Transcript Gaseous Equilibrium – Chapter 12

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.
Equilibrium Systems
2N2 (g) + 3H2 (g) ⇌ 2NH3 (g)
N2O4(g) ⇌ 2NO2 (g)
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
Calculating Keq
Kc = [C]c[D]d
[A]a[B]b
or
Kp = (PC)c(PD)d
(PA)a(PB)b
Kp = Kc(RT)Δng


R = 0.0821 L atm/mol K
Δng = the change in the # of moles of gas in the equation (products
– reactants).
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)
then for….

½ N2O4(g) ⇌ NO2 (g)
K’ = ?
K = 11
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’’ = ?
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
⇌ NO
⇌ SO
SO2 (g)
+ NO2 (g)
SO2 (g)
+ ½ O2 (g)
NO2 (g)
⇌ NO
(g)
(g)
+ SO3 (g)
3 (g)
+ ½ O2 (g)
K3 = ?
K1 = 2.2
K2 = 4.0
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 = ?
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)
I2 (s)
⇌
⇌
CO (g) + H2O (l) K =
I2 (g) K =
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.
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.
Example 4 - An Equilibrium Table

However, it is a bit more difficult when your given only
one partial pressure @ 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) PI at equilibrium
(b) PHI at equilibrium
(c) K
2
Understanding K
 When
K > 1 the reaction favors the products,
the forward reaction (ΔGo < 0).
 When K < 1 the reaction favors the reactants,
the reverse reaction (ΔGo > 0).
 When K = 1 neither the forward nor the
reverse is favored (ΔGo = 0).
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 < K then the reaction will move in the forward
direction
 If Q > K then the reaction will move in the reverse
reaction
 If Q = K then the reaction is already at equilibrium
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.
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.
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?
Le Chatelier’s Principle:
Effects 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.

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.

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
van’t 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 van’t Hoff
equation maybe used:
Equilibrium & Gibbs Free Energy

Relationship between standard Gibbs free
energy and the equilibrium constant (or
reaction quotient):
 ΔGo

= -RT lnK
Gibbs free energy and standard Gibbs free
energy:
 ΔG
= ΔGo + RT lnQ
Example 7

Consider the equilibrium reaction for the
formation of gaseous ammonia from gaseous
nitrogen and hydrogen, where ΔGo = -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
MC #1
CuO(s) + H2(g) <===> Cu(s) + H2O(g) ΔH = - 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
MC #2
HgO(s) + 4 I¯ + H2O  HgI42¯ + 2 OH¯ ΔH < 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
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) <===> CO2(g) + 1/2 N2(g)
(B) N2 (g) + 3 H2 (g) <===> 2 NH3(g)
(C) N2 (g) + 2 O2 (g) <===> 2 NO2(g)
(D) N2O4 (g) <===> 2 NO2(g)
(E) NO(g) + O3 (g) <===> NO2(g) + O2(g)
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
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
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 105°C.
(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 105°C, 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 105°C.
(d) Calculate the mole fraction of F2(g) in the container at equilibrium.
FRQ #2
(a)
(b)
(c)
(d)
(e)
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
Write the expression for the equilibrium constant, Kp for the reaction.
Calculate the number of moles of CO2(g) initially placed in the container. (Assume that the
volume of the solid carbon is negligible.)
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
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.
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.
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 3.72×10–2 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.
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.
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]
[H2O]
= 0.30 mol/L
= [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.