Chemical Kinetics What do we know about chemical reactions? The area of chemistry that concerns reaction rates and reaction mechanisms.

Table 12.1 Concentrations of Reactant and Products as a Function of Time for the Reaction 2NO2(g) 2NO + O2(g)

Figure 12.1 Starting with a Flask of Nitrogen Dioxide at 300 degrees C, Concentrations of Nitrogen Dioxide, Nitric Oxide, and Oxygen are Plotted versus Time

Figure 12.2 Representation of the reaction of: 2NO2 (g) 2NO(g) + O2 (g)

Reaction Rate The change in concentration of a reactant or product per unit of time

Reaction Rate - What is the average rate at which the concentration of NO2 changes over the first 50 seconds of the reaction???? Change in [NO2] Δ [NO2] = - Time elapsed Δ t [.0079]t=50 – [.0100]t-0 [NO2]t=50 – [NO2]t-0 = = 50 s – 0 s 50 s – 0 s = 4.2 x 10-5 mol/L x s

Table 12.2 Average Rate (in mol/L-s) of Decomposition of Nitrogen Dioxide as a Function of Time

Reaction Rates: 1. Can measure disappearance of reactants 2NO2(g)  2NO(g) + O2(g) Reaction Rates: 1. Can measure disappearance of reactants 2. Can measure appearance of products 3. Are proportional stoichiometrically

Reaction Rates: 4. Are equal to the slope tangent to that point 2NO2(g)  2NO(g) + O2(g) Reaction Rates: 4. Are equal to the slope tangent to that point 5. Change as the reaction proceeds, if the rate is dependent upon concentration [NO2] t

Rate Laws Differential rate laws express (reveal) the relationship between the concentration of reactants and the rate of the reaction. The differential rate law is usually just called “the rate law.” Integrated rate laws express (reveal) the relationship between concentration of reactants and time

Writing a (differential) Rate Law Problem - Write the rate law, determine the value of the rate constant, k, and the overall order for the following reaction: 2 NO(g) + Cl2(g)  2 NOCl(g) Experiment [NO] (mol/L) [Cl2] Rate Mol/L·s 1 0.250 1.43 x 10-6 2 0.500 5.72 x 10-6 3 2.86 x 10-6 4 11.4 x 10-6

Writing a Rate Law Part 1 – Determine the values for the exponents in the rate law: R = k[NO]x[Cl2]y Experiment [NO] (mol/L) [Cl2] Rate Mol/L·s 1 0.250 1.43 x 10-6 2 0.500 5.72 x 10-6 3 2.86 x 10-6 4 1.14 x 10-5 In experiment 1 and 2, [Cl2] is constant while [NO] doubles. The rate quadruples, so the reaction is second order with respect to [NO]  R = k[NO]2[Cl2]y

Writing a Rate Law Part 1 – Determine the values for the exponents in the rate law: R = k[NO]2[Cl2]y Experiment [NO] (mol/L) [Cl2] Rate Mol/L·s 1 0.250 1.43 x 10-6 2 0.500 5.72 x 10-6 3 2.86 x 10-6 4 1.14 x 10-5 In experiment 2 and 4, [NO] is constant while [Cl2] doubles. The rate doubles, so the reaction is first order with respect to [Cl2]  R = k[NO]2[Cl2]

Writing a Rate Law Part 2 – Determine the value for k, the rate constant, by using any set of experimental data: R = k[NO]2[Cl2] Experiment [NO] (mol/L) [Cl2] Rate Mol/L·s 1 0.250 1.43 x 10-6

Writing a Rate Law Part 3 – Determine the overall order for the reaction. R = k[NO]2[Cl2] 2 + 1 = 3  The reaction is 3rd order Overall order is the sum of the exponents, or orders, of the reactants

Order =3. Units are L2mol-2s-1 Order =5. Units are L4 mol-4s-1 What is the overall order of each of the following rate laws, and what are the units of the rate constant, K, in each of these rate laws? (Time is in seconds) Rate= k[A] [B] [C] Rate= k[X]2 [Y]3 Rate= k[M]2 [N] Rate= k (e)Rate= k [R] Order =3. Units are L2mol-2s-1 Order =5. Units are L4 mol-4s-1 Order =3. Units are L2 mol-2s-1 Order =0 Units are mol L-1s-1 Order =1 Units are s-1

Analyze: We are given three boxes containing Solution: Analyze: We are given three boxes containing different numbers of spheres representing mixtures containing different reactant concentrations. We are asked to use the given rate law and the compositions of the boxes to rank the mixtures in order of increasing reaction rates. Plan: Because all three boxes have the same volume, we can put the number of spheres of each kind into the rate law and calculate the rate of each box. Solve: next slide

Box 1 contains 5 red spheres and 5 purple spheres, giving the following rate: Box 1: Rate = k(5)(5)2 = 125k Box 2 contains 7 red spheres and 3 purple Spheres: Box 2: Rate = k(7)(3)2 = 63k Box 3 contains 3 red spheres and 7 purple spheres: Box 3: Rate = k(3)(7)2 = 147k

The slowest rate is 63k (box 2), and the Highest is 147k (box 3). Thus, the rates vary in the order 2<1<3 How about if the rate = k[A][B] What would the rank of the mixtures in order of increasing rank be now? Eh? 2=3<1

Order =3. Units are L2mol-2s-1 Order =5. Units are L4 mol-4s-1 What is the overall order of each of the following rate laws, and what are the units of the rate constant, K, in each of these rate laws? (Time is in seconds) Rate= k[A] [B] [C] Rate= k[X]2 [Y]3 Rate= k[M]2 [N] Rate= k (e)Rate= k [R] Order =3. Units are L2mol-2s-1 Order =5. Units are L4 mol-4s-1 Order =3. Units are L2 mol-2s-1 Order =0 Units are mol L-1s-1 Order =1 Units are s-1

Rate Laws Differential rate laws express (reveal) the relationship between the concentration of reactants and the rate of the reaction. The differential rate law is usually just called “the rate law.” Integrated rate laws express (reveal) the relationship between concentration of reactants and time

Determining Order with Concentration vs. Time data (the Integrated Rate Law) Zero Order: First Order: Second Order:

Solving an Integrated Rate Law Time (s) [H2O2] (mol/L) 1.00 120 0.91 300 0.78 600 0.59 1200 0.37 1800 0.22 2400 0.13 3000 0.082 3600 0.050 Problem: Find the integrated rate law and the value for the rate constant, k A graphing calculator with linear regression analysis greatly simplifies this process!!

Time vs. [H2O2] Regression results: y = ax + b a = -2.64 x 10-4 Time (s) [H2O2] 1.00 120 0.91 300 0.78 600 0.59 1200 0.37 1800 0.22 2400 0.13 3000 0.082 3600 0.050 Regression results: y = ax + b a = -2.64 x 10-4 b = 0.841 r2 = 0.8891 r = -0.9429

Time vs. ln[H2O2] Regression results: y = ax + b a = -8.35 x 10-4 Time (s) ln[H2O2] 120 -0.0943 300 -0.2485 600 -0.5276 1200 -0.9943 1800 -1.514 2400 -2.04 3000 -2.501 3600 -2.996 Regression results: y = ax + b a = -8.35 x 10-4 b = -.005 r2 = 0.99978 r = -0.9999

Time vs. 1/[H2O2] Regression results: y = ax + b a = 0.00460 Time (s) 1/[H2O2] 1.00 120 1.0989 300 1.2821 600 1.6949 1200 2.7027 1800 4.5455 2400 7.6923 3000 12.195 3600 20.000 Regression results: y = ax + b a = 0.00460 b = -0.847 r2 = 0.8723 r = 0.9340

And the winner is… Time vs. ln[H2O2] 1. As a result, the reaction is 1st order 2. The (differential) rate law is: 3. The integrated rate law is: 4. But…what is the rate constant, k ?

Finding the Rate Constant, k Method #1: Calculate the slope from the Time vs. ln[H2O2] table. Time (s) ln[H2O2] 120 -0.0943 300 -0.2485 600 -0.5276 1200 -0.9943 1800 -1.514 2400 -2.04 3000 -2.501 3600 -2.996 Now remember:  k = -slope k = 8.32 x 10-4s-1

Finding the Rate Constant, k Method #2: Obtain k from the linear regresssion analysis. Regression results: y = ax + b a = -8.35 x 10-4 b = -.005 r2 = 0.99978 r = -0.9999 Now remember:  k = -slope k = 8.35 x 10-4s-1

Half Life t1/2 Half Life: is defined as the time required for a reactant to reach half its original concentration. 0.693 t1/2 = k

Half Life t1/2 Try this: A certain first-order reaction has a half-life of 20.0 minutes. A. Calculate the rate constant for this reaction. K = 3.47 x 10-2 min-1 What am I NOT dependent on???

Rate Laws Summary Rate = k Rate = k[A] Rate = k[A]2 [A] = -kt + [A]0 Zero Order First Order Second Order Rate Law Rate = k Rate = k[A] Rate = k[A]2 Integrated Rate Law [A] = -kt + [A]0 ln[A] = -kt + ln[A]0 Plot the produces a straight line [A] versus t ln[A] versus t Relationship of rate constant to slope of straight line Slope = -k Slope = k Half-Life

Reaction Mechanism The reaction mechanism is the series of elementary steps by which a chemical reaction occurs. The sum of the elementary steps must give the overall balanced equation for the reaction The mechanism must agree with the experimentally determined rate law

Rate-Determining Step In a multi-step reaction, the slowest step is the rate-determining step. It therefore determines the rate of the reaction. The experimental rate law must agree with the rate-determining step

Identifying the Rate-Determining Step For the reaction: 2H2(g) + 2NO(g)  N2(g) + 2H2O(g) The experimental rate law is: R = k[NO]2[H2] Which step in the reaction mechanism is the rate-determining (slowest) step? Step #1 H2(g) + 2NO(g)  N2O(g) + H2O(g) Step #2 N2O(g) + H2(g)  N2(g) + H2O(g) Step #1 agrees with the experimental rate law

Table 12.7 Examples of Elementary Steps The molecularity of a process tells how many molecules are involved in the process.

Figure 12.9 A Molecular Representation of the Elementary Steps in the Reaction of NO2 and CO

NO2 (g) + CO (g)  NO (g) + CO2 (g) The rate law for this reaction is found experimentally to be Rate = k [NO2]2 CO is necessary for this reaction to occur, but the rate of the reaction does not depend on its concentration. This suggests the reaction occurs in two steps.

A proposed mechanism for this reaction is Rate = k [NO2]2 A proposed mechanism for this reaction is Step 1: NO2 + NO2  NO3 + NO (slow) Step 2: NO3 + CO  NO2 + CO2 (fast) NO2 (g) + CO (g)  NO (g) + CO2 (g)

Identifying Intermediates For the reaction: 2H2(g) + 2NO(g)  N2(g) + 2H2O(g) Which species in the reaction mechanism are intermediates (do not show up in the final, balanced equation?) Step #1 H2(g) + 2NO(g)  N2O(g) + H2O(g) Step #2 N2O(g) + H2(g)  N2(g) + H2O(g) 2H2(g) + 2NO(g)  N2(g) + 2H2O(g)  N2O(g) is an intermediate

2NO (g) + Br2(g)  2 NOBr(g) Challenge Time! 2NO (g) + Br2(g)  2 NOBr(g) Rate = k [NO]2 [Br2] A proposed mechanism for this reaction is: Step 1: NO(g) + Br2(g) NOBr2(g) (fast) Step 2: NOBr2(g) + NO(g) 2NOBr(g) (slow) k1 k-1 k2 dddddddddd dddddddddd dddddddddd

Intermediates are intrinsically UNSTABLE Rate = k [NO]2 [Br2] k1 Step 1: NO(g) + Br2(g) NOBr2(g) (fast) Step 2: NOBr2(g) + NO(g) 2NOBr(g) (slow) k-1 k2 Dynamic equilibrium gets reached. So…. K1[NO][Br2] = k-1[NOBr2] FINALLY

Dynamic equilibrium gets reached. So…. K1[NO][Br2] = k-1[NOBr2] Rate = k [NO]2 [Br2] Dynamic equilibrium gets reached. So…. K1[NO][Br2] = k-1[NOBr2] Solving for [NOBr2] k1 [NOBr2] = [NO][Br2] k-1 Sub into RDS Step 2: NOBr2(g) + NO(g) 2NOBr(g) (slow) Step 2: [NO][Br2] + NO(g) 2NOBr(g) (slow)

Let’s check our understanding of Integrated Rate Laws & Half-life and Reaction Mechanisms Please begin (at your workstations) the following Problems: pg. 571: 49-52

A Model For Chemical Kinetics

Collision Model Key Idea: Molecules must collide to react. However, only a small fraction of collisions produces a reaction. Why?

Collision Model Collisions must have sufficient energy to produce the reaction (must equal or exceed the activation energy). 1. Colliding particles must be correctly oriented to one another in order to produce a reaction. 2.

Factors Affecting Rate Increasing temperature always increases the rate of a reaction. Particles collide more frequently Particles collide more energetically Double the temp…effective collisions increase exponentially! Increasing surface area increases the rate of a reaction Increasing Concentration USUALLY increases the rate of a reaction Presence of Catalysts, which lower the activation energy by providing alternate pathways

Endothermic Reactions

Exothermic Reactions

The Arrhenius Equation k = rate constant at temperature T A = frequency factor Ea = activation energy R = Gas constant, 8.31451 J/K·mol

Catalysis Catalyst: A substance that speeds up a reaction without being consumed Enzyme: A large molecule (usually a protein) that catalyzes biological reactions. Homogeneous catalyst: Present in the same phase as the reacting molecules. Heterogeneous catalyst: Present in a different phase than the reacting molecules.

Lowering of Activation Energy by a Catalyst

Catalysts Increase the Number of Effective Collisions

Heterogeneous Catalysis Carbon monoxide and nitrogen monoxide adsorbed on a platinum surface Step #1: Adsorption and activation of the reactants.

Heterogeneous Catalysis Carbon monoxide and nitrogen monoxide arranged prior to reacting Step #2: Migration of the adsorbed reactants on the surface.

Heterogeneous Catalysis Carbon dioxide and nitrogen form from previous molecules Step #3: Reaction of the adsorbed substances.

Heterogeneous Catalysis Carbon dioxide and nitrogen gases escape (desorb) from the platinum surface Step #4: Escape, or desorption, of the products.