Transcript Ch. 15 Chemical Equilibrium

Ch. 15 Chemical Equilibrium
• Consider colorless N2O4. At room temperature, it
decomposes to brown NO2:
N2O4(g)  2NO2(g).
• At some time, the color stops changing and we have a
mixture of N2O4 and NO2. We now write it like this…
N2O4(g)
2NO2(g)
• Each is constantly being formed at the same rate that it is
being consumed. It is therefore a “dynamic equilibrium”.
• Chemical equilibrium is the point at which the
concentrations of all species are constant.
The Equilibrium Constant
•
Consider the following reaction:
aA + bB
•
cC + dD
There are two reactions going on, forward and reverse. The rate of
each reaction can be expressed separately:
Ratef = kf[A]a[B]b
Rater = kr[C]c[D]d
• At equilibrium, Rf = Rr or…kf[A]a[B]b = kr[C]c[D]d
•
We can rearrange this equation and combine the rate constants into
a single constant. We end up with this…

Cc Dd
Keq 
Aa Bb
where Keq is called the equilibrium constant.
The Equilibrium Constant
• If the substances in the reaction are gases, then their
concentrations are proportional to their partial pressures
according to PV=nRT, so we can express Keq in terms of
pressures as well…
(PC)c (PD)d
Keq = (P )a (P )b
A
B
•
(Reminder… Molarity = n/V =P/RT…and R= 0.0821 L·atm/mol·K)
(Changes in pressure at equilibrium will be explored later in the notes.)
• The actual value for Keq is determined experimentally at a specific
temperature for a specific reaction.
Facts About Keq
•
Keq does not depend on initial concentrations of the products or
reactants since it’s only the equilibrium concentrations that we are
concerned with!
•
The larger Keq , the more products are present at equilibrium. If
Keq>> 1, then products dominate at equilibrium and equilibrium lies
to the right.
•
Conversely, the smaller Keq , the more reactants are present at
equilibrium. If Keq << 1, then reactants dominate at equilibrium
and the equilibrium lies to the left.
The Magnitude of Equilibrium Constants
More Facts About Keq
• The equilibrium constant of a reaction in the reverse direction is
the inverse of the equilibrium constant of the reaction in the
forward direction… Keq(forward) = 1/Keq(reverse)
For example:
N2O4(g)
2NO2(g)
At 100 ºC, Keq(forward) = 6.49
At 100 ºC, Keq(reverse) = 1/6.49 = 0.154
• Keq does not depend on the reaction mechanism.
• The value of Keq varies with temperature. (We will see this later.)
• The stoichiometry of a reaction that has been multiplied by a
number changes the equilibrium constant. Keq gets raised to the
power equal to that number.
For example: 2
4
Keq = (6.49)2 = 42.12
More Facts About Keq
• The equilibrium constant for a net reaction made up of two or more
steps is the product of the equilibrium constants for the individual
steps.
For example: A + B  X + C
Keq(1) = 2.0
X+BD
Keq(2) = 5.0
A + 2B  C + D
Keq = K1 x K2 = 10.0
When this is written in terms of concentrations at equilibrium…
Keq=10.0 = [C][D]/[A][B]2
• Keq usually is not written with units. Why? The Molarity &
Pressures are based on a “standard”: (1 M or 1 atm.)
More Facts About Keq
• If concentrations are used in the equilibrium expression, Keq is
sometimes written as Kc.
• When pressures are used in the expression, Keq it sometimes written
as Kp.
•The equilibrium expression only contains the concentrations of gases
or aqueous substances and NEVER solids or pure liquids. Why?
- Consider the decomposition of calcium carbonate:
CaCO3(s)  CaO(s) + CO2(g)
- The concentrations of solids and pure liquids are constant.
The amount of CO2 formed will not depend greatly on the
amounts of CaO and CaCO3 present.
Keq = [PCO2]
Note: Although the concentrations of these species are not included in
the equilibrium expression, they do participate in the reaction and must
be present for an equilibrium to be established!
Changes in Equilibrium
• Le Chatelier’s Principle: If a stress is applied to a system that is
already at equilibrium, the equilibrium will shift to reduce the effect
of the stress.
• We will now look at changing various things on a system at
equilibrium and discuss how the equilibrium will shift to “relieve the
stress.”
(1) Changing Concentrations:
A+B↔C+D
• If more [A] is added, the rate of the forward reaction increases
to relieve the stress. As this occurs [C] and [D] increase, the rate
of the reverse reaction increases and quickly equilibrium is reestablished. At equilibrium –Rf = Rr. The concentrations of A, B,
C, D have changed but:
[C][D] remains constant at a given temperature.
[A][B]
Changes in Equilibrium
Changing Concentrations Continued..
A+B↔C+D
• If [C] is removed for the system, the rate of the forward reaction
will increase to replace the [C] that was removed. As this occurs
[D] increases, and [A] and [B] decreases the rate of the reverse
reaction increases and quickly equilibrium is re-established. Once
again, Keq is the same even though the concentrations are
different.
(2) Pressure:
ONLY AFFECTS CHEMICAL REACTIONS WHICH INVOLVE GASES
• As the pressure on the gaseous system increases, the gaseous
substances are compressed, their concentrations and # of
molecules/liter increase. The reaction, (forward or reverse),
which favors the reduction of the number of molecules per liter
will be favored. (Remember, if V↓, then this would cause P↑.)
Changes in Equilibrium
Changing pressure continued…
Example: 2H2(g) + O2(g) ↔ 2H2O(g)
• If pressure is increased by decreasing the volume of the gases in the
container at equilibrium, then the forward reaction is favored. Why?
- This reduces the # of gas particles from 3 to 2.
- Note: If the # of gas particles on both sides of the equation is the
same, then changing pressure has NO EFFECT on the equilibrium.
• If an inert gas is added, it WILL NOT change the equilibrium.
(3) Temperature: All chemical reactions either give off heat
(exothermic) or take in heat (endothermic). An increase in
temperature favors the endothermic reaction; a decrease in
temperature favors the exothermic reaction.
• The temperature of a reaction will change the value of Keq, but for
now, let’s just focus on how the equilibrium will shift.
Changes in Equilibrium
Changes in temperature continued…
• Example:
H2 + I2 ↔ 2HI
∆H = - 25 kJ/mol (exothermic)
Another way of looking at the reaction…
H2 + I2 ↔ 2HI + heat
Lowering the temperature favors HI formation. (You can
think of it as though we are removing the “heat” product from
the equation.) Raising the temperature favors the reverse
reaction.
• van Hoff’s Law: In a system at equilibrium, an increase in heat
energy is displaced so that heat is absorbed.
(4) Catalyst: A catalyst increases the rate of both the forward and the
reverse reaction by decreasing Ea. It has no effect on Keq but does
cause equilibrium to be reached more quickly.
Catalysts & Changes in Equilibrium
Predicting the Direction of a Reaction
• When given the concentrations of reactants & products
and the value of Keq, you can easily predict the direction
of a reaction.
- Simply plug the given concentrations into the
equilibrium equation and calculate the “reaction
quotient”, Q.
- If Q > Keq then the reverse reaction must occur to
reach equilibrium (i.e., products are consumed,
reactants are formed, the numerator in the equilibrium
constant expression decreases and Q decreases until it
equals Keq).
- If Q < Keq then the forward reaction must occur to
reach equilibrium. (See Practice Problems for an example.)
Predicting the Direction of a Reaction
Calculating Keq
• Practice Problem: At a certain temperature, a 1.0 L flask initially
contains 0.298 moles of PCl3(g) and 0.0087 moles of PCl5(g). After the
system reaches equilibrium, 0.00200 moles of Cl2(g) was found in the
flask. Calculate the equilibrium concentrations of the gases in the
flask and also the value of Keq. PCl5 decomposes according to the
following reaction…PCl5(g) PCl3(g) + Cl2(g)
• Step 1-- Tabulate initial and equilibrium concentrations (or partial
pressures) given.
Initial
[PCl5]
[PCl3]
[Cl2]
0.0087 M
0.298 M
0
Change
Equilibrium
0.002 M
•This sort of table goes by the nickname “ICEbox”…get it?
Calculating Keq
Step 2- If an initial and equilibrium concentration is given for a
species, calculate the change in concentration.
Step 3- Use stoichiometry on the change in concentration line only to
calculate the changes in concentration of all species…(Since
our reaction occurs in a 1:1:1 ratio, the changes for each
species is the same. Only the sign is different on the reactant.)
Step 4- Deduce the equilibrium concentrations of all species.
Initial
Change
Equilibrium
[PCl5]
[PCl3]
[Cl2]
0.0087 M
0.298 M
0
− 0.002 M
0.0067 M
+ 0.002 M
+ 0.002 M
0.3 M
0.002 M
Step 5- Finally, plug the equilibrium concentrations into the
equilibrium equation and solve!
Calculating Keq
Keq = [ PCl3][Cl2]/[PCl5]
Keq = [0.3][0.002]/[0.0067] = 0.08955≈ 0.09
Initial
Change
Equilibrium
• This
[PCl5]
[PCl3]
[Cl2]
0.0087 M
0.298 M
0
− 0.002 M
0.0067 M
+ 0.002 M
+ 0.002 M
0.3 M
0.002 M
system of tabulating data will allow you to solve for
equilibrium concentrations if you are given Keq…
Calculating Keq
Practice Problem #2-- Given this equation…H2(g)+ I2(g)  2 HI(g)
…calculate all three equilibrium concentrations when:
[H2]o = [I2]o = 0.200 M and Kc = 64.0
Initial
Change
Equilibrium
[H2]
[I2]
[HI]
0.200 M
0.200 M
0
−x
−x
+ 2x
0.200 − x
0.200 − x
+ 2x
• Here’s where we use stoichiometry to calculate the changes in the concentrations…
• Now determine the equilibrium concentrations…
• Now plug them into the equilibrium expression and solve for “x”…
Calculating Keq
Kc = [ HI]2/[H2][I2]
64.0 = [2x]2/[0.200 – x][0.200 – x]
• Solving this takes some good algebra skills & maybe even a quadratic equation will
have to be solved…YUCK!
• Both sides are perfect squares, (done so on purpose), so we square root both sides
to get…
8.00 = (2x) / (0.200 - x)
• From there, the solution should be easier, and so after some cross-multiplying and
dividing, etc… x = 0.160 M
• This is not the end of the solution since the question was asking for the equilibrium
concentrations, so…
[H2] = 0.200 - 0.160 = 0.040 M
[I2] = 0.200 - 0.160 = 0.040 M
[HI] = 2 (0.160) = 0.320 M
• You can check for correctness by plugging back into the equilibrium expression:
Kc = (0.320)2 / (0.040)(0.040) = 64
• Since Kc = 64.0 we know that the problem was correctly solved.
The End
Now we get to work on more practice problems!!