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# Power point chapter 17

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### Power point chapter 17

1. 1. Chapter 17Reaction Energy and Reaction Kinetics
2. 2. THERMOCHEMISTRY• Virtually every chemical reaction is accomplished by a change in energy.• Chemical reactions usually absorb or release energy in the form of heat.
3. 3. THERMOCHEMISTRY• Thermochemistry is the study of the changes in heat energy that accompany chemical reactions and physical changes.• The heat absorbed or released in a chemical or physical change is measured in a calorimeter.
4. 4. THERMOCHEMISTRY• In a calorimeter, known quantities of reactants are sealed in a reaction chamber, which is immersed in a known quantity of water in an insulated vessel.• The heat given off or absorbed during the reaction is equal to the heat absorbed or given off by the known quantity of water.
5. 5. THERMOCHEMISTRY• The amount of heat is determined from the temperature change of the known mass of the surrounding water.• The data collected from calorimetry experiments are temperature changes because heat cannot be measured directly, but temperature can be.
6. 6. THERMOCHEMISTRYWhat is the difference between heat and temperature?• Temperature is the measure of the average kinetic energy of the particles in a sample of matter.• To measure temperature we use the Celsius and Kelvin scales.• Kelvin = °Celsius + 273.15
7. 7. THERMOCHEMISTRY• The ability to measure temperature is based on heat transfer.• The amount of heat transfer is usually measured in joules (J).• A joule is the SI unit of heat energy as well as all other forms of energy.
8. 8. THERMOCHEMISTRY• Heat can be thought of as the sum total of the kinetic energies of the particles in a sample of matter.• Heat always flows spontaneously from matter at a higher temperature to matter at a lower temperature.
9. 9. THERMOCHEMISTRY• The quantity of heat transferred during a temperature change depends on: –The nature of the material changing temperature. –The mass of the material changing temperature. –The size of the temperature change.
10. 10. THERMOCHEMISTRY• A quantity called specific heat can be used to compare heat absorption capacities for different materials.• Specific Heat is the amount of heat energy required to raise the temperature of one gram of a substance by one Celsius degree or one Kelvin.
11. 11. THERMOCHEMISTRY• Specific Heat is measured in units of Joules/gram•°Celsius (J/g•°C) or calories/gram•Kelvin (cal/g•K) or calories/gram•°Celsius (cal/g•°C).• Table 17-1, page 513.
12. 12. THERMOCHEMISTRY• Specific Heat is usually measured under constant pressure conditions, so its symbol, cp, contains a subscripted p as a reminder.• Specific heat can be found using the following equation: cp = q/(m × T)
13. 13. THERMOCHEMISTRY• cp = Specific Heat• q = Heat lost or gained• m = mass of the sample• T = Difference between the initial and final temperatures.
14. 14. THERMOCHEMISTRYEXAMPLE:A 4.0 g sample of glass was heated from 274 K to 314 K, a temperature increase of 40 K, and was found to have absorbed 32 J of heat. What is the specific heat of this type of glass? How much heat will the same glass sample gain when it is heated from 314 K to 344 K?
15. 15. THERMOCHEMISTRYEXAMPLE:• Determine the specific heat of a material is a 35 g sample absorbed 48 J as it was heated from 293 K to 313 K.
16. 16. THERMOCHEMISTRYEXAMPLE:• A piece of copper alloy with a mass of 85.0 g is heated from 30 °C to 45 °C. In the process, it absorbs 523 J of heat. What is the specific heat of this copper alloy? How much heat will the sample lose if it is cooled to 25 °C?
17. 17. THERMOCHEMISTRYEXAMPLE:• The temperature of a 74 g sample of material increases from 15 °C to 45 °C when it absorbs 2.0 kJ of heat. What is the specific heat of this material?
18. 18. THERMOCHEMISTRYEXAMPLE:• How much heat is needed to raise the temperature of 5.0 g of gold by 25 °C?
19. 19. THERMOCHEMISTRYEXAMPLE:• What mass of liquid water at room temperature (25 °C) can be raised to its boiling point with the addition of 24 kJ of heat energy?
20. 20. THERMOCHEMISTRYEXAMPLE:• Heat in the amount of 420 J is added to a 35 g sample of water at a temperature of 10 °C. What will be the final temperature of the water?
21. 21. THERMOCHEMISTRY• The heat of reaction is the quantity of heat released or absorbed during a chemical reaction.• You can think of heat of reaction as the difference between the stored energy of the reactants and the products.
22. 22. THERMOCHEMISTRYEXAMPLE: 2H2(g) + O2(g) --> 2H2O(g)• If a mixture of hydrogen and oxygen is ignited, water will form and heat energy will be released.• The heat released comes from the formation of the product (water).
23. 23. THERMOCHEMISTRY• Chemical equations do not tell you that heat is evolved during the reaction.• Experiments show that 483.6 kJ of heat are evolved when 2 mol of gaseous water are formed at 298.15 K from its elements.
24. 24. THERMOCHEMISTRY• Thermochemical equations include the quantity of heat released or absorbed during the reaction.2H2(g) + O2(g) --> 2H2O(g) + 483.6 kJ• The quantity of heat for any reaction depends on the amounts of reactants and products.
25. 25. THERMOCHEMISTRY• Producing twice as much water vapor would require twice as many moles of reactants, and would release twice as much heat.4H2(g) + 2O2(g) --> 4H2O(g) + 967.2 kJ
26. 26. THERMOCHEMISTRY• Producing one-half as much water would require one-half as many moles of reactants and would release only one-half as much heat.H2(g) + 1/2O2(g) --> H2O(g) + 241.8 kJ
27. 27. THERMOCHEMISTRY• Fractional coefficients are sometimes used in thermochemical equations.• In chemical reactions, heat can either be released or absorbed. –Exothermic = release of heat –Endothermic = absorption of heat
28. 28. THERMOCHEMISTRY• In writing thermochemical equations for endothermic reactions, the situation is reversed because the products have a higher heat content than the reactants.• Decomposition of water:2H2O(g) + 483.6 kJ --> 2H2(g) + O2(g)
29. 29. THERMOCHEMISTRY• The physical states of the reactants and products must always be included in thermochemical equations because they influence the overall amount of heat exchanged (Ex: Ice).
30. 30. THERMOCHEMISTRY• The heat absorbed or released during a chemical reaction at constant pressure is represented by the symbol H.• The H is the symbol for a quantity called enthalpy.• Enthalpy is the heat content of a system at constant pressure.
31. 31. THERMOCHEMISTRY• Most chemical reactions are run in open vessels, so the pressure is constant and equal to atmospheric pressure.• Only changes in enthalpy can be measured ( H = Enthalpy change).
32. 32. THERMOCHEMISTRY• Enthalpy change is the amount of heat absorbed or lost by a system during a process at constant pressure. H = Hproducts - Hreactants
33. 33. THERMOCHEMISTRY• Thermochemical equations are usually written by designating the value for H rather than writing the heat as a reactant or product.• Exothermic reactions have a H that is always negative, because the system loses energy.2H2(g) + O2(g) --> 2H2O(g) H = -483.6 kJ
34. 34. THERMOCHEMISTRY• The H for endothermic reactions is always positive, because the system gains energy.2H2O(g) --> 2H2(g) + O2(g) H = +483.6 kJ
35. 35. THERMOCHEMISTRY• Keep in mind the following concepts when using thermochemical equations:1. The coefficients in a balanced thermochemical equation represent the number of moles, not molecules, which allows us to write these as fractions.
36. 36. THERMOCHEMISTRY2. The physical state of the product/reactant involved in a reaction is an important factor and must be included in the thermochemical equation.
37. 37. THERMOCHEMISTRY3. The change of energy represented by a thermochemical equation is directly proportional to the number of moles of substances undergoing a change.
38. 38. THERMOCHEMISTRY4. The value of the energy change, H, is usually not significantly influenced by changing temperature.
39. 39. THERMOCHEMISTRY• The formation of water from hydrogen and oxygen is a composition reaction - the formation of a compound from its elements.• The amount of heat released or absorbed when one mole of a compound is formed from its elements is called the molar heat of formation.
40. 40. THERMOCHEMISTRY• To make comparisons, heats of formations are given for the standard states of reactants and products.• These states are found at atmospheric pressure and room temperature (298.15 K). –Example: Water is a liquid at room temperature, not a solid.
41. 41. THERMOCHEMISTRY• To signify that a value represents measurements on substances in their standard states, a 0 sign is added to the enthalpy symbol ( H0).• Adding the subscript f, further indicates a standard heat of formation ( H0f).
42. 42. THERMOCHEMISTRY• Heats of Formation, Appendix Table A-14.• Each entry in the table is the heat of formation for the synthesis of one mole of the compound listed from its elements in their standard states.
43. 43. THERMOCHEMISTRY• Elements in their standard states are defined as having a H0f = 0.• Compounds with a high negative heat of formation are very stable, which means that they will not decompose back into their respective elements.
44. 44. THERMOCHEMISTRY• Compounds with relatively positive values, or slightly negative values, are relatively unstable and will spontaneously decompose into their elements if the conditions are appropriate. –Ex: Hydrogen Iodide (HI), ethyne (C2H2), and Mercury Fulminate (HgC2N2O2).
45. 45. THERMOCHEMISTRY• Combustion reactions produce a considerable amount of energy in the form of light and heat when a substance is combined with oxygen.• The heat released by the complete combustion of one mole of a substance is called the heat of combustion of the substance.
46. 46. THERMOCHEMISTRY• Heat of combustion is defined in terms of one mole of reactant (opposite of the heat of formation = one mole of product).• Heat of combustion is given the notation Hc.• Heats of Combustion, Appendix Table A-5.
47. 47. THERMOCHEMISTRY• Remember that CO2 and H2O are the products of the complete combustion of organic compounds.Example:• Combustion of propane. C3H8(g) + 5O2(g) --> 3CO2 + 4 H2O(l) H0c = -2219.2 kJ/mol
48. 48. THERMOCHEMISTRY• Thermochemical equations can be rearranged, terms can be canceled, and the equations can be added to give enthalpy changes for reactions not included in the data tables.• The basis for calculating heats of reaction is known as Hess’s Law.
49. 49. THERMOCHEMISTRY• Hess’s Law –The overall enthalpy change in a reaction is equal to the sum of the enthalpy changes for the individual steps in the process.• Measured heats of reaction can be combined to calculate heats of reaction that are difficult or impossible to actually measure.
50. 50. THERMOCHEMISTRY Calculating Heats of Reaction:• Suppose you want to know the heat of reaction for the decomposition of ice to give oxygen and hydrogen gases. Given the information below: H0f (liquid water) = -285.83 kJHeat of Fusion (melting of ice) = 6.0 kJ/mol
51. 51. THERMOCHEMISTRYGiven:H2(g) + 1/2O2(g) --> H2O(l) H0f = -285.83 H2O(s) --> H2O(l) H = 6.0 kJWant: H2O(s) --> H2(g) + 1/2O2(g) H = ?????
52. 52. THERMOCHEMISTRY General Principles for combining thermochemical equations:1. Reverse the direction of the known equations so that when added together they give the desired thermochemical equation.• When reversing equations, the sign of H is also reversed.
53. 53. THERMOCHEMISTRY2. Multiply the coefficients of the known equations so that when added together they give the desired thermochemical equation.• When multiplying the coefficients, you must also multiply the value for H.
54. 54. THERMOCHEMISTRY• To solve this example, we must reverse the direction of the equation for the formation of H2O as a liquid, then add the equations and heats together.H2O(l) --> H2(g) + 1/2O2(g) H0f = 285.83 + H2O(s) --> H20(l) H = 6.0H2O(s) --> H2(g) + 1/2O2(g) H = 291.83
55. 55. THERMOCHEMISTRYExample:• What is the heat of reaction used to prepare ultra-pure silicon for the electronics industry?Given:Si(s) + 2Cl2(g) --> SiCl4(l) H = -687.01Mg(s) + Cl2(g) --> MgCl2(s) H = -641.32
56. 56. THERMOCHEMISTRYWant: 2Mg(s) + SiCl4(l) --> Si(s) + 2MgCl2(s) H=?• In this example we must reverse the first equation and multiply the second equation by 2, then add them together.
57. 57. THERMOCHEMISTRY SiCl4(l) --> Si(s) + 2Cl2(g) H = 687.01+ 2Mg(s) + 2Cl2(g) --> 2MgCl2(s) H = -1282.642Mg(s) + SiCl4(l) --> Si(s) + 2MgCl2(s) H = -595.63
58. 58. THERMOCHEMISTRYExample:• Calculate the heat of reaction for the combustion of of methane gas, CH4, to form CO2(g) and H2O(l). Write all equations involved.
59. 59. THERMOCHEMISTRY• To find the heat of formation of a compound, you would use the same method as finding the heat of reaction.
60. 60. THERMOCHEMISTRYExample:• When carbon is burned in a limited supply of oxygen, carbon monoxide and is produced. In this reaction, carbon is probably first oxidized to carbon dioxide. Then part of the carbon dioxide is reduced with carbon to give some carbon monoxide.Find the heat of formation of carbon monoxide.
61. 61. THERMOCHEMISTRY• We know the heat of formation of carbon dioxide and the heat of combustion of carbon monoxide from the tables in the book.C(s) + O2(g) --> CO2(g) H0f = -393.5 kJ/molCO(g) + 1/2O2(g) --> CO2(g) H0c = -283.0 kJ/mol
62. 62. THERMOCHEMISTRY• We want to find the heat of formation of carbon monoxide. C(s) + 1/2O2(g) --> CO(g) H0f = ?• From the information given we should now be able to solve this problem by rearranging and combining equations.
63. 63. THERMOCHEMISTRYExample:• Calculate the heat of formation of pentane, C5H12, using the information on heats of formation in Appendix Table A-14 and the information of heats of combustion in Appendix Table A- 5. Solve by combining known thermochemical equaitons.
64. 64. THERMOCHEMISTRY• Given:C(s) + O2(g) --> CO2(g) H0f = -393.5 kJ/mol H2(g) + O2(g) --> H2O(l) H0f = -285.8 kJ/mol C5H12(g) + 8O2(g) --> 5CO2(g) + 6H2O(l) H0c = -3535.6 kJ/mol• Want: 5C(s) + 6H2(g) --> C5H12(g) H0f = ?
65. 65. THERMOCHEMISTRY• Practice Problems, pg. 524, 1-3.
66. 66. THERMOCHEMISTRY• The change in heat content of a reaction system is one of two factors that allow chemists to predict whether a reaction will occur spontaneously and to explain how it occurs.• The randomness of the particles in a system is the second factor affecting spontaneity.
67. 67. THERMOCHEMISTRY• A great majority of chemical reactions in nature are exothermic.• As these reactions proceed, energy is liberated and the products have less energy than the original reactants.• The products are also more resistant to change, more stable, than the original reactants.
68. 68. THERMOCHEMISTRY• The tendency throughout nature is for a reaction to proceed in a direction that leads to a lower energy state.• In endothermic reactions, energy is absorbed and the products are at a higher potential energy and are less stable than the original reactants.
69. 69. THERMOCHEMISTRY• So one might think that endothermic reactions do not take place spontaneously, but they do.
70. 70. THERMOCHEMISTRY• Consider the melting of ice. –An ice cube melts spontaneously as heat is transferred from the warm air to the ice. –Well-ordered crystal arrangement to a less orderly liquid phase. –A system that can go from one state to another without an enthalpy change does so by becoming more disordered.
71. 71. THERMOCHEMISTRY• A disordered system is one that lacks a regular arrangement of its parts.• This factor is called Entropy.• Entropy (S): –A measure of the degree of randomness of the particles, such as molecules.
72. 72. THERMOCHEMISTRY• The more disordered or the more random a system, the higher its entropy.• Entropy increases when a gas expands, a solid or liquid dissolves, or the number of particles in a system increases.
73. 73. THERMOCHEMISTRY• Processes in nature are driven in two directions: –Towards lowest enthalpy. –Towards highest entropy.• When these two oppose each other, the dominant factor determines the direction of change.
74. 74. THERMOCHEMISTRY• To predict which will dominate for a given system, a function has been defined to relate the enthalpy and entropy factors at a given temperature.• Free Energy (G): –The combined enthalpy-entropy function of a system.
75. 75. THERMOCHEMISTRY• Natural processes proceed in the direction that lowers the free energy of a system.• Only the change in free energy can be measured.• The change in free energy can be defined in terms of the changes in enthalpy and entropy.
76. 76. THERMOCHEMISTRY• Free-Energy Change ( G): –The difference between the change in enthalpy, H, and the product of the Kelvin temperature and the entropy change, which is defined as T S. –Equation for Free Energy Change: G 0 = H 0 - T S0
77. 77. THERMOCHEMISTRY• Each of the variables in the equation can have positive or negative values, this leads to four possible combinations of terms.• The more negative H is, the more negative G is likely to be.• The more positive S is, the more negative G is likely to be.
78. 78. THERMOCHEMISTRY• Reactions systems that change from a high enthalpy state to a low one tend to proceed spontaneously.• Also, systems that change from a well-ordered state to a highly disordered state also tend to proceed spontaneously.
79. 79. THERMOCHEMISTRY• G can either be positive or negative, depending on the temperature (T). –If the temperature of the system is the dominant factor that determines the relative importance of the tendencies toward lower energy and higher entropy.
80. 80. THERMOCHEMISTRYEXAMPLE:• Consider the reaction between water and carbon to produce carbon monoxide and hydrogen. H2O(g) + C(s) --> CO(g) + H2(g) H = +131.3 kJ S = +0.134 kJ/(mol•K) T = 298 K
81. 81. THERMOCHEMISTRY• Calculate free energy change and tell whether or not this reaction is spontaneous or not. G = 131.3 - (298)(0.134) G = 131.4 - 39.9 G = +91.4 kJ/mol
82. 82. THERMOCHEMISTRY• Since G has a positive value, the reaction that produces produces hydrogen gas is not spontaneous at a relatively low temperature of 298 K.
83. 83. THERMOCHEMISTRY• Increases in temperature tend to favor increases in entropy.
84. 84. THERMOCHEMISTRYEXAMPLE:• For the reaction NH4Cl(s) --> NH3(g) + HCl(g), at 25°C, H = 176 kJ/mol and S = 0.285 kJ/(mol•K). Calculate G and decide if this reaction will be spontaneous in the forward direction.
85. 85. THERMOCHEMISTRY G = 176 - (298)(0.285) G = 176 - 84.9 G = 91 kJ/mol• The positive value of G shows that this reaction is not spontaneous at 25°C.
86. 86. THERMOCHEMISTRYEXAMPLE:• For the vaporization reaction Br2(l) --> Br2(g), H = 31.0 kJ/mol and S = 93.0 kJ/(mol•K). At what temperature will this process be spontaneous?
87. 87. THERMOCHEMISTRY• No matter how simple a balanced equation, the reaction pathway may be complicated and difficult to determine.• Chemists believe that simple, one step reaction mechanisms are unlikely, and that nearly all reactions are complicated.
88. 88. THERMOCHEMISTRYEXAMPLE:• A reaction between colorless H2 gas and violet colored I2 gas at elevated temperatures produces hydrogen iodide, a colorless gas. HI molecules, in turn, tend to decompose and re-form hydrogen and iodine.
89. 89. THERMOCHEMISTRY H2(g) + I2(g) --> 2HI(g) 2HI(g) --> H2(g) + I2(g)• These types of reactions do not show the reaction pathway by which either reaction proceeds.
90. 90. THERMOCHEMISTRY• Reaction Mechanism: –The step-by-step sequence of reactions by which the overall chemical change occurs.• Most reactions take place in a sequence of simple steps.
91. 91. THERMOCHEMISTRY• This reaction is also a Homogenous reaction.• Homogenous Reaction: –A reaction whose reactants and products exist in a single phase. –In such a reaction, all reactants and products in intermediate steps are in the same phase.
92. 92. THERMOCHEMISTRY• The real reaction has 2 possible mechanisms: First Reaction Mechanism I2 <--> 2I 2I + H2 <--> 2HI I2 + H2 <--> 2HI
93. 93. THERMOCHEMISTRY Second Reaction Mechanism I2 <--> 2I I + H2 <--> H2I H2I + I <--> 2HI I2 + H2 <--> 2HI• Some species that appear in some steps, but not in the net equation are known as intermediates.
94. 94. THERMOCHEMISTRY• Collision Theory: –A set of assumptions regarding collisions and reactions. –In order for reactions to occur between substances, their particles (molecules, atoms, or ions) must collide.
95. 95. THERMOCHEMISTRYEXAMPLE: HI + HI --> H2 + 2I• According to the collision theory, 2 HI molecules must collide in order to react.
96. 96. THERMOCHEMISTRY• The HI molecules must collide while favorably orientated and with enough energy to disrupt the bonds of the molecules.
97. 97. THERMOCHEMISTRY • If the collision is too gentle, the old bonds are not disrupted and new ones are not formed. They simply bounce off each other.
98. 98. THERMOCHEMISTRY • If the reactant molecules are poorly orientated, the collision has little effect.
99. 99. THERMOCHEMISTRY • If the reactants collide with enough energy and proper orientation, a reshuffling of bonds leads to the formation of 1 H2 molecule and 2 I atoms.
100. 100. THERMOCHEMISTRY• The collision theory provides two reasons why a collision between reactants molecules may fail to produce a new chemical species: –Collision is not energetic enough. –Colliding molecules are not orientated properly.
101. 101. THERMOCHEMISTRY• Activation Energy (Ea): –Energy required to transform the reactants into the activated complex.• Activated Complex: –A transitional structure that results from an effective collision and that persists while old bonds are breaking and new bonds are forming.
102. 102. THERMOCHEMISTRY Reading Energy Diagrams:• Symbols: – H = Change in Enthalpy (Heat) H = HPRODUCTS - HREACTANTS –Ea = Activation Energy of the forward reaction. –Ea’ = Activation Energy of the reverse reaction.
103. 103. THERMOCHEMISTRY
104. 104. THERMOCHEMISTRY• Reaction Rate: –The change in concentration of reactants per unit time as a reaction proceeds.• Chemical Kinetics: –The area of chemistry that is concerned with reaction rates and reaction mechanisms.
105. 105. THERMOCHEMISTRY• There are 5 important factors that influence the rate of a chemical reaction:1. Nature of Reactants:• Substances vary greatly in their tendencies to react.• The rate of reaction depends on the particular reactants and bonds involved.
106. 106. THERMOCHEMISTRY2. Surface Area:• The more surface area an object has the faster the reaction will occur.
107. 107. THERMOCHEMISTRY3. Concentration:• Increasing the concentration of a substance could increase the rate of reaction or it could have no effect on the reaction. The effect depends on the particular reaction.
108. 108. THERMOCHEMISTRY• Reactions take place in a series of steps. The step that proceeds the slowest determines the overall rate of reaction.• The slowest step is called the Rate-Determining Step.
109. 109. THERMOCHEMISTRY4. Temperature:• The rates of many reactions roughly double or triple with a 10° C rise in temperature.• The increase in temperature increases the energy of the particles in the reaction, thus increasing the number of collisions between those particles.
110. 110. THERMOCHEMISTRY5. Presence of a Catalyst:• A substance that changes the rate of a chemical reaction without itself being permanently consumed.• The action of a catalyst is called catalysis.• Catalysts do not appear among the final products of reactions the accelerate.
111. 111. THERMOCHEMISTRY• A catalyst may be effective in forming an alternative activated complex that requires a lower activation energy.
112. 112. THERMOCHEMISTRY• A catalyst that is in the same phase as all the reactants and products in a reaction system is called a homogeneous catalyst.• When its phase is different from that of the reactants, it is called a heterogeneous catalyst.
113. 113. THERMOCHEMISTRY• Sometimes chemists need to slow down the rate of reaction, to do so they add an inhibitor.• Inhibitor: –A substance that slows down a process. –A negative catalyst.
114. 114. THERMOCHEMISTRY• The relationship between rate of reaction and the concentration of one reactant is determined experimentally by first keeping the concentrations of other reactants and the temperature of the system constant.
115. 115. THERMOCHEMISTRY• Then the reaction rate is measured for various concentrations of the reactant in question.• A series of such experiments reveals how the concentration for each reactant affects the reaction rate.
116. 116. THERMOCHEMISTRYEXAMPLE:• Hydrogen gas reacts with nitrogen monoxide gas at constant volume and at an elevated constant temperature, according to the following equation:2H2(g) + 2NO(g) --> N2(g) + 2 H2O(g)
117. 117. THERMOCHEMISTRY• Four moles of reactant gases produce three moles of product gases; thus, the pressure of the system diminishes as the reaction proceeds.• The rate of reaction can, therefore, be determined by measuring the change of pressure in the vessel with time.
118. 118. THERMOCHEMISTRY• Suppose a series of experiments is conducted using the same initial concentration of NO but different initial concentrations of H2.
119. 119. THERMOCHEMISTRY• The initial reaction rate is found to vary directly with H2 concentration: –Doubling the concentration of H2 doubles the rate. –Tripling the concentration of H2 triples the rate.
120. 120. THERMOCHEMISTRY• If R represents the reaction rate and [H2] is the concentration of hydrogen in moles per liter, the mathematical relationship between rate and concentration can be written as: R [H2]• = “is proportional to”
121. 121. THERMOCHEMISTRY• Now suppose the same initial concentration of H2 is used but the initial concentration of NO is varied.• The initial reaction rate is found to increase fourfold when the NO concentration is doubled and ninefold when the concentration of NO is tripled.
122. 122. THERMOCHEMISTRY• Thus, the reaction rate varies directly with the square of the NO concentration. R [NO]2
123. 123. THERMOCHEMISTRY• Because R is proportional to [H2] and to [NO]2, it is proportional to their product: R [H2][NO]2
124. 124. THERMOCHEMISTRY• By introduction of an appropriate proportionality constant, k, the expression becomes an equality: R = k [H2][NO]2
125. 125. THERMOCHEMISTRY• An equation that relates reaction rate and concentration of reactants is called the rate law.• It is applicable for a specific reaction at a given temperature.• A rise in temperature increases the reaction rates of most reactions.
126. 126. THERMOCHEMISTRY• The value of k usually increases as the temperature increases, but the relationship between reaction rate and concentration almost always remains unchanged.
127. 127. THERMOCHEMISTRY• The form of rate law depends on the reaction mechanism.• For a reaction that occurs in a single step, the reaction rate of that step is proportional to the product of the reactant concentrations, each of which is raised to its stoichiometric coefficient.
128. 128. THERMOCHEMISTRYEXAMPLE:• Suppose one molecule of a gas A collides with one molecule of gas B to form two molecules of substance C. A + B --> 2C
129. 129. THERMOCHEMISTRY• One particle of each reactant is involved in each collision. Thus, doubling the concentration of either reactant will double the collision frequency.• It also will double the reaction rate for this step.
130. 130. THERMOCHEMISTRY• Therefore, the rate for this step is directly proportional to the concentration of A and B.• The rate law for this one-step reaction follows: R = k[A][B]
131. 131. THERMOCHEMISTRY• Now suppose the reaction is reversible.• In the reverse step, 2 molecules of C must decompose to form one molecule of A and one of B. 2C --> A + B
132. 132. THERMOCHEMISTRY• Thus, the reaction rate for this reverse step is directly proportional to [C] × [C].• The rate law for this step is: R = k[C]2
133. 133. THERMOCHEMISTRY• The power to which the molar concentration of each reactant is raised in the rate laws above corresponds to the coefficient for the reactant in the balanced chemical equation.
134. 134. THERMOCHEMISTRY• Such a relationship holds only if the reaction follows a simple one- step path.• If a chemical reaction proceeds in a series of steps, the rate law is determined from the slowest step because it has the lowest rate. –Rate Determining Step.
135. 135. THERMOCHEMISTRYEXAMPLE:• Consider the following reaction:NO2(g) + CO(g) --> NO(g) + CO2(g)
136. 136. THERMOCHEMISTRY• The reaction is believed to be a two step process represented by the following mechanism:Step 1: NO2 + NO2 --> NO3 + NO slowStep 2: NO3 + CO --> NO2 + CO2 fast
137. 137. THERMOCHEMISTRY• The first step is the slower of the two and is therefore the rate- determining step.• We can write the rate law from this rate-determining step: R = k[NO2]2
138. 138. THERMOCHEMISTRY• CO does not appear in the rate law because it reacts after the rate-determining step, so the reaction rate will not depend on [CO].
139. 139. THERMOCHEMISTRY• The general form for the rate law is given by the following equation: R = k[A]n[B]m….
140. 140. THERMOCHEMISTRY• The reaction rate is represented by R, k is the rate constant, and [A] and [B]… represent the molar concentrations of reactants.• The n and m are the respective powers to which the concentrations are raised. They must be determined from experimental data.