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Chemistry II Chapter 13 Chemical Kinetics The Rate of a Chemical Reaction • kinetics is the study of the factors that affect the speed of a reaction and the mechanism by which a reaction proceeds • Factors that influence the speed of a reaction: physical state of reactants, temperature, catalysts, concentration The Rate of a Chemical Reaction Defining Rate rate is how much a quantity changes in a given period of time, e.g. distance Speed time The Rate of a Chemical Reaction • Chemical reaction - concentration change with time concentrat ion Rate time • for reactants, a negative sign is used to show a decrease in concentration [product] [reactant] Rate time time • as time goes on, the rate of a reaction generally slows down and stops because the concentration of the reactants decreases at t = 0 [A] = 8 [B] = 8 [C] = 0 at t = 0 [X] = 8 [Y] = 8 [Z] = 0 at t = 16 [A] = 4 [B] = 4 [C] = 4 at t = 16 [X] = 7 [Y] = 7 [Z] = 1 C A 1 CA C2A 2 1 Rate t 2 tt21 t1 tt 44 08 0.025.25 Rate 1616 00 1 X 1 Z ZX Z2X 2 Rate t 2 tt21 t1 tt 17 08 0.00625 .0625 Rate 1616 00 at t = 16 [A] = 4 [B] = 4 [C] = 4 at t = 16 [X] = 7 [Y] = 7 [Z] = 1 at t = 32 [A] = 2 [B] = 2 [C] = 6 at t = 32 [X] = 6 [Y] = 6 [Z] = 2 2C1 A 1 CA C2A Rate t 2 tt21 t1 tt 62 44 0.0125 Rate .125 1616 00 1 X 1 Z ZX Z2X 2 Rate t 2 tt21 t1 tt 26 17 0.00625 .0625 Rate 1616 00 at t = 32 [A] = 2 [B] = 2 [C] = 6 at t = 32 [X] = 6 [Y] = 6 [Z] = 2 at t = 48 [A] = 0 [B] = 0 [C] = 8 at t = 48 [X] = 5 [Y] = 5 [Z] = 3 2C1 A 1 CA C2A Rate t 2 tt21 t1 tt 80 62 0.0125 Rate .125 1616 00 1 X 1 Z ZX Z2X 2 Rate t 2 tt21 t1 tt 35 26 0.00625 .0625 Rate 1616 00 Rate overall A2 A1 A t 2 t1 t 0 8 Rate overall 0.174 46 0 Rate overall C C2 C1 t 2 t1 t 8 0 Rate overall 0.174 46 0 8 X X2 X1 Rate overall t 2 t1 t Rate overall Rate overall 5 8 0.065 46 0 Z Z2 Z1 t 2 t1 t 3 0 Rate overall 0.065 46 0 rate of change reactants = rate of change products 9 The Rate of a Chemical Reaction Stoichiometry • e.g. H2 (g) + I2 (g) 2 HI(g) • for the above reaction, for every 1 mole of H2 used, 1 mole of I2 will also be used and 2 moles of HI made therefore the rate of change will be different • in order to be consistent, the change in the concentration of each substance is multiplied by 1/coefficient [H 2 ] [I 2 ] 1 [HI] Rate t t 2 t The Rate of a Chemical Reaction Stoichiometry • e.g. H2 (g) + I2 (g) 2 HI(g) The Rate of a Chemical Reaction Average Rate • the average rate is the change in measured concentrations in any particular time period linear approximation of a curve • the larger the time interval, the more the average rate deviates from the instantaneous rate H2 I2 HI Avg. Rate, M/s Avg. Rate, M/s Time (s) [H2], M [HI], M -[H2]/t 1/2 [HI]/t Stoichiometry tells us that every 1rate mole/L of change H Theforaverage is the in 0.000 1.000 0.000 used, the concentration in a given time 2 10.000 20.000 30.000 0.819 0.670 0.549 0.362 0.660 0.902 40.000 50.000 60.000 70.000 0.449 0.368 0.301 0.247 1.102 1.264 1.398 1.506 Ave. rate slows down as reaction proceeds 2 moles/L of HI are made. period. 0.0181 0.0181 0.0149 0.0149 Assuming a 1 L container, at 10 s, we used 0.181 moles of H . Therefore the amount of HI made is 2(0.181 0.0121 moles) 0.0121 = 0.362 moles 2 In the first 10 s, the Δ[H2] is -0.181 0.0100 M, so the rate0.0100 is 0.0081 0.0081 0.0067 0.0067 At 60 s, we used 0.699 moles of H . Therefore the amount0.0054 of HI made is 2(0.699 moles)0.0054 = 1.398 moles 2 80.000 0.202 1.596 0.0045 90.000 100.000 0.165 0.135 1.670 1.730 0.0037 0.0030 Rate of loss reactant = Rate gain product 0.181 M 0.0045 10.000 s 0.0037 M 0.0181 0.0030 s 2.000 1.800 concentration, (M) 1.600 1.400 Concentration vs. Time for H2 + I2 --> 2HI average rate in a given time period = slope of the line connecting the [H2] points; and ½ +slope of the line for [HI] 1.200 the average rate for the first 80 10 s is 0.0108 40 0.0181 M/s 0.0150 1.000 [H2], M [HI], M 0.800 0.600 0.400 0.200 Ave. rate slows down as 0.000 reaction 0.000 proceeds 10.000 20.000 30.000 40.000 50.000 time, (s) 60.000 70.000 80.000 90.000 100.000 The Rate of a Chemical Reaction Instantaneous Rate • average rate becomes less accurate over longer time spans • the instantaneous rate is the change in concentration at any one particular time slope at one point of a curve • determined by taking the slope of a line tangent to the curve at that particular point first derivative of the function for you calculus fans H2 (g) + I2 (g) 2 HI (g) Using [H2], the instantaneous rate at 50 s is: Rate 0.28 M 40 s Rate 0.0070 M s Using [HI], the instantaneous rate at 50 s is: 1 0.56 M Rate 2 40 s rate reactants = rate products Rate 0.0070 M s Generalized rate law: aA + bB → cC + dD 1 [A] 1 [B] 1 [C] 1 [D] Rate a t b t c t d t Ex 13.1 - For the reaction given, the [I] changes from 1.000 M to 0.868 M in the first 10 s. Calculate the average rate in the first 10 s and the Δ[H+]. H2O2 (aq) + 3 I(aq) + 2 H+(aq) I3(aq) + 2 H2O(l) Solve the equation for the Rate (in terms of the change in concentration of the Given quantity) 1 [I ] 1 0.868 M 1.000 M Rate 10 s 3 t 3 Rate 4.40 10-3 Solve the equation of the Rate (in terms of the change in the concentration for the quantity to find) for the unknown value M s 1 [H ] Rate 2 t [H ] 2Rate t [H ] M M 2 4.40 10-3 8.80 10-3 s s t Factors Affecting Reaction Rate - nature of reactants • nature of the reactants means what kind of reactant molecules and what physical condition they are in. small molecules tend to react faster than large molecules; gases tend to react faster than liquids which react faster than solids; powdered solids are more reactive than “blocks” more surface area for contact with other reactants certain types of chemicals are more reactive than others e.g., the activity series of metals ions react faster than molecules no bonds need to be broken Factors Affecting Reaction Rate - Temperature • increasing temperature increases reaction rate chemist’s rule of thumb - for each 10°C rise in temperature, the speed of the reaction doubles • there is a mathematical relationship between the absolute temperature and the speed of a reaction discovered by Svante Arrhenius which will be examined later Factors Affecting Reaction Rate - Catalysts • catalysts are substances which affect the speed of a reaction without being consumed • most catalysts are used to speed up a reaction, these are called positive catalysts catalysts used to slow a reaction are called negative catalysts • homogeneous = present in same phase • heterogeneous = present in different phase • how catalysts work will be examined later Factors Affecting Reaction Rate - Reactant Concentration • generally, the larger the concentration of reactant molecules, the faster the reaction increases the frequency of reactant molecule contact concentration of gases depends on the partial pressure of the gas higher pressure = higher concentration • concentration of solutions depends on the solute to solution ratio (molarity) The Rate Law: Effect of Concentration On Reaction Rate • Mathematical relationship between the rate of the reaction and the concentrations of the reactants • for the reaction aA + bB products the rate law would have the form given below n and m are called the orders for each reactant k is called the rate constant Rate k[A] [B] n m The Rate Law: Effect of Concentration On Reaction Rate • sum of the exponents is called the order of the reaction • The rate law for the reaction: 2 NO(g) + O2(g) ⇌ 2 NO2(g) Rate = k[NO]2[O2] The reaction is second order with respect to [NO], first order with respect to [O2], and third order overall The Rate Law: Effect of Concentration On Reaction Rate Sample Rate Laws Reaction Rate Law CH3CN CH3NC Rate = k[CH3CN] CH3CHO CH4 + CO Rate = k[CH3CHO] 2 N2O5 4 NO2 + O2 Rate = k[N2O5]2 H2 + I2 2 HI Rate = k[H2][I2] The Rate Law: Effect of Concentration On Reaction Rate Example: A → Products Rate = k[A]1 [A] (M) Initial Rate (M/s) 0.10 0.015 0.20 0.030 0.30 0.060 k = 0.015 / 0.10 = 0.15 s-1 If concentration of A doubles, the new rate, Rate2 = k[2A]1 = 2 k[A]1 = 2 x Rate The Rate Law: Effect of Concentration On Reaction Rate Example: Zero order: Second order: [A] (M) Initial Rate (M/s) [A] (M) Initial Rate (M/s) 0.10 0.015 0.10 0.015 0.20 0.015 0.20 0.060 0.30 0.015 0.30 0.240 Concentration of A doubles, the rate is constant Rate1 = Rate2 = k Conc. A doubles, new rate Rate2 = k[2A]2 = 4 x k[A]2 Rate2 = 4 x Rate Reactant Concentration vs. Time A Products 0: Concentration dec. linearly with time. Rate Is constant, reaction does not slow down as [A] dec. 1 and 2: Rate slows as reaction proceeds since [A] dec. Rate = k[A]2 Rate = k[A] Rate = k The Integrated Rate Law • can only be determined experimentally • graphically rate = slope of curve [A] vs. time if graph [A] vs time is straight line, then exponent on A in rate law is 0, rate constant = -slope if graph ln[A] vs time is straight line, then exponent on A in rate law is 1, rate constant = -slope if graph 1/[A] vs time is straight line, exponent on A in rate law is 2, rate constant = slope The Integrated Rate Law • the half-life, t1/2, of a reaction is the length of time it takes for the concentration of the reactants to fall to ½ its initial value • the half-life of the reaction depends on the order of the reaction Zero Order Reactions (n = 0) Rate = -d[A] = k[A]0 = k dt constant rate reactions Solution: [A] = -kt + [A]0 y = mx + b ∫ -d[A]/dt = ∫ k ∫ d[A] = -∫ k dt [A] = -kt + C, where C = [A]0 (= integrated rate law) graph of [A] vs. time is straight line with slope = -k and y-intercept = [A]0 [A] = [A0]/2, t ½ = [A0]/2k Units: when Rate = M/sec, k = M/sec [A]0 [A] time First Order Reactions (n = 1) ∫ -d[A]/dt = ∫ k[A] ∫ 1 d[A] = -∫ k dt [A] ln[A] = -kt + C, where C = ln[A]0 Rate = -d[A] = k[A] dt Solution: ln[A] = -kt + ln[A]0 graph of ln[A] vs. time gives straight line with slope = -k and y-intercept = ln[A]0 used to determine the rate constant [A] = [A0]/2, t½ = ln 2 k [lna-lnb = ln(a/b)] ln[A]0 the half-life of a first order reaction is constant ln[A] Units: when Rate = M/sec, k = sec-1 (dim. Analysis: M/s = k.M) Rate slows as reaction proceeds since [A] dec. time Rate Data for hydrolysis of C4H9Cl Time (sec) [C4H9Cl], M 0.0 0.1000 Show reaction is first-order and find k 50.0 100.0 150.0 0.0905 0.0820 0.0741 200.0 300.0 400.0 500.0 0.0671 0.0549 0.0448 0.0368 800.0 10000.0 0.0200 0.0000 C4H9Cl + H2O C4H9OH + 2 HCl Concentration vs. Time for the Hydrolysis of C 4H9Cl 0.12 concentration, (M) 0.1 0.08 0.06 0.04 0.02 0 0 200 400 600 time, (s) 800 1000 C4H9Cl + H2O C4H9OH + 2 HCl Rate vs. Time for Hydrolysis of C 4H9Cl 2.5E-04 Rate, (M/s) 2.0E-04 1.5E-04 1.0E-04 5.0E-05 0.0E+00 0 100 200 300 400 time, (s) 500 600 700 800 C4H9Cl + H2O C4H9OH + 2 HCl LN([C4H9Cl]) vs. Time for Hydrolysis of C 4H9Cl 0 slope = -2.01 x 10-3 -0.5 LN(concentration) -1 k= 2.01 x 10-3 s-1 -1.5 -2 -2.5 -3 t1 y = -2.01E-03x - 2.30E+00 2 -3.5 0.693 k 0.693 2.0110 3 s -1 345 s -4 -4.5 0 100 200 300 400 time, (s) 500 600 700 800 Second Order Reactions Rate = -d[A] = k[A] = k[A]2 dt Solution: 1/[A] = kt + 1/[A]0 y = mx + b graph 1/[A] vs. time gives straight line with slope = k and y-intercept = 1/[A]0 used to determine the rate constant 1/[A] t½ = 1 k[A0] when Rate = M/sec, k = M-1∙sec-1 l/[A]0 time Example Second-Order Reaction Show that the reaction: NO2 → NO + O is incorrect according to the data below PNO2, (mmHg) Time (hrs.) ln(PNO2) 1/(PNO2) 0 100.0 4.605 0.01000 30 60 62.5 45.5 4.135 3.817 0.01600 0.02200 90 35.7 3.576 0.02800 120 150 29.4 25.0 3.381 3.219 0.03400 0.04000 180 21.7 3.079 0.04600 210 19.2 2.957 0.05200 240 17.2 2.847 0.05800 Rate Data Graphs For NO2 → NO + O Partial Pressure NO2, mmHg vs. Time 100.0 90.0 Pressure, (mmHg) 80.0 Non-linear so not zero order 70.0 60.0 50.0 40.0 30.0 20.0 10.0 0.0 0 50 100 150 Time, (hr) 200 250 Rate Data Graphs For NO2 ® NO + O ln(PNO2) vs. Time 4.8 4.6 4.4 ln(pressure) 4.2 4 Non-linear so not first order 3.8 3.6 3.4 3.2 3 2.8 2.6 2.4 0 50 100 150 Time (hr) 200 250 Rate Data Graphs For NO2 → NO + O 1/(PNO2) vs Time k = 2 x 10-4 M-1 s-1 0.07000 1/PNO2 = 0.0002(time) + 0.01 -1 Inverse Pressure, (mmHg ) 0.06000 0.05000 0.04000 0.03000 0.02000 We can deduce actual reaction should be: 2NO2 → 2NO + O2 0.01000 0.00000 0 50 100 150 Time, (hr) 200 250 Tro, Chemistry: A Molecular Approach 47 Ex. 13.4 – The reaction SO2Cl2(g) SO2(g) + Cl2(g) is first order with a rate constant of 2.90 x 10-4 s-1 at a given set of conditions. Find the [SO2Cl2] at 865 s when [SO2Cl2]0 = 0.0225 M Given: Find: [SO2Cl2]0 = 0.0225 M, t = 865, k = 2.90 x 10-4 s-1 [SO2Cl2] Concept Plan: [SO2Cl2]0, t, k [SO2Cl2] Relationships: for a 1st order process : ln[A] kt ln[A] 0 Solution: ln[SO 2Cl2 ] kt ln[SO 2Cl2 ]0 ln[SO 2Cl2 ] 2.90 10- 4 s -1 865 s ln 0.0225 ln[SO 2Cl2 ] 0.251 3.79 4.04 [SO 2Cl2 ] e(-4.04) 0.0175 M Check: the new concentration is less than the original, as expected Ex 13.2 – Determine the rate law and rate constant for the reaction NO2(g) + CO(g) NO(g) + CO2(g) given the data below. Write a general rate law including all reactants Examine the data and find two experiments in which the concentration of one reactant changes, but the other concentrations are the same Expt. Initial Initial Rate Initial Number (M) (M/s) Number [NO22], (M) [CO], [CO], (M) (M/s) 1. 0.10 0.10 0.0021 2. 0.20 0.10 0.0082 3. 0.20 0.20 0.0083 0.033 4. 0.40 0.10 Rate k[NO2 ] [CO] n m Comparing Expt #1 and Expt #2, the [NO2] changes but the [CO] does not 49 Ex 13.2 – Determine the rate law and rate constant for the reaction NO2(g) + CO(g) NO(g) + CO2(g) given the data below. Determine by what factor the concentrations and rates change in these two experiments. [NO2 ]expt 2 [NO2 ]expt 1 Expt. Initial Initial Rate Initial Number [NO2], (M) [CO], (M) (M/s) 1. 2. 0.10 0.20 0.10 0.10 0.0021 0.0082 3. 0.20 0.20 0.0083 4. 0.40 0.10 0.033 0.20 M 2 0.10 M Rate expt 2 Rate expt 1 0.0082 M s 4 M 0.0021 s Ex 13.2 – Determine the rate law and rate constant for the reaction NO2(g) + CO(g) NO(g) + CO2(g) given the data below. Determine to what power the concentration factor must be raised to equal the rate factor. [NO2 ]expt 2 [NO2 ]expt 1 Expt. Initial Initial Rate Initial Number [NO2], (M) [CO], (M) (M/s) 1. 0.10 0.10 0.0021 2. 0.20 0.10 0.0082 3. 0.20 0.20 0.0083 4. 0.40 0.10 0.033 0.20 M n M Rate 0 . 0082 expt 2 2 s [NO 2 ]expt 2 Rate expt 2 4 0.10 M M Rate expt 1 0.0021 s [NO 2 ]expt 1 Rate expt 1 2n 4 n2 Ex 13.2 – Determine the rate law and rate constant for the reaction NO2(g) + CO(g) NO(g) + CO2(g) given the data below. Repeat for the other reactants Expt. Initial Initial Rate Initial Number [NO2], (M) [CO], (M) (M/s) 1. 0.10 0.10 0.0021 2. 0.20 0.10 0.0082 3. 0.20 0.20 0.0083 4. 0.40 0.10 0.033 m 0[CO] M Rate Rate 0 . 0083 expt 3 expt 3 . 20 M expt 3 s [CO] 2 Rate 1 expt 2 Rateexpt 0.0082 M s [CO]expt 2 0.10 M expt2 2 2m 1 m0 [CO]expt 3 Ex 13.2 – Determine the rate law and rate constant for the reaction NO2(g) + CO(g) NO(g) + CO2(g) given the data below. Substitute the exponents into the general rate law to get the rate law for the reaction Expt. Initial Initial Rate Initial Number [NO2], (M) [CO], (M) (M/s) 1. 0.10 0.10 0.0021 2. 0.20 0.10 0.0082 3. 0.20 0.20 0.0083 4. 0.40 0.10 0.033 n = 2, mRate =0 Rate n0 m 2 2k][NO Rate k[NO [CO] 2 ] [CO] 2 k[NO2 ] Ex 13.2 – Determine the rate law and rate constant for the reaction NO2(g) + CO(g) NO(g) + CO2(g) given the data below. Substitute the concentrations and rate for any experiment into the rate law and solve for k Expt. Initial Initial Rate Initial Number [NO2], (M) [CO], (M) (M/s) 1. 0.10 0.10 0.0021 2. 0.20 0.10 0.0082 3. 0.20 0.20 0.0083 4. 0.40 0.10 0.033 Rate k[NO2 ]2 for expt 1 0.0021 M s k 0.10 M 2 0.0021 M s -1 -1 k 0 . 21 M s 0.01 M 2 The Effect of Temperature on Reaction Rate • Rate constant k is temperature dependent • Arrhenius investigated this relationship and showed that: k A e Ea RT where T is the temperature in kelvin R is the gas constant in energy units, 8.314 J/(mol∙K) A is a constant called the frequency factor Ea is the activation energy, the extra energy needed to start the molecules reacting As x (temperature) increases e-1/x will increase up to a maximum value of 1, k increases 10 8 ex 6 e-x y e1/x e-1/x 4 2 0 0 10 x 20 As Ea increases k will decrease (follows e-x graph) The Effect of Temperature on Reaction Rate Activation Energy and the Activated Complex • Ea is an energy barrier to the reaction • amount of energy needed to convert reactants into the activated complex aka transition state • the activated complex is a chemical species with partially broken and partially formed bonds always very high in energy because partial bonds The Effect of Temperature on Reaction Rate The Exponential Factor • e-Ea/RT is a number between 0 and 1 • it represents the fraction of reactant molecules with sufficient energy to make it over the energy barrier • that extra energy comes from converting the KE of motion to PE in the molecule when the molecules collide • e-Ea/RT decreases as Ea increases Reaction rate inc.: -Increasing T increases the Ave. KE of the molecules -Increases no. of molecules with sufficient energy to overcome the energy barrier The Effect of Temperature on Reaction Rate Arrhenius Plots • the Arrhenius Equation can be algebraically solved: Ea ln( k ) R 1 ln A T y = mx + b where y = ln(k) and x = (1/T) a graph of ln(k) vs. (1/T) is a straight line slope of the line = -Ea/R so Ea = -mR ey-intercept = A, (unit is the same as k …why?) k = Ae-Ea/RT lnk = ln(Ae-Ea/RT) lnk = lnA + lne-Ea/RT lnk = lnA – Ea/RT Ex. 13.7 Determine the activation energy and frequency factor for the reaction O3(g) O2(g) + O(g) given the following data: Temp, K 600 700 k, M-1∙s-1 3.37 x 103 4.83 x 104 Temp, K 1300 1400 k, M-1∙s-1 7.83 x 107 1.45 x 108 800 900 1000 1100 3.58 x 105 1.70 x 106 5.90 x 106 1.63 x 107 1500 1600 1700 1800 2.46 x 108 3.93 x 108 5.93 x 108 8.55 x 108 1200 3.81 x 107 1900 1.19 x 109 Ex. 13.7 Determine the activation energy and frequency factor for the reaction O3(g) O2(g) + O(g) given the following data: use a spreadsheet to graph ln(k) vs. (1/T) Ex. 13.7 Determine the activation energy and frequency factor for the reaction O3(g) O2(g) + O(g) given the following data: Ea = m∙(-R) solve for Ea J 4 J Ea 1.12 104 K 8.314 9.3110 mol K mol kJ Ea 93.1 mol A = ey-intercept solve for A A e 26 .8 4.36 1011 A 4.36 1011 M -1 s 1 The Effect of Temperature on Reaction Rate The Collision Model • • for most reactions, in order for a reaction to take place, the reacting molecules must collide into each other. once molecules collide they may react together or they may not, depending on two factors – 1. whether the collision has enough energy to "break the bonds holding reactant molecules together"; 2. whether the reacting molecules collide in the proper orientation for new bonds to form. The Effect of Temperature on Reaction Rate The Collision Model Effective Collisions • collisions in which these two conditions are met (and therefore lead to reaction) are called effective collisions • the higher the A value (frequency of effective collisions), the higher k value and the faster the reaction rate • when two molecules have an effective collision, a temporary, high energy (unstable) chemical species is formed - called an activated complex or transition state The Effect of Temperature on Reaction Rate The Collision Model Orientation Effect The Effect of Temperature on Reaction Rate The Collision Model • A is the factor called the frequency factor and is the number of molecules that can approach overcoming the energy barrier • there are two factors that make up the frequency factor – the orientation factor (p) and the collision frequency factor (z) RTEa k A e Ea pze RT The Effect of Temperature on Reaction Rate The Collision Model Orientation Factor • proper orientation is when the atoms are aligned so that old bonds can break and the new bonds can form • the more complex the reactants, the less frequently they will collide with the proper orientation reactions between atoms generally have p = 1 reactions where symmetry results in multiple orientations leading to reaction have p slightly less than 1 • for most reactions, the orientation factor is less than 1 For many, p << 1 e.g. H(g) + I(g) → HI(g) H2(g) + I2(g) → 2HI(g) HCl(g) + HCl(g) → H2(g) + Cl2(g) Smallest p Reaction Mechanisms • we generally describe chemical reactions with an equation listing all the reactant molecules and product molecules • but the probability of more than 3 molecules colliding at the same instant with the proper orientation and sufficient energy to overcome the energy barrier is negligible • most reactions occur in a series of small reactions involving 1, 2, or at most 3 molecules • describing the series of steps that occur to produce the overall observed reaction is called a reaction mechanism • knowing the rate law of the reaction helps us understand the sequence of steps in the mechanism Reaction Mechanisms • • • Overall reaction: H2(g) + 2 ICl(g) 2 HCl(g) + I2(g) Mechanism: 1) H2(g) + ICl(g) HCl(g) + HI(g) 2) HI(g) + ICl(g) HCl(g) + I2(g) the steps in this mechanism are elementary steps, meaning that they cannot be broken down into simpler steps and that the molecules actually interact directly in this manner without any other steps Reaction Mechanisms Intermediates 1) H2(g) + ICl(g) HCl(g) + HI(g) 2) HI(g) + ICl(g) HCl(g) + I2(g) H2(g) + 2 ICl(g) 2 HCl(g) + I2(g) • • • notice that the HI is a product in Step 1, but then a reactant in Step 2 since HI is made but then consumed, HI does not show up in the overall reaction materials that are products in an early step, but then a reactant in a later step are called intermediates Reaction Mechanisms Molecularity • the number of reactant particles in an elementary step is called its molecularity • a unimolecular step involves 1 reactant particle • a bimolecular step involves 2 reactant particles though they may be the same kind of particle • a termolecular step involves 3 reactant particles though these are exceedingly rare in elementary steps Reaction Mechanisms Rate Laws for Elementary Steps • each step in the mechanism is like its own little reaction – with its own activation energy and own rate law • the rate law for an overall reaction must be determined experimentally • but the rate law of an elementary step can be deduced from the equation of the step H2(g) + 2 ICl(g) 2 HCl(g) + I2(g) 1) H2(g) + ICl(g) HCl(g) + HI(g) 2) HI(g) + ICl(g) HCl(g) + I2(g) Rate = k1[H2][ICl] Rate = k2[HI][ICl] Reaction Mechanisms Rate Laws for Elementary Steps Reaction Mechanisms Rate Determining Step • in most mechanisms, one step occurs slower than the other steps • the result is that product production cannot occur any faster than the slowest step – the step determines the rate of the overall reaction • we call the slowest step in the mechanism the rate determining step the slowest step has the largest activation energy • the rate law of the rate determining step determines the rate law of the overall reaction Another Reaction Mechanism NO2(g) + NO2(g) NO3(g) + NO(g) NO3(g) + CO(g) NO2(g) + CO2(g) NO2(g) + CO(g) NO(g) + CO2(g) Rate = k1[NO2]2 slow Rate = k2[NO3][CO] fast Rateobs = k[NO2]2 The first step is slower than the second step because its activation energy is larger. The first step in this mechanism is the rate determining step. The rate law of the first step is the same as the rate law of the overall reaction. Reaction Mechanisms Validating a Mechanism in order to validate (not prove) a mechanism, two conditions must be met: 1. the elementary steps must sum to the overall reaction 2. the rate law predicted by the mechanism must be consistent with the experimentally observed rate law Reaction Mechanisms Mechanisms with a Fast Initial Step • when a mechanism contains a slow initial step, the rate law will not contain • • • • intermediates when a mechanism contains a fast initial step, the rate limiting step, and hence the rate law may contain intermediates We can express[intermediate] in terms of [reactant] If first step is fast, intermediate products build up (limited by slower step down the line) as they build up they react to re-form reactants reaches equilibrium, the forward and reverse reaction rates are equal – so the concentrations of reactants and products of the step are related substituting into the rate law of the RDS will produce a rate law in terms of just reactants An Example 2 H2(g) + 2 NO(g) 2 H2O(g) + N2(g) Experimentally observed Rateobs = k [H2][NO]2 Proposed mechanism: k1 2 NO(g) ⇌ N2O2(g) Fast k-1 H2(g) + N2O2(g) H2O(g) + N2O(g) H2(g) + N2O(g) H2O(g) + N2(g) Slow Fast 2 H2(g) + 2 NO(g) 2 H2O(g) + N2(g) (rate limiting) Reaction Mechanisms Mechanisms with a Fast Initial Step • Is the mechanism valid? 1. steps must sum to the over all reaction 2. rate law predicted by mechanism must be consistent with exp. observation Since 2nd step is rate limiting, Rate = k2[H2][N2O2] BUT! Contains intermediate [N2O2] , not consistent with observation Since 1st step is in equilibrium we can express [intermediate] in terms of reactants An Example k1 2 NO(g) ⇌ N2O2(g) k-1 H2(g) + N2O2(g) H2O(g) + N2O(g) H2(g) + N2O(g) H2O(g) + N2(g) Fast Slow Rate = k2[H2][N2O2] Fast 2 H2(g) + 2 NO(g) 2 H2O(g) + N2(g) Rateobs = k [H2][NO]2 for Step 1 Rateforward = Ratereverse k1[NO]2 k1[N 2O 2 ] [N 2O 2 ] k1 [NO]2 k1 Rate k 2 [H 2 ][N 2 O 2 ] Rate k 2 [H 2 ] Rate k1 [NO]2 k 1 k 2 k1 [H 2 ][NO]2 k 1 Let k2k1/k-1 = k, we have obs rate law Ex 13.9 Show that the proposed mechanism for the reaction 2 O3(g) 3 O2(g) matches the observed rate law Rate = k[O3]2[O2]-1 k1 O3(g) ⇌ O2(g) + O(g) k-1 O3(g) + O(g) 2 O2(g) Fast Slow Rate = k2[O3][O] (Slow rate has intermediate) for Step 1 Rateforward = Ratereverse k1[O3 ] k1[O 2 ][O] k1 [O] [O3 ][O 2 ]1 k1 Rate k2[O3 ][O] k1 Rate k2[O3 ] [O3 ][O 2 ]-1 k1 k2k1 Rate [O3 ]2[O 2 ]-1 k1 Catalysts • catalysts are substances that affect the rate of a reaction without being consumed • catalysts work by providing an alternative mechanism for the reaction with a lower activation energy • catalysts are consumed in an early mechanism step, then made in a later step mechanism without catalyst mechanism with catalyst O3(g) + O(g) 2 O2(g) Cl(g) + O3(g) ⇌ O2(g) + ClO(g) V. Slow ClO(g) + O(g) O2(g) + Cl(g) O3(g) + O(g) 2 O2(g) Fast Slow Catalysts Demo 2CH3OH(l) + 3O2(g) → 2CO2(g) + 4H2O(g) http://academics.rmu.edu/~short/chem1215/1215demos/oscillating-methanol-7mins.mov Ozone Depletion over the Antarctic Catalysts polar stratospheric clouds contain ice crystals that catalyze reactions that release Cl from atmospheric chemicals Catalysts Homorgeneous and Heterogeneous Catalysts • homogeneous catalysts are in the same phase as the reactant particles Cl(g) in the destruction of O3(g) • heterogeneous catalysts are in a different phase than the reactant particles solid catalytic converter in a car’s exhaust system Catalysts Enzymes • because many of the molecules are large and complex, most biological reactions require a catalyst to proceed at a reasonable rate • protein molecules that catalyze biological reactions are called enzymes • enzymes work by adsorbing the substrate reactant onto an active site that orients it for reaction Enzymatic Hydrolysis of Sucrose