  and III only(E) I, II, and III

29
FRQ 1

• Answer the following questions regarding the
  decomposition of arsenic pentafluoride, AsF5(g).
• (a) A 55.8 g sample of AsF5(g) is introduced into
  an evacuated 10.5 L container at 105C.
• (i) What is the initial molar concentration of
  AsF5(g) in the container?
• (ii) What is the initial pressure, in
  atmospheres, of the AsF5(g) in the container?
• At 105C, AsF5(g) decomposes into AsF3(g) and
  F2(g) according to the following chemical
  equation.
• AsF5(g) ? AsF3(g) F2(g)
• (b) In terms of molar concentrations, write the
  equilibrium-constant expression for the
  decomposition of AsF5(g).
• (c) When equilibrium is established, 27.7 percent
  of the original number of moles of AsF5(g) has
  decomposed.
• (i) Calculate the molar concentration of AsF5(g)
  at equilibrium.
• (ii) Using molar concentrations, calculate the
  value of the equilibrium constant, Keq, at 105C.
  
• (d) Calculate the mole fraction of F2(g) in the
  container at equilibrium.

30
FRQ 2

• C(s) CO2 (g) ? 2 CO(g)
• Solid carbon and carbon dioxide gas at 1,160 K
  were placed in a rigid 2.00 L container, and the
  reaction represented above occurred. As the
  reaction proceeded, the total pressure in the
  container was monitored. When equilibrium was
  reached, there was still some C(s) remaining in
  the container. Results are given below.
• Time (hours) Total Pressure of Gases in
  Container at 1,160 K (atm)
• 0.0 5.00
• 2.0 6.26
• 4.0 7.09
• 6.0 7.75
• 8.0 8.37
• 10.0 8.37
• (a) Write the expression for the equilibrium
  constant, Kp for the reaction.
• (b) Calculate the number of moles of CO2(g)
  initially placed in the container. (Assume that
  the volume of the solid carbon is negligible.)
• (c) For the reaction mixture at equilibrium at
  1,160 K, the partial pressure of the CO2(g) is
  1.63 atm. Calculate
• (i) the partial pressure of CO(g) (ii) the
  value of the equilibrium constant, Kp
• (d) If a suitable solid catalyst were placed in
  the reaction vessel, would the final total
  pressure of the gases at equilibrium be greater
  than, less than, or equal to the final total
  pressure of the gases at equilibrium without the
  catalyst? Justify your answer. (Assume that the
  volume of the solid catalyst is negligible.)
• In another experiment involving the same
  reaction, a rigid 2.00 L container initially
  contains 10.0 g of C(s), plus CO(g) and CO2(g),
  each at a partial pressure of 2.00 atm at 1,160
  K.
• (e) Predict whether the partial pressure of
  CO2(g) will increase, decrease, or remain the
  same as this system approaches equilibrium.
  Justify your prediction with a calculation.

31
FRQ 3

• 2 H2S(g) ? 2 H2(g) S2(g)
• When heated, hydrogen sulfide gas decomposes
  according to the equation above. A 3.40 g sample
  of H2S(g) is introduced into an evacuated rigid
  1.25 L container. The sealed container is heated
  to 483 K, and 3.72102 mol of S2(g) is present
  at equilibrium.
• (a) Write the expression for the equilibrium
  constant, Kc, for the decomposition reaction
  represented above.
• (b) Calculate the equilibrium concentration, in
  mol?L-1, of the following gases in the container
  at 483 K.
• (i) H2(g)
• (ii) H2S(g)
• (c) Calculate the value of the equilibrium
  constant, Kc, for the decomposition reaction at
  483 K.
• (d) Calculate the partial pressure of S2(g) in
  the container at equilibrium at 483 K.
• (e) For the reaction H2(g) S2(g) ? H2S(g) at
  483 K, calculate the value of the equilibrium
  constant, Kc.

32
FRQ 4

• C(s) H2O(g) ? CO(g) H2(g) ?Hº 131kJ
• A rigid container holds a mixture of graphite
  pellets (C(s)), H2O(g), CO(g), and H2(g) at
  equilibrium. State whether the number of moles of
  CO(g) in the container will increase, decrease,
  or remain the same after each of the following
  disturbances is applied to the original mixture.
  For each case, assume that all other variables
  remain constant except for the given disturbance.
  Explain each answer with a short statement.
• (a) Additional H2(g) is added to the equilibrium
  mixture at constant volume.
• (b) The temperature of the equilibrium mixture is
  increased at constant volume.
• (c) The volume of the container is decreased at
  constant temperature.
• (d) The graphite pellets are pulverized.

33
FRQ 5

• CO2(g) H2(g) ? H2O(g) CO(g)
• When H2(g) is mixed with CO2(g) at 2,000 K,
  equilibrium is achieved according to the equation
  above. In one experiment, the following
  equilibrium concentrations were measured.
• H2 0.20 mol/L
• CO2 0.30 mol/L
• H2O CO 0.55 mol/L
• (a) What is the mole fraction of CO(g) in the
  equilibrium mixture?
• (b) Using the equilibrium concentrations given
  above, calculate the value of Kc, the equilibrium
  constant for the reaction.
• (c) Determine Kp in terms of Kc for this system.
• (d) When the system is cooled from 2,000 K to a
  lower temperature, 30.0 percent of the CO(g) is
  converted back to CO2(g). Calculate the value of
  Kc at this lower temperature.
• (e) In a different experiment, 0.50 mole of H2(g)
  is mixed with 0.50 mole of CO2(g) in a 3.0-liter
  reaction vessel at 2,000 K. Calculate the
  equilibrium concentration, in moles per liter, of
  CO(g) at this temperature.