CHEMISTRY 12 - UNIT 3 : Dynamic Equilibrium Notes
I. Introduction
·
evidence shows that not all reactions go to
completion (ie. there are both products and -reactants present)
·
while studying PE diagrams, we have predicted
the reversible pathways of a chemical reaction
·
this makes sense if we assume that there is a
competition between collisions of reactants to form products and collisions of
products to reform reactants.
·
a closed
system is required for this to occur - ie. a system in which no substances
can enter or leave.
·
when the competition described is occurring, an
equilibrium may be set up.
Equilibrium
- A reversible reaction that has the rate
of the forward reaction equal to the rate of the reverse reaction.
There are 3 types of Equilibrium
i. phase equilibrium Go over 3-5 page 39 of
Hebden
·
if a beaker of water is left open it will
evaporate H20(l) «H20(g)
if it is in a closed system, evaporation is soon in equilibrium with
·
( the rate of evaporation is equal to the rate
of condensation)
ii. chemical reaction equilibrium (the topic of
this unit)
iii.
solubility equilibrium (will discuss in next unit)
Equilibrium
is described by:
·
constant macroscopic properties (visible changes)
·
rate of forward reaction = rate of reverse
reaction
·
can be achieved from either direction (see
overhead)
·
concentration of reactants and products is
constant
·
it is dynamic not static. (at the molecular
level there is a constant back-and-forth reaction between reactants and
products)
·
[reactants] differs from the [products]
·
[reactants] is constant in time and [products]
is constant in time
·
the forward and reverse rates do not change as
time passes
·
a system which is not at equilibrium will tend
to move toward a position of equilibrium.
Question 1-2 page 37
Question 6-7 page 40 of Hebden (need
graph paper - give the range for the scale - concentration from 0 - 1.6 M and
time from 0 -18 min)
(the
changes in rates of the forward and reverse reactions relates to the changing
concentrations of the reactants and products as equilibrium is established)
Questions 8-13 page 42 of Hebden
II. Enthalpy and Entropy - How to Predict
if a Chemical Reaction will Occur.
Spontaneous
Change - a change that occurs by itself, without outside assistance
Enthalpy
(H)
·
PE diagrams easily show us which side has the
lowest energy (or Enthalpy)
·
the side of the PE diagram having lowest
enthalpy is said to be "favoured" because molecules will tend to go
or remain at minimum energy. (the
exothermic direction)
·
BUT - not all exothermic reactions are spontaneous
and some endothermic ones are, there is another factor to consider.
Entropy
(S)
·
the amount of randomness or disorder of a system
·
highly random states are favoured because there
are MORE random states possible
·
only a perfect crystal at O K has an entropy of
zero (no motion of molecules)
·
the more disorder, the higher the entropy
·
If entropy is the only factor considered,
systems tend to favor a state of maximum entropy
SUMMARY:
systems tend toward a
position of minimum enthalpy and maximum randomness (entropy)
in
a system at equilibrium, there is a compromise between the tendency to achieve
a state of maximum disorder and a state of minimum energy.
RESULTS :
H = increases S = increases equilibrium reached
H = increases S = decreases reactants are favored (NR)
H = decreases S = increases products are favored (SP)
H = decreases S = decreases equilibrium is reached
Enthalpy / Entropy Questions
1. Zn(s)
+ 2HCI(aq) ®
ZnCl2(aq) + H2(g) AH = -152 KJ
· enthalpy decreases ® favor products
· entropy increases ® favor products
· forward reaction only
2. 3C(s)
+ 3H2(g) ®
C3H6(g) AH
20.4 KJ
· H
increases®
favor reactants
· S
decreases ®
favor reactants
· reverse reaction only
3. 2Pb(NO3)2(s)
+ 597 KJ ®
2PbO(s) + 4NO2(g) + O2(g)
· H increases ®
favor reactants
· S increases ®
favor products both occur
· get equilibrium
Do
questions 14-16 page 48 of Hebden
HINTS
FOR ENTROPY
1. gases
>> solutions > liquids >> solids
2. if
more than one phase is present :
·
the side having the most random phase is the
side with the most entropy
·
if both side contain particles having equally
random phases, then the side having the most number of particles has the most
entropy
Do Questions 14-16 page 48 of Hebden
III. Le
Chatelier's Principle
"When a stress is applied to a
system at equilibrium, the system readjusts so as to counteract the
stress"
·
This principle can be summed up "Whatever
we do, Nature tries to undo.
·
It allows us to quickly and easily predict the
effect that any change of conditions will have on equilibrium
·
the equilibrium eventually readjusts so that the
forward and reverse rates become equal again
These
3 factors may be varied in order to disturb a gaseous system at equilibrium
a) temperature
b) concentration
c) total pressure - brought about by volume
changes
the
addition of a catalyst speeds up both forward and reverse reactions and causes
the system to reach equilibrium in a shorter period of time. It does not
shift an equilibrium one way or the other.
a)
temperature (see pie graph overhead)
·
when the amount of heat energy is decreased the
reaction shifts so as to produce more heat and vice versa.
·
an increase in temperature favors endothermic
reactions, a decrease in temperature favors exothermic reactions
·
the energy term in a chemical equilibrium can be
treated as though it were a reactant or product
reaction kinetics explanation:
·
cooling a system such as this exothermic one
2SO2(g)
+ O2(1) «
2SO3(g) + 198 KJ
causes
both forward and reverse reactions rates to slow, but the reverse reaction rate
decreases more than the forward rate.
The shift causes concentration changes which will increase the reverse
rate and decrease the forward rate until they become equal again.
·
See graph page 51
b) concentration
·
increase concentration of reactant favors
forward reaction
·
decrease concentration of reactant favors
reverse reaction
·
increase concentration of products favors
reverse reaction
·
decrease concentration of products favors forward
reaction
reaction kinetics explanation:
2NO(g) + Cl2(g)
«
2NOCl(g) + 76 KJ
·
when reactant is added, more reactant particles
can have successful collisions so the forward rate increases (hence more
products are formed).
·
If the concentration of a reactant increases,
the rate of the forward reaction increases (rxn rate increases when
concentration increases.)
·
See graph page 52
c) pressure
·
an increase in pressure is like an increase in
concentration
·
increase in pressure favors the reaction that
produces the fewest number of particles per unit volume
·
decrease pressure favors the reaction that
produces the most number of particles per unit volume
reaction kinetics explanation
2SO2(g)
+ O2(g) «
2SO3(g) + 198KJ
·
imbalance of reaction rates occurs because both
forward and reverse reaction increase, but the forward one increases more
because there are more particles 'involved ' in the forward reaction (more
collisions)
·
while rates remain unequal, the observed result
is the production of more product.
· the
shift causes concentration changes which increase the reverse rate and decrease
the forward rate until they become equal.
·
See graph page 53
SUMMARY OF CHANGES THAT CAN BE MADE TO AN
EQUILIBRIUM
1) CHANGING THE TEMPERATURE :
· the
concentrations shown start to SLOWLY change to a new value
· there
are NO sudden changes in the graph.
2) CHANGING THE CONCENTRATION :
· the
concentration of the species being added SUDDENLY jumps up or down.
3) CHANGING THE PRESSURE :
· the
concentration of ALL SPECIES simultaneously jump up or down
4) ADDING A CATALYST
· does
not affect graph
· it
just SPEEDS up the REACTION RATE
Do questions 17 - 28 pages 54 - 55
· odd numbers in class
· even numbers for homework
IV.
Industrial Application -
· industry
strives to design processes where reactants are added continuously and products
are continuously removed, so that an equilibrium is never allowed to establish.
The
Haber Process for the Production of NH3
N2(g) + 3H2(g) « 2NH3(g)+
92.4KJ
high
P - more
product
low
T -
more product
could
only get 30% reactants consumed
*
Haber: pull ammonia out of the system by liquefying it while keeping hydrogen
and nitrogen gaseous.
*
Won 1918 Nobel prize in chemistry.
*
Important because nearly all Nitrogen compounds used are made from NH3
Do questions 29 - 30 page 56 in Hebden
V. The Equilibrium Constant Keq
· empirical evidence reveals a mathematical
relationship that provides a constant value for a chemical system over a range
of concentrations it is called the equilibrium constant Keq
· the equilibrium law states :
for
the reaction
aA + bB « cC + dD
the
equilibrium law is Keq = [C]c[D]d
[A]a [B]b
where: A,B,C,D represent chemical entities and
a,b,c,d
represent their coefficients in the balanced chemical equation
Keq = [products]
[reactants]
The
following types of substances are NOT included in the equilibrium expression
because they have constant concentrations :
· Solids
· Pure
Liquids - a liquid that exists on either side of the reaction.
example
: Write the equilibrium law for the reaction of nitrogen monoxide gas with
oxygen gas to form nitrogen dioxide gas
DO EXAMPLES page 59 of HEBDEN
Do Questions 31-35 page 60 of HEBDEN
THINGS TO REMEMBER ABOUT Keq :
1. the
higher the value of K, the greater the tendency of the system to favor the
forward direction (the more products formed)
2. K
is dependent on temperature
(a temperature must always be specified)
· an
increase in temperature will increase Keq if the reaction is
endothermic
· an
increase in temperature will decrease Keq if the reaction is
exothermic
3. K
is only dependent on very large changes in concentration, it is not affected by
small or moderate changes in concentration of reactants or products
4. K
is not affected by pressure
or volume of gases, surface area, or the addition of catalysts
5. K
provides only info on the position of equilibrium, none on reaction rate!
6. solids
and liquids have a constant concentration value, they become part of the Keq
constant - ie. they are not shown in the equilibrium expression
7. since
gases and dissolved substances have variable concentrations they are always
shown in the equilibrium expression.
8. ions
in solution, must be represented as individual entities (always use the net
ionic form of reactions)
Do
Questions 36-45 page 62 Hebden
The
Size of Keq
Keq =
[products]
[reactants]
· A
LARGE Keq indicates a LARGE amount of products is present at
equilibrium
· A
SMALL value for Keq implies that a SMALL amount of products is
present at equilibrium
Example: write an equilibrium expression for the decomposition of
solid ammonium chloride to gaseous ammonia and gaseous hydrogen chloride.
NH4CI(l) « NH3(g) + HCI(g)
Keq = [NH3(g)][HCI(g)]
Example : write an
equilibrium expression for the reaction of zinc in copper(II)chloride solution
Zn(s) + Cu 2+(aq)
+ 2CI-(aq) «
Cu(s) + Zn2+(aq) + 2CI-(aq)
take
out spectator ions
Zn(s)
+ Cu 2+(aq) « Cu(s)
+ Zn2+(aq)
Keq =
[Zn 2+(aq)]
[Cu2+(aq)]
***********
Quiz
***********
VII. Quantitative Changes in Equilibrium Systems
A. Calculate the value of Keq given
the equilibrium concentration of all species
example
: C02(g) + H2(g)
n
CO(g) + H20(g)
at
equilibrium, [CO21 = 0.648 mol/L, [H2] = 0. 148mol/L,
[CO] = 0.352 mol/L, [H20] = 0.352 mol/L. Find the value of Keq
Keq
= [CO][H2O] = (0.352)(0.352) =
1.29
[CO2][H2] (0.648)(0.148)
B. Calculate the value of Keq given
the initial concentrations of all species and one equilibrium concentration
example: Into a l.00 L container is placed 0.500 mol of
SO2(g) and 0.500 mol of O2(g). The following reaction
2SO2(g)
+ 02(g) n 2SO3(g)
reaches
equilibrium. At equilibrium [SO3(g)]
= 0.150 M,
Calculate
Keq
Solution
*
use ice table
*
record initial conc of SO2 = O2 = 0.500 M assume
[SO3] = 0
* record equilibrium concentration of SO3
= 0.150 mol/L
* calculate change in concentration of S03
= 0.150 M
* refer to the molar relationship given by the
coefficients in the equation to determine the change in concentration of SO2
and O2. DS02
= 0.150 M, DO2 =
0.075 M
* Calculate equilm concentration of
S02 and O2:
[SO2] = 0.350 M, [O21
= 0.425 M
|
Concentration |
[SO2] |
[O2] |
[SO3] |
Initial |
0.500 |
0.500 |
0.00 |
|
Change |
-2/2(0.150) |
-1/2(0.150) |
+ 0.150 |
|
Equilibrium |
0.350 |
0.425 |
0.150 |
write the equilibrium expression
substitute
values and solve for Keq
all
concentrations in mol/L
Keq
= [SO3]2 = (0.150)2 =
0.432
[SO2]2[02] (0.350)-(0.425)
C. Calculate the equilibrium concentration of
all species given the value of Keq and the initial concentrations
example
: CO(g) + H2O(g) n
CO2(g) + H2(g)
Keq = 0.80
If
4.0 mol CO(g) and 4.0 mol H2O(g) are placed into a 2.0 L container, find the
equilibrium concentration of all species.
Solution
:
in
a problem where Keq is given, you will probably use 'x' to represent
some unknown quantity
· write
the equation
· construct
an 'Ice' table
· calculate
the initial of all materials
CO = 2.0 M, H20 = 2.0 M, H2 = 0.0 M, CO2
= 0.0
· let
x be the equilibrium concentration of H2
· determine
the change in concentration of H2 = (+x)
· determine
the change in concentration of all other species based on change in H2
: CO = -x, H2O = -x, CO2 = +x
· determine
equilibrium concentration of all species in terms of x: CO = H20 =
(2.0 - x), H2 = CO2 = x
Concentration
|
[CO] |
[H2O] |
[H2] |
[CO2] |
Initial
|
2.0 |
2.0 |
0.0 |
0.0 |
|
Change |
-x |
-x |
+x |
+x |
|
Equilibrium |
2.0-x |
2.0-x |
x |
x |
· write
equilibrium expression
· put
values into expression and solve for x
Keq =
[CO2] [H2] =
(x)(x) = 0.80
[CO] [H20] (2.0-x)(2.0-x)
(square root each side to make it easier to solve)
x = 0.89
(2.0-x)
(cross
multiply to solve for x)
x = 0.89(2.0-x) = 1.78 - 0.98x
1.87
x = 1.78
x = 0.95 mol/L
[CO]
= [H20]= 2.0 - 0.95 = 1. I M
[H2]
= [CO2]= 0.95 M
Note
- if the initial reactant concentrations are not identical, you must use the
quadratic formula to solve.
D. Determine whether a system is at equilibrium, and if not, in which direction it will shift to reach equilibrium when given a set of concentrations for reactants and products.
Because
the size of Keq varies with the extent (position) of the
equilibrium, if the concentrations of the substances in any closed system are
known, we can find out if it is at equilibrium, if not, we can determine in
what direction the system will shift to reach equilibrium.
We
produce a trial value (reaction quotient, ktrial – sometimes called
Q) by substituting in the concentrations
· if
ktrial is equal to Keq, the system is at equilibrium
· if
ktrial is greater than Keq, the system is shifting to the
left to reach equilibrium because the product to reactant ratio is too high
· if
ktrial is less than Keq, the system is shifting to the
right to reach equilibrium, because the product to reactant ratio is too low.
Example
: if the concentration of the N2= 0.82 M, H2 = 0.74 M,
and NH3= 1.7 M, is the reaction at equilibrium and if not, in which
direction will it proceed to establish equilibrium?
N2(g)
+ 3H2(g) n
2NH3(g) Keq
= 5.3
· write
expression
· substitute
and determine trial Keq
· if value is greater than Keq, rxn shifts to form reactants, if less, it shifts to form more products.
Ktrial =
[NH3]2
= (1.7)2
= 9.0
[N2 ][H2]3 (0.82)(0.74)3
since
9.0 > 5.3 the reaction is not at equilibrium and it proceeds to the left to
form more reactants.