ADVANCED PLACEMENT CHEMISTRY EQUILIBRIUM

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ADVANCED PLACEMENT CHEMISTRYEQUILIBRIUM
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Chemical equilibrium- state where
concentrations of products and reactants remain
constant-equilibrium is dynamic-any chemical
reaction in a closed vessel will reach
equilibrium-at equilibrium, forward reaction
rate reverse reaction rate
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Law of Mass Action for jA kB ? lC
mDKc ClDm concentration in
-------- mol/L Molarity
AjBk
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Kc K Keq equilibrium constant (used
interchangeably)
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Ex. 4 NH3 7 O2 ? 4 NO2 6 H2O
NO24H2O6 Kc -----------------
NH34O27 If we know
equilibrium concentrations, we can calculate the
equilibrium constant, Kc.
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K changes with temperature (not with
concentration or pressure).
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For the reverse reaction lC mD ? jA kB
Kc' AjBk

ClDm Kc' 1/ Kc
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If the original reaction is multiplied by some
factor to give njA nkB ? nlC nmD
CnlDnm Kc" --------------
Kcn AnjBnk
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For a 2-step reaction with Kc1 and Kc2 as Kc
values for each step, the Kc for the overall
reaction is Kc3 Kc1 x Kc2.
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The units for K depend on the reaction. They are
usually not used.
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Equilibrium position - a set of equilibrium
concentrations. - depends on initial
concentrations. (Kc doesn't)
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Pressures can be used in equilibrium expressions.
The equilibrium constant is called Kp. Using
the same mass action equation as above, the Kp
expression becomes (Pc)l(PD)m Kp
---------- (PA)j(PB)k P partial
pressure at equilibrium in atm.
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Kc involves concentrations while Kp involves
pressures.Kp and Kc can be interconverted using
the following relationshipKp Kc(RT) ?n
R 0.0821 mol/L atm
T Kelvin temperature?n moles gaseous
product - moles gaseous reactantKc
Kp if moles gaseous product moles gaseous
reactant
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Concentrations of pure solids and pure liquids
are not included in equilibrium expressions
because they are constant. CaCO3(s) ? CaO (s)
CO2(g) Kc CO2
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A value for K greater than one means that the
equilibrium is far to the right (mostly
products).A value for K less than one means
that the equilibrium is far to the left (mostly
reactants). The size of K and the time needed
to reach equilibrium are not directly related.
K values can not always be directly compared
because stoichiometry differs.
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If the value for K is very small, reactants are
present in great excess at equilibrium. If the
value for K is very large, products are present
in great excess at equilibrium. Values for K in
the range of 0.001 to 1000 describe reactions
where reactants and products are both present in
significant quantities at equilibrium.
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Consider the reaction 2NOCl(g) ? 2NO(g) Cl2(g)
at 35oC, when 3.00 mol NOCl(g), 1.00 mol NO(g),
and 2.00 mol Cl2(g) are mixed in a 10.0 L flask.
After the system has reached equilibrium the
concentrations are observed to be
Cl2 1.52 x 10-1 M NO 4.00 x 10-3
M NOCl 3.96 x 10-1 MCalculate the value of
K for this system at 35oC.

• K NO2Cl2 NOCl2
• (4 x 10-3)2 (1.52 x 10-1)
• (3.96 x 10-1)2 1.55 x 10-5

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Calculate the value of K for the reaction 2NO(g)
Cl2(g) ? 2NOCl(g).

• K' 1/K
• K' 1/1.55 x 10-5
• 6.45 x 104

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Calculate the value of K for the reaction
4NOCl(g) ? 4NO(g) 2Cl2(g).

• K" Kn
• K" (1.55 x 10-5)2
• 2.40 x 10-10

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Calculate Kp for the first reaction.

• Kp Kc(RT)? n
• Kp (1.55 x 10-5)(0.0821)(308)1
• 3.92 x 10-4

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When working equilibrium problems, it is not
always obvious which direction that equilibrium
is going to shift. To determine this, solve for
the reaction quotient, Q. Q only needs to be
calculated when there is some of each reactant
and product present.
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Q reaction quotient - calculated like Kc, but
using initial concentrations instead of
equilibrium concentrations1. If K Q, then
system is at equilibrium (no shift occurs)2.
If Q gt K, product/reactant is too large
(system shifts to left)3. If Q lt K,
product/reactant is too small (system shifts
to the right)
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Example 2. For the synthesis of ammonia, the
value of K is 6 x 10-2 at 500oC. In an
experiment, 0.50 mol of N2(g), 1.0 x 10-2 mol of
H2(g), and 1.0 x 10-4 mol of NH3(g) are mixed at
500oC in a 1.0 L flask. In which direction will
the system proceed to reach equilibrium?

• N2 3 H2
  ? 2NH3
• Initial 0.50 mol/L 0.010 mol/L 1.0 x 10-4
  mol/L
• Q NH32 (1.0 x 10-4)2
  2.0 x 10-2 N2H23 (0.50)(1.0
  x 10-2)3
• Since Q lt K, the reaction will shift to the right
  to reach equilibrium.

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When solving equilibrium problems, it is very
important to follow a series of steps. Skipping
these can lead to problems (and fewer points on
your AP exam).
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STEPS FOR SOLVING EQUILIBRIUM PROBLEMS1.
Write a balanced equation. If a chemical
reaction occurs, work out the stoichiometry and
then write a second equation for the equilibrium
reaction. Always do stoichiometry(in moles)
first!2. Set up the equilibrium expression.
(No numbers yet!)
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3. If you can't tell which way the reaction is
going to shift, solve for Q.4. Set up a chart
that includes the equation, initial
concentrations, changes in concentration in terms
of X, and final concentrations.
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5. Substitute these final concentrations into
the equilibrium expression and solve for X. 6.
Check your answer to make sure that it is
logical!
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When solving an equilibrium problem, some x and
-x values can be treated as negligible. X is
considered negligible if it is less that 5 of
the number that it was to be subtracted from or
added to. If X is not negligible, the quadratic
equation or the method of successive
approximation must be used.
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At 700 K, carbon monoxide reacts with water to
form CO2 and H2 CO(g) H2O(g) ? CO2(g)
H2(g)The equilibrium constant for this reaction
at 700 K is 5.10. Consider an experiment in
which 1.00 mol of CO(g) and 1.00 mol of H2O(g)
are mixed together in a 1.00 L flask at 700 K.
Calculate the concentrations of all species at
equilibrium.

• Reaction CO H2O ? CO2
  H2
• Initial 1.00 M 1.00 M 0
  0
• Change -x -x
  x x
• Equil. 1.00-x 1.00-x
  x x
• Keq CO2H2 5.10 x2
  COH2O
  (1.00-x)2
• x/(1.00-x) ?5.10
• x 0.69M
• H2OCO 1.00-0.69 0.31 M CO2H2
  0.69 M

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Example 4. Calculate the number of moles of Cl2
produced at equilibrium in a 10.0 L vessel when
1.00 mol of PCl5 is heated to 250oC. K 0.041
mol/L

• 1.00 mol/10.0 L 0.100 M
• R PCl5(g) ? PCl3(g) Cl2(g)
• I 0.100 0 0
• C -x x x
• E 0.100-x x x
• 0.041 x2 ? x2
• 0.100-x 0.100
• x 0.064M
• 0.064 is more than 5 of 0.100 so this is not a
  valid approximation.

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Use of the quadratic equation x2
0.0041-0.041xx2 0.041x 0.0041 0
x 0.047 and 0.088(not possible)
Cl2 0.047 moles Cl2 0.047M x
10.0 L 0.47 moles Cl2
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Using successive approximation When we assumed
that x was negligible, we got x 0.064.1
x 0.0382
x 0.0503
x 0.0454
x 0.047
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5 x
0.0466
x 0.0477
x 0.047
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or/ with a graphing calculator ?0.041 x
0.100 0.06403 (not negl) ?
((0.041)(0.100-ans)) 0.038
0.050 0.045 0.047
0.046
0.047
0.047
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Example 5. Consider the reaction 2HF(g) ? H2(g)
F2(g) where K 1.0 x 10-2 at some very high
temperature. In an experiment, 5.00 mol of
HF(g), 0.500 mol of H2(g) , and 0.750 mol of
F2(g) are mixed in a 5.00 L flask and allowed to
react to equilibrium. Solve for the equilibrium
concentrations.

• 5.00 mol/5 L 1.00 M HF
• 0.500 mol/5 L 0.100 M H2
• 0.750 mol/5 L 0.150 M F2
• Q H2F2 (0.100)(0.150) 0.015 HF2
  (1.00)Q gt K so reaction shifts left

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• Reaction 2HF ? H2 F2
• Initial 1.00 M 0.100 M 0.150 M
• Change 2x -x
  -x
• Equil. 1.002x 0.100-x 0.150-x
• K 1.0 x 10-2 (0.100-x)(0.150-x)
• (1.00
  2x)2

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Using the quadratic equation0.960x2 - 0.290x
(5.00 x 10-3) 0

• x 0.284 M and 0.0180 M
• Since 0.284 gt 0.100, it doesn't work!
• x 0.0180 M
• H2 0.10 - 0.0180 0.082 M
• F2 0.150 - 0.0180 0.132 M
• HF 1.00 2(0.0180) 1.04 M

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When K is very large, it is easier mathematically
to work from completion backward to equilibrium
rather than from initial concentrations forward
to equilibrium. It is best to work with very
large equilibria by determining what the
concentrations would be at completion and then
defining x to be the difference between
completion and equilibrium.
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The equilibrium position of the system lies very
near the completion point of the reaction, so x
is relatively small. Solving for x leads to the
true equilibrium concentrations. In building a
concentration table for a reaction with a large
K, we add an additional row of information, the
concentrations(or pressures) at completion. We
can do this, even though the actual reaction
never reaches completion, because the equilibrium
position is the same no matter how it is
approached.
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Example 6 Nitrogen oxide reacts with oxygen to
form nitrogen dioxide 2NO(g) O2(g)
? 2NO2(g)The Kc for this reaction is 3 x 106 at
200oC. Assume initial concentrations of 0.100M
for NO and 0.050M for O2. Calculate the
concentrations of all species present at
equilibrium.

• Since K is very large, this problem would be very
  difficult to solve in the conventional manner.
  The reaction will proceed almost to completion
  and x would be very large. We can take the
  reaction to completion and work backwards to find
  the equilibrium concentrations.

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Watch out for a limiting reagent!

• Reaction 2NO(g) O2(g) ? 2NO2(g)
• Initial Conc. 0.100 0.050
  0
• Change to Comp. 0.100 -0.050 0.100
  
• New Initial 0 0
  0.100
• Change to Equil 2x x
  -2x
• Equil Conc. 2x x
  0.100-2x RIC NICE!

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• Kc NO22 _ 3 x 106 (0.100-2x)2
• NO 2O2
  (2x)2x
• ? (0.100)2 4x3
• x ? 9 x 10-4M
• NO2 ? 0.098
• NO ? 1.8 x 10-3
• O2 ? 9 x 10-4

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LeChatelier's PrincipleWhen a stress is applied
to a system, the equilibrium will shift in the
direction that will relieve the stress.
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Changes in concentration An increase in
concentration of a reactant will cause
equilibrium to shift to the right to form more
products. An increase in concentration of a
product will cause equilibrium to shift to the
left to form more reactants.
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A decrease in concentration of a product will
cause equilibrium to shift to the right to form
more products. A decrease in the concentration
of a reactant will cause equilibrium to shift to
the left to make more reactants.A change in
concentration of reactant or product will not
affect the value of K.
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If CO is increased, the forward reaction
increases to reestablish equilibrium. Therefore
the quantity of H2 will decrease and the quantity
of product will increase. The value for the
equilibrium constant (K) is unchanged.
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If product is added to this system at
equilibrium, the reverse reaction will increase
to reestablish the equilibrium. Therefore
quantities of both reactants (CO and H2) will
increase. The value for the equilibrium constant
(K) again remains unchanged.
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Continuous removal of product from a reaction
forces more of it to be produced, according to
LeChatelier's Principle. Metabolic reactions as
well as industrial processes make use of this
effect to continuously make products in
equilibrium reactions.
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A B ? C DAdd A or B --------gtRemove A
or B lt-------Add C or D lt-------Remove
C or D --------gt Example N2(g) 3H2(g)
?2NH3(g)a. addition of N2b. addition of
NH3c. addition of H2d. removal of NH3
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Changes in temperature Changes in
temperature may easily be treated as changes in
concentration if you think of heat as a product
(exothermic rxn) or a reactant (endothermic rxn).
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An increase in temperature of an exothermic
reaction will cause equilibrium to shift to the
left. K will decrease. A decrease in
temperature of an exothermic reaction will cause
equilibrium to shift to the right. K will
increase.An increase in temperature of an
endothermic reaction will cause equilibrium to
shift to the right. K will increase. A decrease
in temperature of an endothermic reaction will
cause equilibrium to shift to the left. K will
decrease.
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Changes in pressureChanges in pressure only
affect equilibrium systems having gaseous
products and/or reactants. Increasing the
pressure of a gaseous system will cause
equilibrium to shift to the side with fewer gas
particles. Decreasing the pressure of a
gaseous system will cause equilibrium to shift to
the side with more gas particles.
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If the system has the same number of moles of gas
on each side, changes in pressure do not affect
equilibrium.Adding an inert gas does not affect
equilibrium since the partial pressures of the
gases in the reaction are not affected.
Changing pressure does not affect the value of
the equilibrium constant.
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Example P4(s) 6Cl2(g) ? 4PCl3(l)a. increase
container volumeb. decrease container volumec.
add argon gasExample PCl3(g) Cl2(g) ?
PCl5(g)a. decrease container volumeb. add
helium gasExample PCl3(g) 3NH3(g) ?
P(NH2)3(g) 3HCl(g)a. increase container
volume
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In the production of ammonia from nitrogen and
hydrogen, raising the temperature favors the
reverse reaction, which absorbs heat. Temperature
and pressure are carefully produced in the
industrial production of ammonia, exploiting
LeChatelier's Principle to maximize the amount of
product obtained. The production of ammonia is of
tremendous importance in feeding the world--since
ammonia is used as fertilizer.
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In the production of ammonia from nitrogen and
hydrogen, raising the pressure favors the forward
reaction because 4 moles of gas is converted to 2
moles of gas.
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Addition of a catalyst Adding a catalyst
does not affect equilibrium.
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Example 6. Consider the reaction 2NO2(g) ?
N2(g) 2O2(g) which is exothermic. A vessel
contains NO2(g), N2(g), and O2(g) at equilibrium.
Predict how each of the following stresses will
affect the concentration of O2 and the value of
K.A. NO2 is addedB. N2 is removedC. The
volume is halved
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D. He(g) is addedE. The temperature is
increasedF. A catalyst is added