Chemical Kinetics Chapter 12

Reaction Rates 01 2 HI(g)  H2(g) + I2(g) Reaction Rate: The change in the concentration of a reactant or a product with time (M/s). change in concentration divided by the change in time Reactant  Products A  B 2 HI(g)  H2(g) + I2(g) Chapter 12

Reaction Rates and Stoichiometry What is the general rate of the following reaction ? 2 HI(g)  H2(g) + I2(g) Rate = − 1 2 [HI] t = [I2] Chapter 12

Reaction Rates and Stoichiometry To generalize, for the reaction aA + bB cC + dD Rate* = − 1 a [A] t = − b [B] = c [C] d [D] *: General rate of reaction Chapter 12

Which of the expressions below, corresponding to the reaction of bromine with formic acid, is incorrect?

Which of the expressions below, corresponding to the reaction of bromine with formic acid, is incorrect?

How Do we study Rate of a reaction? Consider the decomposition of N2O5 to give NO2 and O2: 2 N2O5(g) 4 NO2(g) + O2(g) Brown Colorless Chapter 12

Reaction Rates: concentration versus time curve 03 2 N2O5(g) 4 NO2(g) + O2(g) Reaction Rates: concentration versus time curve 03 Average Rate = Rate between two points in time The slope of each triangle Between two points Chapter 12

Reaction Rates 2N2O5(g) 4NO2(g) + O2(g) The rate of change for both NO2 and O2 is positive (increasing) and the rate of change for N2O5 is negative (decreasing). The calculated rate is dependent on the time points taken.

Instantaneous rate: Rate for specific instance in time Slope of the tangent to a concentration versus time curve Initial Rate Chapter 12

Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g) time 393 nm light Detector 393 nm Br2 (aq) D[Br2] a DAbsorption Chapter 12

Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g) slope of tangent slope of tangent slope of tangent average rate = - D[Br2] Dt = - [Br2]final – [Br2]initial tfinal - tinitial instantaneous rate = rate for specific instance in time The slope of a line tangent to the curve at any point is the instantaneous rate at that time Chapter 12

slope of tangent slope of tangent Chapter 12

The Rate Law; rate = 3.50 x 10-3 s-1 [Br2] rate a [Br2] rate = k [Br2] k = rate [Br2] = rate constant The Rate Law; rate = 3.50 x 10-3 s-1 [Br2] Chapter 12

k = 3.50 x 10-3 s-1 Chapter 12

The Rate Law and Reaction Order The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers. aA + bB cC + dD Rate = k [A]x[B]y reaction is xth order in A reaction is yth order in B reaction is (x +y)th order overall Chapter 12

The Rate Law and Reaction Order are Experimentally Determined Chapter 12

Determine the reaction order for: F2 (g) + 2ClO2 (g) 2FClO2 (g) ---- rate = k [F2]x[ClO2]y 1 vs 3 Double [F2] with [ClO2] constant Rate doubles x = 1 1 vs 2 rate = k [F2][ClO2] Quadruple [ClO2] with [F2] constant Rate quadruples y = 1 The instantaneous rate at the beginning of a reaction is called initial rate Chapter 12

Rate Laws Rate laws are always determined experimentally. Reaction order is always defined in terms of reactant (not product) concentrations. The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation. F2 (g) + 2ClO2 (g) 2FClO2 (g) 1 rate = k [F2][ClO2] Chapter 12

S2O82- (aq) + 3I- (aq) 2SO42- (aq) + I3- (aq) Determine the rate law and calculate the rate constant for the following reaction from the following data: S2O82- (aq) + 3I- (aq) 2SO42- (aq) + I3- (aq) Experiment [S2O82-] [I-] Initial Rate (M/s) 1 0.08 0.034 2.2 x 10-4 2 0.017 1.1 x 10-4 3 0.16 rate = k [S2O82-]x[I-]y y = 1 x = 1 rate = k [S2O82-][I-] Double [I-], rate doubles (experiment 1 & 2) Double [S2O82-], rate doubles (experiment 2 & 3) k = rate [S2O82-][I-] = 2.2 x 10-4 M/s (0.08 M)(0.034 M) = 0.08/M•s Chapter 12

Determine the Rate Law and Reaction Order NH4+(aq) + NO2−(aq) N2(g) + 2 H2O(l) Comparing Experiments 1 and 2, when [NH4+] doubles, the initial rate doubles. Chapter 12

NH4+(aq) + NO2−(aq) N2(g) + 2 H2O(l) Likewise, comparing Experiments 5 and 6, when [NO2−] doubles, the initial rate doubles. Chapter 12

This equation is called the rate law, and k is the rate constant. This means Rate  [NH4+] Rate  [NO2−] Rate  [NH+] [NO2−] or Rate = k [NH4+] [NO2−] This equation is called the rate law, and k is the rate constant. Chapter 12

Rate Laws The exponents tell the order of the reaction with respect to each reactant. This reaction is First-order in [NH4+] First-order in [NO2−] Chapter 12

Rate Law & Reaction Order The reaction of nitric oxide with hydrogen at 1280°C is: 2 NO(g) + 2 H2(g)  N2(g) + 2 H2O(g) From the following data determine the rate law and rate constant. E x p e ri m n t [ NO ] H 2 I iti a l Ra (M/ s ) 1 5 . x 1 – 3 4 Second order in NO, First order in H2 k = 1/3(250+250+260) = 250 M-2.s-1 Chapter 12

First-Order Reactions Concentration and Time Equation 01 First Order: Reaction rate depends on the reactant concentration raised to first power. Rate = k[A] A product D[A] Dt = k [A] - Chapter 12

-(ln[A] -ln[A]0) = kt Concentration and Time Equation For A First-Order Reactions D[A] = K Δt - -(ln[A] -ln[A]0) = kt [A] ln[A] = ln[A]0 - kt See next slide for proof of the formula ln [A]0 [A] = k t [A] = [A]0exp(-kt) [A] is the concentration of A at any time t rate [A] M/s M = [A]0 is the concentration of A at time t = 0 k = = 1/s or s-1 [A] = [A]0exp(-kt) Chapter 12

Integration:     Chapter 12

First-Order Reactions ln [A]0 [A] = k t First-Order Reactions ln[A] = ln[A]0 - kt [A] is the concentration of A at any time t ln[A] = ln[A]0 - kt [A]0 is the concentration of A at time t = 0 Chapter 12

The reaction 2A B is first order in A with a rate constant of 2 The reaction 2A B is first order in A with a rate constant of 2.8 x 10-2 s-1 at 800C. How long will it take for A to decrease from 0.88 M to 0.14 M ? [A]0 = 0.88 M ln [A]0 [A] = k t [A] = 0.14 M t = ? ln [A]0 [A] k ln 0.88 M 0.14 M 2.8 x 10-2 s-1 = t = = 66 s Chapter 12

What is Half- Life ? The half-life, t½, is the time required for the concentration of a reactant to decrease to half of its initial concentration. t½ = t when [A] = [A]0/2 Half Life For the First Order Reaction ln [A]0 [A]0/2 k = t½ ln [A]0 [A] = k t ln2 k = 0.693 Chapter 12

Units of Rate Constants vs Reaction Orders Zeroth Order Reaction: Rate = K [A]0 = K Units of Rate Constants vs Reaction Orders Overall order of this reaction is third. Chapter 12

What is the order of decomposition of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1? What is the half life of decomposition of N2O5 ? 2N2O5(g)  4NO2(g) + O2(g) units of k (s-1) Therefore, decomposition is first order? t½ ln2 k = 0.693 5.7 x 10-4 s-1 = = 1200 s = 20 minutes Chapter 12

Half life of a First Order Reaction Chapter 12

First-order reaction A product # of half-lives [A] 1 ½ [A]0 2 1/4 [A]0 3 1/8 [A]0 4 1/16 [A]0 [A] = [A]0 x (1/2)n Chapter 12

First-Order Reaction 2N2O5(g)  4NO2(g) + O2(g) Show that the decomposition of N2O5 is first order and calculate the rate constant and Half life. k = 1.7 x 10-3 s-1 t1/2 = 408 S Chapter 12

Second-Order Reactions rate = - D[A] Dt A product rate = k [A]2 What is Unit of k ? rate [A]2 M/s M2 = k = = 1/M•s or M-1 s-1 What is Conc. Vs time equation? D[A] Dt = k [A]2 - 1 [A] = [A]0 + kt [A] is the concentration of A at any time t [A]0 is the concentration of A at time t = 0 Chapter 12

Second-Order Reactions So if a process is second-order in A, a plot of 1/[A] vs. t will yield a straight line, and the slope of that line is k. Drive the formula for half life of a second order reaction t = t1/2 = t1/2 [A] 2 [A]0 Chapter 12

Half-life for a second-order reaction t = t1/2 = t1/2 [A] 2 [A]0 = kt + [A]0 1 [A]t [A]0 1 = kt1/2 + 2 = t1/2 k[A]0 1 Chapter 12

Second-Order Reactions = t1/2 k[A]0 1 For a second-order reaction, the half-life is dependent on the initial concentration. Each successive half-life is twice as long as the preceding one. Chapter 12

2 NO2(g)  2NO(g) + O2(g) Example: Is the following reaction first or second order ? What is the value of k? 2 NO2(g)  2NO(g) + O2(g) Figure: Table 12.4.1UN Title: Caption: Chapter 12

Figure: Table 12.4.2UN Title: Caption: Chapter 12

Second-Order Reactions Figure: 12-08-01UN Title: Worked Example 12.8 Caption: Concentration–time data are given for the decomposition of nitrogen dioxide to nitric oxide and molecular oxygen. Plot the data using the relationships from the integrated rate laws to determine if the reaction is first- or second-order. k = 0.54 M-1 . S-1 Second-Order Reactions Chapter 12

Zero Order Reaction: Rate = k Example of Zeroth Order Reaction: Figure: Table 12.3.1UN Title: Caption: Example of Zeroth Order Reaction: Decomposition of N2O on hot platinum surface:        N2O →  N2  +  1/2 O2        Rate  [N2O]0 =  k[N2O]0 = k        d[N2 O]/dt = k Chapter 12

Reaction Mechanisms 01 A reaction mechanism is a sequence of molecular events, or reaction steps, that defines the pathway from reactants to products. Chapter 12

Reaction Mechanisms 02 Single steps in a mechanism are called elementary steps (reactions). An elementary step describes the behavior of individual molecules. An overall reaction describes the reaction stoichiometry. Chapter 12

Reaction Mechanisms NO2(g) + CO(g)  NO(g) + CO2(g) NO2(g) + NO2(g)  NO(g) + NO3(g) Elementary NO3(g) + CO(g)  NO2(g) + CO2(g) Elementary NO2(g) + CO(g)  NO(g) + CO2(g) Overall The chemical equation for an elementary reaction is a description of an individual molecular event that involves the breaking and/or making of chemical bonds. NO3(g) is called reaction intermediate. Chapter 12

Reaction Mechanisms 04 Molecularity: is the number of molecules (or atoms) on the reactant side of the chemical equation. Unimolecular: Single reactant molecule. Chapter 12

Reaction Mechanisms 05 Bimolecular: Two reactant molecules. Termolecular: Three reactant molecules. Chapter 12

Reaction Mechanisms 06 Determine individual steps , the reaction intermediates, and the molecularity of each individual step. 2N2 + O2 2N2O Chapter 12

Rate Laws and Reaction Mechanisms 01 Rate law for an overall reaction must be determined experimentally. Rate law for elementary step follows from its molecularity. Chapter 12

Rate Laws and Reaction Mechanisms 02 The rate law of each elementary step follows its molecularity. The overall reaction is a sequence of elementary steps called the reaction mechanism. Chapter 12

Rate-Determining Step The slowest elementary step in a multistep reaction is called the rate-determining step. The overall reaction cannot occur faster than the speed of the rate-determining step. The rate of the overall reaction is therefore determined by the rate of the rate-determining step. Chapter 12

What is the equation for the overall reaction? The experimental rate law for the reaction between NO2 and CO to produce NO and CO2 is rate = k[NO2]2. The reaction is believed to occur via two steps: Step 1: NO2 + NO2 NO + NO3 Step 2: NO3 + CO NO2 + CO2 What is the equation for the overall reaction? NO2+ CO NO + CO2 What is the intermediate? NO3 What can you say about the relative rates of steps 1 and 2? rate = k[NO2]2 is the rate law for step 1 so step 1 must be slower than step 2 Chapter 12

Determining Reaction Mechanism From The Rate Law

Activation Energy (Ea): Chapter 12

The Arrhenius Equation Typically, as the temperature increases, the rate of reaction increases. 2N2O5(g) 4NO2(g) + O2(g) rate = k[N2O5] Where is temperature dependence? Is it hidden in k? Chapter 12

The Arrhenius Equation 01 Collision Theory: A bimolecular reaction occurs when two correctly oriented molecules collide with sufficient energy. Activation Energy (Ea): The potential energy barrier that must be surmounted before reactants can be converted to products. Chapter 12

Temperature dependence of Rate Constat Adding more of the reactants speeds up a reaction by increasing the number of collisions that occur. Collision rate = Z [A][B] The fraction of collisions with an energy equal or more than activation energy( Ea) : f = e-Ea/RT Raising the temperature speeds up a reaction by providing the energy of activation to more colliding molecules. Z is a constant related to collision frequency Chapter 12

The Arrhenius Equation 02 Only the fraction of collisions having proper orientation can result to products. This is called steric factor, p., In the above example p = 0.5 Chapter 12

The Arrhenius Equation Collision rate = Z [A][B] Where Z is a constant, related to the collision frequency . Reaction rate = p.f.Z [A][B] Reaction rate = k [A][B] k = p.f.Z, Assume p.Z = A, frequency factor A = frequency factor k = A. f, f = e-Ea/RT k = Ae-Ea/RT (p.Z )= A Chapter 12

The Arrhenius Equation K = Ae-Ea/RT A = pZ Figure: 12-15-03UN Title: The Arrhenius equation Caption: Relationship between the rate constant and the activation energy of the reaction. The term pZ is usually represented with the symbol A and is called the frequency factor. (steric factor) Chapter 12

Ea = -R . (slope) Calculating Activation Energy k = A .e-Ea/RT Figure: 12-15-05UN Title: The Arrhenius Plot Caption: According to the rearranged Arrhenius equation, the activation energy for a reaction can be determined by plotting ln k versus (1/T) which yields a line whose slope is equal to (–Ea/R). Ln k Ea = -R . (slope) 1/T Chapter 12

2HI(g) + H2(g)  I2(g) + H2(g) Find the activation energy for the following reaction 2HI(g) + H2(g)  I2(g) + H2(g) Figure: Table 12.5.1UN Title: Caption: Chapter 12

Calculating Activation Energy

Slope = -2.24 x 10 4 K Ea = -R . (slope) Ea = - (8.314 j/K.mol) (-2.24 x 10 4 K) Ea = 190 kj/mol Figure: 12-15-06UN Title: Worked Example 12.11 Caption: Plot of ln k vs. (1/T) from the data for the gas-phase decomposition of hydrogen iodide. Chapter 12

Effect of Temperature on Fraction of Collisions with Activation energy f = e-Ea/RT Chapter 12

f = e-Ea/RT Effect of Temperature Collision Theory: As the average kinetic energy increases, the average molecular speed increases, and thus the collision rate increases.

Change of Rate Constant with temperature If the Ea is known , we can calculate the Rate Constant when temperature is changed: The above formula could be used to determine the rate constant at a different temperature. Chapter 12

Homework: Determination the Activation Energy The second-order rate constant for the decomposition of nitrous oxide (N2O) into nitrogen molecule and oxygen atom has been measured at different temperatures: Determine (graphically) the activation energy for the reaction. k (M - 1 s ) t ( ° C .8 7 x 3 6 .0 5 9 .2 4 Ea = 241 KJ/mole Chapter 12

Catalysis 01 Chapter 12

Catalysis 01 A catalyst is a substance that increases the rate of a reaction without being consumed in the reaction. Chapter 12

Catalysis Note that the presence of a catalyst does not affect the energy difference between the reactants and the products

Catalysis 02 The relative rates of the reaction A + B  AB in vessels a–d are 1:2:1:2. Red = A, blue = B, green = third substance C. (a) What is the order of reaction in A, B, and C? (b) Write the rate law. (c) Write a mechanism that agrees with the rate law. (d) Why doesn’t C appear in the overall reaction? 1 Chapter 12 2 1 2

Catalysis Catalyst: A substance that increases the rate of a reaction without itself being consumed in the reaction. A catalyst is used in one step and regenerated in a later step. rate-determining step H2O2(aq) + I1-(aq) H2O(l) + IO1-(aq) H2O2(aq) + IO1-(aq) H2O(l) + O2(g) + I1-(aq) fast step 2H2O2(aq) 2H2O(l) + O2(g) overall reaction Chapter 12

Catalysis 03 Homogeneous Catalyst: Exists in the same phase as the reactants. Heterogeneous Catalyst: Exists in different phase to the reactants. Chapter 12

Catalysis 04 Catalytic Hydrogenation: Chapter 12

Mechanism of Catalytic Hydrogenation: B X Y H Chapter 12 8

Mechanism of Catalytic Hydrogenation: B Y C C A X H H H H Chapter 12 9

Mechanism of Catalytic Hydrogenation: B X Y H H H H Chapter 12 9

Mechanism of Catalytic Hydrogenation: B Y A X H H H C C H Chapter 12 9

Mechanism of Catalytic Hydrogenation: B X Y H H Chapter 12 9

Mechanism of Catalytic Hydrogenation: B X Y H H Chapter 12 9