Transcript Gaseous Equilibrium – Chapter 12

Gaseous Equilibrium – Chapter 12

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)
*Reference Table 12.1 – page 322
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 12.1 – Page 326

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 12.2 – Page 328

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.
Determining K

When given the partial pressures, fairly simple,
apply the equation previously discussed.
 Example
12.3: 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.
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.
Example: 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.
 When K < 1 the reaction favors the reactants,
the reverse reaction.
 When K = 1 neither the forward nor the
reverse is favored.
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
Example 12.5 – page 332
Example 12.6 – Page 333

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.
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 – an equilibrium table is suggested.
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.
Effects of Concentration

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.
Effects of Pressure

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 away from the
side with the most moles.
Effects
of
Temperature
 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: (see page 338)

Example 12.8 – page 339
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