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17/11/2016 Chemical kinetics studies: Lecture 7 Kinetics rates of chemical reactions the factors that affect rate reactions the mechanisms (the series of steps) by which reactions occur 1 Kinetics, rate of the reaction, reaction mechanism 2 Rate of reaction The speed with which the reactants disappear and the products form is called the rate of the reaction. A study of the rate of reaction can give detailed information about how reactants change into products. The series of individual steps that add up to the overall observed reaction is called the reaction mechanism. We are all familiar with processes in which some quantity change with time Car travels at 40 km/hour (miles/hour) Faucet delivers water at 30 l/min (gallons/minute) Factory produces 32,000 tires/day Each of these ratios is called a rate 3 4 1 17/11/2016 A spectroscopic method for determining reaction rates. Light of the wavelength that is absorbed by the investigated substance is passed through a reaction chamber. The changes in the reactant or product concentration as reaction progresses cause the decrease or increase of the light intensity. Blue dye is reacting with bleach, which converts it into a colourless products. The colour decreases and eventually disappears. The rate of the reaction could be determined by repeatedly measuring both the colour intensity and the elapsed time. The concentration of the dye could be calculated from the intensity of the blue colour. 5 Quiz 6 Experience tells us that different chemical reactions occur at very different rates, e.g. Identify which of the following are rates: A) 15 cm B) 30 m / s2 C) 25oC D) 5oC/min E) 1.2 mol /min F) 45 min 1. combustion reactions – burning methane (component of natural gas) or combustion of isooctane (C8H18)in gasoline – proceed very rapidly, even explosively. 2. Erosion of the rocks – proceed very slowly. 3. Similarly the iron rusting is slow process. 7 8 2 17/11/2016 The reactions of strong acids with strong bases are thermodynamically favoured (spontaneous) and occur at very rapid rates. 3) C(diamond) + O2(g)→CO2(g) 1) 2HCl + Mg(OH)2 → MgCl2 + 2H2O ΔG0rxn = -591.8 kJ/mol Similarly reaction of some compound with oxygen (e.g. This reaction does not occur at an observable rate. 4) C(graphite) + O2(g) →CO2(g) burning) is thermodynamically favoured e.g. ΔG0rxn = - 397 kJ/mol ΔG0rxn = - 394 kJ/mol. This reaction occurs rapidly. 2) CH4 + 2O2 → CO2 + 2H2O The difference in the reaction speed of 3 and 4 reactions is ΔG0rxn = -800.8 kJ/mol explained by kinetics, not thermodynamics. 9 10 Rate, formula Rate of reaction, v, The rate of reaction describes how fast reactants are used up and products are formed. Knowledge of the rate of a reaction can be an invaluable tool in helping us to understand how chemical compounds behave when they interact. 11 A rate, is always expressed as a ratio. One way to describe a reaction rate is to select one component of the reaction and describe the change in its concentration per unit of time: 12 3 17/11/2016 rate with respect to X Quiz (conc. of X at time t2 conc. of X at time t1 ) (t 2 t1 ) An 8.00 g piece of magnesium was placed into 6.0 M HCl . After 25 s. 3.50 g of unreacted magnesium remained. The average rate at which magnesium was consumed is: A. 0.14 g/s B. 0.18 g/s C. 0.32 g/s D. 4.50 g/s (conc. of X ) t Molarity (mol/L) is normally presented as the concentration unit whereas the second (s) is the most often used unit of time. Typically, the reaction rate has the unit mol/L or mol L-1 s-1 s 13 14 Consider the following reaction: Consider the following reaction at a constant temperature in an open system: MgCO3(s) + 2HCl(aq) → CO2(g) + H2O(l) + MgCl2(aq) Which of the following properties could be used to determine reaction rate? A. Mass of the system B. Pressure of the gas C. Concentration of H2O D. Concentration of MgCO3 15 2CaCrO4(s) +2H+(aq) 2Ca2+(aq) + H2O(l) + Cr2O72- (aq) (orange) The progress of the reaction could be followed by observing the rate of A. mass loss B. decrease in pH C. precipitate formation D. formation of orange colour in the solution 16 4 17/11/2016 a A+ b B → c C+ d D If no other reaction takes place, the changes in rate= -Δ[A]/ Δt; concentration are related to one another. rate = -Δ[B]/ Δt, or For every a mol/L that described decrease of [A], [B] rate = Δ[C]/ Δt; must decrease by b mol/L, [C] must increase by c rate= Δ[D]/ Δt mol/L and so on… The reaction rate must be positive because it describes the The number of moles of reaction that occur per litter forward reaction, which consumes A and B. in a given time describe the rate of reaction. The concentration of reactants A and B decrease in time interval Δt. Δ[A]/ Δt and Δ[B]/ Δt are negative quantities. 17 18 19 20 a A+ b B → c C+ d D 5 17/11/2016 Quiz 2A + B →3C +D If the rate of disappearance of A is equal to -0.084 mol/L s at the start of the reaction what are the rates of change for B, C and D at this time? Rate of change of B = Rate of change of C = Rate of change of D = a) B= 0.042 M/s; C= 0.056 M/s; D= - 0.042 mol/L s b) B = -0.042M/s; C = 0.112 M/s; D = 0.042 mol/L s c) B= -0.042 M/s; C= - 0.112 M/s; D= 0.042 mol/L s 21 22 Consider the following reaction: N2H4(l) + 2H2O2(l) C3 H8 ( g ) 5O2 ( g ) 3CO2 ( g ) 4H 2O( g ) N2(g) + 4H20(l) Compared to the rate with respect to propane: ◦ Rate with respect to oxygen is five times faster ◦ Rate with respect to carbon dioxide is three times faster ◦ Rate with respect to water is four times faster In 1.0 seconds, 0.015 mol of H2O2 is consumed. The rate of production of N2 is A.1.5 x 10-3 mol/s B. 7.5 x 10-3 mol/s C. 6.0 x 10-3 mol/s Since the rates are all related any may be monitored to determine the reaction rate D. 1.5 x 10-2 mol/s 23 24 6 17/11/2016 Quiz Determine relative reaction rates of the four substrates involved in the following chemical reaction. Give the appropriate numbers instead w, x, y and z letters: 2C2H2(g) + 5O2(g) → 4CO2 + 2H2O(l) A reaction rate is generally not constant throughout the reaction. Since the most of reactions depend on the concentration of reactants, the rate changes as they are used up. The rate at any particular moment is called the instantaneous rate. It can be calculated from a concentration versus time plot. 25 26 Plot of [H2] vs time for the reaction of 1.000 M H2 with 2.000 M ICl. The instantaneous rate of reaction at any time, t, equals the negative slope of the tangent to this curve at time t. The initial ratio of the reaction is equal to the negative of the initial slope (t=0). The determination of the instantaneous rate at t=2 second is illustrated. A plot of the concentration of HI versus time for the reaction: 2HI(g) H2(g) + I2(g). The slope is negative because we are measuring the disappearance of HI. When used to express the rate it is used as a positive number. 27 28 7 17/11/2016 A rate law is a mathematical equation that describes the concentration Quiz Based on the graph below determine the instantaneous rate of change of [X] at 8 seconds [X] =….. progress of the reaction. In general, rate laws must be determined experimentally. Unless a reaction is an elementary reaction, it is not M/s possible to predict the rate law from the overall chemical equation. 0.10 There are two forms of a rate law for chemical kinetics: 0.06 0.04 the differential rate law and the integrated rate law. 2 8 24 s time 29 30 Rate constant, k For a generic reaction: rate = k [reactant 1]n [reactant 2]m reactant 1 + reactant 2 → product where k is the constant of proportionality with no intermediate steps in this reaction named rate constant. Its value is generally mechanism, the rate law is given by constant provided that reaction is performed at Rate = k [reactant 1]n [reactant 2]m constant temperature T. Values of k are always positive, although it may be an exception of this rule. 31 32 8 17/11/2016 Consider the following reaction: Order of reaction H 2 SeO3 6 I 4 H Se 2 I 3 3H 2O From experiment, the rate law (determined from initial rates) is rate k[ H 2 SeO3 ]1[ I ]3[ H ]2 At 0oC, k equals 5.0 x 105 L5 mol-5 s-1 Thus, at 0oC The exponents in the rate law are generally unrelated to the chemical equation’s coefficients. The exponent in a rate law is called the order of reaction with respect to the corresponding reactant. ◦ Never simply assume that the exponents and coefficients are the same. ◦ The exponents must be determined from the results of experiments. rate (5.0 105 L5 mol -5 s 1 ) [ H 2 SeO3 ][I ]3[ H ]2 33 34 Reaction order quiz For the rate law: Analyse the following rate equations, and determine the orders of reactions with respect to reactant and overall reaction order: 1. rate = k [Cu2+] [NH3] 2. rate = k [OH-] 3. rate = k [NO]2[O2] 4. rate = k [A]3[B]2 rate k[ H 2 SeO3 ]1[ I ]3[ H ]2 We can say ◦ ◦ ◦ ◦ The The The The reaction is reaction is reaction is reaction is first order with respect to H2SeO3 third order with respect to Isecond order with respect to H+ sixth order overall Exponents (orders of reactions) in a rate law can be fractional, negative, and even zero. 35 36 9 17/11/2016 Looking for patterns in experimental data provide way to determine the exponents in a rate law. One of the easiest ways to reveal patterns in data is to form ratios of results using different sets of conditions. This technique is generally applicable. Consider the hypothetical reaction: Suppose the experimental concentration-rate data for five experiments is: Inital Conc. [ A] [ B] Expt 1 2 3 4 5 A B products rate k[ A]m [ B]n Initial Rate (mol L-1 ) (mol L-1 ) (mol L-1 s -1 ) 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 37 For experiments 1, 2, and 3 [B] is held constant, so any change in rate must be due to changes in [A] The rate law says that at constant [B] the rate is proportional to [A]m m rate2 [ A]2 rate1 [ A]1 rate2 0.40 mol L-1 s -1 2 rate1 0.20 mol L-1 s -1 m 38 For experiments 3, 4, and 5 [A] is held constant, so any change must be due to changes in [B] The rate law says that at constant [A] the rate is proportional to [B]n Using the results from experiment 3 and 4: m [ A]2 0.20 mol L-1 2m -1 [ A ] 0.10 mol L 1 Thus m=1 39 40 10 17/11/2016 n rate 4 [ B]4 rate3 [ B]3 rate 4 2.40 mol L-1 s -1 4 rate3 0.60 mol L-1 s -1 n Thus n=2 k n [ B]4 0.20 mol L-1 2n -1 [ B]3 0.10 mol L The rate constant (k) can be determined using data from any experiment Using experiment 1: rate 0.20 mol L-1 s -1 2 [ A][B] (0.10 mol L-1 )(0.10 mol L-1 ) 2 2.0 102 L2 mol -2 s -1 The reaction is second order in respect to B and rate=k[A]1[B]2 Using data from a different experiment might give a slightly different value of k 41 42 Quiz Quiz For 2NO + O2 --> 2NO2 , initial rate data are: The reaction has the rate law: rate = k[C][D]2. What will happen to the reaction rate when the following change in conditions is performed? doubling [C] tripling [D] Experiment [NO] M/L [O2 ] M/L Rate [mM/ s] 1 0.010 0.010 2.5 2 0.010 0.020 5.0 3 0.030 0.020 45.0 Determine the reaction rates in terms of [NO], [O2] and k 43 44 11 17/11/2016 The relationship between concentration and time can be derived from the rate law and calculus. Integration of the rate laws gives the integrated rate laws, which present concentration as a function of time. Integrated laws can be very complicated, so only a few simple forms will be considered. In this reaction the speed of the reaction does not depend on the reagent concentrations. 𝑣= − ∆𝑐 = 𝑘𝑐 0 = 𝑘 ∆𝑡 45 46 v k c0 k v v •photochemical reactions •heterophaseous reaction in which the slowest process t C c is connected with the phase change •burning ethanol by living organism C0 • decomposition of ammonia tg a - k • synthesis of HCl from hydrogen and chlorine performed on the sun light t 47 48 12 17/11/2016 First order reaction examples First order reaction ◦ Formula for the rate law is: rate = k [A] ◦ The integrated rate law can be expressed as: ln [ A]0 kt or [ A]t [ A]0 e kt [ A]t [A]0 is [A] at t (time) = 0 [A]t is [A] at t = t e = base of natural logarithms = 2.71828… Absorption, distribution, elimination rates Microbial death kinetics Photo dissociation of ozone with UV light Decomposition of hydrogen peroxide at room temperature Hydrolysis of sucrose (sugar) to glucose and fructose e.g. SO2Cl2→ SO2 + Cl2 at 320 oC k = 2 x 10-5 kJ/ s C2H6→ 2 CH3• at 700 oC k = 5.36 x 10-4 kJ/ s 49 Example of calculation 50 Solution For certain first-order reaction the initial [A]6 =[A]0 e-kt concentration of reactant A is equal 2M/L and the constant of this reaction k = 0.15M/ L s. What is the concentration of A after time 6s ? [A]6 = [2] (2.718-0.15 x6) = 0.813M/L 51 52 13 17/11/2016 Quiz If a reaction is first order with a rate constant of 5.48 x 10-2 sec-1, how long is required for 3/4 of the initial concentration of reactant to be used up? Graphical methods can be used to determine the first-order rate constant, note ln [ A]0 kt [ A]t ln[ A]0 ln[ A]t kt ln[ A]t ln[ A]0 kt ln[ A]t kt ln[ A]0 y mx b 53 54 A plot of ln[A]t versus t gives a straight line with a slope of -k Determination of reaction constant from plot The decomposition of N2O5. (a) A graph of concentration versus time for the decomposition at 45oC. (b) A straight line is obtained from a logarithm versus time plot. The slope is negative the rate constant. 55 56 14 17/11/2016 ln of initial Decreasing function, abtuse angle reactant concentration 1.4 Plot ln [A] vs. time 1.4 1.2 1.2 1 1 0.8 0.8 0.6 0.6 a 0.4 ln CA ln CA 0.4 0.2 0 -0.2 tg α = a/b a = 1.2- (-0.28) = 1.48 = b= 125 tg α = - 0.0118 = -k k = 0.0118 0 25 50 75 100 125 150 175 0.2 0 200 0 25 50 75 100 125 150 175 200 -0.2 -0.4 -0.4 -0.6 -0.6 -0.8 -0.8 -1 -1 t [min] b t [min] 57 58 Second order reaction Second order reaction examples The simplest second-order rate law has the form 2 rate k[ B] The integrated form of this equation is 1 1 kt [ B]t [ B]0 [ B]0 the initial concentration of B [ B]t the concentration of B at time t 63 Decomposition of HI without catalyst Decomposition of NO2 to NO and O2 ClO- + Br-→ BrO- + Clat 25 oC k = 4.2 x 10-7 kJ/l mol-1s-1 H+ + OH-→ H2O at 25 oC k = 1.35 x 1011 kJ/l mol-1s-1 64 15 17/11/2016 Half life t1/2 Graphical methods can also be applied to second-order reactions A plot of 1/[B]t versus t gives a straight line with a slope of k The amount of time required for half of a reactant concentration to disappear is called the half-life, t1/2 Second-order kinetics. A plot of 1/[HI] versus time 65 66 ◦The half-life of a first-order 𝑡1 2 = It is not affected by the initial concentration 𝐴0 2𝑘 First - order rate law : ln [ A]0 kt [ A]t 1 at t t1/ 2 , [ A]t [ A]0 , substituti ng 2 [ A]0 ln 2 ln 1 kt1/ 2 or t1/ 2 k [ A ] 0 2 For zero-order reactions, the half-life depends on the initial concentration of reactant and the rate constant. 67 68 16 17/11/2016 ◦ The half-life of a second-order reactions does depend on the initial concentration The equation for the half-life of a second order reaction: 𝑡1/2 = 1 𝑘 ∙ 𝐵0 This inverse relationship suggests that as the initial concentration of reactant is increased, there is a higher probability of the two reactant molecules interacting to form product. Consequently, the reactant will be consumed in a shorter amount of time, i.e. the reaction will have a shorter half-life. 1 1 kt [ B]t [ B]0 Second - order rate law : 1 at t t1/ 2 , [ B]t [ B]0 , substituti ng 2 1 1 kt1/ 2 1 [ B]0 [ B]0 2 1 kt1/ 2 [ B]0 or t1/ 2 ln 2 k[ B]0 69 70 (2) Graphical method Reagent concentration to t1 t2 t3 t4 ... co c1 c2 c3 c4 ... n=0 n=1 c co tg a - k ln co tg a - k c c k t 1 c t 71 ln c n=2 t ln c ln c k t 1 c tg a k t 1 1 k t c c 72 17 17/11/2016 Quiz Substance A decomposes by a first-order reaction. Starting initially with [A] = 2.00 M, after 150 min [A] = 0.50 M. For this reaction what is t1/2 ? First-order radioactive decay of iodine-131. The initial concentration is represented by [I]0. 73 74 Quiz 120 100 The table below presents plote of concentration of biologically active metabolite T-IDA vs time. Using of these data determine graphically the half-time life (t1/2) of this metabolite. Concentra tion 0 10 20 30 40 50 60 70 100 50 25 12,5 6.25 3.13 1.56 0.781 concentration [mol /L] Time [min] 80 [mol /L] 60 40 20 0 0 10 20 30 40 50 Time [min] 60 70 80 T1/2 = 10 min 75 76 18 17/11/2016 There are five principle factors that influence reaction rates: 1) Chemical nature of the reactants 2) Ability of the reactants to come in contact with each other 3) Concentration of the reactants 4) Temperature The progress of the reaction A B. The number of A molecules (in red) decreases with time while the number of B molecules (in blue) increases. The steeper the concentration versus time curve, the faster the reaction rate is. The film strip represents the relative number of A and B molecules at each time. 5) Availability of rate-accelerating agents called catalysts 77 78 Chemical nature of the reactants ◦ Bonds break and form during reactions The most fundamental difference in reaction rates lies in the molecules, or ions) collide before the reaction can Some reactions are fast by nature whereas others are slow occur. ΔG0rxn = -397 kJ/mol ◦ This depends on the phase in which the reactants This reaction one does not occur. C(graphite) + O2(g) →CO2(g) Ability of the reactants to meet ◦ Most reactions require that particles (atoms, reactants themselves. C(diamond) + O2(g)→CO2(g) are. ΔG0rxn = -394 kJ/mol. This reaction occurs rapidly. 79 80 19 17/11/2016 ◦ In a homogeneous reaction the reactants are in the same phase: Effect of crushing a solid. When a single solid is subdivided into much smaller pieces, the total surface area on all of the pieces becomes very large. For example both reactants in the gas (vapour) phase. ◦ In a heterogeneous reaction the reactants are in different phases: For example one reactant is present in the liquid whereas the second is in the solid phase. ◦ In heterogeneous reactions the reactants meet only at the intersection between the phases. ◦ Thus the area of contact between the phases determines the rate of the reaction. 81 82 Concentration of the reactants H2O2(aq) → H2O(l) + ½ O2 (g) Volume O2 [cm3] ◦ Both homogeneous and heterogeneous reaction rates are affected by reactant concentration Concentration of H2O2 0,5 mol/l of 0,4 mol/l For example, red hot steel wool bursts into flames in the presence of pure oxygen 0,3 mol/l 0,2 mol/l 0,1 mol/ l Time [s] 83 84 20 17/11/2016 Temperature Example ◦ The rates for almost all chemical reactions Consider this reaction: enhance as the temperature is increased Cold-blooded creatures, such as insects and Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) reptiles, become sluggish at lower temperatures as their metabolism slows down. http://www.ausetute.com.au 85 86 Condition Condition Concentration Temperature Affect on Rate Explanation Increasing the More HCl particles means there concentration of HCl will will be more collisions between increase the reaCtion rate. HCl and Zn. Increasing temperature increases the reaction rate. HCl particles will gain more kinetic energy increasing the number of collisions with Zn atoms. More Zn and HCl particles will have sufficient energy to react resulting in more successful collisions. Affect on Rate Explanation Particle Size Reducing the size of the Zn Reducing the size of particles increases the surface area Zn particles will available for reaction with HCl increase the rate of molecules resulting in more reaction. collisions. Stirring Rate Stirring will keep small Zn particles Increasing the in suspension, increasing the stirring rate of this surface area available for mixture will increase collisions, resulting in an increased the reaction rate. reaction rate. http://www.ausetute.com.au/reactrate.html http://www.ausetute.com.au/reactrate.html 87 88 21 17/11/2016 Quiz Catalyst Student performed experiment on dissolution of Mg in HCl in four test tubes according to the following conditions. Test tube Mg size HCl concentration 1 cube 1.0 M 2 cube 0.5 M 3 powder 1.0 M 4 powder 0.5 M A catalyst is a substance that changes the rate of a chemical reaction without itself being used up. ◦ Positive catalysts speed up reactions ◦ Negative catalysts or inhibitors slow reactions Determine the order of test tubes according to the reaction time decreasing (Positive) catalysts speed reactions by allowing the rate-limiting step to proceed with a lower activation energy. Thus a larger fraction of the collisions are effective. 89 91 Catalysts can be divided into two groups ◦ Homogeneous catalysts exist in the same phase as the reactants. ◦ Heterogeneous catalysts exist in a separate phase. NO2 is a homogeneous catalyst for the production of sulfuric acid in the lead chamber process. The mechanism is: (a) The catalyst provides an alternate, low-energy path from the reactants to the products. (b) A larger fraction of molecules have sufficient energy to react when the catalyzed path is available. 92 93 22 17/11/2016 S O 2 SO 2 SO 2 12 O 2 SO 3 SO3 H 2 O H 2SO 4 Heterogeneous catalysts are typically solids Consider the synthesis of ammonia from hydrogen and nitrogen by the Haber process 3H2 N 2 2NH3 The second step is slow, but is catalyzed by NO2: NO 2 SO 2 NO SO3 NO 12 O2 NO2 94 95 Mechanism theories The Haber process. Catalytic formation of ammonia molecules from hydrogen and nitrogen on the surface of a catalyst. 96 97 23 17/11/2016 The reaction’s mechanism is the series of simple reactions called elementary processes. The rate law of an elementary process can be written from its chemical equation. One of the simplest models explaining reaction rates is collision theory. According to collision theory, the rate of reaction is proportional to the effective number of collisions per second among the reacting molecules. An effective collision is one that actually gives product molecules. The number of all types of collisions increase with concentration, including effective collisions. 98 99 There are a number of reasons why only a small fraction of all the collisions leads to the formation of product: ◦ Only a small fraction of the collisions are energetic enough to lead to products. ◦ Molecular orientation is important because a collision on the “wrong side” of a reacting species cannot produce any product. This becomes more important as the complexity of The key step in the decomposition of NO2Cl to NO2 and Cl2 is the collision of a Cl atom with a NO2Cl molecules. (a) A poorly orientated collision. (b) An effectively orientated collision. the reactants increases. 10 0 10 1 24 17/11/2016 ◦ The minimum of kinetic energy of the colliding particles must have is called the activation energy, Ea . ◦ In a successful collision, the activation energy changes to potential energy as the bonds rearrange to form products. ◦ Activation energies can be large, so only a small fraction of the well-orientated, colliding molecules have it. ◦ When temperature increases the average kinetic energy of the reacting particle also increases. 10 2 Kinetic energy distribution for a reaction at two different temperatures. At the higher temperature, a larger fraction of the collisions have sufficient energy for reaction to occur. The shaded area under the curves represent the reacting fraction of the collisions. 10 3 Activation energies and heats of reactions can be determined from potential-energy diagrams Potential-energy diagram for an endothermic reaction. The heat of reaction and activation energy are labeled. The potential-energy diagram for an exothermic reaction. The extent of reaction is represented as the reaction coordinate. 10 4 10 5 25 17/11/2016 Reactions generally have different activation energies in the forward and reverse direction A unsuccessful (a) and successful (b) collision for an exothermic reaction. Activation energy barrier for the forward and reverse reactions. 10 6 Transition state theory explains what happens when reactant particles come together. 10 7 The brief moment during a successful collision that the reactant bonds are partially broken and the product bonds are partially formed is called the transition state. 10 8 10 9 26 17/11/2016 The overall rate law determined for the mechanism must agree with the observed rate law. The exponents in the rate law for an elementary process are equal to the coefficients of the reactants in chemical equation Elementary process : 2NO 2 NO 3 NO Formation of the activated complex in the reaction between NO2Cl and Cl. NO2Cl+ClNO2+Cl2 rate k[NO 2 ]2 11 0 Multistep reactions are common. The sum of the element processes must give the overall reaction. Slow set in a multistep reaction limits how fast the final products can be formed and is called the rate-determining or ratelimiting step. Simultaneous collisions between three or more particles are extremely rare. 11 1 A reaction that depended on a three-body collision would be extremely slow. Thus, reaction mechanism seldom includes elementary process that involves more than two-body or bimolecular collisions. Consider the reaction 2NO 2H 2 N 2 2H 2 O rate k[NO] 2 [H 2 ] (experimental) 11 2 The mechanism is thought to be 11 3 27 17/11/2016 2NO N 2O 2 (fast) N 2O 2 H 2 N 2O H 2O (slow) N 2O H 2 N 2 H 2O (fast) rate(forward) k f [ NO]2 The second step is the rate-limiting step, which gives rate(reverse) k r [ N 2 O 2 ] thus rate k[ N 2O2 ][H2 ] The first step reaches a fast equilibrium At equilibrium, the rates of the forward and reverse reaction are equal k f [ NO]2 k r [ N 2 O 2 ] or N2O2 is a reactive intermediate, and can be eliminated from the expression. [ N 2O 2 ] kf kr [ NO]2 11 4 Substituting, the rate law becomes rate k[ N 2 O 2 ][H 2 ] The activation energy is related to the rate constant by the Arrhenius equation k Ae Ea / RT kf rate k [ NO]2 [H 2 ] or kr rate k '[ NO]2 [H 2 ] 11 5 k = rate constant Ea = activation energy e = base of the natural logarithm R = gas constant = 8.314 J mol-1 K-1 T = Kelvin temperature A = frequency factor or pre-exponential factor Which is consistent with the experimental rate law. 11 6 11 7 28 17/11/2016 The Arrhenius equation can be put in standard slope-intercept form by taking the natural logarithm ln k ln A Ea / RT k E 1 1 ln 2 a k1 R T2 T1 or ln k ln A ( Ea / R) (1 / T ) y b m The activation energy can be related to the rate constant at two temperatures x A plot of ln k versus (1/T) gives a straight line with slope = -Ea/RT 11 8 11 9 12 0 121 Quiz The activation energy of the first order reaction is 50,2 kJ/mol at 25oC. At what temperature will the rate constant double? 29 17/11/2016 12 2 12 3 True or false quiz 1. A catalyst alters the rate of a chemical reaction by: a) always providing a surface on which molecules react b) changing the products formed in the reaction c) inducing an alternate pathway for the reaction with generally lower activation energy d) changing the frequency of collisions between molecules 2. The Rate of a Chemical Reaction a) usually is increased when the concentration of one of the reactants is increased b) is dependent on temperature c) may be inhibited by certain catalytic agents d) will be very rapid if the activation energy is large 125 126 30