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Chapter 7
Energy, Rate, and Equilibrium
Denniston
Topping
Caret
5th Edition
7.1 Thermodynamics
• Thermodynamics - the study of energy,
work, and heat
– Applied to chemical change
• Calculate the quantity of heat obtained from
combustions of one gallon of fuel oil
– Applied to physical change
• Determine the energy released by boiling water
• The laws of thermodynamics help us to
understand why some chemical reactions
occur and others do not
7.1 Thermodynamics
The Chemical Reaction and Energy
Basic Concepts – from Kinetic Molecular
Theory
– Molecules and atoms in a reaction mixture are
in constant, random motion
– Molecules and atoms frequently collide with
each other
– Only some collisions, those with sufficient
energy, will break bonds in molecules
– When reactant bonds are broken, new bonds
may be formed and products result
7.1 Thermodynamics
Change in Energy and
Surroundings
• Absolute value for energy stored in a
chemical system cannot be measured
• Can measure the change in energy during
these chemical changes
• System - contains the process under study
• Surroundings - the rest of the universe
7.1 Thermodynamics
Changes in the System
• Energy can be lost from the system to the
surroundings
• Energy may be gained by the system at the
expense of the surroundings
– This energy change is usually in the form of heat
– This change can be measured
7.1 Thermodynamics
Law of Conservation of Energy
• The first law of thermodynamics - energy of
the universe is constant
• This law is also called the Law of
Conservation of Energy
• Where does the reaction energy come from
that is released and where does the energy
go when it is absorbed?
7.1 Thermodynamics
Changes in Chemical Energy
A-B + C-D  A-D + C-B
• Consider the reaction converting AB and
CD to AD and CB
• Each chemical bond is stored chemical
energy
• If a reaction will occur:
– Bonds must break
– Breaking bonds requires energy
7.1 Thermodynamics
Exothermic Reactions
If the energy required to break the bonds is
less than the energy released when the bonds
are formed, there is a net release of energy
– This is called an exothermic reaction
– Energy is a product in this reaction
A-B + C-D  A-D + C-B
These bonds must
be broken in the
reaction, requiring
energy
These bonds are
formed, releasing
energy
7.1 Thermodynamics
Endothermic Reactions
If the energy required to break the bonds is
larger than the energy released when the
bonds are formed, there will need to be an
external supply of energy
– This is called an endothermic reaction
A-B + C-D  A-D + C-B
These bonds must
be broken in the
reaction, requiring
energy
These bonds are
formed, releasing
energy
7.1 Thermodynamics
Exothermic Reaction
Combustion
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g) + 211 kcal
Endothermic Reaction
Decomposition
22 kcal + 2NH3(g) 
N2(g) + 3H2(g)
7.1 Thermodynamics
Enthalpy
• Enthalpy - represents heat energy
• Change in Enthalpy (D Ho) - energy difference
between the products and reactants of a chemical
reaction
• Energy released, exothermic reaction, enthalpy
change is negative
– In the combustion of CH4, D Ho = –211 kcal
• Energy absorbed, endothermic, enthalpy change is
positive
– In the decomposition of NH3, D Ho = +22 kcal
7.1 Thermodynamics
Spontaneous and
Nonspontaneous Reactions
• Spontaneous reaction - occurs without
any external energy input
• Most, but not all, exothermic reactions
are spontaneous
• Thermodynamics is used to help predict
if a reaction will occur
• Another factor is needed, Entropy
7.1 Thermodynamics
Spontaneous and
Nonspontaneous Reactions
D S o is positive
D So is negative
7.1 Thermodynamics
Spontaneous and
Nonspontaneous Reactions
Are the following processes exothermic or
endothermic?
– Fuel oil is burned in a furnace
– C6H12O6(s)
2C2H5OH(l) + 2CO2(g)
D H = –16 kcal
– N2O5(g) + H2O(l)
2HNO3(l) + 18.3 kcal
7.1 Thermodynamics
Entropy
• The second law of thermodynamics - the
universe spontaneously tends toward increasing
disorder or randomness
• Entropy (So) – a measure of the randomness of
a chemical system
• High entropy – highly disordered system, the
absence of a regular, repeating pattern
• Low entropy – well organized system such as a
crystalline structure
• No such thing as negative entropy
7.1 Thermodynamics
Entropy of Reactions
D So of a reaction = So(products) - So(reactants)
• A positive D So means an increase in
disorder for the reaction
• A negative D So means a decrease in
disorder for the reaction
7.1 Thermodynamics
Processes Having Positive
Entropy
Phase change
Melting
Vaporization
Dissolution
All of these processes have a positive D So
7.1 Thermodynamics
Entropy and Reaction
Spontaneity
• If exothermic and positive D So…
SPONTANEOUS
• If endothermic and negative D So…
NONSPONTANEOUS
• For any other situations, it depends on the
relative size of D Ho and D So
7.1 Thermodynamics
Greatest Entropy
• Which substance has the greatest
entropy?
– He(g) or Na(s)
– H2O(l) or H2O(g)
7.1 Thermodynamics
Free Energy
• Free energy (D Go) - represents the
combined contribution of the enthalpy and
entropy values for a chemical reaction
• Free energy predicts spontaneity of
chemical reactions
DGo = DHo - TDSo
– Negative D Go…Always Spontaneous
– Positive D Go…Never Spontaneous T in Kelvin
7.1 Thermodynamics
Free Energy and Reaction
Spontaneity
• Need to know both D H and D S to predict the
sign of D G, making a statement on reaction
spontaneity
• Temperature also may determine direction of
spontaneity




D
D
D
D
H +, D S - : D G always +, regardless of T
H -, D S + : D G always -, regardless of T
H +, D S + : D G sign depends on T
H -, D S - : D G sign depends on T
7.2 Experimental Determination
of Energy Change in Reactions
• Calorimetry - the
measurement of heat energy
changes in a chemical reaction
• Calorimeter - device which
measures heat changes in
calories
• The change in temperature
is used to measure the loss
or gain of heat
7.2 Determination of Energy
Change in Reactions
Heat Energy in Reactions
• Change in temperature of a solution, caused by a
chemical reaction, can be used to calculate the
gain or loss of heat energy for the reaction
– Exothermic reaction – heat released is absorbed
– Endothermic reaction – reactants absorb heat from
the solution
• Specific heat (SH) - the number of calories of
heat needed to raise the temperature of 1 g of a
substance 1oC
Q  ms  DTs  SHs
7.2 Determination of Energy
Change in Reactions
Heat Energy in Reactions
• Specific heat of the solution along with the total
number of grams of solution and the
temperature change, permits calculation of heat
released or absorbed during the reaction
• SH for water is 1.0 cal/goC
• To determine heat released or absorbed, need:
– specific heat
– total number of grams of solution
– temperature change (increase or decrease)
7.2 Determination of Energy
Change in Reactions
Calculation of Heat Energy
in Reactions
• Q is the product
– ms is the mass of solution in the calorimeter
 D Ts is the change in temperature of the solution
from initial to final state
– SHs is the specific heat of the solution
• Calculate with this equation
Q  ms  DTs  SHs
– Units are: calories = gram x ºC x calories/gram - ºC
7.2 Determination of Energy
Change in Reactions
Calculating Energy Involved in
Calorimeter Reactions
If 0.10 mol HCl is mixed with 0.10 mol KOH in a
“coffee cup” calorimeter, the temperature of 1.50 x
102 g of the solution increases from 25.0oC to
29.4oC. If the specific heat of the solution is 1.00
cal/goC, calculate the quantity of energy evolved in
the reaction.
D Ts = 29.4oC - 25.0oC = 4.4oC
Q = ms x D Ts x SHs
= 1.50 x 102 g solution x 4.4oC x 1.00 cal/goC
= 6.6 x 102 cal
7.2 Determination of Energy
Change in Reactions
Calculating Energy Involved in
Calorimeter Reactions
Is the reaction endothermic or exothermic?
– 0.66 kcal of heat energy was released to the
surroundings—the solution
– The reaction is exothermic
What would be the energy evolved for
each mole of HCl reacted?
– 0.10 mol HCl used in the original reaction
– [6.6 x 102 cal / 0.10 mol HCl] x 10 = 6.6 kcal
7.2 Determination of Energy
Change in Reactions
Bomb Calorimeter and
Measurement of Calories in Foods
Nutritional Calorie
(large “C” Calorie) =
– 1 kilocalorie (1kcal)
– 1000 calories
• The fuel value of
food
• Bomb calorimeter
is used to measure
nutritional Calories
7.2 Determination of Energy
Change in Reactions
Calculating the Fuel Value
of Foods
1 g of glucose was burned in a bomb calorimeter. 1.25 x
103 g H2O was warmed from 24.5oC to 31.5oC.
Calculate the fuel value of the glucose (in Kcal/g).
 D Ts = 31.5oC - 24.5oC = 6.1oC
Surroundings of calorimeter is water with specific heat
capacity = 1.00 cal/g H2OoC
Fuel Value = Q  ms  DTs  SHs
= 1.25 x 103 g H2O x 6.1oC x 1.00 cal/g H2OoC
= 7.6 x 103 cal
7.6 x 103 cal x 1 Calorie / 103 cal = 7.6 nutritional
Calories
7.3 Kinetics
• Thermodynamics determines if a reaction
will occur spontaneously, but tells us
nothing about the amount of time the
reaction will take
• Kinetics - the study of the rate (or speed) of
chemical reactions
– Also supplies an indication of the mechanism –
step-by-step description of how reactants
become products
7.3 Kinetics
Kinetic Information
• Kinetic information represents changes over
time, seen here:
– Disappearance of reactant, A
– Appearance of product, B
7.3 Kinetics
Alternative Presentation of
Kinetic Data
Rather than the graph shown before, this figure
demonstrates the change from purple reactant to
green product over time from the molecular
perspective
7.3 Kinetics
Kinetic Data Assessed by
Color Change
• Change in color over time can be used to monitor
the progress of a chemical reaction
• The rate of color change can aid in calculating the
rate of the chemical reaction
The Chemical Reaction
7.3 Kinetics
CH4(g) + 2O2(g)  CO2(g) +2H2O(g) + 211 kcal
• C-H and O=O bonds must be broken
• C=O and O-H bonds must be formed
• Energy is required to break the bonds
– This energy comes from the collision of the molecules
– If sufficient energy available, bonds break and atoms
recombine in a lower energy arrangement
– Effective collision is one that produces product
molecules
– Leads to a chemical reaction
7.3 Kinetics
Activation Energy and the
Activated Complex
• Activation energy - the minimum amount of
energy required to initiate a chemical reaction
• Picture a chemical reaction in terms of changes
in potential energy occurring during the reaction
– Activated complex - an extremely unstable,
short-lived intermediate complex
– Formation of this activated complex requires
energy (Ea) to overcome the energy barrier to
start the reaction
7.3 Kinetics
Activation Energy and the
Activated Complex
• Reaction proceeds
from reactants to
products via the
activated complex
• Activated complex - can’t
be isolated from the
reaction mixture
• Activation energy (Ea) is
the difference between the
energy of the reactants and
that of the activated
complex
•To be an Exothermic reaction requires a net release of energy (D Ho)
7.3 Kinetics
Activation Energy in the
Endothermic Reaction
• This figure diagrams an
endothermic reaction
• Reaction takes place
slowly due to the large
activation energy
required
• The energy of the
products is greater than
that of the reactants
7.3 Kinetics
Factors That Affect
Reaction Rate
1. Structure of the reacting species
2. Concentration of reactants
3. Temperature of reactants
4. Physical state of reactants
5. Presence of a catalyst
Structure of Reacting Species
7.3 Kinetics
• Oppositely charged species react more rapidly
• Dissociated ions in solution whose bonds are already
broken have a very low activation energy
• Ions with the same charge do not react
• Bond strength plays a role
• Covalent molecules bonds must be broken with the
activation energy before new bonds can be formed
• Magnitude of the activation energy is related to bond
strength
• Size and shape influence the rate
• Large molecules may obstruct the reactive part of the
molecule
• Only molecular collisions with correct orientation lead to
product formation
7.3 Kinetics
The Concentration of Reactants
• Rate is related to the concentration of one or
more of the reacting substances
• Rate will generally increase as concentration
increases
– Higher concentration means more reactant
molecules per unit volume
– More reactant molecules means more collisions
per unit time
The Temperature of Reactants
7.3 Kinetics
• Rate increases as the temperature increases
– Increased temperature relates directly to
increased average kinetic energy
– Greater kinetic energy increases the speed of
particles
– Faster particles increases likelihood of collision
• Higher kinetic energy means a higher
percentage of these collisions will result in
product formation
The Physical State of Reactants
7.3 Kinetics
Reactions occur when reactants can collide
frequently with sufficient energy to react
•Solid state: atoms, ions, compounds are close together
but restricted in motion
•Gaseous state: particles are free to move but often are
far apart causing collisions to be relatively infrequent
•Liquid state: particles are free to move and are in close
proximity
•Reactions to be fastest in the liquid state and
slowest in the solid state
• Liquid > Gas > Solid
The Presence of a Catalyst
7.3 Kinetics
• Catalyst - a substance that increases the reaction
rate
– Undergoes no net change
– Does not alter the final product of the reaction
– Interacts with the reactants to create an alternative
pathway for product production
7.3 Kinetics
Use of a Solid Phase Catalyst
N2+3H2  2NH3
Haber Process is a synthesis of ammonia facilitated
by a solid phase catalyst
–
–
–
–
Diatomic gases bind to the surface
Bonds are weakened
Dissociation of diatomic gases and reformation of NH3
Newly formed NH3 leaves the solid surface with the
catalyst unchanged
7.3 Kinetics
Mathematical Representation of
Reaction Rate
• Consider a decomposition reaction with the
following balanced chemical equation:
D
2N2O5 ( g ) 
 4NO2 ( g )  O2 ( g )
• When heated N2O5 decomposes to 2
products – NO2 and O2
• When holding all factors constant, except
concentration, rate of reaction is
proportional to the concentration
7.3 Kinetics
Mathematical Representation of
Reaction Rate
D
2N2O5 ( g ) 

4NO2 ( g )  O2 ( g )
• Reaction rate is proportional to reactant
concentration – rate [N2O5 ]
– Concentration of N2O5 is denoted as [N2O5]
– Replace proportionality symbol with (=) and
proportionality constant k
– k is called the rate constant
rate k[N2O5 ]
Rate Equation
7.3 Kinetics
• For a reaction Aproducts we write the equation:
rate = k[A]n
• This is called the rate equation (or rate law)
• The exponent n is the order of the reaction
– If n=1, first order
– If n=2, second order
– n must be determined experimentally
– This exponent is not the same as the coefficient of the
reactant in the balanced equation
Rate Equation
8.3 Kinetics
• For the equation A + B  products
the rate equation is:
– rate = k[A]n[B]m
• What would be the general form of the
rate equation for the reaction:
CH4+2O2CO2+2H2O
– Rate = k[CH4]n[O2]m
• Knowing the rate equation and the order
helps industrial chemists determine the
optimum conditions for preparing a
product
8.3 Kinetics
Writing Rate Equations
• Write the form of the rate equation for the
oxidation of ethanol (C2H5OH)
• The reaction has been experimentally determined to
be first order in ethanol and third order in oxygen
– Rate expression involves only the reactants
– Concentrations: [C2H5OH][O2]
– Raise each to exponent corresponding to its order
rate = k [C2H5OH][O2]3
– Remember that 1 as an exponent is understood and
NOT written
– Knowing the rate equation and the order helps
industrial chemists determine the optimum conditions
for preparing a product
7.4 Equilibrium
Rate and Reversibility of Reactions
• Equilibrium reactions - chemical reactions
that do not go to completion
– Completion – all reactants have been converted
to products
– Equilibrium reactions are also called incomplete
reactions
– Seen with both physical and chemical processes
• After no further observable change,
measurable quantities of reactants and
products remain
Physical Equilibrium
7.4 Equilibrium
• Physical equilibria are reversible reactions
– Dissolved oxygen in lake water
– Stalactite and stalagmite formation
– Sugar dissolved in water
• Reversible reaction - a process that can
occur in both directions
– Use the double arrow symbol
• Dynamic equilibrium - the rate of the
forward process in a reversible reaction is
exactly balanced by the rate of the reverse
process
7.4 Equilibrium
Sugar in Water
1. If you add 2-3 g of sugar into 100 mL water
•
•
All will dissolve with stirring in a short time
No residual solid sugar, sugar dissolved completely
Sugar(s)  Sugar(aq)
2. If add 100 g of sugar in 100 mL of water
•
•
•
Not all of it will dissolve even with much stirring
Over time, you observe no further change in the
amount of dissolved sugar
Appears nothing is happening – Incorrect!
Sugar in Water
7.4 Equilibrium
Appears nothing is happening is incorrect!
–
Individual sugar molecules are constantly going into
and out of solution
–
Both happen at the same rate
–
Over time the amount of sugar dissolved in the
measured volume of water does not change
•
An equilibrium situation has been established
Sugar(s)
Sugar(aq)
•
Some molecules dissolve and others return to the
solid state – the rate of each process is equal
Dynamic Equilibrium
7.4 Equilibrium
sugar(s)
sugar(aq)
• The double arrow serves as an indicator of:
– a reversible process
– an equilibrium process
– the dynamic nature of the process
• Continuous change is taking place without
observable change in the amount of sugar
in either the solid or the dissolved form
7.4 Equilibrium
Equilibrium Constant
• ratef = forward rate
rater = reverse rate
• at equilibrium: ratef = rater
• ratef = kf[sugar(s)]
• rater = kr[sugar(aq)]
• kf[sugar(s)]=kr[sugar(aq)]
• Equilibrium constant (Keq) - ratio of the two rate
constants
kf
[sugar(aq)]
K eq 

kr
[sugar(s)]
7.4 Equilibrium
Chemical Equilibrium
The Reaction of N2 and H2
N2(g) + 3H2(g)
2NH3(g)
• Mix components at elevated temperature
• Some molecules will collide with
sufficient energy to break N-N and H-H
bonds
• Rearrangement of the atoms will produce
the product NH3
Chemical Equilibrium
7.4 Equilibrium
N2(g) + 3H2(g)
2NH3(g)
• Initially the forward
reaction is rapid
– Reactant concentrations are
high
– Product concentration
negligible
• Forward reaction rate
decreases with time
– Concentrations of reactants
are decreasing
– Product concentration
increasing
Equilibrium occurs when the rate of reactant depletion is
equal to the rate of product depletion
Rates of forward and reverse reactions are Equal
7.4 Equilibrium
Chemical Equilibrium
N2(g) + 3H2(g) 2NH3(g)
Basic equation divides into 2 parts:
• forward rxn: N2(g) + 3H2(g)  2NH3(g)
• reverse rxn: 2NH3(g) N2(g) + 3H2(g)
• ratef = kf[N2]n[H2]m
• rater = kr[NH3]p
• ratef = rater
[NH3 ]p
K eq 

kr [N2 ]n [H 2 ]m
kf
7.4 Equilibrium
Chemical Equilibrium
[NH3 ]p
K eq 

kr [N2 ]n [H 2 ]m
kf
• The exponents in the rate expression are
numerically equal to the coefficients
[NH3 ]2
K eq 
[N2 ][H 2 ]3
• Keq is a constant at constant temperature
7.4 Equilibrium
The Generalized EquilibriumConstant Expression for a
Chemical Reaction
aA + bB
cC + dD
• A and B are reactants
• C and D are products
• a, b, c, and d are the coefficients of the
balanced equation
[C]c [D]d
K eq 
a
b
[A] [B]
7.4 Equilibrium
Writing Equilibrium-Constant
Expressions
• Equilibrium constant expressions can only be
written after a correct, balanced chemical equation
• Each chemical reaction has a unique equilibrium
constant value at a specified temperature
• The brackets represent molar concentration
• All equilibrium constants are shown as unitless
• Only the concentration of gases and substances in
solution are shown
• Concentration for pure liquids and solids are not shown
7.4 Equilibrium
Writing an Equilibrium-Constant
Expression
Write an equilibrium-constant expression for
the reversible reaction:
H2(g) + F2(g)
2HF(g)
• No solids or liquids are present
–
–
–
–
All reactants and products appear in the expression
Numerator term is the product term [HF]2
Denominator term is the reactants [H2] and [F2]
Each term contains an exponent identical to the
corresponding coefficient in the balanced equation
Keq = [HF]2
[H2][F2]
7.4 Equilibrium
Writing an Equilibrium-Constant
Expression
Write an equilibrium-constant expression for the
reversible reaction:
MnO2(s) + 4H+(aq) + 2Cl-(aq) Mn2+(aq) + Cl2(g) + H2O(l)
• MnO2 is a solid
• H2O(l) is a product, but negligible compared to solvent water
– Numerator term is the product terms [Mn2+] and [Cl2]
– Denominator term is the reactants [H+]4 and [Cl-]2
– Each term contains an exponent identical to the corresponding
coefficient in the balanced equation
Keq = [Mn2+] [Cl2]
[H+]4 [Cl-]2
7.4 Equilibrium
Interpreting Equilibrium Constants
•
Reversible arrow in chemical equation
indicates equilibrium exists
• The numerical value of the equilibrium
constant tells us the extent to which
reactants have converted to products
1. Keq greater than 1 x 102
•
•
Large value of Keq indicates numerator
(product term) >>> denominator (reactant term)
At equilibrium mostly product present
7.4 Equilibrium
Interpreting Equilibrium Constants
2. Keq less than 1 x 10-2
•
•
Small value of Keq indicates numerator
(product term) <<< denominator (reactant term)
At equilibrium mostly reactant present
3. Keq between 1 x 10-2 and 1 x 102
•
Equilibrium mixture contains significant
concentration of both reactants and products
7.4 Equilibrium
Calculating Equilibrium Constants
• A reversible reaction
is allowed to proceed
until the system
reaches equilibrium
• Amount of reactants
and products no longer
changes
• Analyze reaction
mixture to determine
molar concentrations
of each product and
reactant
2NO2(g)
N2O4(g)
7.4 Equilibrium
Calculating an Equilibrium
Constant
HI placed in a sealed container and comes to
equilibrium; equilibrium reaction is:
2HI(g)
H2(g) + I2(g)
•Equilibrium concentrations:
– [HI] = 0.54 M
Keq = [H2] [I2]
– [H2] = 1.72 M
2
[HI]
– [I2] = 1.72 M
•Substitute concentrations:Keq= [1.72] [1.72] = 2.96
[0.54]2
0.29
= 10.1 or 1.0 x 101
7.4 Equilibrium
Le Chateleir’s Principle
• Le Chateleir’s principle - if a stress is
placed on a system at equilibrium, the
system will respond by altering the
equilibrium composition in such a way as
to minimize the stress
N2(g) + 3H2(g) 2NH3(g)
• If reactants and products are present in a
fixed volume and more NH3 is added into
the container, the system will be stressed
– Stressed = the equilibrium will be disturbed
7.4 Equilibrium
Le Chateleir’s Principle
• Adding NH3 to the system causes stress
– To relieve stress, remove as much of added
material as possible by converting it to
reactants
• Adding N2 or H2 to the system causes
stress also
– To relieve stress, remove as much of added
material as possible by converting it to
product
N2(g) + 3H2(g) 2NH3(g)
Equilibrium shifted
Product introduced:
Reactant introduced:
Effect of Concentration
7.4 Equilibrium
N2(g) + 3H2(g)
2NH3(g)
• Adding or removing either reactants or products at a
fixed volume is saying that the concentration is
changed
• Removing material decreases concentration
• System will react to this stress to return
concentrations to the appropriate ratio
A
B
C
A: Reaction at
equilibrium
B: Shift to reactant
with more red color
C: Shift to product
with loss of red color
7.4 Equilibrium
Effect of Heat
• Exothermic reactions: treat heat as a product
N2(g) + 3H2(g)
2NH3(g) + 22 kcal
• Addition of heat is
treated as increasing
the amount of product
• More product shifts
equilibrium to the left
– Increases amount of
reactants
– Decreases amount of
product
Heat favors the blue species
while cold favors the pink
7.4 Equilibrium
Effect of Heat
• Endothermic Reaction – treat heat as a
reactant
39 kcal + 2N2(g) + O2(g)
2NH3(g)
• This reaction shift will shift to the right if heat
is added by increasing the temperature
7.4 Equilibrium
Effect of Pressure
• Pressure affects the equilibrium only if one
or more substances in the reaction are gases
• Relative number of gas moles on reactant
and product side must differ
• When pressure goes up…shift to side with
less moles of gas
• When pressure goes downs…shifts to side
with more moles of gas
7.4 Equilibrium
Effect of Pressure
N2(g) + 3H2(g)
2NH3(g)
• If increase pressure, which way will the
equilibrium shift?
– Increased pressure favors decreased volume with more
product (2 moles) formed and less reactant (4 moles)
2HI(g)
H2(g) + I2(g)
• If pressure increases in this reaction, which way
will the equilibrium shift?
– No shift in equilibrium as both reactant and product
have 2 moles of gas
7.4 Equilibrium
Effect of a Catalyst
• A catalyst has no effect on the
equilibrium composition
• It increases the rate of both the forward
and reverse reaction to the same extent
• While equilibrium composition and
concentration do not change, equilibrium
is reached in a shorter time