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1/14/2014
CHEMICAL
EQUILIBRIUM
Unit #11
Chapter 13
Equilibrium
• Equilibrium occurs when the rate of the
forward reaction is equal to the rate of the
reverse reaction
• Equilibrium is a dynamic state, this means that
even though it seems that the reaction has
stopped, in fact the products are still being
made and used up-but at the same speed (rate).
• Equilibrium is symbolized by the use of a
double arrow (
) or an equals sign (=)
Equilibrium, continued
• At the point when equilibrium occurs in a chemical
reaction, some reactants has been converted to
products.
• The amounts depend on the reaction and are most often
described in Molarity (mol/L)
• Because of this use of measurement; solids, liquids and
water are not considered in calculations.
• Solids and liquids are pure and have no concentration
and of course water is a liquid.
• Each reaction has its own set of conditions that must be
met and it is the same for the reaction (always) based
on temperature.
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How Equilibrium is Measured
• Each reaction that is reversible and
therefore, can reach equilibrium, is assigned
a constant, Kc.
• Kc = equilibrium constant has no units and
tells the ratio of concentrations (in M) at
equilibrium. That is what the “c” stands
for.
• Sometimes, just “K” is used and this is the
same as Kc or Keq which also is molarity
dependent.
Equilibrium is not for everyone!
• Some reactions go to completion and are
not reversible.
• These reactions show only a forward arrow.
• Equilibrium (reversible) reactions must
happen in a closed container.
• All gases are considered because their
concentration changes by the size of the
container.
How is Kc Calculated
• When equilibrium occurs, the concentration of the
products and reactants are used to determine the
equilibrium constant (Kc).
• Let’s look at a reaction at equilibrium …
H2(g) + I2(g)
2 HI(g)
The equilibrium expression is …
Kc= [HI]2
[H2][I2]
it is measured in the concentrations of the products
(raised to the coefficient power) divided by the
concentration of the reactants (raised to the coefficient
power) – excluding liquids and solids
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What are the equilibrium expressions for these
reactions?
1. AgCl(s)
Ag+(aq) + Cl-(aq)
answer = Kc = [Ag+][Cl-]
2. 2 NH3(g)
N2(g) + 3 H2(g)
answer Kc = [N2][H2]3
[NH3]2
3. Pb(l) + H2O(g)
PbO(s) + H2 (g)
Kc = [H2]
[H2O]
In this case since water is a gas it is included!
Kc, Kp and Q
• Kc = equilibrium constant using concentrations
• Kp = equilibrium constant using partial pressures
and is related to Kc but not always equal to Kc
• Kp is always calculated using Atmospheres (atm)
• Q = the ratio of products and reactants using the
same formula as Kc but means that the reaction
is not necessarily at equilibrium. (Instantaneous)
• If Q<Kc the reaction still needs to proceed
forward to achieve equilibrium and if Q> Kc
then the reaction must go in reverse to achieve
equilibrium if Q=Kc the reaction is at
equilibrium
Calculating Kc
What is the equilibrium constant for this reaction
at 250 oC? 2 HI(g)
H2(g) + I2(g)
[HI] = 1.5 x 10-2 M and [H2] = [I2]= 3.4 x 10-3 M
Answer
Kc = [H2][I2] = (3.4 x 10-3 M)2 = 0.051
[HI]2
(1.5 x 10-2 M)2
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Calculating Kp
• Kp is calculated the same way if all your data is
given in partial pressures at 25 o C.
• For example: What is the Kp for the reaction of
N2(g) + 3 H2(g)
2 NH3(g) at equilibrium the
partial pressures are as follows; pN2 = .0223 atm,
pH2 =.0669 atm and pNH3 = 1.97 atm.
• Answer
• Kp = (pNH3)2 = (1.97 atm)2 = 5.81 x 105
(pN2)(pH2)3 (.0223)(.0669)3
Calculating Kp, continued
• If you are trying to determine the relationship
between Kc and Kp you must use a factor called
∆n.
• ∆n = total moles of gaseous products – total
moles of gaseous reactants
• Kp = Kc(RT) ∆n
• Kc = 3.48 x 108
Comparing Q and K
3 H2(g) + N2(g) = 2 NH3(g) Kc= 5.9
At a certain time the following data was
gathered…[H2] = 1.19 M, [N2] = .0515 M and
[NH3] = 2.27 M…
Is the reaction at equilibrium? If not which
direction must the reaction shift to get to
equilibrium?
ANSWER: Q = 59 so it is NOT at equilibrium
Reaction needs to shift reverse or
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