Download Chemical Equilibrium Stress? What stress? 1

Chemical Equilibrium
Equilibrium Expression
• Chemist’s can express the equilibrium
position in terms of a numerical
constant.
 The equilibrium constant shows the
relationship between the amount of
product and reactant at equilibrium.
• Consider this hypothetical reaction:
aA + bB
Equilibrium Expression
• We can write an expression to show the
ratio of product concentrations to
reactant concentrations called a mass
action expression:
[C]c [D]d
[A]a [B]b
•
•
The concentration of each substance is
raised to a power equal to the # of mols
of that substance in the balanced
equation.
The square brackets indicate
concentration in Molarity (mol/L).
cC + dD
Equilibrium Expression
o
The resulting ratio of the equilibrium is
called the equilibrium constant or Keq.
• When the reactants and products are
in molarity the constant is called a Kc.
• When the reactants and
[C]c [D]d
products amounts are in
pressure units is called Keq=
[A]a [B]b
a Kp.
o The constant is dependent on the
temperature.
• If the temperature changes so does the
constant.
NOTE: water and solid materials are not
included in mass action expressions.
Water, a special case…
2 H2O (l)  2 H2 (g) + O2 (g)
 The concentration of a pure
liquid cannot change, it is
fixed and equal to the liquid’s
K = [H2]2 [O2]
density.
[H2O]2
 We know that K remains
constant for all combinations
of reactant and product
K[H2O]2 = [H2]2 [O2] = K
concentrations at equilibrium.
 Therefore, since K is
constant and the
concentration of water is
constant, we can combine
these two values.
Stress? What stress?
In general, the concentrations of
entities in a condensed state are not
included as variables in the
equilibrium law expression; they are
incorporated into the constant K.
1
Chemical Equilibrium
For example:
1. Write the mass action expression for
each of the following reactions:
A.2SO2(g) + O2(g) <==> 2SO3(g)
B. Bi2S3(s) <==> 2Bi3+ (aq) + 3S2- (aq)
Class work:
Write the mass action expression for each
of the following reactions:
2C4H10(g) + 13O2(g) <==> 8CO2(g) +10H2O(g)
4Al(s) + 3O2(g) <==> 2Al2O3(s)
Equilibrium Constant
 Equilibrium constants provide a measure
of the extent to which a reaction has
gone to completion when equilibrium is
reached.
 K gives no information about the rate of
a reaction.
 They show whether the products or the
reactants are favoured in a reaction.
A value of:
 Keq > 1 means that products are favored.
 Keq < 1 means that reactants are favored.
 Keq = 1 means [product] is approximately
equal to [reactant].
K values for formation and decomposition
are reciprocals of each other
Forward reaction:
Reverse reaction:
N2 (g) + 3 H2 (g)  2 NH3 (g)
2 NH3 (g)  N2 (g) + 3 H2 (g)
K = _[NH3]2 = 6.49 x 10-2
[N2] [H2]3
K΄ = [N2] [H2]3 = 15.4
[NH3]2
Therefore: K = 1 = 1
= 6.49 x 10-2
K΄ 15.4
Stress? What stress?
o
Molarity is a measure of how much
“stuff” is dissolved in water.
̵ The more stuff dissolved,
the more concentrated the solution.
̵ The higher the Molarity.
2
Chemical Equilibrium
Class work:
1. What is the molarity of a solution
that was made by dissolving 0.100
mols of sugar in 100 ml of water?
2. How many moles of salt are
contained in 300.0 mL of a 0.40 M
NaCl solution?
3. A chemist dissolves 98.4 g of FeSO4
in enough water to make 2.00 L of
solution. What is the molarity of the
solution?
Sample Problem 1
Dinitrogen tetroxide (N2O4), a colorless
gas, and nitrogen dioxide (NO2), a brown
gas, exist in equilibrium with each other
according to the following equation:
N2O4(g)
Analyze: list what we know
 Known:
 [N2O4] = .0045 mol/1.0 L
 [NO2] = .030 mol/1.0 L
 Unknown:
 Mass action expression = ?
 Kc = ?
o
At equilibrium, there is no net
change in the amount of N2O4 or NO2
at any given instant.
Class work:
1. Find the equilibrium constant if [SO2] =
1.0 M; [O2]=1.0 M; [SO3]=2.0 M;
2SO2(g) + O2(g) <==> 2SO3(g)
2. Find the equilibrium constant if [Bi3+] =
0.00058 M; [S2-] = 0.00087 M
Bi2S3(s) <==> 2Bi3+ (aq) + 3S2- (aq)
Stress? What stress?
2NO2(g)
A 1.0 liter of gas mixture at 10C at
equilibrium contains 0.0045 mol N2O4
& 0.030 mol NO2. Write the mass action
expression and calculate Kc for the
reaction.
Calculate: solve for unknowns
 The only product of the reaction is NO2,
which has a coefficient of 2 in the balanced
equation.
 The only reactant N2O4 has a coefficient of
1 in the balanced equation.
 The mass action expression is:
Kc =
o
o
[NO2]2
Kc =
[.030M]2
[.0045M]1
[N2O4]1
Kc is equal to: Kc = 0.20
Kc < 1, therefore reaction does not
favor products
Reaction Quotient Q
 We can also determine if a reaction has
reached equilibrium by calculating a
reaction quotient (Q).
 It’s like taking a snapshot of a reaction
at a given time and interpreting how
far along the reaction is.
 Once the reaction quotient is solved, it is
compared to the equilibrium constant
 The following picture helps us decide
how to interpret the direction the
reaction will continue.
3
Chemical Equilibrium
Sample Problem
At a certain temperature Keq = 55 and a
reaction vessel contains a mixture with
the following concentrations :
[SO3] = 0.85 M, [NO] = 1.2 M ,
[SO2] = 1.5 M and [NO2] = 2.0 M.
SO3(g) + NO(g)
SO2(g) + NO2(g)
Is the reaction at equilibrium and if
not which direction will the reaction
proceed?
 Solve just as if you were solving for
the equilibrium constant.
 Then analyze the resulting quotient
with the given Keq.
o
o
o
Q = 2.94 which is < Keq (55)
The reaction is not at equilibrium
If Q < Keq then the numerator
of our quotient must increase
•
Therefore the reaction continues in
order to increase [products] until it
reaches Keq.
Class work:
1. 0.035 moles of SO2, 0.500 moles of
SO2Cl2, and 0.080 moles of Cl2 are
combined in an evacuated 5.00 L
flask and heated to 100°C.
a) What is Q before the reaction
begins?
b) Which direction will the reaction
proceed in order to establish
equilibrium?
SO2Cl2(g) <==> SO2(g) + Cl2(g)
Kc = 0.078 at 100°C
Manipulating the Equilibrium…
 There is a principle that can be studied
to govern changes in equilibrium (Le
Chatelier’s Principle).
 Le Chatelier’s Principle states:
 “If a stress is applied to a system
in dynamic equilibrium, the system
changes to relieve the stress.”
 Stresses are changes in temperature,
pressure, concentration of reactants,
or concentration of products.
Stress? What stress?
4
Chemical Equilibrium
Pressure & Equilibrium
 If A, B, and C are all gases, then the
equilibrium they establish is pressure
dependent.
 When the pressure is increased, the
system relieves the pressure by favoring
the direction that produces fewer gas
molecules.
 Pressure is # of particles dependent, the
more particles the higher the pressure.
 Fewer gas molecules will exert less pressure.
 So, more product is formed, which
overall reduces the pressure, this is a
shift right.
Pressure & Equilibrium
 Conversely, a decrease in pressure will
favor the reaction that produces the
most molecules.
Class work
Predict the effect of the following changes
on the reaction in which SO3 decomposes to
form SO2 and O2.
2SO3(g) <=> 2SO2 (g) + O2 (g)
Ho = 197.78 kJ
(a) Increasing the temperature of the
reaction.
(b) Increasing the pressure on the reaction.
(c) Adding more O2 when the reaction is at
equilibrium.
(d) Removing O2 from the system when the
reaction is at equilibrium.
Stress? What stress?
5