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

UNIT 3 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. TO PREVIOUS SLIDEPREVIOUS The light produced by a firefly depends on the speed of a particular chemical reaction that occurs in its abdomen. Chapter 6: Rates of Reaction

UNIT 3 Section Chemical Reaction Rates TO PREVIOUS SLIDEPREVIOUS Chemical kinetics is the study of the rate at which chemical reactions occur. Chapter 6: Rates of Reaction 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

UNIT 3 Section 6.1 Determining Reaction Rates TO PREVIOUS SLIDEPREVIOUS 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. Chapter 6: Rates of Reaction

UNIT 3 Section 6.1 Determining Reaction Rates TO PREVIOUS SLIDEPREVIOUS 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. Chapter 6: Rates of Reaction

UNIT 3 Section 6.1 Average and Instantaneous Reaction Rates TO PREVIOUS SLIDEPREVIOUS 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. Chapter 6: Rates of Reaction

UNIT 3 Section 6.1 Expressing Reaction Rates in Terms of Reactants or Products TO PREVIOUS SLIDEPREVIOUS 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. Chapter 6: Rates of Reaction

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

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

UNIT 3 Section 6.1 Methods for Measuring Rates of Reaction TO PREVIOUS SLIDEPREVIOUS Chapter 6: Rates of Reaction

Section 6.1 Review UNIT 3 Section 6.1 TO PREVIOUS SLIDEPREVIOUS Chapter 6: Rates of Reaction

UNIT 3 Section & 6.3: Collision Theory and Factors Affecting Rates of Reaction TO PREVIOUS SLIDEPREVIOUS 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. Chapter 6: Rates of Reaction

UNIT 3 Section 6.2 Effective Collision Criteria 1: The Correct Orientation of Reactants TO PREVIOUS SLIDEPREVIOUS 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 NO 3 (g) are shown. Only one has the correct collision geometry for reaction to occur. Chapter 6: Rates of Reaction

UNIT 3 Section 6.2 Effective Collision Criteria 2: Sufficient Activation Energy TO PREVIOUS SLIDEPREVIOUS 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 ≥ E a ). Chapter 6: Rates of Reaction For a chemical reaction, reactant molecules must also collide with sufficient energy. Activation energy, E a, is the minimum amount of collision energy required to initiate a chemical reaction. Collision energy depends on the kinetic energy of the colliding particles.

UNIT 3 Section 6.2 Representing the Progress of a Chemical Reaction TO PREVIOUS SLIDEPREVIOUS Chapter 6: Rates of 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) ExothermicEndothermic

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

UNIT 3 Section 6.2 Activation Energy for Reversible Reactions TO PREVIOUS SLIDEPREVIOUS 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 Chapter 6: Rates of Reaction

UNIT 3 Section 6.2 Analyzing Reactions Using Potential Energy Diagrams TO PREVIOUS SLIDEPREVIOUS The BrCH 3 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. Chapter 6: Rates of Reaction

Describe the relative values of E a(fwd) and E a(rev) for an exothermic reaction UNIT 3 Answer on the next slide TO PREVIOUS SLIDEPREVIOUS Chapter 6: Rates of Reaction L EARNING C HECK

E a(rev) is greater than E a(fwd) Section 6.2 UNIT 3 TO PREVIOUS SLIDEPREVIOUS Chapter 6: Rates of Reaction L EARNING C HECK

UNIT 3 Section 6.2 Factors Affecting Reaction Rate TO PREVIOUS SLIDEPREVIOUS 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 ≥ E a ) Chapter 6: Rates of Reaction

UNIT 3 Section 6.2 Factors Affecting Reaction Rate TO PREVIOUS SLIDEPREVIOUS Chapter 6: Rates of Reaction 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

UNIT 3 Section 6.2 A Catalyst Influences the Reaction Rate TO PREVIOUS SLIDEPREVIOUS A catalyst lowers the E a 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 E a(fwd) and E a(rev). Chapter 6: Rates of Reaction

UNIT 3 Section 6.2 Catalysts in Industry TO PREVIOUS SLIDEPREVIOUS 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. Chapter 6: Rates of Reaction A catalyst (V 2 O 5 ) is used for industrial-scale production of sulfuric acid from sulfur.

UNIT 3 Section 6.2 Catalysts in Industry TO PREVIOUS SLIDEPREVIOUS The Ostwald process uses a platinum-rhodium catalyst for the industrial production of nitric acid. Chapter 6: Rates of Reaction 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.

Section 6.2 & 6.3 Review UNIT 3 Section 6.2 TO PREVIOUS SLIDEPREVIOUS Chapter 6: Rates of Reaction

UNIT 3 Section : Rate Law TO PREVIOUS SLIDEPREVIOUS Initial rate is found by determining the slope of a line tangent to the curve at time zero. Chapter 6: Rates of Reaction 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

UNIT 3 Section 6.3 The Rate Law TO PREVIOUS SLIDEPREVIOUS The rate law shows the relationship between reaction rates and concentration of reactants for the overall reaction. Chapter 6: Rates of 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

UNIT 3 Section 6.3 Graphing Reaction Rate in Terms of Concentration TO PREVIOUS SLIDEPREVIOUS 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 Chapter 6: Rates of Reaction Initial rates are determined (A) and these are plotted against concentration (B).

UNIT 3 Section 6.3 First-order Reactions TO PREVIOUS SLIDEPREVIOUS The initial rate vs starting concentration graph on the previous slide is a straight line. Chapter 6: Rates of Reaction 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.”

Example1: The empirically determined rate law equation for the reaction between nitrogen dioxide and fluorine gas : 2NO 2 + F 2 → 2NO 2 F r = K[NO 2 ] 1 [F 2 ] 1 The order of reaction with respect to NO 2, F 2 is 1 respectively. the overall order of reaction is (1 + 1) = 2 ( The reaction is second order overall)

Example 2: The discomposition of dinitrogen pentoxide is first order with respect to N 2 O 5. If the initial rate of consumption is 2.1 x mol/(L*s) when the initial concentration of N 2 O 5 is 0.40 mol/L. Predict what the rate would be if another experiment were performed in which the initial concentration of N 2 O 5 were 0.80 mol/L 2N 2 O 5 → 2NO 2 +O 2

Solution: First: write the rate law equation for the system: r = K[N 2 O 5 ] 1 (first order reaction, so exponent is 1) r = 0.8 mol/L = x 0.4 mol/L 2.1 x mol/(L*s) Since it is first order, any change in [N 2 O 5 ] will have the same effect on the rate. The [N 2 O 5 ] is doubled from 0.4 to 0.8 mol/L, so the rate of consumption will double from 2x (2.1 x ) to 4.2 x mol/(L*s)

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

Example 3: The dimerization reaction of 1,3-butadiene (C 4 H 6 ) is a second order with respect to C 4 H 6. If the initial rate of reaction were 32 mmol/(L*min) C 4 H 6 at a given initial concentration of C 4 H 6,. What would be the initial rate of reaction if the initial concentration of C 4 H 6 were doubled 2C 4 H 6 → C 8 H 12

Solution: The rate equation is r = K[C 4 H 6 ] 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 2 2, or 4. The new rate is 4 x 32 mmol C 4 H 6 /(L*min) or 0.13 mol /(L*min) C 4 H 6

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

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] 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 mol/L.s k(0.1mol/L) m 2.0 = (2.0) m

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 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 2 2

[A] [B] Rates 1) ) ) ) ) rate 5 = 5.4 mol/L.s = k(0.30 mol/L) n rate mol/L.s k(0.10 mol/L) n 9 = 3

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 -1 s -1 (0.10 mol L -1 )(0.010 mol 2 L -2 ) K=0.2 mol L -1 s mol 2 L -2 K=2.0 x 10 2 L 2 mol -2 s -1 [A] [B] Rates

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) +2H 2(g) →N 2(g) +2H 2 O (g) Initial Concentrations Initial Rate of Disappearance of NO (mol L -1 s -1 ) [NO] [H 2 ] 1) X ) X ) X What is the rate law for this reaction? 2.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

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

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

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

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

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: 1.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.

UNIT 3 Section 6.3 The Rate-determining Step the slowest of the elementary steps in a complex reaction is called the rate-determining step. TO PREVIOUS SLIDEPREVIOUS This reaction occurs in three elementary steps: Chapter 6: Rates of Reaction Step 2 is the rate-determining step: it is the slowest elementary step the overall rate of the reaction is dependent on this step the E a for this step is higher than E a for each of the other steps

Potential Energy graph of Reaction Mechanism A B C D E

UNIT 3 Section 6.3 A Proposed Reaction Mechanism TO PREVIOUS SLIDEPREVIOUS Experiments show that this reaction is zero order with respect to OH – (i.e. its rate does not depend on [OH – ]) This can be explained by a two-step mechanism Step 2 is very fast and depends on completion of Step 1, not on the concentration of OH –. Chapter 6: Rates of Reaction

Potential Energy graph of Reaction Mechanism

Example: The balanced equation for a reaction of the gases nitrogen dioxide and fluorine is 2NO 2(g) + F 2(g) → 2NO 2 F (g) Experimentally determined rate law is Rate = k[NO 2 ][F 2 ] The suggested mechanism for this reaction is: NO 2 + F 2 → NO 2 F + F slow F + NO 2 → NO 2 F fast Is this an acceptable mechanism? Does it satisfy the requirements for reaction mechanism? 1. 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. 3. The reactant (s) of the slowest reaction gives the rate law equation.

Solution: First:The sum of the steps should give the balanced equation NO 2 + F 2 → NO 2 F + F F + NO 2 → NO 2 F ___________________ NO 2 + F 2 + F + NO 2 →NO 2 F + F + NO 2 F Overall reaction : 2NO 2 + F 2 → 2NO 2 F Second: Mechanism must agree with the experimentally determine rate. Since the first step is the slowest (the rate determining step), it must determine the reaction rate Rate = k[NO 2 ][F 2 ]