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UNIT 3: Energy Changes and Rates of Reaction

UNIT 3: Energy Changes and Rates of Reaction. Chapter 6: Rates of Reaction. UNIT 3. Chapter 6: Rates of Reaction. Chapter 6: Rates of Reaction.

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UNIT 3: Energy Changes and Rates of Reaction

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  1. UNIT 3: Energy Changes and Rates of Reaction Chapter 6: Rates of Reaction

  2. UNIT 3 Chapter 6: Rates of Reaction Chapter 6: Rates of Reaction An important part of studying chemical reactions is to monitor the speed at which they occur. Chemists look at how quickly, or slowly, reactions take place and how these rates of reaction are affected by different factors. The light produced by a firefly depends on the speed of a particular chemical reaction that occurs in its abdomen. TO PREVIOUS SLIDE

  3. UNIT 3 Chapter 6: Rates of Reaction Section 6.1 6.1 Chemical Reaction Rates Chemical kinetics is the study of the rate at which chemical reactions occur. • The term reaction rate, or rate of reaction refers to: • the speed that a chemical reaction occurs at, or • the change in amount of reactants consumed or products formed over a specific time interval TO PREVIOUS SLIDE

  4. UNIT 3 Chapter 6: Rates of Reaction Section 6.1 Determining Reaction Rates The reaction rate is often given in terms of the change in concentration of a reactant or product per unit of time. The change in concentration of reactant A was monitored over time. TO PREVIOUS SLIDE

  5. UNIT 3 Chapter 6: Rates of Reaction Section 6.1 Determining Reaction Rates The change in concentration of reactant or product over time is often graphed. For the reaction A → B, over time, the concentration of A decreases, and the concentration of B increases. TO PREVIOUS SLIDE

  6. UNIT 3 Chapter 6: Rates of Reaction Section 6.1 Average and Instantaneous Reaction Rates • Average rate of reaction: • change in [reactant] or [product] over a given time period (slope between two points) • Instantaneous rate of reaction: • the rate of a reaction at a particular point in time (slope of the tangent line) Average rate of reaction and instantaneous rate of reaction can be determined from a graph of concentration vs. time. TO PREVIOUS SLIDE

  7. UNIT 3 Chapter 6: Rates of Reaction Section 6.1 Expressing Reaction Rates in Terms of Reactants or Products A known change in concentration of one reactant or product and coefficients of a chemical equation allows determination of changes in concentration of other reactants or products. TO PREVIOUS SLIDE

  8. UNIT 3 Chapter 6: Rates of Reaction Section 6.1 Express the rate of formation of ammonia relative to hydrazine, for the reaction on the previous slide. Answer on the next slide TO PREVIOUS SLIDE

  9. UNIT 3 Chapter 6: Rates of Reaction Section 6.1 The mole ratio of ammonia to hydrazine is 4:3 TO PREVIOUS SLIDE

  10. UNIT 3 Chapter 6: Rates of Reaction Section 6.1 Methods for Measuring Rates of Reaction TO PREVIOUS SLIDE

  11. UNIT 3 Chapter 6: Rates of Reaction Section 6.1 Calculating Reaction Rates from Experimental Data The following data were collected in order to calculate the rate of a reaction. TO PREVIOUS SLIDE

  12. UNIT 3 Chapter 6: Rates of Reaction Section 6.1 Calculating Reaction Rates from Experimental Data TO PREVIOUS SLIDE

  13. UNIT 3 Chapter 6: Rates of Reaction Section 6.1 Calculating Reaction Rates from Experimental Data TO PREVIOUS SLIDE

  14. UNIT 3 Chapter 6: Rates of Reaction Section 6.1 Section 6.1 Review TO PREVIOUS SLIDE

  15. UNIT 3 Chapter 6: Rates of Reaction Section 6.2 6.2 & 6.3: Collision Theory and Factors Affecting Rates of Reaction According to collision theory, a chemical reaction occurs when the reacting particles collide with one another. Only a fraction of collisions between particles result in a chemical reaction because certain criteria must be met. TO PREVIOUS SLIDE

  16. UNIT 3 Chapter 6: Rates of Reaction Section 6.2 Effective Collision Criteria 1:The Correct Orientation of Reactants For a chemical reaction to occur, reactant molecules must collide with the correct orientation relative to each other (collision geometry). Five of many possible ways that NO(g) can collide with NO3(g) are shown. Only one has the correct collision geometry for reaction to occur. TO PREVIOUS SLIDE

  17. UNIT 3 Chapter 6: Rates of Reaction Section 6.2 Effective Collision Criteria 2:Sufficient Activation Energy For a chemical reaction, reactant molecules must also collide with sufficient energy. Activation energy, Ea, is the minimum amount of collision energy required to initiate a chemical reaction. Collision energy depends on the kinetic energy of the colliding particles. The shaded part of the Maxwell-Boltzmann distribution curve represents the fraction of particles that have enough collision energy for a reaction (ie the energy is ≥ Ea). TO PREVIOUS SLIDE

  18. UNIT 3 Chapter 6: Rates of Reaction Section 6.2 Representing the Progress of a Chemical Reaction • From left to right on a potential energy curve for a reaction: • potential energy increases as reactants become closer • when collision energy is ≥ maximum potential energy, reactants will transform to a transition state • products then form (or reactants re-form if ineffective) Exothermic Endothermic TO PREVIOUS SLIDE

  19. UNIT 3 Chapter 6: Rates of Reaction Section 6.2 Activation Energy and Enthalpy • The Ea for a reaction cannot be predicted from ∆H. • ∆H is determined only by the difference in potential energy between reactants and products. • Ea is determined by analyzing rates of reaction at differing temperatures. • Reactions with low Ea occur quickly. Reactions with high Ea occur slowly. Potential energy diagram for the combustion of octane. TO PREVIOUS SLIDE

  20. UNIT 3 Chapter 6: Rates of Reaction Section 6.2 Activation Energy for Reversible Reactions • Potential energy diagrams can represent both forward and reverse reactions. • follow left to right for the forward reaction • follow right to left for the reverse reaction TO PREVIOUS SLIDE

  21. UNIT 3 Chapter 6: Rates of Reaction Section 6.2 Analyzing Reactions Using Potential Energy Diagrams The BrCH3 molecule and OH- collide with the correct orientation and sufficient energy and an activated complex forms. When chemical bonds reform, potential energy decreases and kinetic energy increases as the particles move apart. TO PREVIOUS SLIDE

  22. Chapter 6: Rates of Reaction UNIT 3 Learning Check Describe the relative values of Ea(fwd) and Ea(rev) for an exothermic reaction Answer on the next slide TO PREVIOUS SLIDE

  23. UNIT 3 Chapter 6: Rates of Reaction Section 6.2 Learning Check Ea(rev) is greater than Ea(fwd) TO PREVIOUS SLIDE

  24. UNIT 3 Chapter 6: Rates of Reaction Section 6.2 Factors Affecting Reaction Rate • 1. Nature of reactants • reactions of ions tend to be faster than those of molecules • 2. Concentration • a greater number of effective collisions are more likely with a higher concentration of reactant particles • 3. Temperature • with an increase in temperature, there are more particles with sufficient energy needed for a reaction (energy is ≥ Ea) TO PREVIOUS SLIDE

  25. UNIT 3 Chapter 6: Rates of Reaction Section 6.2 Factors Affecting Reaction Rate • 4. Pressure • for gaseous reactants, the number of collisions in a certain time interval increases with increased pressure • 5. Surface area • a greater exposed surface area of solid reactant means a greater chance of effective collisions • 6. Presence of a catalyst • a catalyst is a substance that increases a reaction rate without being consumed by the reaction TO PREVIOUS SLIDE

  26. UNIT 3 Chapter 6: Rates of Reaction Section 6.2 A Catalyst Influences the Reaction Rate • A catalyst lowers the Ea of a reaction. • this increases the fraction of reactants that have enough kinetic energy to overcome the activation energy barrier • a catalyzed reaction has the same reactants, products, and enthalpy change as the uncatalyzed reaction A catalyst decreases both Ea(fwd) and Ea(rev). TO PREVIOUS SLIDE

  27. UNIT 3 Chapter 6: Rates of Reaction Section 6.2 Catalysts in Industry A metal catalyst is used for industrial-scale production of ammonia from nitrogen and hydrogen. Hydrogen and nitrogen molecules break apart when in contact with the catalyst. These highly reactive species then recombine to form ammonia. A catalyst (V2O5) is used for industrial-scale production of sulfuric acid from sulfur. TO PREVIOUS SLIDE

  28. UNIT 3 Chapter 6: Rates of Reaction Section 6.2 Catalysts in Industry The Ostwald process uses a platinum-rhodium catalyst for the industrial production of nitric acid. Many industries use biological catalysts, called enzymes, which are most often proteins. For example: the use of enzymes decreases the amount of bleach (an environmental hazard) needed to whiten fibres used in paper production. TO PREVIOUS SLIDE

  29. UNIT 3 Chapter 6: Rates of Reaction Section 6.2 Section 6.2 Review TO PREVIOUS SLIDE

  30. UNIT 3 Chapter 6: Rates of Reaction Section 6.3 6.5: Rate Law Initial rate is the rate of a chemical reaction at time zero. • products of the reaction are not present, so the reverse reaction cannot occur • it is a more accurate method for studying the relationship between concentration of reactant and reaction rate Initial rate is found by determining the slope of a line tangent to the curve at time zero. TO PREVIOUS SLIDE

  31. UNIT 3 Chapter 6: Rates of Reaction Section 6.3 The Rate Law The rate law shows the relationship between reaction rates and concentration of reactants for the overall reaction. rate = k[A]m[B]n m: order of the reaction for reactant A n: order of the reaction for reactant B k: rate constant m + n: order of the overall reaction TO PREVIOUS SLIDE

  32. UNIT 3 Chapter 6: Rates of Reaction Section 6.3 Graphing Reaction Rate in Terms of Concentration • To study the effects of concentration on reaction rate: • different starting concentrations of reactant are used • initial rates are calculated using the slopes of the tangent lines from concentration vs time curves • initial rates are plotted against starting concentration Initial rates are determined (A) and these are plotted against concentration (B). TO PREVIOUS SLIDE

  33. UNIT 3 Chapter 6: Rates of Reaction Section 6.3 First-order Reactions The initial rate vs starting concentration graph on the previous slide is a straight line. • the equation of the line can be expressed as: rate = k[A] • This represents a first-order reaction • For reactions with more than one reactant (e.g. A and B): • if experiments for each reactant produce straight lines, the rate is “first order with respect to reactant A and first order with respect to reactant B.” TO PREVIOUS SLIDE

  34. Example1: • The empirically determined rate law equation for the reaction • between nitrogen dioxide and fluorine gas : • 2NO2 + F2 → 2NO2F • r = K[NO2]1[F2]1 • The order of reaction with respect to NO2, F2 is 1 respectively. • the overall order of reaction is (1 + 1) = 2 ( The reaction is second order overall)

  35. Example 2: • The discomposition of dinitrogen pentoxide is first order with respect to N2O5. If the initial rate of consumption is 2.1 x 10-4 mol/(L*s) • when the initial concentration of N2O5 is 0.40 mol/L. Predict what the rate would be if another experiment were performed in which the initial concentration of N2O5 were 0.80 mol/L • 2N2O5 → 2NO2 +O2

  36. Solution: • First: write the rate law equation for the system: • r = K[N2O5]1 (first order reaction, so exponent is 1) • r = 0.8 mol/L = x • 0.4 mol/L 2.1 x 10-4mol/(L*s) • Since it is first order, any change in [N2O5] will have the same effect • on the rate. The [N2O5] is doubled from 0.4 to 0.8 mol/L, so the rate • of consumption will double from 2x (2.1 x 10-4) to 4.2 x 10-4 • mol/(L*s)

  37. UNIT 3 Chapter 6: Rates of Reaction Section 6.3 Second-order Reactions • For chlorine dioxide in this reaction: • the initial rate vs concentration curve is parabolic • the reaction is proportional to the square of [ClO2] • it is a second order reaction with respect to this reactant rate = k[A]2 TO PREVIOUS SLIDE

  38. Example 3: • The dimerization reaction of 1,3-butadiene (C4H6) is a second • order with respect to C4H6. If the initial rate of reaction were • 32 mmol/(L*min) C4H6 at a given initial concentration of C4H6,. • What would be the initial rate of reaction if the initial • concentration of C4H6 were doubled • 2C4H6 → C8H12

  39. Solution: • The rate equation is r = K[C4H6]2 (let x be the initial concentration) • x = 32 • 2x2 • If the initial concentration is doubled (multiplied by 2), the initial • rate will be multiplied by 22, or 4. The new rate is 4 x 32 mmol • C4H6/(L*min) or 0.13 mol /(L*min) C4H6

  40. DETERMINATION OF EXPONENT OF THE RATE LAW • The exponents of the Rate Law must be determined • experimentally. To find the exponent of the rate law, we must • study how changes in concentration affect the rate of reaction. • Example: - • [A]+ B → products •  Rate = k[A]m[B]n

  41. DETERMINATION OF EXPONENT OF THE RATE LAW • The exponents of the Rate • Law must be determined • experimentally by study • how changes in • Concentration affect the • Rate of reaction. • Example: - • A + B → products •  Rate = k[A]m[B]n • Example: Experimental Data • Initial Concentration Initial rate of formation • of products • [A] [B] • 1. 0.10 0.10 0.20 • 2. 0.20 0.10 0.40 • 3. 0.30 0.10 0.60 • 4. 0.30 0.20 2.40 • 5. 0.30 0.30 5.40 To determine the value of m, we use the results from Exp.# 1 and 2 or 3, in which only [A] changes: rate 2 = 0.4 mol/L.s =k(0.2 mol/L)m rate 1 0.2 mol/L.s k(0.1mol/L)m 2.0 = (2.0)m

  42. Meaning, if [A] is doubled the rate doubles (#1 & 2).  Also, if [A] is tripled (#1 &3), the rate is tripled .Therefore, the exponent of A must be 1 ie [A]1 To determine n, we use the results from Exp. #3 and 4 or 5, in which only [B] changes: This time it is the concentration of B that affects the rate rate 4 = 2.4 mol/L.s = k(0.20 mol/L)n rate 3 0.6 mol/L.s K(0.10 mol/L)n n 4 = 2 When [B] is doubled the rate increases by factor of 4 which equals 22

  43. [A] [B] Rates • 1) 0.10 0.10 0.20 • 2) 0.20 0.10 0.40 • 3) 0.30 0.10 0.60 • 4) 0.30 0.20 2.40 • 5) 0.30 0.30 5.40 • rate 5 = 5.4 mol/L.s = k(0.30 mol/L)n • rate 3 0.60 mol/L.s k(0.10 mol/L)n • 9 = 3

  44. Calculating Constant, K, from the Rate Law: • Using the data from set # 1, K can be calculated from the • r= k [A]1 [B]2 • K=Rate • [A][B] • K =0.20 mol L-1s-1 • (0.10 mol L-1)(0.10 mol L-1) • K=0.2 mol L-1s-1 • 0.010 mol2 L-2 • K=2.0 x 101 L mol-1 s-1 • Let’s check other sample on P.378-379 [A] [B] Rates 0.10 0.10 0.20 0.20 0.10 0.40 0.30 0.10 0.60 0.30 0.20 2.40 0.30 0.30 5.40

  45. Try this Example: -Determine the exponents of the Rate Law when the following data were measured for the reduction of Nitric Oxide with Hydrogen 2NO(g)+2H2(g)→N2(g)+2H2O(g) • Initial Concentrations Initial Rate of Disappearance of NO • (mol L-1s-1) • [NO] [H2] • 1) 0.10 0.10 1.37 X 10-3 • 2) 0.10 0.20 2.75 X 10-3 • 3) 0.20 0.10 5.47 X 10-3 • What is the rate law for this reaction? • What is the unit of the constant? Solution: 1. Find the order of the reaction 2. Solve for K value and unit 3. Then write the rate law Equation

  46. Solution:The general form of the rate law for this reaction:Rate = K[NO]m[H2]n To determine m, use result where only NO changes: Rate 3 = 5.47x10-3 mol/L*s= 0.20 mol/L Rate 1 1.37 x 10-3mol/L 0.10 mol/L 4 = [2]n, ie [2]2 , Thus m = 2 To determine n, use result where only [H2] changes: Rate 2 = 2.75x10-3 mol/L*s= 0.20 mol/L Rate 1 1.37 x 10-3mol/L 0.10 mol/L 2.0 = [2]n, ie [2]1 , Thus m = 1 Therefore, r = k[NO]2[H2]1

  47. Using trial # 1, find k 0.00137 mol/L.s = k x 0.100 mol/L x (0.100 mol/L)2 k = 0.00137 mol/L.s = 1.37 L2/mol2.s 0.100 mol/L x (0.100 mol/L)2 r = 1.37 L2/(mol2.s)[NO]2[H2] Note: Monitor the Unit of the rate constant, K, closely as it varies with different order of reactions

  48. UNIT 3 Chapter 6: Rates of Reaction Section 6.3 6.6: Reaction Mechanisms A reaction mechanism is the series of elementary steps that occur as reactants are converted to products. For example, oxygen and nitrogen are not formed directly from the decomposition of nitrogen dioxide: It occurs in two elementary steps: TO PREVIOUS SLIDE

  49. A balanced equation does not tell us how the reactants become product, it only tells us the reactant, the product, and the stoichiometry, but gives no information about the reaction mechanism. O is an intermediate, it is neither a reactant nor a product, but formed and consumed during the reaction sequence. Each of the two reactions is called an elementary step, a reaction whose rate law can be written from its molecules. A reaction with one molecule is called a unimolecular step, two or three molecules are called bimolecular and termolecular steps respectively

  50. A unimolecular step is always first order,bimolecular, second order Termolecular steps are uncommon • Elementary step Molecularity Rate law • A → Product Unimolecular Rate = [A] • A + A → Products Bimolecular Rate = [A]2 • A + B → Products Bimolecular Rate = [A][B] • a reaction mechanism must satisfy two requirements: • The sum of the elementary steps must give the overall • balanced equation for the reaction • 2. The mechanism must agree with the experimentally • determined rate law.

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