CHAPTER 14 Chemical kinetics Introduction Spontaneity refers to the inherent tendency for the process to occur; however, it implies nothing about speed. Who might be interested in the rate of a chemical reaction? • Would the grocer be interested in fast or slow chemical reactions? Introduction Spontaneity refers to the inherent tendency for the process to occur; however, it implies nothing about speed. Who might be interested in the rate of a chemical reaction? • Would the grocer be interested in fast or slow chemical reactions? Slow reactions; to preserve food. Introduction • How about the doctor and the patient? Introduction • How about the doctor and the patient? Fast or slow-depends? It would be nice to take an aspirin and have the head ache go away in a second or two, but if the issue is cancer, then slow! Introduction What would be a word to describe extremely fast chemical reactions that occur in a nanosecond? Introduction What would be a word to describe extremely fast chemical reactions? Explosion! Introduction What would be a word to describe extremely fast chemical reactions that occur in a nanosecond? Explosion! There really is no word for very slow reactions, but some examples are in order. • Rusting of Iron • Rotting of wood Factors Affecting Rate During our study of Kinetics the following will be shown to affect the rate of chemical reactions • Reactant concentration: The closer reactants are to each other the higher the probability of combining to make products • Reaction temperature: Increased temperature increases kinetic energy to break reactant bonds to form products • Catalyst: Lowers the activation energy. • Surface Area: Exposes more reactants to the other reactant. • Nature of Reactants How is Rate Measured All rates are inversely proportional to time, which means time must be in the denominator. In a chemical reaction a change in concentration of reactants or products is measured per unit time. To measure rate, then concentration and time must be measured. Stoichiometry and Rate Consider the Stoichiometry of reaction: 2 H2 + O2 → 2 H2O • Hydrogen disappears twice as fast as oxygen. • Water appears at the same rate as hydrogen disappears. • Oxygen disappears ½ as fast as hydrogen appears General Rate Relations Note: - means decreasing, while + means increasing. Consider the equation: aA + bB → cC + dD Δ[A] = - 1 Δ[B] = 1 Δ[C] = 1 [∆D] a Δt b Δt c Δt d ∆t Rate = - 1 Special Note about Δ Δ means change. To be consistent change in chemistry is always final - initial Sample Problem Consider the decomposition of dinitrogenpentoxide 2 N2O5 (g) → 4 NO2 (g) + O2 (g) If the rate of appearance of O2 is 0.23 M/min, then what would the rate of appearance of NO2? ¼ Δ[NO2]/Δt = Δ[O2]/Δt Δ[NO2]/Δt = 4 Δ[O2]/Δt = 4 [0.23M/min] = 0.92 M/min Average vs. Instantaneous Rate Consider a Road Trip: What does the speedometer tell us? Using the road signs how can the average rate be calculated? Average vs. Instantaneous Rate Consider a Road Trip: What does the speedometer tell us? Instantaneous Rate Using the road signs how can the average rate be calculated? Average vs. Instantaneous Rate Consider a Road Trip: What does the speedometer tell us? Instantaneous Rate Using the road signs how can the average rate be calculated? Rateave = ΔD/ Δt Average vs. Instantaneous Rate Consider a the following chemical reaction: CV+ (aq) + OH-(aq) → CVOH (aq) Where CV is crystal violet calculate the average rate for the given time frames from the data on the next slide. Crystal Violet Data 1. Graph [CV+] vs. time 2. Calculate average rate for a. t=0.0 to 10.0 seconds b. t=10 to 20 seconds 3. Calculate the instantaneous rate at t= 10.0 seconds Rate = Rate = ∆[CV+] ∆t = (3.68-5.00)(10-5) 10.0-0.00 -1.32X10-5 M/s Time (s) 0 10.0 20.0 30.0 40.0 50.0 60.0 80.0 90.0 [CV+] 5.00 X 10-5 3.68 X 10-5 2.71 X 10-5 1.99 X 10-5 1.46 X 10-5 1.08 X 10-5 0.793 X 10-5 0.429 X 10-5 0.232 X 10-5 About Graphs Central Graphing Guidelines • Graph done on graph paper; or Excell (possibly Logger-Pro) • Scales clearly and axes clearly labeled • Allowed scales – 1,2,3,4 etc.; 10, 20, 30,40 etc.; 100, 200, 300 etc. – 2,4,6 etc; 20,40, 60 etc. – 5, 10, 15, 20, etc; 50, 100, 150, 200, etc • Smoothest curve between data points, or best fit line • If drawing the line, then the same number of data points should be above and below the smoothest curve • Large graphs required; full sheet of graph paper, no smaller than half sheet • Design graph to maximize the number of significant figures. Graphing With Excel 1. 2. 3. 4. Place ordered pairs in Excell Highlight the ordered pairs Go to insert, chart, scatter Right click on line and choose trend line, options display equation. 2NO2 → 2NO + O2 Rate Laws • Most all reactions are reversible, this means that products can make reactants. • The overall rate would be the rate of the product formation minus rate of reactant formation. • In order not to deal with this difference we determine rates shortly after the reaction starts, so that the rate of the reverse reaction is negligible. • The previous graphs were not linear, but exponential in nature, due to reactant concentration decreasing. • The rate of a reaction can be more easily stated as Rate = k[A]n , where k is the rate constant, [A] the initial concentration of the reactants, and n the order of the reaction. This expression is called the rate law. Rate Law cont. • The value of n cannot be determined from the balanced equation, but must be determined experimentally. • Concentrations of products can be ignored, since they are minimal, thus producing a negligible rate in the reverse direction. • There are two different types of rate laws a. The differential rate law (rate law), describes how the rate depends on concentration. b. The integrated rate law, which shows how the concentrations of species depend on time. The Differential Rate Law Consider the following reaction: CH3CO2CH3 + OH- → CH3CO2- + CH3OH Exp# 1 [CH3CO2CH3] 0.040 [OH-] 0.040 Rate M/s 0.00022 2 0.040 0.080 0.00048 3 0.080 0.080 0.00090 Simplify the Rate Data Exp# [CH3CO2CH3] [OH-] Rate M/s assume k =1 Method of initial rates 1 2 3 R = k[CH3CO2CH3]X[OH-]y R = 1[1]1[1]1 = 1 for exp #1 R = 1[1]1[2]1 = 2 for exp #2 R = 1[1]1[2]1 = 4 for exp #3 Conclusion X=1, y=1 1 1 2 1 2 2 1 2 4 The reaction must be first order in methylacetate and first order in hydroxide, with an overall order of two NH4+ + NO2- → N2 + 2 HOH R1/R2 =? R2/R3 =? R1 = k[NH4+]x [NH2-]y R1/R2 = R2/R3 = R2 = k[NH4+]x[NH2-]y x y 1.35X10-7 = [0.100] [0.005] 2.70x10-7 [0.100]x[0.010]y 2.70X10-7 5.40X10-7 = [0.100]x[0.010]y [0.200]x[0.010]y 0.500 = [0.500]y y=1 0.500 = [0.500]x x=1 Log Review Log(X)(Y) = Log(X) + Log(Y) Log X Y = Log(X) - Log(Y) Log (Y)x = X Log (Y) Example: 1.37 = (4.39)x Log 1.37 = X Log 4.39 0.136 = X (4.39) 0.136 X= = 0.212 4.39 Checking on Calculator: [4.39] 0.212 = 1.37 Reaction Rate and Order Order is predicted from the exponents in the rate equation. • Zero order in the reactant, means that there is no rate effect on the initial concentration of that reactant, even if it is increased. • First order in the reactant, means that doubling the initial concentration will double the rate. • Second order in a reactant, means that doubling the initial concentration will make the rate quadruple. • Third order in a reactant, means that doubling the initial concentration increases the rate eightfold. Sample Problems Consider the following equation: N2 (g) + 3 H2 (g) 2 NH3 (g) Experiments show R = k[N2][H2]3 1. What is the order with respect to N2? 2. What is the order with respect to H2? 3. What is the overall order? Sample Problems Consider the following equation: N2 (g) + 3 H2 (g) 2 NH3 (g) Experiments show R = k[N2][H2]3 1. What is the order with respect to N2? First 2. What is the order with respect to H2? 3. What is the overall order? Sample Problems Consider the following equation: N2 (g) + 3 H2 (g) 2 NH3 (g) Experiments show R = k[N2][H2]3 1. What is the order with respect to N2? First 2. What is the order with respect to H2? Third 3. What is the overall order? Sample Problems Consider the following equation: N2 (g) + 3 H2 (g) 2 NH3 (g) Experiments show R = k[N2][H2]3 1. What is the order with respect to N2? First 2. What is the order with respect to H2? Third 3. What is the overall order? Fourth Sample Problems Consider the following equation: N2 (g) + 3 H2 (g) 2 NH3 (g) Experiments show R = k[N2][H2]3 1. How many times faster if N2 is doubled? Sample Problems Consider the following equation: N2 (g) + 3 H2 (g) 2 NH3 (g) Experiments show R = k[N2][H2]3 1. How many times faster if N2 is doubled? 2 Sample Problems Consider the following equation: N2 (g) + 3 H2 (g) 2 NH3 (g) Experiments show R = k[N2][H2]3 1. How many times faster if N2 is doubled? 2 2. How many times faster if H2 is doubled? Sample Problems Consider the following equation: N2 (g) + 3 H2 (g) 2 NH3 (g) Experiments show R = k[N2][H2]3 1. How many times faster if N2 is doubled? 2 2. How many times faster if H2 is doubled? 8 Sample Problems Consider the following equation: N2 (g) + 3 H2 (g) 2 NH3 (g) Experiments show R = k[N2][H2]3 1. How many times faster if N2 is doubled? 2 2. How many times faster if H2 is doubled? 8 3. How many times faster if both reactants are doubled? Sample Problems Consider the following equation: N2 (g) + 3 H2 (g) 2 NH3 (g) Experiments show R = k[N2][H2]3 1. How many times faster if N2 is doubled? 2 2. How many times faster if H2 is doubled? 8 3. How many times faster if both reactants are doubled? 16 Sample Problems Consider the following equation: N2 (g) + 3 H2 (g) 2 NH3 (g) 1. Experiments show R = k[N2][H2]3 2. What is the overall order? 4 3. How many times faster if N2 is doubled? 2 4. How many times faster if H2 is doubled? 8 5. How many times faster if both are doubled? 16 6. How many times faster if H2 is reduced by half of its original concentration? Sample Problems Consider the following equation: N2 (g) + 3 H2 (g) 2 NH3 (g) 1. Experiments show R = k[N2][H2]3 2. What is the overall order? 4 3. How many times faster if N2 is doubled? 2 4. How many times faster if H2 is doubled? 8 5. How many times faster if both are doubled? 16 6. How many times faster if H2 is reduced by half of its original concentration? 1/8 Zero-Order Reactions A →B • R = k[A]0 • Reactant concentration has no effect on initial rate of reaction. • [A] vs. time is linear with a negative slope. • The rate is equal to k, since [A]0 =1 • The units of k are M/s • Equation y = mx + B → [A]t = -kt + [A]0 • This is called the integrated rate law (derived on the next slid), which shows how concentration is related to time. • The integrated rate law can give instantaneous rates. Kinetics and Integral Calculus Since rate is the change in concentration per unit time (Δ[A]/Δt) then we can apply calculus principles of extremely small changes of concentrations per extremely small changes of time d[A]/dt. Consider a zeroth order reaction R = d [A] 0/dt = - k[A0]0 where A0 = initial concentration at time zero Rearranging this expression gives: d [A]0 = -k[A0]0 dt Integrating from t=0 to t=t gives the integrated rate law. t t ∫ d [A]0 = ∫ -k[A0]0 dt 0 0 gives [A]t = -kt + [A]0 First-Order Reactions A →B • R = k[A]1 • Reactant concentration is directly proportional to the rate of a reaction. • [A] vs. time not linear with a negative slope. • A graph of the ln[A] vs. time is linear. • The units of k are 1/s Calculus and Kinetics; First-Order Integrated Rate Law Since R = Δ[A]/Δt = -k[A]1 (first order) and integral calculus uses d for Δ, the equation has a calculus form of d[A]/[A] 1 dt = -k Applying the integration process we get; d[A] /[A] = -k dt ln[A]t / [A]0 = - kt → ln[A]t = - kt + ln[A]0 Second-Order Reactions A →B • R = k[A]2 • Reactant concentration is exponentially proportional to the rate of a reaction. • [A] vs. time not linear with a negative slope. • A graph of the 1/[A] vs. time is linear. • The units of k are 1/Ms Second-Order Integrated Rate Law Since R = Δ[A]/Δt = -k[A]1 (first order) and integral calculus uses d for Δ, the equation has a calculus form of d[A]/dt = -k[A]2, or rearranging d[A]/ [A]2= -k dt Applying the integration process we get; d[A]/[A]2 = -k dt 1/[A]t = kt + 1/[A]0 Reaction Half Lives • The half life of a reaction is the amount of time required for one half of the initial reactant concentration to disappear. •The half life for each type of integrated equation can be calculated from the equation itself. Zero-Order Half Life [A]t = ½[A]0 (half life definition) [A]t = - kt + [A]0 → ½[A]0 = -kt1/2 + [A]0 - [A]0/2 = - kt1/2 → t1/2 = [A]0/2k Reaction Half Lives First Order Half Life ln[A]t = -kt + ln[A]0 ln[A]t - ln[A]0 = - kt1/2 → ln[A]0 /[A]t = - kt ln[A]t /[A]0 = - kt → ln[A]0 /[A]t = kt [A]t = ½[A]0 (definition) ln[A]0 /1/2[A]0 = kt1/2 → ln2[A]0 / [A]0 = kt1/2 ln2 = kt1/2 → t1/2 = 0.693/k Note: the half life does not depend on initial concentration Reaction Half Lives Second-Order Half Life 1/[A]t = kt + 1/[A]0 → 1/[A]t - 1/[A]0 = kt [A]t = ½[A]0 (definition) → 2/[A]0 - 1/[A]0 = kt1/2 1/[A]0 = kt1/2 → t1/2 = 1/k[A]0 Decomposition of N2O5 0.0500 0.0250 0.0125 Is this a zero order reaction? Decomposition of N2O5 No, the graph should be linear! Decomposition of N2O5 Is this a first order reaction? Decomposition of N2O5 Note the half times are the same Decomposition of N2O5 T1/2=0.693/k ( a constant divided by a constant would always give same time) Consider a Second Order t1/2= 1/k[A]0 for a first order reaction. The first half life would be [A]0 [A]0/2 The second half life would be [A]0/2 [A]0/4 Consider a Second Order t1/2= 1/k[A]0 for a first order reaction. The first half life would be [A]0 [A]0/2 The second half life would be [A]0/2 [A]0/4 Now do the math: Consider a Second Order t1/2= 1/k[A]0 for a first order reaction. The first half life would be [A]0 [A]0/2 The second half life would be [A]0/2 [A]0/4 Now do the math: Assume [A]0=1 and k=1 t1/2=1/(1)(1)=1 for the first half life Consider a Second Order t1/2= 1/k[A]0 for a first order reaction. The first half life would be [A]0 [A]0/2 The second half life would be [A]0/2 [A]0/4 Now do the math: Assume [A]0=1 and k=1 t1/2=1/(1)(1)=1 first half life t1/2=1/(1)(1/2)=2 second half life twice as long Consider a Second Order t1/2= 1/k[A]0 for a first order reaction. The first half life would be [A]0 [A]0/2 The second half life would be [A]0/2 [A]0/4 Now do the math: Assume [A]0=1 and k=1 t1/2=1/(1)(1)=1 first half life t1/2=1/(1)(1/2)=2 second half life twice as long The third half life would be 4 times longer than the first. So the graph must be for a first order reaction since the time increments for the half-lives are the same. Order Determination From the information below, determine the order of the reaction. Time (s) [A] ln[A] 1/[A] 0 0.01000 -4.605 100 1000 0.00625 -5.075 160 1800 0.00476 -5.348 210 2800 0.00370 -5.599 270 3600 0.00313 -5.767 320 4400 0.00270 -5.915 370 5200 0.00241 -6.028 415 6200 0.00208 -6.175 481 Graphs of the Data Graphs of the Data ln[A]t = - kt + ln[A]0 1/[A]t = kt + 1/[A]0 Graphs of the Data ln[A]t = - kt + ln[A]0 1/[A]t = kt + 1/[A]0 Second Order Pseudo-Ordered Reactions If a reaction is first order in each reactant and second order over all how can we calculate the rate constant? Pseudo-Ordered Reactions If a reaction is first order in each reactant and second order over all how can we calculate the rate constant? Do we use the first order rate law, the second order rate law, or neither of these? Pseudo-Ordered Reactions If a reaction is first order in each reactant and second order over all how can we calculate the rate constant? Do we use the first order rate law, the second order rate law, or neither of these? Neither; use the pseudoorder technique Pseudo-Ordered Reactions The pseduo-order technique uses very large concentration differences between the reactants. Always keep one reactant at low concentrations and the other reactants at higher concentrations by 4-5 orders of magnitude. Pseudo-Ordered Reactions The pseudo-order technique uses very large concentration differences between the reactants. Always keep one reactant at low concentrations and the other reactants at higher concentrations by 4-5 orders of magnitude. Consider the following reaction: A + B C 1.0 M 1.0X10-4 M When half of B disappears its concentration will be 5.0X10-5. The concentration of A should then be what? Pseudo-Ordered Reactions The pseudo-order technique uses very large concentration differences between the reactants. Always keep one reactant at low concentrations and the other reactants at higher concentrations by 4-5 orders of magnitude. Consider the following reaction: A + B C 1.0 M 1.0X10-4 M When half of B disappears its concentration will be 5.0X10-5. The concentration of A should then be what? 1.0 – 0.00005 = ? Pseudo-Ordered Reactions The pseudo-order technique uses very large concentration differences between the reactants. Always keep one reactant at low concentrations and the other reactants at higher concentrations by 4-5 orders of magnitude. Consider the following reaction: A + B C 1.0 M 1.0X10-4 M When half of B disappears its concentration will be 5.0X10-5. The concentration of A should then be what? 1.0 – 0.00005 = 1.0, right? Rounded to the correct number of significant figures. Pseudo-Ordered Reactions Since the concentration did not change for A it can be considered a constant and incorporated into the constant (that is why we use one, since 1 X k = k), then any rate information is relative to B, thus allowing calculation of the B order. After the order of B is determined, then an experiment with [B]=1.0 and [A]=1.0X10-4 can be done to calculate the order of [A], thus giving the rate law. Then the experimental data can be substituted into the rate equation to provide the value of k. Pseudo-Ordered Example Experiment [A] [B] Rate 1 0.1000 0.0050 1.35X10-7 2 0.1000 0.0100 2.70X10-7 3 0.0050 0.1000 2.7X10-7 4 0.0100 0.1000 1.08X10-6 Rate = k[A]x[B]y R1/R2 = R3/R4 = 1.35X10-7 2.70x10-7 [0.100]x[0.005]y = [0.100]x[0.010]y y x 2.70X10-7 = [0.0050] [0.1000] 1.08X10-6 [0.0100]y[0.1000]x 0.500 = [0.500]y y=1 0.250 = [0.500]y x=2 Pseudo-Ordered Example Experiment [A] [B] Rate 1 0.1000 0.0050 1.35X10-7 2 0.1000 0.0100 2.70X10-7 3 0.0050 0.1000 2.7X10-7 4 0.0100 0.1000 1.08X10-6 Rate = k[A]x[B]y R1/R2 = R3/R4 = 1.35X10-7 2.70x10-7 [0.100]x[0.005]y = [0.100]x[0.010]y y x 2.70X10-7 = [0.0050] [0.1000] 1.08X10-6 [0.0100]y[0.1000]x Rate = k[A]1[B]2 0.500 = [0.500]y y=1 0.250 = [0.500]y x=2 Multiple Reactants Use pseudo-order technique A + B + C → D R = k[A]x[B]y[C]z The experimental conditions, will use high concentrations of two of the three reactants relative to the third reactant. This will cause relatively little change in concentration of the large ones and significant changes in concentration of the more dilute reactant. Thus the order to the dilute reactant can be determined by the method of initial concentrations. Then another set of experiments will be carried out by keeping two reactants at high concentrations and the other at dilute concentration, to determine its order. A Sample Problem • Consider the reaction: 3A + B + C → D + E, where the rate law is defined as • Δ[A]/Δt = – k[A]2[B][C] experiment is carried out where [B]0 = [C]0 = 1.00 M and [A]0 = 1.00 x 10–4 M. After 3.00 minutes, • [A] = 3.26 x 10–5 M. – Calculate the value of k using integrated second order equation. 1 1 = kt + [A]0 [A]t 1 1 = kt + 1.00X10-4 -5 3.26X10 30675 = k (3.00 min) + 1.00 X 104 K = 6.89 x 103 min-1 A Sample Problem • • Consider the reaction: 3A + B + C → D + E, where the rate law is defined as Δ[A]/Δt = – k[A]2[B][C] experiment is carried out where [B]0 = [C]0 = 1.00 M and [A]0 = 1.00 x 10–4 M. After 3.00 minutes, [A] = 3.26 x 10–5 M. – Find the concentration of [B] and [C] after 3.00 minutes [B] = [C] = 1.00 – 0.0000225 = 1.00 M A Sample Problem • • • Consider the reaction: 3A + B + C → D + E, where the rate law is defined as Δ[A]/Δt = – k[A]2[B][C] experiment is carried out where [B]0 = [C]0 = 1.00 M and [A]0 = 1.00 x 10–4 M. After 3.00 minutes, [A] = 3.26 x 10–5 M. – Describe, in your own words, an experiment that could be carried out to determine the order of reactants B and C. A Sample Problem • • • Consider the reaction: 3A + B + C → D + E, where the rate law is defined as Δ[A]/Δt = – k[A]2[B][C] experiment is carried out where [B]0 = [C]0 = 1.00 M and [A]0 = 1.00 x 10–4 M. After 3.00 minutes, [A] = 3.26 x 10–5 M. – Describe, in your own words, an experiment that could be carried out to determine the order of reactants B and C. Same way, keep A and C at 1.00 molar and do rate expt. with B to find order Reaction Kinetics: A Summary 1. To calculate rate of reaction when the rate law is known, use this expression: R = k[A]x[B]y 2. To determine a rate of reaction when the rate law is not given, use • • The slope of an appropriate tangent line to the graph of [A] versus t The expression –Δ[A]/ Δt, with a short time interval Δt 3. To determine the order of a reaction, use one of the following methods. • • • Find the graph of rate data that yields a straight line Test for constancy of the half-life (only first order) Substitute rate data into integrated rate laws to find the one that gives a constant value of k. Reaction Kinetics: A Summary 1. To find the rate constant k for a reaction, use one of the following methods. • • • Obtain k from the slope of a straight-line graph. Substitute concentration-time data into the appropriate integrated rate law. Obtain k from the half-life of the reaction (first order only) t1/2 = 0.693/k 2. To relate reactant concentrations and times, use the appropriate integrated rate law after first determining k. Reaction Mechanisms 1. A reaction mechanism is a step-by-step description of how chemical changes occur. 2. Each step is called an elementary process 3. A reaction mechanism must meet the following: • Be consistent with the overall Stoichiometry • The slowest step must account for the experimentally determined rate law • The individual equations must add up to equal the total balanced equation. Like Hess’s Law Elementary Processes 1. Are either unimolecular or bimolecular collisions. termolecular are not probable since they require simultaneous collision of three particles. 2. The exponents of concentration terms for the rate law of an elementary step are the coefficients of the equation. This is not the case for the overall equation for the chemical change. 3. Elementary processes are reversible and some may have the forward and reverse reaction rates equal (equillibrium) Elementary Processes 4. Certain species are produced in one elementary process and consumed in another are called reactive intermediates. 5. Of all of the elementary steps, only one is the slowest and is called the rate determining step Mechanism Examples Slow Step Followed by a Fast Step. H2(g) + 2 ICl(g) → I2(g) + 2 HCl(g) Mechanism: (Note: HI is a reactive intermediate) H2 + ICl → HI + HCl slow HI + ICl → I2 + HCl fast H2 (g) + 2 ICl (g) → I2 (g) + 2 HCl (g) Overall Rx bimolecular bimolecular The rate law for the overall reaction is just the rate of the slowest step. R = k[H2][ICl] A reactive intermediate is produced in one reaction and consumed in the next reaction Reaction Profile HCl + ICl Mechanism Examples Fast followed by slow step 2 NO (g) + O2 (g) 2 NO (g) N2O2 (g) + O2 (g) N2O4 (g) overall k1 k-1 K2 N2O2 (g) fast 2 NO (g) + O2 (g) R= 2 k[NO] [O N2O4 (g) slow N2O4 (g) experiment ] 2 The Fast Slow Mechanism • In this mechanism there is a rapid equilibrium in the first step. • Some of the product of the first step is slowly drawn off and consumed in the second step. • Since N2O2 is a reactive intermediate it cannot be in the rate law. Rf = K[N2O2][O2] • Rf = k1[N O]2 and Rr = k-1[N2O2] • Equilibrium means that Rf = Rr ; k1[NO]2= k-1[N2 O2] • [N2O2] = k1/k-1[NO]2 The Fast Slow Mechanism • • • • • Rf = K[N2O2][O2] but [N2O2] = k1/k-1[NO2]2 Substituting Rf =Kk1/k-1[NO2]2[O2] Kk1/k-1 = K Rf = K[NO2]2[O2] (The rate law from experiment) This is a very common mechanism when the overall reaction suggests a termolecular collision Hydration Mechanism CH2=CH2 + HOH Step #1 H+ CH3CH2OH CH2=CH2 + H+ CH3CH2+ fast CH3CH2+ + HOH CH3CH2OH2+ slow CH3CH2OH2+ CH3CH2OH + H+ slow Effect of Temperature on Rates • In 1889 Arrhenius demonstrated that the rate constants of many chemical reactions vary with temperature according to the Arrhenius equation below: K = Ae-Ea/RT • This form is not useful, since it is exponential. • A more useful form is found by taking the natural logarithm of both sides, also below: -Ea + ln A ln k = RT An equation of a straight line where –Ea/R is the slope and ln A is the y-intercept About the Arrhenius Equation K = A e-Ea/RT 1. A is the pre-exponential factor having the same units as k a For a first order reaction the units are s-1, since these are units of frequency, then sometimes A is called the frequency factor 2. Ea is the energy of activation 3. T is the absolute temperature 4. R is the ideal gas constant, 8.314 j/mole-K Rearrangement of Arrhenius Equation 1.Taking the natural logarithm of both sides of the equation lnK = -Ea/RT + lnA Fits the equation of a straight line, where –Ea/R is the slope of the line Two Point Equation K1 is constant at a constant T1 temperature K2 is constant at a constant T2 temperature lnK1 = -Ea/RT1 + lnA lnK2 = -Ea/RT2 + lnA On the next slide the second equation will be subtracted from the first equation, thus eliminating the y-intercept lnA The Two Point Equation Consider the the equation for different temperatures ln k1 = ln k2 = -Ea RT1 -Ea RT2 Lnk1 - ln k2 = + ln A + ln A -Ea RT1 - Subtract the bottom equation from the top equation -Ea RT2 Lnk1/lnk2 = Ea/R(1/T2 – 1/T1) + ln k1 – lnk2 Collision Theory • The number of collisions per second is called the collision frequency. • Typically gases have a frequency of 1030 collisions per second. • If each collision made product then the rate would be 106M/s. (explosive) • Gas reactions are normally around 10-4M/s • We must conclude then, that most collisions do not end up as products. Collision Theory 1. Collision theory views reaction rate as a process of reactants colliding with a certain frequency and minimum energy. 2. The number of collisions per unit time is the upper limit of reaction rate. 3. This model is restricted to A + B →C a. If there are 2 A and 2 B reactants then there are possible 4 collisions, which both multiplication and addition will predict. b. Add another A particle, thus 3 A particles and two B particles, each a can collide with either of the two B particles, thus a total of 6 collisions, not 5 that addition would indicate. c. Add another B particle, so that there are 3 A particles and 3 B particles; each A can collide with three B particles, for a total of 9 collisions and not 6. Collision Theory Increasing the temperature of the reaction increases the speed of the reacting particles and therefore their collision frequency. a.Collision frequency cannot be the only factor affecting rate, or every gaseous reaction would be over instantaneously. b.Since at 1 atm at 20°C the molecules in a milliliter of gas experience about 1027 collisions per second (Total collisions) c. Each molecule, on the other hand, experiences 4X107 collisions Collision Theory d. This means that if every collision resulted in a reaction the reaction would be over in 10-7 seconds e. Obviously most collisions do not result in a reaction. Arrhenius suggested that every reaction had an energy threshold (Ea) that reactants must meet or exceed in order to react. This is the energy require to break the bonds, so that the atoms can rearrange into other molecules Collision Theory Usually a ten degree increase in reaction temperature doubles or triples the reaction rate a. Is this increase due to increased frequency? b. An increase of 10°only increases the speed by about 2%, which we would expect a 4% increase in reaction rate, which does not explain the two or three fold increase in reaction rate. c. Far more important is the increase of the fraction of particles having the correct kinetic energy to form products. This is shown on the next slide. Kinetic Energy Distribution Energy of activation About Colliding Particles Orientation a Colliding particles must collide with a particle that can make a product, called effective collisions b In Arrhenius equation the frequency factor A contains information on orientation 1. A = pz 2. P is the steric factor (always less than one) and represents the fraction of collisions with effective orientations. 3. Z is the collision frequency 4. e-Ea/RT is the fraction of collisions with sufficient energy to produce products About Colliding Particles As molecules become more complex the probability factor p (steric factor) decreases. – When NO collides with NO3 to make NO2, the probability factor is 0.006, or six out of a thousand are effective – In nucleic acids the factor is about 10-6 Transition State Theory • Transition State was first proposed by Henry Eyring (1901-19810). • The hypothetical species that exists between the reactants and products is called the activated complex. • The state where the activated complex is found is called the transition state. • The activated complex is formed from collisions between molecules. Transition State Theory • Once the reactants have collided, they can go back to reactants, or produce products. • Meaning they roll, like a ball, down one side of the hill or the other. • Consider the collision of dinitrogen oxide with nitrogen monoxide as illustrated below: N N O + N=O O N=O N N activated complex Transition State Theory Activated Complex The Role of a Catalyst – A substance that speeds up a reaction with out being consumed by the reaction. – Biological catalyst is called an enzyme. – Catalysts work by lowering the energy of activation. – The enthalpy does not change. – Homogeneous catalyst, same phase as the reactants – Heterogeneous catalyst, not the same phase as the reactants Heterogeneous Catalyst • Has a different phase than the rest of the reaction mixture Common example is an adsorption mechanism Difference between adsorption and absorption • • – – – – Adsorption means sticks to the surface like adhesive tape Absorption means penetrates into the substance itself. Metals are common choice for non-metal reactions When a gas for example is adsorbed on to the surface of a metal, the bonds of the gas are weakened, thus allowing the other gas, also with weaker bonds to react. Heterogeneous Catalyst • Hydrogenation of an unsaturated hydrocarbon – For ethylene reacting with hydrogen, they are both adsorbed on the surface of a metal such as Pt, Pd, Ni, Cu, etc. The metal hydrogen interactions weaken the covalent bond between the two hydrogen atoms. The steps are typically – – • • • • Adsorption and activation of the reactants Migration of the absorbed reactants on the surface Reaction among the absorbed substances Escape, or desorption of the products Homogeneous Catalysis •The catalyst is in the same phase as the reaction mixture •Homogeneous catalysts have advantages over heterogeneous catalysts •Cheaper than transition metals •Can catalyze a specific portion of a molecule •Reactions can be carried out under atmospheric conditions Heterogeneous Catalyst Example The hydrogenation of ethylene requires the use of a heterogeneous catalyst Pt CH2=CH2 + H2 CH3CH3 The hydrogen molecules are adsorbed on the surface of the platinum metal, thus weakening the bonds holding the hydrogen atoms together. The exposed protons are then attracted to the pi bonding electrons from the ethylene molecule forming new sigma bonds between the hydrogen atoms and the carbon atoms of the ethylene. Homogeneous Catalysis 1. The catalyst is in the same phase as the reaction mixture 2. Homogeneous catalysts have advantages over heterogeneous catalysts a. Cheaper than transition metals b. Can catalyze a specific portion of a molecule c. Reactions can be carried out under atmospheric conditions 3. Example the hydration of ethene into ethyl alcohol • R = K [CH3COOC2H5] uncatalyzed • R = Kc[CH3COOC2H5][H+] catalyzed Review Predict the Order N2O N2 + 1/2O2 The radioactive isotope 32P decays by firstorder kinetics and has a half-life of 14.3 days. a. Predict the order. The radioactive isotope 32P decays and has a half-life of 14.3 days. a. Predict the order. Since the half-life of 14.3 days is given, and the concentration of 32P is constant, then the rate cannot be dependent on the initial concentration, leaving first order rate law as the only choice. The radioactive isotope 32P decays and has a half-life of 14.3 days. a. Predict the order. Since the half-life of 14.3 days is given, and the concentration of 32P is constant, then the rate cannot be dependent on the initial concentration, leaving first order rate law as the only choice. Why is the concentration constant? The radioactive isotope 32P decays and has a half-life of 14.3 days. a. Predict the order. Since the half-life of 14.3 days is given, and the concentration of 32P is constant, then the rate cannot be dependent on the initial concentration, leaving first order rate law as the only choice. Why is the concentration constant, because the density is constant, right? About Phosphorus • Phosphorus, P, is a non-metallic main group element belonging to Group 15 of the periodic table. Atomic Number : 15 • Relative Atomic Mass : 31 • Melting Point : 41 degC • Boiling Point : 280 degC • Relative Density : 1.8 g/mL (White Phosphorus) • Relative Density : 2.2 g/mL (Red Phosphorus) • White phosphorus is the most common allotrope. Density to Molarity 1.8 g mole mL 31 g Density to Molarity 1.8 g mole mL mL 31 g 10-3 L Density to Molarity 1.8 g mole mL = 58 M mL 31 g 10-3 L Sample Problem The radioactive isotope 32P decays and has a half-life of 14.3 days. a. Predict the order. b. How long does it take for 95.0% of a sample of 32P to decay? Sample Problem The radioactive isotope 32P decays and has a half-life of 14.3 days. a. Predict the order. b. How long does it take for 95.0% of a sample of 32P to decay? Since this is first order, we can use the first order half-life equation to calculate the rate constant k. Sample Problem The radioactive isotope 32P decays by firstorder kinetics and has a half-life of 14.3 days. a. Predict the order. b. How long does it take for 95.0% of a sample of 32P to decay? 95% means [A] =100.0 [A] =5.0 0 t t1/ 2 = 0.693 k [A]t ln [A] = -kt 0 Sample Problem The radioactive isotope 32P decays by firstorder kinetics and has a half-life of 14.3 days. a. Predict the order. b. How long does it take for 95.0% of a sample of 32P to decay? [A] =100.0 [A] =5.0 0 t t1/ 2 = 0.693 k T1/2 = 0.693 0.693 = k 14.3 Sample Problem The radioactive isotope 32P decays by firstorder kinetics and has a half-life of 14.3 days. a. Predict the order. b. How long does it take for 95.0% of a sample of 32P to decay? [A] =100.0 [A] =5.0 0 t t1/ 2 = 0.693 k 0.693 0.693 = k 14.3 = 0.04846 1/day T1/2 = The radioactive isotope 32P decays by firstorder kinetics and has a half-life of 14.3 days. a. Predict the order. b. How long does it take for 95.0% of a sample of 32P to decay? [A] =100.0 [A] =5.0 0 t t1/ 2 = 0.693 k [A]t ln [A] = -kt 0 0.693 0.693 = k 14.3 = 0.04846 1/day T1/2 = The radioactive isotope 32P decays by firstorder kinetics and has a half-life of 14.3 days. a. Predict the order. b. How long does it take for 95.0% of a sample of 32P to decay? [A] =100.0 [A] =5.0 0 t t1/ 2 = 0.693 k [A]t ln [A] = -kt 0 0.693 0.693 = k 14.3 = 0.04846 1/day T1/2 = 5.0 ln 100.0 The radioactive isotope 32P decays by firstorder kinetics and has a half-life of 14.3 days. a. Predict the order. b. How long does it take for 95.0% of a sample of 32P to decay? [A] =100.0 [A] =5.0 0 t t1/ 2 = 0.693 k [A]t ln [A] = -kt 0 0.693 0.693 = k 14.3 = 0.04846 1/day T1/2 = 5.0 ln 100.0 = - 0.04846t The radioactive isotope 32P decays by firstorder kinetics and has a half-life of 14.3 days. a. Predict the order. b. How long does it take for 95.0% of a sample of 32P to decay? [A] =100.0 [A] =5.0 0 t t1/ 2 = 0.693 k [A]t ln [A] = -kt 0 0.693 0.693 = k 14.3 = 0.04846 1/day T1/2 = 5.0 ln 100.0 = - 0.04846t - 2.996 = - 0.04846t The radioactive isotope 32P decays by firstorder kinetics and has a half-life of 14.3 days. a. Predict the order. b. How long does it take for 95.0% of a sample of 32P to decay? [A] =100.0 [A] =5.0 0 t t1/ 2 = 0.693 k [A]t ln [A] = -kt 0 0.693 0.693 = k 14.3 = 0.04846 1/day T1/2 = 5.0 ln 100.0 = - 0.04846t - 2.996 = - 0.04846t t = 61.8 days Sample Problem The gas-phase reaction between methane and diatomic sulfur is given by the equation. Calculate Ea CH4 (g) 2S2 (g) K (L/mole-s) K1 = 1.1 K2 = 6.4 k2 ln k1 = Ea R [ CS2 (g) + 2 H2S (g) T (°C) 550 T (K) 823 625 898 1 - 1 ] T1 T2 Sample Problem The gas-phase reaction between methane and diatomic sulfur is given by the equation. Calculate Ea CH4 (g) 2S2 (g) K (L/mole-s) K1 = 1.1 K2 = 6.4 k1 ln k2 6.4 ln 1.1 = = Ea R Ea 8.314 CS2 (g) + 2 H2S (g) T (°C) 550 T (K) 823 625 898 [ 1 - 1 ] T1 T2 [ 1 - 1 823 898 ] Sample Problem The gas-phase reaction between methane and diatomic sulfur is given by the equation. Calculate Ea CH4 (g) 2S2 (g) K (L/mole-s) K1 = 1.1 K2 = 6.4 k1 ln k2 6.4 ln 1.1 = = Ea R Ea 8.314 [ [ CS2 (g) + 2 H2S (g) T (°C) 550 T (K) 823 625 898 1 - 1 ] T1 T2 1 - 1 823 898 ] Ea= 1.4 X 102 Kj/mole Sample Problem The decomposition of ammonia to nitrogen and hydrogen was studied on two surfaces. Ea with tungsten is 163 kj/mole and 197 kj/mole using Os. Without catalyst the activation energy is 335 kj/mole. a. Which surface is the better catalyst? b. How many times faster is the reaction at 298 K on tungsten compared to no catalyst? Sample Problem The decomposition of ammonia to nitrogen and hydrogen was studied on two surfaces. Ea with tungsten is 163 kj/mole and 197 kj/mole using Os. Without catalyst the activation energy is 335 kj/mole. a. Which surface is the better catalyst? Tungsten b. How many times faster is the reaction at 298 K on tungsten compared to no catalyst? Sample Problem The decomposition of ammonia to nitrogen and hydrogen was studied on two surfaces. Ea with tungsten is 163 kj/mole and 197 kj/mole using Os. Without catalyst the activation energy is 335 kj/mole. a. Which surface is the better catalyst? b. How many times faster is the reaction at 298 K on tungsten compared to no catalyst? 335 kj 1000 j mole-K Ea/RT = mole =135.2 8.3145 j kj 298 K K = e-135.2 Sample Problem The decomposition of ammonia to nitrogen and hydrogen was studied on two surfaces. Ea with tungsten is 163 kj/mole and 197 kj/mole using Os. Without catalyst the activation energy is 335 kj/mole. a. Which surface is the better catalyst? b. How many times faster is the reaction at 298 K on tungsten compared to no catalyst? 335 kj 1000 j mole-K Ea/RT = mole =135.2 8.3145 j kj 298 K K = e-135.2 K = 1.920X10-59 Sample Problem The decomposition of ammonia to nitrogen and hydrogen was studied on two surfaces. Ea with tungsten is 163 kj/mole and 197 kj/mole using Os. Without catalyst the activation energy is 335 kj/mole. a. Which surface is the better catalyst? b. How many times faster is the reaction at 298 K on tungsten compared to no catalyst? 335 kj 1000 j mole-K Ea/RT = mole =135.2 8.3145 j kj 298 K 163 kj 1000 j mole-K Ea/RT= mole =65.78 kj 8.3145j 298 K K = e-135.2 K = 1.920X10-59 Sample Problem The decomposition of ammonia to nitrogen and hydrogen was studied on two surfaces. Ea with tungsten is 163 kj/mole and 197 kj/mole using Os. Without catalyst the activation energy is 335 kj/mole. a. Which surface is the better catalyst? b. How many times faster is the reaction at 298 K on tungsten compared to no catalyst? 335 kj 1000 j mole-K Ea/RT = mole =135.2 8.3145 j kj 298 K K = e-135.2 K = 1.920X10-59 -29 K = 2.705X10 1000 j 163 kj mole-K Ea/RT= mole =65.78 kj 8.3145j 298 K Sample Problem The decomposition of ammonia to nitrogen and hydrogen was studied on two surfaces. Ea with tungsten is 163 kj/mole and 197 kj/mole using Os. Without catalyst the activation energy is 335 kj/mole. a. Which surface is the better catalyst? b. How many times faster is the reaction at 298 K on tungsten compared to no catalyst? 335 kj 1000 j mole-K Ea/RT = mole =135.2 8.3145 j kj 298 K K = e-135.2 K = 1.920X10-59 -29 K = 2.705X10 1000 j 163 kj mole-K Ea/RT= mole =65.78 kj 8.3145j 298 K K/K = 1.409X10-30 Sample Problem The decomposition of ammonia to nitrogen and hydrogen was studied on two surfaces. Ea with tungsten is 163 kj/mole and 197 kj/mole using Os. Without catalyst the activation energy is 335 kj/mole. a. Which surface is the better catalyst? b. How many times faster is the reaction at 298 K on tungsten compared to no catalyst? 335 kj 1000 j mole-K Ea/RT = mole =135.2 8.3145 j kj 298 K K = e-135.2 K = 1.920X10-59 -29 K = 2.705X10 1000 j 163 kj mole-K Ea/RT= mole =65.78 kj 8.3145j 298 K K/K = 1.409X10-30 1.409X1030 times faster Sample Problem The experimental rate law for ozone decomposition has 2 [O ] 3 been determined to be rate = k Propose a two [O2] step mechanism consisting of a fast reversible first step and a slow second step. 2 O3 3 O2 Remember mechanism steps must add to give overall reaction. O3 k1 K-1 O + O3 O2 + O fast 2 O2 slow Sample Problem The experimental rate law for ozone decompositin has 2 [O ] 3 been determined to be rate = k Propose a two [O2] step mechanism consisting of a fast reversible first step and a slow second step. 2 O3 3 O2 Remember mechanism steps must add to give overall reaction. O3 k1 O + O3 2 O3 O2 + O fast K-1 k2 2 O2 slow 3 O2 Sample Problem The experimental rate law for ozone decompositin has 2 [O ] 3 been determined to be rate = k Propose a two [O2] step mechanism consisting of a fast reversible first step and a slow second step. 2 O3 3 O2 Remember mechanism steps must add to give overall reaction. O3 k1 O + O3 2 O3 O2 + O fast K-1 k2 2 O2 slow 3 O2 Sample Problem The experimental rate law for ozone decomposition has 2 [O ] 3 been determined to be rate = k Propose a two [O2] step mechanism consisting of a fast reversible first step and a slow second step. 2 O3 3 O2 Remember mechanism steps must add to give overall reaction. R =R O3 k1 K-1 O + O3 2 O3 f O2 + O fast k2 2 O2 slow 3 O2 r K1[O3] = k-1[O2][O] [O] = k1/k-1[O3]/O2 Sample Problem O3 Rf = Rr k1 O2 + O fast K-1 O + O3 k2 2 O3 2 O2 slow [O] = k1/k-1[O3]/O2 3 O2 R = k2[O][O3] R = k2 k1/k-1[O3]/O2 [O3] R= K1[O3] = k-1[O2][O] K[O3]2 [O2] [O] = k1/k-1[O3]/O2 Sample Problem O3 Rf = Rr k1 O2 + O fast K-1 O + O3 k2 2 O3 2 O2 slow 3 O2 R = k2[O][O3] R = k2 k1/k-1[O3]/O2 [O3] R= K[O3]2 [O2] K1[O3] = k-1[O2][O] [O] = k1/k-1[O3]/O2 Since the elementary steps add up to equal the overall reaction and the slowest step of the mechanism yields a rate law consistent with experimental evidence this must be an acceptable mechanism. Other mechanisms are possible. Review Continued Conceptual Questions If the rate of a reaction is directly proportional to the concentration of a reactant, what does this tell you about? a. The order of the reaction with respect to the reactant? b. The overall order? Conceptual Questions If the rate of a reaction is directly proportional to the concentration of a reactant, what does this tell you about? a. The order of the reaction with respect to the reactant? The reaction is first order in that reactant. b. The overall order? Conceptual Questions If the rate of a reaction is directly proportional to the concentration of a reactant, what does this tell you about? a. The order of the reaction with respect to the reactant? The reaction is first order in that reactant. b. The overall order? The overall order is ≥ 1 Conceptual Questions What is the relationship between the order of a rate law and the molecularity of a reaction? Conceptual Questions What is the relationship between the order of a rate law and the molecularity of a reaction? The order of a reaction depends upon the molecularity of the rate determining step Conceptual Questions What is the relationship between the rate law and the balanced chemical equation? Conceptual Questions What is the relationship between the rate law and the balanced chemical equation? None! Conceptual Questions For any given reaction, what is the relationship between the energy of activation and a. electrostatic repulsions? Conceptual Questions For any given reaction, what is the relationship between the energy of activation and a. electrostatic repulsions? An increase in repulsions between reactants will increase the activation energy. Conceptual Questions For any given reaction, what is the relationship between the energy of activation and a. electrostatic repulsions? An increase in repulsions between reactants will increase the activation energy. b. bond formation? Conceptual Questions For any given reaction, what is the relationship between the energy of activation and a. electrostatic repulsions? An increase in repulsions between reactants will increase the activation energy. b. bond formation? Bond formation tends to tends to lower the activation energy Conceptual Questions For any given reaction, what is the relationship between the energy of activation and a. electrostatic repulsions? An increase in repulsions between reactants will increase the activation energy. b. bond formation? Bond formation tends to tends to lower the activation energy c. The nature of the activated complex? Conceptual Questions For any given reaction, what is the relationship between the energy of activation and a. electrostatic repulsions? An increase in repulsions between reactants will increase the activation energy. b. bond formation? Bond formation tends to tends to lower the activation energy c. The nature of the activated complex? An increase in the energy of the activated complex increases the activation energy of the reaction. Conceptual Questions What is the relationship between A, activation energy, and temperature? Conceptual Questions What is the relationship between the frequency factor (A), activation energy, and temperature? Conceptual Questions What is the relationship between the frequency factor (A), activation energy, and temperature? The activation energy is not related to either the temperature or the frequency factor A, but the frequency factor is usually proportional to the square root of the temperature. How does an increase in frequency factor affect the rate of reaction? Conceptual Questions What is the relationship between the frequency factor (A), activation energy, and temperature? The activation energy is not related to either the temperature or the frequency factor A, but the frequency factor is usually proportional to the square root of the temperature. How does an increase in frequency factor affect the rate of reaction? An increase in frequency factor increases the rate. Conceptual Questions What effect does a catalyst have on the energy of activation of a reaction? Conceptual Questions What effect does a catalyst have on the energy of activation of a reaction? A catalyst lowers the activation energy of a reaction, thus increasing the fraction of particles having the necessary activation energy to react. Conceptual Questions What effect does a catalyst have on the energy of activation of a reaction? A catalyst lowers the activation energy of a reaction, thus increasing the fraction of particles having the necessary activation energy to react. What effect does it have on the frequency factor, A? Conceptual Questions What effect does a catalyst have on the energy of activation of a reaction? A catalyst lowers the activation energy of a reaction, thus increasing the fraction of particles having the necessary activation energy to react. What effect does it have on the frequency factor, A? Some catalysts can also orient the reactants and thereby increase the frequency factor. Catalysts have no effect on the change in potential energy for the reaction, ΔH. Conceptual Questions Consider the reaction between cerium(IV) and thallium(I). 2Ce4+ + Tl+ 2Ce3+ + Tl3+ This reaction is slow, but it is catalyzed by Mn2+, as shown in the following mechanism: Ce4+ + Mn2+ Ce3+ + Mn3+ Ce4+ + Mn3+ Mn4+ + Tl+ Ce3+ + Mn2+ Tl3+ + Mn2+ In what way does Mn2+ increase the reaction rate? Conceptual Questions Consider the reaction between cerium(IV) and thallium(I). 2Ce4+ + Tl+ 2Ce3+ + Tl3+ This reaction is slow, but it is catalyzed by Mn2+, as shown in the following mechanism: Ce4+ + Mn2+ Ce3+ + Mn3+ Ce4+ + Mn3+ Mn4+ + Tl+ Ce3+ + Mn4+ Tl3+ + Mn2+ In what way does Mn2+ increase the reaction rate? The Mn2+ ion donates an electron to Ce4+ and then accepts and electron from Tl+. It functions as an electron conduit. Numerical Questions Half-lives for the reaction A + B C were calculated at three values of [A]0, and the concentrations of B was the same in all cases. The data are listed bellow. [A]0, M t1/2, s 0.50 420 0.75 280 1.0 210 Does this reaction follow first-order kinetics? Numerical Questions Half-lives for the reaction A + B C were calculated at three values of [A]0, and the concentrations of B was the same in all cases. The data are listed bellow. [A]0, M t1/2, s 0.50 420 0.75 280 1.0 210 Does this reaction follow first-order kinetics? No. The reaction is second order: the half-life decreases with increasing reactant concentration according to t1/2 = 1/k[A]0 Numerical Questions Nitramide (O2NNH2) decomposes in solution to N2O and H2O . Give the rate law for this process. O2NNH2 O2NNH H+ + OH - K1 K-1 K2 K3 O2NNH2 [O2NNH2] Rate = K [H+] O2NNH - + H+ fast N2O + OH - slow H2O fast N2O + H2O Note: H+ is not a reactive intermediate, since it is not consumed in the next step The End