Transcript Equilibrium - Warren County Public Schools

Equilibrium
Chapter 15
Equilibrium
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Have you ever tried to
maintain your balance
as you walked across
a narrow ledge?
In a chemical reaction
balance or equilibrium
is also maintained.
You can think of the
yields  sign as the
ledge
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Equilibrium systems
exist in ocean water,
blood, urine, and
many other biological
systems
Chemical reactions for
the most part are
reversible
You can think of the
yield sign  as the
ledge in a chem
system
I. Chemical Equilibrium Concept
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Chemical Equilibrium occurs when opposing
reactions are proceeding at equal rates
At equil rate forward = rate reverse
AB
[] indicates a molar conc.
Fr AB rate kf [A] kf [A] = kr [B]
Rr BA rate kr [B]
[B] = kf cons= Kc
Rearranging formula
[A] kr
A dynamic equilibrium
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Individual molecules are
undergoing change but there is
no net exchange in the
concentration of reactants and
products
Does not mean that the
concentrations are not
changing just that the ratio
equals a definite value
Look at Habber reaction figure
15.6 text.
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Dihydrogen monoxide:
is also known as hydric acid,
and is the major component of
acid rain.
contributes to the "greenhouse
effect."
may cause severe burns.
contributes to the erosion of
our natural landscape.
accelerates corrosion and
rusting of many metals.
may cause electrical failures
and decreased effectiveness of
automobile brakes.
has been found in excised
tumors of terminal cancer
patients.
Law of Mass Action
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Equilibrium can be reached from either direction
Concentrations of reactants and products are
expressed as
aA + bB  pP + qQ
equil constant Kc = [P]p [Q]q
products
[A]a [Q]b
reactants
Equilibrium Constant mass action
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Kc depends on the stoichiometry not on
mechanics
Doesn’t depend on the initial conc of
reactants and products
Doesn’t depend on other added sub as
long as they do not react
Varies with tempt
Catalyst do not effect just speeds –
reaching eq.
Putting it Together
Question?
Main Ideas Details
Monitor
Writing equilibrium expressions
2O3 (g)  3O2 (g)
 2NO (g) + Cl2 (g)  2NOCl (aq)
 AgCl (s)  Ag+ (aq) + Cl- (aq)
Kc = [O2]3
Kc = [NOCl]2
[O3]2
[NO]2 [Cl2]
Kc = [Ag+] [Cl-]
Pure solids and liquids do not effect the
equilibrium because their conc remain
unchanged
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Eq expressed as pressure
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C3H8(g) + O2(g)  CO2(g) + H2O(l)
Kp = CO2p3
C3H8p O2p5
P = partial pressure of the gas
Kp = kc(RT)delta n
What does kc tell you?
Ex CO (g) + Cl2 (g)  COCl2 (g)
 Kc = [COCl2]
= 4.57 X 10^9
[CO] [Cl2]
Kc>>>1 larger numerator reaction goes
almost totally to products – eq lies right –
favors products
Kc<<<1 larger denominator
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Le Chatelier’s Principle
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If a system at eq is
disturbed by a change
in temperature,
pressure, or the
concentration of one
of the components,
the system will shift
it’s eq pos so as to
counter the effects of
the distrubance
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Henri-Louis Le
Chatelier (1858-1936)
Change in Concentration
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Le Châtelier's principle states that if the
concentration of one of the components of the
reaction (either product or reactant) is changed,
the system will respond in such a way as to
counteract the effect
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If a substance (either reactant or product) is
removed from a system, the equilibrium will shift
so as to produce more of that component (and
once again achieve equilibrium)
Change in concentration
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If a substance (either reactant or product)
is added to a system, the equilibrium will
shift so as to consume more of that
component (and once again achieve
equilibrium)
Putting it Together
N2(g) + 3H2(g) <=> 2NH3(g)
Question?
How would
inceasing H2
change eq
Predict the
relative
conc of
each
reactant
and product
Main Ideas
Details
Monitor
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The reaction is driven "to
the right" by the effects
of added H2
The eq conc’s will not be
identical to the original
state. However, Kc will
be the same. The new
equilibrium state contains
a slightly higher
concentration of NH3(g),
and slightly lower
concentration of N2(g) (as
well as a slightly higher
concentration of H2(g).
Change in Volume and Pressure
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A chemical system in equilibrium can respond to
the effects of pressure also. According to Le
Châtelier's Rule, if the pressure is increased on a
system, it will respond by trying to reduce the
pressure. How does it do this?
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We are primarily concerned with homogeneous
gaseous reactions
The stoichiometry of the reaction may lead to a
greater number of molecules on one side of the
equation.
For example, in the Haber reaction, N2(g) +
3H2(g) <=> 2NH3(g) there are twice as many
moles of reactants as products
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If the Haber reaction were in equilibrium,
and the pressure was increased, the
reaction would respond to oppose the
increase in pressure. It could accomplish
this by shifting the equilibrium to the right
(producing NH3(g))
This would reduce the overall number of
moles in the reaction, and therefore, lower
the pressure
Systems shift to the side with the fewest
number of moles if both are the same
then no change in eq con’s occurs
Changes in Temperature
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The intrinsic value of K does not change when
we increase concentrations or pressures of
components in a reaction. However, almost
every equilibrium constant (K) changes in
response to changes in temperature.
We will consider reaction conditions under which
no work is done, and therefore all energy
changes associated with reactions will be
manifested by temperature changes)
Temperature Changes
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Exothermic reactions are associated with
heat release when the reaction proceeds
in the forward direction
Endothermic reactions are associated with
heat release when the reaction proceeds
in the reverse direction (i.e. heat is
absorbed in the forward direction)
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These two types of reactions and their
associated heat changes can be written as:
Exothermic: Reactants yield Products + Heat
Endothermic: Reactants + Heat yield Products
If temperature is increased, the equilibrium will
shift so as to minimize the effect of the added
heat
The reaction will shift in the appropriate
direction such that the added heat is
absorbed
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When heat is added to exothermic
reactions at equilibrium, products will be
consumed to produce reactants (shift to
the LEFT) May also be written delta t is
negative.
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When heat is added to endothermic
reactions at equilibrium, reactants will be
consumed to produce products (shift to
the RIGHT) May also be written delta t is
positive.
Based on this behavior, what is the
effect of T upon K?
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Assume K = 1.0 for an exothermic reaction at
equilibrium.
Added heat causes the reaction to shift to
the left. Reactants <= Products + Heat
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Thus, 1.0 must represent a reaction quotient, Q,
that is too large in comparison to the new value
of K.
Thus, the effect of increasing temperature on
an exothermic reaction is to lower the value of
K.
Conversely, the effect of increasing
temperature on an endothermic reaction is to
increase the value of K
Putting it Together
Calc Delta H of formation for
C3H8(g) + O2(g)  CO2(g) + H2O(l)
Question? Main Ideas Details
Is the value
Exo,or
endotherm
How would
inc. temp
effect eq
How would
dec temp
eff eq k
Monitor
Calculations with eq K
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Example calculating unknown
concentrations using the eq constant
CO(g) + 3H2(g) <-> CH4(g) + H20(g)
At eq 0.3 mol of CO, 0.1 mol H2 and 0.02
mol of H20 are in 1.0 liter of a vessel at
1200 k kC is 3.92 what is the conc of
CH4?
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Kc = [CH4] [H2O]
[CO] [H2]3
3.93 = [CH4] (.020)
(0.30) (0.10)3
[CH4] = (0.30)(0.10)3 3.93
(0.020)
0.059 mol/l
Learning Check
PCl5(g)<-> PCl3(g) + Cl2(g)
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A l.0 liter vessel has a unknown amount of
PCl5 at eq Kc at 250 0C is 0.0415. Calc the
unknow conc. if 0.02moles of PCl3 and Cl2
are in the container. (0.0096)
Solving linear eq equasions
CO(g) + H20(g) <-> C02(g) + H2(g)
Given 1.0 mol of CO2 and H20 in a 50.0 l
vessel. How many moles are in an eq mix
at 1000 oC Ec = 0.58 at 1000oC
CO(g) + H20(g) <-> C02(g) + H2(g
I 0.02
0.02
0
0
C
-x
-x
+x
+x
E 0.02-x 0.02-x
x
x
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0.58 = [CO2][H2]
[CO] [H2O]
X2
(0.02-X)2
0.76 = X2
(0.02-X)
(the neg one gives a neg answer x can’t be neg)
+0.76(0.02-X)=X
0.0152-0.76X=X
0.0152= 1.76X
X= 0.0086
+,-
_
=
H2(g) + I2(g) <-> 2HI(g)
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What is the eq comp of a reaction mixture
starting with 0.500 mol each of H2 and I2
in a 1 l vessel? Kc = 49.7 at 458 oC (H2
&I2 = 0.11 mol/l HI = 0.78 mol/l
Equil with quadratic expressions
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Calc the conc. of the previous problem
with 1.00 molar H2 and 2.00 molar I2 as
the starting concentrations.
H2(g) +
I 1.00
C -x
E 1.00-x
I2(g) <-> 2HI(g)
2.00
0
-x
2x 49.9 = (2x)2
2.00-x 2x
(1.00-x)(2.00-x)
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(1.00-x)(2.00-x)= (2x)2
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