AP CHEMISTRY
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Reactions are reversible
A + B → C + D ( forward)
C + D → A + B (reverse)
Initially, only the forward reaction is possible
As C and D build up, the reverse reaction can
occur
Eventually the rates of the forward and
reverse reactions are equal: equilibrium
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Rates are equal
Concentrations are not.
The net concentrations do not change at
equilibrium
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Given the following reversible reaction:
N2O4 ↔ 2NO2
Both the forward and reverse reactions are
elementary
The rate law can be written for both
equations
◦
◦
◦
◦
◦
◦
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Forward Rxn: N2O4 → 2NO2
Rate = kf [N2O4]
Reverse Rxn: 2NO2 → N2O4
Rate = kr [NO2]2
At equilibrium
kr [NO2]2 = Kf [N2O4]
Therefore
[𝑁2𝑂4]
𝑘𝑓
2=
𝑁𝑂2
𝑘𝑟
= constant, kc
Once equilibrium is reached, the
concentrations of the products and reactants
no longer change. No Net Change.
 Reaction still occurs, it is in a state of
dynamic equilibrium
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For any reaction
aA + bB ⇆ cC + dD
Kc =
𝐶
𝐴
𝑐
𝐷
𝑎
𝐵
𝑑
𝑏
This relationship is called the equilibriumconsant expression
Kc is the numerical value obtained when we
substitute equilibrium concentrations into the
equilibrium-constant expression.
The subscripts indicates that concentrations
expressed in molarity are used to evaluate the
constant.
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Write the equilibrium expression for the
following reactions:
2O3(g) ⇆ 3 O2(g)
𝑂2
𝑂3
3
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Kc =
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2NO (g) + Cl2 (g) ⇆ 2 NOCl (g)
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Kc =
2
𝑁𝑂𝐶𝑙 2
𝑁𝑂 2[𝐶𝑙2]
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When reactants and products are gases, we can
formulate the equilibrium constant expression in
terms of partial pressures.
aA + bB⇆ cC + dD
𝑃𝐷
𝑃𝐴
𝑑
𝑃𝐶
𝑃𝐵
𝑐
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Kp =
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Converting between Kp and Kc
𝑎
𝑏
◦ Kp = Kc (RT)∆n
◦ ∆n = moles of gaseous product – moles of gaseous reactant
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Equilibrium constants are related not only to
the kinetics of a reaction but also its
thermodynamic measurements.
Measured in activities, not concentration or
pressure.
Beyond the scope of our course…
IT IS ACCEPTED PRACTICE NOT TO INCLUDE
UNITS
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Range from very large to very small
If K>>1: Equilibrium lies to the right;
products predominate
If K<<1: Equilibrium lies to the left; reactants
predominate
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The equilibrium constant expression for a
reaction written in one direction is the
reciprocal of the one for the reaction
written in the reverse direction.
For the reaction:
N2 + 3H2 ⇆ 2NH3,
Kc = 4.34 x 10-3 at 300⁰C. What is the value
of Kc for the reverse reaction?
◦ Kc(r) = 1/4.34 x 10-3 = 2.30 x 102
◦ If an equation is expressed in steps, then the Kc for
the overall equation is the product of the Kc for
each of the steps.
◦ The equilibrium constant of a reaction in the
reverse direction is the inverse of the equilibrium
constant in the forward direction.
◦ The equilibrium constant of a reaction that has
been multiplied by a number is the equilibrium
constant raised to a power equal to that number.
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Given the following information
HF(aq) ⇆ H+(aq) + F-(aq)
H2C2O4(aq) ⇆ 2H+(aq) + C2O42-(aq)
Kc = 6.8 x 10-4
Kc = 3.8 x 10-6
determine the value of the equilibrium constant for the
following reaction:
2HF(aq) + C2O42-(aq) ⇆ 2F-(aq) + H2C2O4(aq)
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Double rxn 1 and reverse rxn 2
◦ Kc1= (6.8 x 10-4)2 = 4.6 x 10-7
◦ Kc2 = 1/3.8 x 10-6 = 2.6 x 105
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Kc = (Kc1)(Kc2) = (4.6 x 10-7)(2.6 x 105)
◦ = 0.12
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When determining equilibrium constants for
heterogeneous equations, it is not necessary
to include solids or liquids.
Example
◦ Write the equilibrium-constant expression for the
following reaction:
SnO2(s) + 2CO(g) ⇆ Sn(s) + 2CO2(g)
◦ Kp =
◦ Kc =
(PCO2)2
(PCO)2
𝐶𝑂2 2
[𝐶𝑂]2
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We often don’t know the equilibrium concentrations
of all chemical species in an equation. If we know the
equilibrium concentration of at least one species, we
can use stoichiometry to deduce the others.
◦ Tabulate the known initial and equilibrium concentrations
of all species in the equilibrium-constant expression.
◦ For those species for which both the initial and equilibrium
concentrations are known, calculate the change in
concentration
◦ Use the stoichiometry of the reactions to calculate the
changes in concentration for other species
◦ From initial concentrations and the changes, calculate the
equilibrium concentrations. These are used to evaluate the
equilibrium constant.
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Enough ammonia is dissolved in 5.00 liters of
water at 25⁰C to produce a solution that is
0.0124 M in ammonia. The solution is then
allowed to come to equilibrium. Analysis of
the equilibrium mixture shows that the
concentration of OH- is 4.64 x 10-4 M.
Calculate Kc at 25⁰C for the reaction
NH3(aq) + H2O(l) ⇆ NH4+(aq) + OH-(aq)
NH3
Initial
+ H2 O
⇆
0.0124 M
NH4+
+
0M
OH0M
Change
4.64 x 10-4 M
Equilibrium
NH3
+ H2O ⇆
NH4+
0M
+
OH-
Initial
0.0124 M
0M
Change
-(4.64 x 10-4 M)
+4.64 x 10-4 M
+4.64 x 10-4 M
Equilibrium
0.0119 M
4.64 x 10-4 M
4.64 x 10-4 M
Kc = [NH4+] [OH-] = (4.64 x 10-4 M) (4.64 x 10-4 M) = 1.81 x 10-5
[NH3]
(0.0119)
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The equilibrium constant allows us to
◦ Predict the extent to which a reaction will proceed.
(If Kc is very large, the reaction to the right
(products) and if Kc is very small, the reaction will
proceed to left (reactants)).
◦ Predict the direction in which a reaction mixture will
proceed to achieve equilibrium
◦ Calculate the concentrations of reactants and
products when equilibrium has been reached.
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If we know the concentrations or the partial
pressures of gases in an reaction, we can
substitute those into the equilibrium gas
constant expression.
The result is know as the reaction quotient
and is represented by Q.
◦ Q = Kc only if the system is at equilibrium
◦ Q > Kc, the reaction moves from right to left
(toward reactants)
◦ Q < Kc, the reaction will move left to right (toward
products)
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At 448⁰C, Kc = 51 for the reaction:
H2(g) + I2(g) ⇆ 2HI(g)
Predict how the reaction will proceed to reach equilibrium at
448⁰C if we start with 0.020 mol of HI, 0.010 mol H2, and
0.030 mol I2 in a 2.00-L containier.
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Solve for initial partial pressures:
◦ [HI] = mol/L= 0.020mol/2.00L = 0.010 M
◦ [H2] = mol/L = 0.010mol/2.00L = 0.0050 M
◦ [I2] = mol/L = 0.030mol/2.00L = 0.015 M
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Q=
[𝐻𝐼]2
𝐻2 [𝐼2]
=
0.010 2
(0.0050)(0.015)
= 1.3
Because Q < Kc, the partial pressure of HI must increase and
those of H2 and I2 must decrease to reach equilibrium; the
reaction will proceed left to right.
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A 1.000-L flask is filled with 1.000 mol of H2 and 2.000
mol of I2 at 448⁰C. The value of Kc is 50.5 for the
reaction: H2(g) + I2(g) ⇆ 2HI(g)
What are the partial pressures of H2, I2, and HI in the
flask at equilibrium?
First, solve for initial partial pressures of H2 and I2.
◦ PH2 = nRT/V = (1.000mol)().08206)(721K)/1.000L = 59.19 atm
◦ PI2 = nRT/V = (2.000mol)().08206)(721K)/1.000L = 118.4 atm
H2
I2
HI
59.19 atm
118.4 atm
0 atm
H2
I2
HI
Initial
59.19 atm
118.4 atm
0 atm
Change
- x atm
- x atm
+ 2x atm
Equilibrium
59.19 – x
118.4 – x
Initial
Change
Equilibrium
Kc=
𝑃𝐻𝐼 2
(𝑃𝐻2 𝑃𝐼2
)(
)
=
2𝑥 2
(59.19 −𝑥)(118.4 −𝑥)
0 + 2x
= 50.5
4x2 = 50.5(x2-177.6x + 0.00701
46.5x2 – 8970x + 35400 = 0
x=
−𝑏± 𝑏2 −4𝑎𝑐
2𝑎
=
−(−8970)± (−8970)2 −4(46.5)(35400)
2(46.5)
= 137.5 or 55.3
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When 137.5 gets substituted in, we get a
negative value so the answer must be 55.3
PH2 = 59.29-55.3 = 3.9 atm
PI2 = 118.4 – 55.3 = 63.1 atm
PHI = 2(55.3) = 110.6 atm
Kc =
𝑃𝐻𝐼 2
(𝑃𝐻2 𝑃𝐼2
)(
=
)
110.6 2
(3.9)(63.1)
= 50
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If a system at equilibrium is disturbed by a
change in temperature, pressure, or the
concentration of one of the components, the
system will shift its equilibrium position so as
to counteract the effect of the disturbance.
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If a system is at equilibrium and we add a
substance (either a reactant or product), the
reaction will shift so as to reestablish
equilibrium by consuming part of the added
substance. Removing a substance will cause
the reaction to move in the direction that
forms more of that substance.
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Reducing the volume of a gaseous
equilibrium mixture causes the system to
shift in the direction that reduces the number
of moles of gas. (Conversely, increasing
volume causes a shift in the direction that
produces more gas molecules.)
Example
◦ N2O4(g) ⇆ 2NO2(g)
◦ When volume is decreased, we expect a shift to the
side that reduces the total number of moles of gas,
which is to the reactant side. (NO2 will be converted
to N2O4.)
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Endothermic: Reactants + heat ⇆ products
Exothermic: Reactants ⇆ products + heat
When applying Le Chatelier’s principle, heat is
treated as if it were a chemical reagent.
When the temperature is increased, it is as if we
have added a reactant (or product) to the system
at equilibrium. The equilibrium shifts in the
direction that consumes the excess reactant.
◦ Endothermic: Increasing T results in equilibrium shift to
right (products formed) and Keq increases
◦ Exothermic: Increasing T results in equilibrium shift to left
(reactants formed) and Keq decreases
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A catalyst lowers the activation barrier
between reactants and products.
The activation energy of a forward reaction is
lowered to the same extent as the reverse
reaction
As a result, a catalyst increase the rate at
which equilibrium is achieved, but it does not
change the composition of the equilibrium
mixture.
◦ The value of Keq is also not affected by a catalyst.