1. Energy, Rate, and Equilibrium
Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Chapter 7
Energy, Rate, and Equilibrium
Denniston Topping Caret
5th Edition

2. 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

3. 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

4. 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

5. Changes in the System 7.1 Thermodynamics
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

6. 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

7. 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

8. Exothermic Reactions 7.1 Thermodynamics
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
7.1 Thermodynamics
A-B + C-D  A-D + C-B
These bonds must be broken in the reaction, requiring energy
These bonds are formed, releasing energy

9. 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
7.1 Thermodynamics
A-B + C-D  A-D + C-B
These bonds must be broken in the reaction, requiring energy
These bonds are formed, releasing energy

10. 7.1 Thermodynamics Exothermic Reaction Combustion CH4(g) + 2O2(g)
CO2(g) + 2H2O(g) kcal
7.1 Thermodynamics
Endothermic Reaction
Decomposition
22 kcal + 2NH3(g) 
N2(g) + 3H2(g)

11. Enthalpy 7.1 Thermodynamics 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

12. 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

13. Spontaneous and Nonspontaneous Reactions
7.1 Thermodynamics
DS o is positive
DSo is negative

14. Spontaneous and Nonspontaneous Reactions
Are the following processes exothermic or endothermic?
Fuel oil is burned in a furnace
C6H12O6(s) C2H5OH(l) + 2CO2(g)
DH = –16 kcal
N2O5(g) + H2O(l) HNO3(l) kcal
7.1 Thermodynamics

15. Entropy 7.1 Thermodynamics
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

16. Entropy of Reactions 7.1 Thermodynamics
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

17. Processes Having Positive Entropy
Phase change
Melting
Vaporization
Dissolution
7.1 Thermodynamics
All of these processes have a positive DSo

18. 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

19. Greatest Entropy 7.1 Thermodynamics
Which substance has the greatest entropy?
He(g) or Na(s)
H2O(l) or H2O(g)
7.1 Thermodynamics

20. Free Energy 7.1 Thermodynamics
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
Negative DGo…Always Spontaneous
Positive DGo…Never Spontaneous
7.1 Thermodynamics
DGo = DHo - TDSo
T in Kelvin

21. 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H +, DS - : DG always +, regardless of T
DH -, DS + : DG always -, regardless of T
DH +, DS + : DG sign depends on T
DH -, DS - : DG sign depends on T
7.1 Thermodynamics

22. 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

23. 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
7.2 Determination of Energy Change in Reactions

24. 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

25. 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
Units are: calories = gram x ºC x calories/gram - ºC
7.2 Determination of Energy Change in Reactions

26. 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 oC = 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

27. 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

28. 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

29. Calculating the Fuel Value of Foods
1 g of glucose was burned in a bomb calorimeter 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 oC = 6.1oC
Surroundings of calorimeter is water with specific heat capacity = 1.00 cal/g H2OoC
Fuel Value =
= 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
7.2 Determination of Energy Change in Reactions

30. 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

31. Kinetic Information 7.3 Kinetics
Kinetic information represents changes over time, seen here:
Disappearance of reactant, A
Appearance of product, B
7.3 Kinetics

32. 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

33. 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
7.3 Kinetics

34. CH4(g) + 2O2(g)  CO2(g) +2H2O(g) + 211 kcal
The Chemical Reaction
CH4(g) + 2O2(g)  CO2(g) +2H2O(g) 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

35. 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

36. 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
7.3 Kinetics
To be an Exothermic reaction requires a net release of energy (DHo)

37. 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

38. Factors That Affect Reaction Rate
Structure of the reacting species
Concentration of reactants
Temperature of reactants
Physical state of reactants
Presence of a catalyst
7.3 Kinetics

39. Structure of Reacting Species
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

40. 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
7.3 Kinetics

41. The Temperature of Reactants
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
7.3 Kinetics

42. The Physical State of Reactants
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
7.3 Kinetics

43. The Presence of a Catalyst
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

44. Use of a Solid Phase Catalyst
N2+3H2  2NH3
7.3 Kinetics
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

45. Mathematical Representation of Reaction Rate
Consider a decomposition reaction with the following balanced chemical equation:
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

46. Mathematical Representation of Reaction Rate
Reaction rate is proportional to reactant concentration –
Concentration of N2O5 is denoted as [N2O5]
Replace proportionality symbol with (=) and proportionality constant k
k is called the rate constant
7.3 Kinetics

47. 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
7.3 Kinetics

48. 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

49. 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
8.3 Kinetics

50. 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

51. 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

52. Sugar in Water 7.4 Equilibrium
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)
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!
7.4 Equilibrium

53. 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
Some molecules dissolve and others return to the solid state – the rate of each process is equal
7.4 Equilibrium
Sugar(s) Sugar(aq)

54. 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

55. Equilibrium Constant 7.4 Equilibrium
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
7.4 Equilibrium

56. Chemical Equilibrium 7.4 Equilibrium The Reaction of N2 and H2
N2(g) + 3H2(g) NH3(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
7.4 Equilibrium

57. 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
7.4 Equilibrium
Equilibrium occurs when the rate of reactant depletion is
equal to the rate of product depletion
Rates of forward and reverse reactions are Equal

58. 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
7.4 Equilibrium

59. 7.4 Equilibrium Chemical Equilibrium
The exponents in the rate expression are numerically equal to the coefficients
7.4 Equilibrium
Keq is a constant at constant temperature

60. The Generalized Equilibrium-Constant 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
7.4 Equilibrium

61. 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

62. Writing an Equilibrium-Constant Expression
Write an equilibrium-constant expression for the reversible reaction:
H2(g) + F2(g) HF(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

63. 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

64. 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
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

65. Interpreting Equilibrium Constants
Keq less than 1 x 10-2
Small value of Keq indicates numerator (product term) <<< denominator (reactant term)
At equilibrium mostly reactant present
Keq between 1 x 10-2 and 1 x 102
Equilibrium mixture contains significant concentration of both reactants and products
7.4 Equilibrium

66. 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

67. 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
[H2] = 1.72 M
[I2] = M
Substitute concentrations:
7.4 Equilibrium
Keq = [H2] [I2]
[HI]2
Keq= [1.72] [1.72] = 2.96
[0.54]
= or 1.0 x 101
2 significant figures

68. 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
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
N2(g) + 3H2(g) NH3(g)

69. 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
7.4 Equilibrium
N2(g) + 3H2(g) NH3(g)
Equilibrium shifted
Product introduced:
Reactant introduced:

70. Effect of Concentration
N2(g) + 3H2(g) NH3(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
7.4 Equilibrium
A: Reaction at equilibrium
B: Shift to reactant with more red color
C: Shift to product with loss of red color
A B C

71. Effect of Heat 7.4 Equilibrium
Exothermic reactions: treat heat as a product
N2(g) + 3H2(g) NH3(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
7.4 Equilibrium
Heat favors the blue species
while cold favors the pink

72. Effect of Heat 7.4 Equilibrium
Endothermic Reaction – treat heat as a reactant
39 kcal + 2N2(g) + O2(g) NH3(g)
This reaction shift will shift to the right if heat is added by increasing the temperature
7.4 Equilibrium

73. Effect of Pressure 7.4 Equilibrium
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

74. Effect of Pressure 7.4 Equilibrium 2HI(g) H2(g) + I2(g)
N2(g) + 3H2(g) NH3(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

75. Effect of a Catalyst 7.4 Equilibrium
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
7.4 Equilibrium