Download Reaction Rates/Chemical Kinetics

Reaction Rates
and Equilibrium
What is meant by the rate of a
chemical reaction?
• Can also be explained as the
speed of he reaction, it is the
amount of time required for a
chemical reaction to come to
completion.
• Different reactions take
different times
– Burning
– Aging
– Ripening
– Rusting
Collision Theory
• Atoms, ions and molecules can react to form
products when they collide provided that the
particles are orientated correctly and have enough
kinetic energy.
• The relative orientations of the molecules during
their collisions determine whether the atoms are
suitably positioned to form new bonds.
• Particles lacking necessary kinetic energy to react
bounce apart when they collide.
• Imagine clay balls…
• The minimum amount of energy that particles must
have in order to react is called the activation energy
Activated Complex
Activation Energy
• Swedish Chemist Svante Arrhenius explored
activation energy, Ea
• He found that most reaction rate data obeyed an
equation based on three factors:
– The fraction of molecules possessing the necessary
activation energy or greater
– The number of collisions occurring per second
– The fraction of collisions that have the appropriate
orientation
• Arrhenius equation
k = Ae-Ea/RT
Factors that Affect Reaction Rates
• Temperature (Alkaseltzer activity)
– Raise temperature, faster reaction rate
– Lower temperature, slower reaction rate
– Higher temperatures make molecules move faster
because they have more kinetic energy so reaction is
more likely
• Particle Size/Surface Area (lycopodium demo)
– The smaller the particle size the larger the surface area.
– An increase in surface area, increases the amount of
reactant exposed, which increases the collision frequency.
– Can be dangerous… coal powder or gas particles reacting
to our lungs.
• Catalyst (demo)
– A substance that increases the rate of a reaction
without being used up itself during the reaction.
– They help reactions to proceed at a lower energy
than is normally required.
– Enzymes catalyze reactions in our body
– An inhibitor interferes with the action of a catalyst
• Concentration
– The number of reacting particles in a given
volume also affects the rate at which
reactions occur.
– Cramming more particles into a fixed volume
increases the concentration of reactants, the
collision frequency and therefore, reaction
rate.
Rate Law: Effect of
Concentration on Rate
• One way of studying the effect of concentration
on reaction rate is to determine the way in which
the rate at the beginning of a reaction depends
on the starting concentrations.
aA+bBcC+dD
Rate = k[A]m[B]n
• The exponents m and n are called reaction
orders and depend on how the concentration of
that reactant affects the rate of reaction.
Concentration and Rate
NH4+(aq) + NO2−(aq)
N2(g) + 2 H2O(l)
If we compare Experiments 1 and 2, we see
that when [NH4+] doubles, the initial rate
doubles.
Concentration and Rate
NH4+(aq) + NO2−(aq)
N2(g) + 2 H2O(l)
Likewise, when we compare Experiments 5
and 6, we see that when [NO2−] doubles, the
initial rate doubles.
Concentration and Rate
• This means
Rate  [NH4+]
Rate  [NO2−]
−
+
Therefore, Rate  [NH4 ] [NO2 ]
which, when written as an equation, becomes
−
+
Rate = k [NH4 ] [NO2 ]
• This equation is called the rate law, and k is the
rate constant.
Rate Laws
• A rate law shows the relationship between
the reaction rate and the concentrations of
reactants.
• The exponents tell the order of the reaction
with respect to each reactant.
• Since the rate law is
Rate = k [NH4+] [NO2−]
the reaction is
First-order in [NH4+] and
−
First-order in [NO2 ].
Rate Laws
Rate = k
[NH4+]
−
[NO2 ]
• The overall reaction order can be found by
adding the exponents on the reactants in
the rate law.
• This reaction is second-order overall.
Multistep Mechanisms
• In a multistep process, one of the steps will
be slower than all others.
• The overall reaction cannot occur faster than
this slowest, rate-determining step.
Slow Initial Step
NO2 (g) + CO (g)  NO (g) + CO2 (g)
• The rate law for this reaction is found
experimentally to be
Rate = k [NO2]2
• CO is necessary for this reaction to occur, but the
rate of the reaction does not depend on its
concentration.
• This suggests the reaction occurs in two steps.
Slow Initial Step
• A proposed mechanism for this reaction is
Step 1: NO2 + NO2  NO3 + NO (slow)
Step 2: NO3 + CO  NO2 + CO2 (fast)
• The NO3 intermediate is consumed in the second
step.
• As CO is not involved in the slow, rate-determining
step, it does not appear in the rate law.
Fast Initial Step
2 NO (g) + Br2 (g)  2 NOBr (g)
• The rate law for this reaction is found to be
Rate = k [NO]2 [Br2]
• Because termolecular processes are rare,
this rate law suggests a two-step
mechanism.
Fast Initial Step
• A proposed mechanism is
Step 1: NO + Br2
NOBr2
Step 2: NOBr2 + NO  2 NOBr
(fast)
(slow)
Step 1 includes the forward and reverse reactions.
Fast Initial Step
• The rate of the overall reaction depends
upon the rate of the slow step.
• The rate law for that step would be
Rate = k2 [NOBr2] [NO]
• But how can we find [NOBr2]?
Fast Initial Step
• NOBr2 can react two ways:
– With NO to form NOBr
– By decomposition to reform NO and Br2
• The reactants and products of the first
step are in equilibrium with each other.
• Therefore,
Ratef = Rater
Fast Initial Step
• Because Ratef = Rater ,
k1 [NO] [Br2] = k−1 [NOBr2]
• Solving for [NOBr2] gives us
k1
[NO]
[Br
]
=
[NOBr
]
2
2
k−1
Fast Initial Step
Substituting this expression for [NOBr2] in
the rate law for the rate-determining step
gives
Rate =
k 2k 1
k−1 [NO] [Br2] [NO]
= k [NO]2 [Br2]
Integrated Rate Laws
Using calculus to integrate the rate law
for a first-order process gives us
[A]t
ln
= −kt
[A]0
Where
[A]0 is the initial concentration of A, and
[A]t is the concentration of A at some time, t,
during the course of the reaction.
First-Order Processes
ln [A]t = -kt + ln [A]0
Therefore, if a reaction is first-order, a plot
of ln [A] vs. t will yield a straight line, and
the slope of the line will be -k.
First-Order Processes
Consider the process in
which methyl isonitrile is
converted to acetonitrile.
CH3NC
CH3CN
First-Order Processes
CH3NC
This data was
collected for this
reaction at 198.9 °C.
CH3CN
First-Order Processes
• When ln P is plotted as a function of time, a
straight line results.
• Therefore,
– The process is first-order.
– k is the negative of the slope: 5.1  10-5 s−1.
Second-Order Processes
Similarly, integrating the rate law for a
process that is second-order in reactant A,
we get
1
1
= kt +
[A]t
[A]0
y = mx + b
also in the form
Second-Order Processes
1
1
= kt +
[A]t
[A]0
So if a process is second-order in A, a plot
1
of [A] vs. t will yield a straight line, and the
slope of that line is k.
Second-Order Processes
The decomposition of NO2 at 300°C is described by
the equation
1
NO2 (g)
NO (g) + 2 O2 (g)
and yields data comparable to this:
Time (s)
[NO2], M
0.0
0.01000
50.0
0.00787
100.0
0.00649
200.0
0.00481
300.0
0.00380
Second-Order Processes
• Plotting ln [NO2] vs. t yields
the graph at the right.
• The plot is not a straight
line, so the process is not
first-order in [A].
Time (s)
[NO2], M
ln [NO2]
0.0
0.01000
−4.610
50.0
0.00787
−4.845
100.0
0.00649
−5.038
200.0
0.00481
−5.337
300.0
0.00380
−5.573
Second-Order Processes
1
[NO2]
• Graphing ln
vs. t, however,
gives this plot.
Time (s)
[NO2], M
1/[NO2]
0.0
0.01000
100
50.0
0.00787
127
100.0
0.00649
154
200.0
0.00481
208
300.0
0.00380
263
• Because this is a
straight line, the
process is secondorder in [A].
Half-Life
• Half-life is defined as
the time required for
one-half of a reactant
to react.
• Because [A] at t1/2 is
one-half of the
original [A],
[A]t = 0.5 [A]0.
Half-Life
For a first-order process, this becomes
0.5 [A]0
ln
=
−kt
1/2
[A]0
ln 0.5 = −kt1/2
−0.693 = −kt1/2
NOTE: For a first-order
process, then, the half-life
does not depend on [A]0.
0.693
= t1/2
k
Half-Life
For a second-order process,
1
1
= kt1/2 +
0.5 [A]0
[A]0
2
1
= kt1/2 +
[A]0
[A]0
2 − 1 = 1 = kt
1/2
[A]
[A]0
0
1
= t1/2
k[A]0
Temperature and Rate
• Generally, as temperature
increases, so does the
reaction rate.
• This is because k is
temperature dependent.
Reversible Reactions
• Until now, most of the reactions we have examined have gone
completely to products. Some reactions are reversible meaning the
reaction from reactants to products also goes from products to reactants
at the same time.
• Consider the following reaction:
2SO2 + O2
2SO3
In this reaction, SO2 and O2 are placed in a container. Initially, the
forward reaction proceeds and SO3 is produced. The rate of the forward
reaction is much greater than the rate of the reverse reaction. As SO3
builds up, it starts to decompose into SO2 and O2. The rate of the
forward reaction is decreasing and the rate of the reverse reaction is
increasing. Eventually, SO3 decomposes to SO2 and O2 as fast as SO2
and O2 combine to form SO3 (see Fig. 19.10 from text). When this
happens, the reaction has achieved chemical equilibrium.
• Chemical equilibrium is when the rate of the forward reaction = the
rate of the reverse reaction. It says nothing about the amounts of
reactants and products at equilibrium.
Simulation
Equilibrium Constants
• When a system reaches equilibrium there is a
mathematical relationship between the
concentrations of the products and the
concentrations of the reactants.
• Pure solids and pure liquids (including water) do
not appear in the equilibrium expression
• When a reactant or product is preceded by a
coefficient, its concentration is raised to the power
of that coefficient in the equilibrium expression.
• aA + bB cC + dD
• Kc =
[C]c * [D]d
[A]a * [B]b
Sample Exercise 15.1 Writing Equilibrium-Constant Expressions
Write the equilibrium expression for Kc for the following reactions:
Solution
Analyze: We are given three equations and are asked to write an equilibrium-constant expression for each.
Plan: Using the law of mass action, we write each expression as a quotient having the product concentration
terms in the numerator and the reactant concentration terms in the denominator. Each concentration term is
raised to the power of its coefficient in the balanced chemical equation.
Practice Exercise
The Equilibrium Constant
Since pressure is proportional to
concentration for gases in a closed
system, the equilibrium expression can
also be written
(PCc) (PDd)
Kp =
(PAa) (PBb)
Relationship Between Kc and Kp
Plugging this into the expression for Kp
for each substance, the relationship
between Kc and Kp becomes
Kp = Kc (RT)n
where
n = (moles of gaseous product) - (moles of gaseous reactant)
Sample Exercise 15.2 Converting between Kc and Kp
In the synthesis of ammonia from nitrogen and hydrogen,
Kc = 9.60 at 300 °C. Calculate Kp for this reaction at this temperature.
Solution
Analyze: We are given Kc for a reaction and asked to calculate Kp.
Plan: The relationship between Kc and Kp is given by Equation 15.14. To apply that equation, we must
determine Δn by comparing the number of moles of product with the number of moles of reactants
(Equation 15.15).
Solve: There are two moles of gaseous products (2 NH3) and four moles of gaseous reactants (1 N2 + 3 H2).
Therefore, Δn = 2 – 4 = –2. (Remember that Δ functions are always based on products minus reactants.) The
temperature, T, is 273 + 300 = 573 K. The value for the ideal-gas constant, R, is 0.0821 L-atm/mol-K. Using
Kc = 9.60, we therefore have
Practice Exercise
Equilibrium Can Be Reached
from Either Direction
As you can see, the ratio of [NO2]2 to [N2O4] remains
constant at this temperature no matter what the initial
concentrations of NO2 and N2O4 are.
Equilibrium Can Be Reached
from Either Direction
This is the data from
the last two trials from
the table on the
previous slide.
Equilibrium Can Be Reached
from Either Direction
For the equation, N2 + H2  NH3
It doesn’t matter whether we start with N2 and H2
or whether we start with NH3: we will have the
same proportions of all three substances at
equilibrium.
What Does the Value of K
Mean?
• If K>>1, the reaction is
product-favored; since
products are on top of
the expression when
they are bigger than
the reactants the
constant will be greater
than 1.
What Does the Value of K
Mean?
• If K<<1, the reaction is
reactant-favored; since
reactants are on the
bottom of the
expression when they
are the constant will be
less than 1. .
If K= 1, then the
reactants and products
are in equal amounts.
Manipulating Equilibrium
Constants
The equilibrium constant of a reaction in
the reverse reaction is the reciprocal of
the equilibrium constant of the forward
reaction.
N2O4 (g)
2 NO2 (g)
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2 NO2 (g) Kc =
N2O4 (g)
Kc =
[NO2]2
= 0.212 at 100 C
[N2O4]
[N2O4]
= 4.72 at 100 C
2
[NO2]
Sample Exercise 15.4 Evaluating an Equilibrium Constant When an Equation is
Reversed
The equilibrium constant for the reaction of N2 with O2 to form NO equals Kc = 1 × 10–30 at 25 °C:
Using this information, write the equilibrium constant expression and calculate the equilibrium constant for
the following reaction:
Solution
Analyze: We are asked to write the equilibrium-constant expression for a reaction and to determine the
value of Kc given the chemical equation and equilibrium constant for the reverse reaction.
Plan: The equilibrium-constant expression is a quotient of products over reactants, each raised to a power
equal to its coefficient in the balanced equation. The value of the equilibrium constant is the reciprocal of
that for the reverse reaction.
Solve:
Writing products over reactants, we have
Both the equilibrium-constant expression and the
numerical value of the equilibrium constant are the
reciprocals of those for the formation of NO from N2
and O2:
Comment: Regardless of the way we express the equilibrium among NO, N2, and O2, at 25 °C it lies on the side
that favors N2 and O2. Thus, the equilibrium mixture will contain mostly N2 and O2, with very little NO present.
The Reaction Quotient (Q)
• Q gives the same ratio the equilibrium
expression gives, but for a system that is
not at equilibrium.
• To calculate Q, one substitutes the initial
concentrations on reactants and products
into the equilibrium expression.
If Q = K,
the system is at equilibrium.
If Q > K,
there is too much product, and the
equilibrium shifts to the left.
If Q < K,
there is too much reactant, and the
equilibrium shifts to the right.
Sample Exercise 15.10 Predicting the Direction of Approach to Equilibrium
At 448 °C the equilibrium constant Kc for the reaction
is 50.5. Predict in which direction the reaction will proceed to reach equilibrium at 448 °C if we start with
2.0 × 10–2 mol of HI, 1.0 × 10–2 mol of H2, and 3.0 × 10–2 mol of I2 in a 2.00-L container.
Solution
Analyze: We are given a volume and initial molar amounts of the species in a reaction and asked to
determine in which direction the reaction must proceed to achieve equilibrium.
Plan: We can determine the starting concentration of each species in the reaction mixture. We can then
substitute the starting concentrations into the equilibrium-constant expression to calculate the reaction
quotient, Qc. Comparing the magnitudes of the equilibrium constant, which is given, and the reaction
quotient will tell us in which direction the reaction will proceed.
Solve: The initial concentrations are
The reaction quotient is therefore
Because Qc < Kc, the concentration of HI must increase and the concentrations of H2 and I2 must decrease to
reach equilibrium; the reaction will proceed from left to right as it moves toward equilibrium.
Sample Exercise 15.12 Calculating Equilibrium Concentrations from
Initial Concentrations
A 1.000-L flask is filled with 1.000 mol of H2 and 2.000 mol of I2 at 448 °C. The value of the equilibrium
constant Kc for the reaction
at 448 °C is 50.5. What are the equilibrium concentrations of H2, I2, and HI in moles per liter?
Solution
Analyze: We are given the volume of a container, an equilibrium constant, and starting amounts of reactants
in the container and are asked to calculate the equilibrium concentrations of all species.
Plan: In this case we are not given any of the equilibrium concentrations. We must develop some
relationships that relate the initial concentrations to those at equilibrium. The procedure is similar in many
regards to that outlined in Sample Exercise 15.9, where we calculated an equilibrium constant using initial
concentrations.
Solve: First, we note the initial
concentrations of H2 and I2 in the
1.000-L flask:
Second, we construct a table in which
we tabulate the initial concentrations:
Sample Exercise 15.12 Calculating Equilibrium Concentrations from
Initial Concentrations
Solution (continued)
Third, we use the stoichiometry of the
reaction to determine the changes in
concentration that occur as the reaction
proceeds to equilibrium. The
concentrations of H2 and I2 will
decrease as equilibrium is established
and that of HI will increase. Let’s
represent the change in concentration of
H2 by the variable x. The balanced
chemical equation tells us the
relationship between the changes in the
concentrations of the three gases:
Fourth, we use the initial concentrations
and the changes in concentrations, as
dictated by stoichiometry, to express
the equilibrium concentrations. With all
our entries, our table now looks like
this:
Sample Exercise 15.12 Calculating Equilibrium Concentrations from
Initial Concentrations
Solution (continued)
Fifth, we substitute the equilibrium
concentrations into the equilibriumconstant expression and solve for the
unknown, x:
If you have an equation-solving
calculator, you can solve this equation
directly for x. If not, expand this
expression to obtain a quadratic
equation in x:
Solving the quadratic equation
(Appendix A.3) leads to two solutions
for x:
When we substitute x = 2.323 into the
expressions for the equilibrium
concentrations, we find negative
concentrations of H2 and I2. Because a
negative concentration is not
chemically meaningful, we reject this
solution. We then use x = 0.935 to find
the equilibrium concentrations:
Sample Exercise 15.12 Calculating Equilibrium Concentrations from
Initial Concentrations
Solution (continued)
Check: We can check our solution by
putting these numbers into the
equilibrium-constant expression to
assure that we correctly calculate the
equilibrium constant:
Comment: Whenever you use a quadratic equation to solve an equilibrium problem, one of the solutions will
not be chemically meaningful and should be rejected.
Practice Exercise
The Concentrations of Solids and
Liquids Are Essentially Constant
The concentrations of solids and liquids do
not appear in the equilibrium expression.
Therefore, when a solid is dissolved in water,
the expression for the solubility product
constant is as follows…
PbCl
Pb2+
+ 2 Cl2 (s)
(aq)
(aq)
Ksp = [Pb2+] [Cl-]2
Low Ksp means low solubility, high Ksp means good solubility
CaCO3 (s)
CO2 (g) + CaO(s)
As long as some CaCO3 or CaO remain in
the system, the amount of CO2 above the
solid will remain the same.
Sample Exercise 15.6 Writing Equilibrium-Constant Expressions for
Heterogeneous Reactions
Write the equilibrium-constant expression for Kc for each of the following reactions:
Solution
Analyze: We are given two chemical equations, both for heterogeneous equilibria, and asked to write the
corresponding equilibrium-constant expressions.
Plan: We use the law of mass action, remembering to omit any pure solids, pure liquids, and solvents from
the expressions.
Solve:
(a) The equilibrium-constant expression is
Because H2O appears in the reaction as a pure liquid, its concentration does not appear in the equilibriumconstant expression.
(b) The equilibrium-constant expression is
Because SnO2 and Sn are both pure solids, their concentrations do not appear in the equilibrium-constant
expression.
Equilibrium Constants for
Reactions with more than one step
• The equilibrium constant of a net reaction
that has two or more steps is found by the
product of the equilibrium constants for
each of the steps…
K=(K1)(K2)(K3)…
Sample Exercise 15.5 Combining Equilibrium Expressions
Given the following information,
determine the value of Kc for the reaction
Solution
Analyze: We are given two equilibrium equations and the corresponding equilibrium constants and are
asked to determine the equilibrium constant for a third equation, which is related to the first two.
Plan: We cannot simply add the first two equations to get the third. Instead, we need to determine how to
manipulate the equations to come up with the steps that will add to give us the desired equation.
Solve: If we multiply the first equation
by 2 and make the corresponding
change to its equilibrium constant
(raising to the power 2), we get
Sample Exercise 15.5 Combining Equilibrium Expressions
Solution (continued)
Reversing the second equation and
again making the corresponding change
to its equilibrium constant (taking the
reciprocal) gives
Now we have two equations that sum to
give the net equation, and we can
multiply the individual Kc values to get
the desired equilibrium constant.
Practice Exercise
Sample Exercise 15.8 Calculating K When All Equilibrium Concentrations
Are Known
A mixture of hydrogen and nitrogen in a reaction vessel is allowed to attain equilibrium at 472 °C. The
equilibrium mixture of gases was analyzed and found to contain 7.38 atm H 2, 2.46 atm N2, and 0.166 atm
NH3. From these data, calculate the equilibrium constant Kp for the reaction
Solution
Analyze: We are given a balanced equation and equilibrium partial pressures and are asked to calculate the
value of the equilibrium constant.
Plan: Using the balanced equation, we write the equilibrium-constant expression. We then substitute the
equilibrium partial pressures into the expression and solve for Kp.
Solve:
Practice Exercise
An aqueous solution of acetic acid is found to have the following equilibrium concentrations at 25 °C:
[HC2H3O2] = 1.65 × 10–2 M; [H+] = 5.44 × 10–4 M; and [C2H3O2–] = 5.44 × 10–4 M. Calculate the
equilibrium constant Kc for the ionization of acetic acid at 25 °C. The reaction is
Answer: 1.79 × 10–5
Sample Exercise 15.9 Calculating K from Initial and Equilibrium Concentrations
A closed system initially containing 1.000 × 10–3 M H2 and 2.000 × 10–3 M I2 at 448 °C is allowed to reach
equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 × 10–3 M.
Calculate Kc at 448 °C for the reaction taking place, which is
Solution
Analyze: We are given the initial concentrations of H2 and l2 and the equilibrium concentration of HI. We
are asked to calculate the equilibrium constant Kc for
Plan: We construct a table to find equilibrium concentrations of all species and then use the equilibrium
concentrations to calculate the equilibrium constant.
Solve: First, we tabulate the initial and
equilibrium concentrations of as many
species as we can. We also provide
space in our table for listing the
changes in concentrations. As shown, it
is convenient to use the chemical
equation as the heading for the table.
Second, we calculate the change in
concentration of HI, which is the
difference between the equilibrium
values and the initial values:
Sample Exercise 15.9 Calculating K from Initial and Equilibrium Concentrations
Solution (continued)
Third, we use the coefficients in the
balanced equation to relate the change
in [HI] to the changes in [H2] and [I2]:
Fourth, we calculate the equilibrium
concentrations of H2 and I2, using the
initial concentrations and the changes.
The equilibrium concentration equals
the initial concentration minus that
consumed:
The completed table now looks like this
(with equilibrium concentrations in blue
for emphasis):
Notice that the entries for the changes are negative when a reactant is consumed and positive when a product
is formed.
Sample Exercise 15.9 Calculating K from Initial and Equilibrium Concentrations
Solution (continued)
Finally, now that we know the
equilibrium concentration of each
reactant and product, we can use the
equilibrium-constant expression to
calculate the equilibrium constant.
Comment: The same method can be applied to gaseous equilibrium problems to calculate Kp, in which case
partial pressures are used as table entries in place of molar concentrations.
Practice Exercise
Sulfur trioxide decomposes at high temperature in a sealed container:
Initially, the vessel is charged at 1000 K with SO3(g) at a partial pressure of 0.500 atm. At equilibrium the SO3
partial pressure is 0.200 atm. Calculate the value of Kp at 1000 K.
Answer: 0.338
Factors Affecting Equilibrium:
Le Chatelier’s Principle
• If a stress is applied to a system at
equilibrium, the system changes to relieve
the stress.
– Concentration
– Temperature
– Pressure
Concentration
• Increasing the concentration of a substance causes the reaction to
shift to remove some of the extra substance.
• Decreasing the concentration of a substance causes the reaction to
shift to produce some of the substance that was removed.
• For example:
2SO2 + O2
2SO3
• a) If [SO2] increases by adding more, what happens the
concentrations of all of the substances?
[SO2] goes up; [O2] goes down; [SO3] goes up.
• Adding SO2 causes that to go up. The system tries to get rid of that
extra SO2 so some of the additional will react with O2 causing the
O2 to decrease. This speeding up of the forward reaction causes
more SO3 to be produced.
• b) If [SO3] decreases by removing some, what happens the
concentrations of all of the substances?
[SO2] goes down; [O2] goes down; [SO3] goes down.
• Removing SO3 causes the reaction to shift to the right to try to
produce more.
Temperature
• Increasing the temp. causes the equilibrium
position to shift in the direction that absorbs heat
(favors the endothermic reaction). Decreasing
the temp. causes the equilibrium position to shift
in the direction that produces heat (favors the
exothermic reaction).
• For example: 2SO2 + O2
2SO3 + heat
• a) If temp increases it will cause the reaction to
shift left to try to remove the extra heat so:
[SO2] goes up; [O2] goes up; [SO3] goes down.
Pressure
• Affects systems containing gas particles. If
pressure is increased, the reaction will shift
to the side that contains the fewest number
of gas particles and vice-versa.
• For example: 2SO2(g) + O2 (g) 2SO3(g) +
heat
• a) If pressure is increased, the reaction will
shift to the side with the fewest gas particles
so it will shift to the right and:
[SO2] goes down; [O2] goes down; [SO3] goes up.
Sample Exercise 15.13 Using Le Châtelier’s Principal to Predict shifts in
Equilibrium
Consider the equilibrium
In which direction will the equilibrium shift when (a) N2O4 is added, (b) NO2 is removed, (c) the total
pressure is increased by addition of N2(g), (d) the volume is increased, (e) the temperature is decreased?
Solution
Analyze: We are given a series of changes to be made to a system at equilibrium and are asked to predict
what effect each change will have on the position of the equilibrium.
Plan: Le Châtelier’s principle can be used to determine the effects of each of these changes.
Solve:
(a) The system will adjust to decrease the concentration of the added N 2O4, so the equilibrium shifts to the
right, in the direction of products.
(b) The system will adjust to the removal of NO2 by shifting to the side that produces more NO2; thus, the
equilibrium shifts to the right.
(c) Adding N2 will increase the total pressure of the system, but N2 is not involved in the reaction. The
partial pressures of NO2 and N2O4 are therefore unchanged, and there is no shift in the position of the
equilibrium.
(d) If the volume is increased, the system will shift in the direction that occupies a larger volume (more gas
molecules); thus, the equilibrium shifts to the right. (This is the opposite of the effect observed in Figure
15.13, where the volume was decreased.)
(e) The reaction is endothermic, so we can imagine heat as a reagent on the reactant side of the equation.
Decreasing the temperature will shift the equilibrium in the direction that produces heat, so the equilibrium
shifts to the left, toward the formation of more N2O4. Note that only this last change also affects the value of
the equilibrium constant, K.
Sample Exercise 15.13 Using Le Châtelier’s Principal to Predict shifts in
Equilibrium
Practice Exercise
For the reaction
in which direction will the equilibrium shift when (a) Cl2(g) is removed, (b) the temperature is decreased, (c)
the volume of the reaction system is increased, (d) PCl3(g) is added?
Answer: (a) right, (b) left, (c) right, (d) left
Free Energy and Equilibrium
Constants
• The standard state free energy of a
reaction can relates to the free energy of a
reaction at any moment in time according
to the following equations…
ΔG°= -RT ln K
ΔG = ΔG°+ RT ln Q
where ΔG = free energy at any moment, ΔG° = standardstate free energy, R=ideal gas constant (8.314 J/molK),
T=temperature in Kelvin