# lec09 ```CHEMICAL EQUILIBRIUM
Chapter 16
• equilibrium vs. completed reactions
• equilibrium constant expressions
• Reaction quotient
• computing positions of equilibria: examples
• Le Chatelier’s principle - effect on equilibria of:
• addition of reactant or product
• pressure
• temperature
YOU ARE NOT RESPONSIBLE for section 16.7 (relation to kinetics)
3 Nov 97
Entropy &amp; Free Energy (Ch 20)
1
THE EQUILIBRIUM CONSTANT
For any type of chemical equilibrium of the type
aA + bB
cC + dD
the following is a CONSTANT (at a given T) :
conc. of products
K =
[C]c [D]d
[A]a [B]b
equilibrium constant
conc. of reactants
If K is known, then we can predict
concentrations of products or reactants.
3 Nov 97
Entropy &amp; Free Energy (Ch 20)
2
Q - the reaction quotient
All reacting chemical systems can be characterized by
their REACTION QUOTIENT, Q.
Q has the same form as K,
. . . but uses existing concentrations
If Q = K, then system is at equilibrium.
0.35
n-Butane
iso-Butane Q = [iso] =
= 1.40
0.25
0.35
[n]
0.25
Since K =2.5, system NOT AT EQUIL.
To reach EQUILIBRIUM
[Iso] must INCREASE and [n] must DECREASE.
3 Nov 97
Entropy &amp; Free Energy (Ch 20)
3
Typical EQUILIBRIUM Calculations
2 general types:
a. Given set of concentrations, Calculate Q
is system at equilibrium ? compare to K
IF:
Q &gt; K or Q/K &gt; 1
 REACTANTS
Q/K
1
Q &lt; K or Q/K &lt; 1
 PRODUCTS
Q
Q=K
3 Nov 97
Q=K
Entropy &amp; Free Energy (Ch 20)
at EQUILIBRIUM
4
H2(g) + I2(g)
2 HI(g)
Kc = 55.3
Step 1 Define equilibrium condition in terms of initial
condition and a change variable
[H2]
[I2]
[HI]
At equilibrium
1.00-x
1.00-x
2x
[2x]2
= 55.3
Put equilibrium [ ] K c =
[1.00 - x][1.00 - x]
Step 2
into Kc .
Step 3. Solve for x.
55.3 = (2x)2/(1-x)2
Square root of both sides &amp; solve gives: x = 0.79
At equilibrium
3 Nov 97
[H2] = [I2] = 1.00 - x = 0.21 M
[HI] = 2x = 1.58 M
Entropy &amp; Free Energy (Ch 20)
5
EQUILIBRIUM AND EXTERNAL EFFECTS
• The position of equilibrium is changed
when there is a change in:
– pressure
– changes in concentration
– temperature
• The outcome is governed by
LE CHATELIER’S PRINCIPLE
“...if a system at equilibrium is disturbed, the
system tends to shift its equilibrium position
to counter the effect of the disturbance.”
3 Nov 97
Entropy &amp; Free Energy (Ch 20)
Henri Le Chatelier
1850-1936
- Studied mining
engineering
- specialized in glass
and ceramics.
6
Shifts in EQUILIBRIUM : Concentration
• If concentration of one species changes,
concentrations of other species CHANGES
to keep the value of K the same (at constant T)
• no change in K - only position of equilibrium changes.
- equilibrium shifts to REACTANTS
- equilibrium shifts to PRODUCTS
- GAS-FORMING; PRECIPITATION
REMOVING PRODUCTS
- often used to DRIVE REACTION TO COMPLETION
3 Nov 97
Entropy &amp; Free Energy (Ch 20)
7
Effect of changed [ ] on an equilibrium
K =
n-Butane
Isobutane
[iso]
[n]
= 2.5
INITIALLY: [n] = 0.50 M
[iso] = 1.25 M
What happens ?
Solution
A. Calculate Q with extra 1.50 M n-butane.
16_butane.mov
Q = [iso] / [n] = 1.25 / (0.50 + 1.50) = 0.63(16m13an1.mov)
Q &lt; K . Therefore, reaction shifts to PRODUCT
3 Nov 97
Entropy &amp; Free Energy (Ch 20)
8
Butane/Isobutane
Solution
B. Solve for NEW EQUILIBRIUM
- set up concentration table
[n-butane]
[isobutane]
Initial
0.50 + 1.50
1.25
Change
-x
+x
Equilibrium 2.00 - x
1.25 + x
B
A
[isobutane]
1.25 + x
K = 2.50 =

[butane]
2.00 - x
x = 1.07 M. At new equilibrium position,
[n-butane] = 0.93 M [isobutane] = 2.32 M.
Equilibrium has shifted toward isobutane.
3 Nov 97
Entropy &amp; Free Energy (Ch 20)
9
Effect of Pressure (gas equilibrium)
N2O4(g)
2 NO2(g)
[NO2 ]2
Kc =
= 0.0059 at 298 K
[N2O4 ]
Increase P in the system
by reducing the volume.
Increasing P shifts equilibrium to side
with fewer molecules (to try to reduce P).
Here, reaction shifts LEFT
PN2O4 increases
16_NO2.mov
(16m14an1.mov)
PNO2 decreases
See Ass#2 - question #6
3 Nov 97
Entropy &amp; Free Energy (Ch 20)
10
EQUILIBRIUM AND EXTERNAL EFFECTS
• Temperature change
 change in K
• Consider the fizz in a soft drink
CO2(g) + H2O(liq)
CO2(aq) + heat
LOWER T
Kc = [CO2(aq)]/[CO2(g)]
HIGHER T
• Change T: New equilib. position? New value of K?
• Increase T
Equilibrium shifts left: [CO2(g)]  [CO2 (aq)] 
K decreases as T goes up.
•Decrease T
[CO2 (aq)] increases and [CO2(g)] decreases.
K increases as T goes down
3 Nov 97
Entropy &amp; Free Energy (Ch 20)
11
Temperature Effects on
Chemical Equilibrium
N2O4 + heat
(colorless)
Kc 
[NO2 ]2
[N2O 4 ]
2 NO2
(brown)
Kc = 0.00077 at 273 K
Kc = 0.00590 at 298 K
Horxn = + 57.2 kJ
Increasing T changes K so as to
shift equilibrium in
ENDOTHERMIC direction
3 Nov 97
Entropy &amp; Free Energy (Ch 20)
16_NO2RX.mov
(16m14an1.mov)
12
EQUILIBRIUM AND EXTERNAL EFFECTS
Catalytic exhaust system
no change in K
• A catalyst only affects the RATE of
approach to equilibrium.
3 Nov 97
Entropy &amp; Free Energy (Ch 20)
13
CHEMICAL EQUILIBRIUM
Chapter 16
• equilibrium vs. completed reactions
• equilibrium constant expressions
• Reaction quotient
• computing positions of equilibria: examples
• Le Chatelier’s principle - effect on equilibria of:
• addition of reactant or product
• pressure
• temperature
YOU ARE NOT RESPONSIBLE for section 16.7 (relation to kinetics)
3 Nov 97
Entropy &amp; Free Energy (Ch 20)
14
Entropy and Free Energy
(Kotz Ch 20)
• Spontaneous vs. non-spontaneous
• thermodynamics vs. kinetics
• entropy = randomness (So)
• Gibbs free energy (Go)
• Go for reactions - predicting spontaneous direction
• Grxn versus Gorxn
• predicting equilibrium constants from Gorxn
3 Nov 97
Entropy &amp; Free Energy (Ch 20)
15
Entropy and Free Energy
( Kotz Ch 20 )
• some processes are spontaneous;
others never occur. WHY ?
• How can we predict if a reaction
can occur, given enough time?
THERMODYNAMICS
9-paper.mov
20m02vd1.mov
• Note: Thermodynamics DOES NOT say how
quickly (or slowly) a reaction will occur.
• To predict if a reaction can occur at a
reasonable rate, one needs to consider: KINETICS
3 Nov 97
Entropy &amp; Free Energy (Ch 20)
16
Thermodynamics
• state of a chemical system (P, T, composition)
• From a given state, would a chemical
reaction decrease the energy of the system?
• If yes, system is favored to react — a
product-favored system which will have
a spontaneous reaction.
• Most product-favored reactions are
exothermic.
• Spontaneous does not imply anything about
time for reaction to occur. (kinetics)
3 Nov 97
Entropy &amp; Free Energy (Ch 20)
17
Thermodynamics versus Kinetics
• Diamond to Graphite
– spontaneous from thermodynamics
– but not kinetically favored.
• Paper burns.
- product - favored reaction.
- Also kinetically favored once
reaction is begun.
3 Nov 97
Entropy &amp; Free Energy (Ch 20)
18
Product-Favored Reactions
In general, productfavored reactions
are exothermic.
E.g. thermite reaction
Fe2O3(s) + 2 Al(s)
 2 Fe(s) + Al2O3(s)
H = - 848 kJ
3 Nov 97
Entropy &amp; Free Energy (Ch 20)
19
Non-exothermic spontaneous reactions
But many spontaneous reactions or
processes are endothermic . . .
NH4NO3(s) + heat  NH4+ (aq) + NO3- (aq)
Hsol = +25.7 kJ/mol
or have H = 0 . . .
3 Nov 97
Entropy &amp; Free Energy (Ch 20)
20
Entropy, S
One property common to
product-favored processes
is that the final state is
more DISORDERED or
RANDOM than the original.
Spontaneity is related to
an increase in
randomness.
The thermodynamic property
related to randomness is
ENTROPY, S.
3 Nov 97
Entropy &amp; Free Energy (Ch 20)
Reaction of K
with water
21
PROBABILITY - predictor of most stable state
WHY DO PROCESSES with  H = 0 occur ?
Consider expansion of gases to equal pressure:
This is spontaneous because the final state,
with equal # molecules in each flask,
is much more probable than the initial state,
SYSTEM CHANGES to state of HIGHER PROBABILITY
THIS IS USUALLY the more RANDOM state.
3 Nov 97
Entropy &amp; Free Energy (Ch 20)
22
Gas expansion - spontaneity from greater probability
Consider distribution of 4 molecules in 2 flasks
P1 &lt; P2
P1 = P2
P1 &gt; P2
With more molecules (&gt;1020) P1=P2 is most probable by far
3 Nov 97
Entropy &amp; Free Energy (Ch 20)
23
Consider
2 H2 (g) + O2 (g)  2 H2O (l)
• HIGHLY EXOTHERMIC
• final state (liquid) is MUCH MORE ORDERED
(less random arrangement) than initial state (2 gases)
• BUT PROBABILITY of final state is higher when one
considers change in the surroundings. WHY ?
• Heat evolved increases motion of molecules in the
surroundings
Increased disorder of
surroundings
&gt;
decreased disorder of
system
MUST consider change in disorder in
BOTH SYSTEM and SURROUNDINGS
to predict DIRECTION of SPONTANEITY
3 Nov 97
Entropy &amp; Free Energy (Ch 20)
24
Directionality of Reactions
How probable is it that reactant
molecules will react?
PROBABILITY suggests that a
product-favored reaction will
result in the dispersal of energy
or dispersal of matter or both.
3 Nov 97
Entropy &amp; Free Energy (Ch 20)
25
Directionality of Reactions
Probability suggests that a product-favored
reaction will result in the dispersal of
energy or of matter or both.
Matter Dispersal
Energy Dispersal
9_gasmix.mov
20m03an1.mov
9_exorxn.mov
20m03an2.mov
3 Nov 97
Entropy &amp; Free Energy (Ch 20)
26
Directionality of Reactions
Energy Dispersal
• Exothermic reactions involve a release of
stored chemical potential energy to the
surroundings.
• The stored potential energy starts out in a
few molecules but is finally dispersed over
a great many molecules.
• The final state—with energy dispersed—is
more probable and makes a reaction
product-favored.
3 Nov 97
Entropy &amp; Free Energy (Ch 20)
27
Standard Entropies, So
• Every substance at a given temperature and in
a specific phase has a well-defined Entropy
• At 298o the entropy of a substance is called
So - with UNITS of J.K-1.mol-1
• The larger the value of So, the greater the
degree of disorder or randomness
e.g. So (in J.K-1mol-1) :
Br2 (liq) = 152.2
Br2 (gas) = 245.5
For any process:
So =  So(final) -  So(initial)
So(vap., Br2) = (245.5-152.2) = 93.3 J.K-1mol-1
3 Nov 97
Entropy &amp; Free Energy (Ch 20)
28
Entropy, S
Vapour
So (J/K•mol)
H2O(gas) 188.8
Ice
Water
H2O(liq)
69.9
H2O (s)
47.9
S (gases) &gt; S (liquids) &gt; S (solids)
3 Nov 97
Entropy &amp; Free Energy (Ch 20)
29
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