Chapter 12 – Chemical Kinetics

Chapter 12 – Chemical Kinetics
Reaction Rates Rate Laws: Intro Determining the Form of the Rate Law Integrated Rate Law Rate Law: Summary Reaction Mechanism Model for Chemical Kinetics Catalysis

Chemical Kinetics Kinetics is the study of how fast chemical reactions occur. There are 4 important factors which affect rates of reactions: reactant concentration, temperature, action of catalysts, and surface area. Goal: to understand chemical reactions at the molecular level.

Chemical Kinetics Reaction Rates
Speed of a reaction is measured by the change in concentration with time. For a reaction A  B Let us begin with 1.00 mole A, represented by 100 red spheres. (B will be shown as blue spheres)

Chemical Kinetics Reaction Rates

Chemical Kinetics Reaction Rates At t = 0 (time zero) there is 1.00 mol A (100 red spheres) and no B present. At t = 20 min, there is 0.54 mol A and 0.46 mol B. At t = 40 min, there is 0.30 mol A and 0.70 mol B.

For the reaction A  B there are two ways of measuring rate: the speed at which the products appear (i.e. change in moles of B per unit time), or the speed at which the reactants disappear (i.e. the change in moles of A per unit time). A plot of number of moles versus time shows that as the reactants (red A) disappear, the products (blue B) appear.

Chemical Kinetics Reaction Rates

Reaction Rate Change in concentration (conc) of a reactant or product per unit time.

At t = 0 (time zero) there is 1.00 mol A (100 red spheres) and no B present. At t = 10 min, there is 0.74 mol A and 0.26 mol B. Calculating, This average rate assumes that there is a constant rate of reaction during the time period being measured.

C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
Chemical Kinetics Rates in Terms of Concentrations Most useful units for rates are to look at molarity. Since volume is constant, molarity and moles are directly proportional. Consider: C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)

Chemical Kinetics Rates in Terms of Concentrations C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)

Chemical Kinetics Rates in Terms of Concentrations C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq) We can calculate the average rate in terms of the disappearance of C4H9Cl. The units for average rate are mol/L•s or M/s. The average rate decreases with time. We plot [C4H9Cl] versus time. The rate at any instant in time (instantaneous rate) is the slope of the tangent to the curve. Instantaneous rate is different from average rate. We usually call the instantaneous rate the rate.

Chemical Kinetics Rates in Terms of Concentrations

Determining a Reaction Rate

Chemical Kinetics Reaction Rates and Stoichiometry For the reaction C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq) we know In general for aA + bB  cC + dD

Reaction Rates and Stoichiometry
The molar ratios between reactants and products correspond to the relative rates of the reaction. Relative rates – relationship between rates of reactant disappearance and product appearance at a given time. 2 HI(g) → H2(g) + I2(g)

Reaction Rate

-(slope of tangent line)
2 NO2(g) → 2 NO(g) + O2(g) Rate = 2.4 x 10-5 M/s Rate = -(slope of tangent line) Rate = 8.6 x 10-5 M/s Rate = 4.3 x 10-5 M/s

Relative Rates Practice At a given time, the rate of C2H4 reaction is 0.23 M/s and CO2 is .115 M/s. What are the rates of the other reaction components? C2H4(g) + 3 O2(g) → 2 CO2(g) H2O(g) M/s ? M/s ? .076 M/s M/s

Rate Law Introduction

Rate Laws 2NO2  2NO + O2 Chemical Reactions are reversible
2NO + O2  2NO2 Initially the change in concentration of NO2 is only dependent on the forward reaction. Over time product accumulate so that the reverse reaction becomes more important. D[NO2] depends on the difference in the rates of the forward and reverse reaction. Typically avoided by studying initial rate Prior to significant build up of products

Rate Laws Rate = k[NO2]n k = rate constant n = order of reactant
2NO2  2NO + O2 Chemical Reactions are reversible 2NO + O2  2NO2 If we choose conditions to neglect reverse reaction then we can write the following rate law: Rate = k[NO2]n k = rate constant n = order of reactant -Both must be determined by experiment -No Products

Two Types of Rate Laws

Types of Rate Laws Ln[A] = -kt + ln[A]o
Differential Rate Law: expresses how rate depends on concentration. Integrated Rate Law: expresses how concentration depends on time. Ln[A] = -kt + ln[A]o

Why Rate Law? Work backwards from rate law to infer the steps of the reaction. (multiple step reactions) Chemist often try to interrupt biological processes (disease) series of reactions with the introduction of a small molecule. Industrial chemist may be trying to speed the rate of reaction

Concentration & Rate aA + bB → cC + dD General form of rate law:
[A], [B] – conc. in M or Pressure rate = k[A]m[B]n k – rate constant; units vary m, n – reaction orders Note the values of the exponents (orders) have to be determined experimentally. They are not simply related to stoichiometry. Note: m ≠ a and n ≠ b Overall order = sum of individual orders Rate constant is independent of concentration. The rate constant, k , is calculated by substituting data from any trial into the rate law expression and solving for the k value. UNITS are very important!

Dependence of Rate on Concentration
The overall rate order of the reaction can be determined by the UNITS of the rate constant, k. A zero order reaction has k units of M/time or M•time-1 A 1st order reaction has k units = 1/time or time-1 A 2nd order reaction has k units = 1/M•time or M-1•time-1 A 3rd order reaction has k units = 1/M2•time or M-2•time-1 and so on…….

Form of the Rate Law 2N2O5  4NO2 + O2(g)
Determine the exponent of the rate law from data Reverse reaction is negligible

Form of the Rate Law - Exponent
2N2O5  4NO2 + O2(g) Evaluate rate at 0.90 M and 0.45 M. Slope of tangents [N2O5] Rate (mol/Ls) 0.90 M m = 5.4 x 10-4 mol/Ls 0.45 M m = 2.7 x 10-4 mol/Ls When N2O5 is ½ the rate is ½ Means: n = 1 The reaction is first order in N2O5

Reaction Orders For the reaction: A →B, the rate law is: rate = k[A]m
Order (m) Δ[A] by a factor of: Effect on rate Zero (0) 2, 4, 15, ½, etc. None 1st (1) 2 2X 3 3X 2nd (2) 4X 9X ¼X

2 NO(g) + 2 H2(g) → N2(g) + 2 H2O(g) rate = k[NO]2[H2]
What is the order with respect to NO? What is the order with respect to H2? What is the overall order? If [NO] is doubled, what is the effect on the reaction rate? If [H2] is halved, what is the effect on the reaction rate? What are the units of k?

What is the rate of Cl¯ production under these conditions?
PtCl2(NH3)2 + H2O → PtCl(H2O)(NH3)2 + Cl¯ rate = k[PtCl2(NH3)2] k = h-1 Calculate the rate of reaction when the concentration of PtCl2(NH3)2 is 0.020M. k=1.8 x M/h What is the rate of Cl¯ production under these conditions?

Practice problem #21 in your Book
Calculate average rate of decomposition of H2O2 between 0 and 21600s Use this rate to calculate the average rate of production of O2 over the same time period. Average Rate H2O2= 2.31 x 10-5 M/s Average Rate O2= 1.16 x 10-5 M/s

Rate Law From Initial Rates

Method of Initial Rates
Initial Rate: the “instantaneous rate” just after the reaction begins. The initial rate is determined in several experiments using different initial concentrations.

NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l) Determine m by observation: m = 1 Reaction is first order in relation to NO2-

NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l) Determine n by observation: n = 1 Reaction is first order in relation to NH4+

NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l) Overall reaction order is the sum of n and m: So the overall reaction order for this reaction is n + m = 2 Evaluate k

Determining Rate Laws Initial Rates Method
Find two experiments in which all but one reactant’s concentration is constant. Observe the relationship between concentration change and rate change to determine the order for that reactant. Repeat for other reactant(s).

Determination of Rate Law– Homework Prob 26
The reaction was studied at 25oC the following results were obtained: I-(aq) + S2O82-(aq)  I2(aq) + SO42-(aq) [I-]0 [S2O82-]0 Initial Rate (mol/L) (mol/L) (mol/L.s) x 10-6 x 10-6 x 10-6 x 10-6 x 10-6 A) Determine the rate law. B) Calculate the rate constant and average rate constant

Determination of Rate Law
[I-]0 [S2O82-]0 Initial Rate (mol/L) (mol/L) (mol/L.s) x 10-6 x 10-6 x 10-6 x 10-6 x 10-6 Rate = k[I]x[S2O82]y = = 2.0x, x = 1 = 2.00 = 2.0y, y = 1; Rate = k[I][S2O82] b. For the First Experiment K=3.9×10-3 L/mol.s All others are the same so the average is the same.

Integrated Rate Laws

The Change of Concentration with Time
First-Order Reactions Goal: convert rate law into a convenient equation to give concentrations as a function of time. For a first order reaction, the rate doubles as the concentration of a reactant doubles. A plot of ln[A]t versus t is a straight line with slope -k and intercept ln[A]0. We can show that the best-fit linear equation is:

The Change of Concentration with Time aA → products
1st order y = mx + b Plot: ln[A] vs. t slope = − k (straight line) integrate

Integrated Rate Law – First Order
2N2O5  4NO2 + O2(g) Integrated first-order rate law is ln[N2O5] = kt + ln[N2O5]o Concentration dependence on time Equation is y = mx + b Can be expressed as:

Integrated Rate Law – First Order
2N2O5  4NO2 + O2(g) [N2O5] Time 0.1000 0.0707 50 0.0500 100 0.0250 200 0.0125 300 400 ln[N2O5] Time -2.303 -2.649 50 -2.996 100 -3.689 200 -4.382 300 -5.075 400

Example Problem #31 Time [H2O2] ln H2O2] (s) (mol/L) The plot of ln [H2O2] vs. time is linear. Thus, the reaction is first order. The rate law and integrated rate law are: Rate = k[H2O2] and ln [H2O2] = -kt + ln [H2O2]o. b. We determine the rate constant k by determining the slope of the ln [H2O2] vs time plot (slope = -k). Using two points on the line gives: slope = -k = = -8.3 ×10-4 , k = 8.3×10-4

Example Problem #31 To determine [H2O2] at s, use the integrated rate law where at t = 0, [H2O2]o = 1.00 M. ln [H2O2] = -kt + ln [H2O2]o or = kt = -8.3 ×10-4 × s, ln [H2O2] = -3.3 [H2O2] = e -3.3 [H2O2] = = M

Nuclear Radiation 1st Order Half Life

Half-Life of a First-Order Reaction
The time required for a reactant to reach half its original concentration [N2O5] Time 0.1000 0.0500 100 0.0250 200 0.0125 300 400 t1/2 = half-life of the reaction k = rate constant For a first-order reaction, the half-life does not depend on concentration. Or ln ([A0]/[A])=kt where [A]=[A0 ]/2

Half-Life Sugar is fermented in a 1st order process (using an enzyme as a catalyst). sugar + enzyme  products Rate of disappear of sugar = k[sugar] k = 3.3 x 10-4 sec-1 What is the half-life of this reaction?

Half-Life Solution [A] / [A]0 = fraction remaining
Rate = k[sugar] and k = 3.3 x 10-4 sec-1. What is the half-life of this reaction? Solution [A] / [A]0 = fraction remaining when t = t1/2 then fraction remaining = 1/2 Therefore, ln (1/2) = - k · t1/2 = - k · t1/2 t1/2 = / k So, for sugar, t1/2 = / k = sec = 35 min

Half-Life Rate = k[sugar] and k = 3.3 x 10-4 sec-1. Half-life is 35 min. Start with 5.00 g sugar. How much is left after 2 hr and 20 min (140 min)? Solution 2 hr and 20 min = 4 half-lives Half-life Time Elapsed Mass Left 1st 35 min g 2nd g 3rd g 4th g

Half-Life Radioactive decay is a first order process.
Tritium  electron helium 3H 0-1e He t1/2 = years If you have 1.50 mg of tritium, how much is left after 49.2 years?

Half-Life Solution ln [A] / [A]0 = -kt
Start with 1.50 mg of tritium, how much is left after 49.2 years? t1/2 = years Solution ln [A] / [A]0 = -kt [A] = ? [A]0 = mg t = y Need k, so we calc k from: k = / t1/2 Obtain k = y-1 Now ln [A] / [A]0 = -kt = - ( y-1)(49.2 y) = Take antilog: [A] / [A]0 = e-2.77 = = fraction remaining

Half-Life Solution [A] / [A]0 = 0.0627
Start with 1.50 mg of tritium, how much is left after 49.2 years? t1/2 = years Solution [A] / [A]0 = is the fraction remaining! Because [A]0 = 1.50 mg, [A] = mg But notice that 49.2 y = 4.00 half-lives 1.50 mg 0.750 mg after 1 half-life 0.375 mg after 2 0.188 mg after 3 0.094 mg after 4

Half-Life of a First-Order Reaction
Calculation: A first-order reaction has a half life of 20.0 min. Calculate rate constant How much time is required for the reaction to be 75% complete?

Second Order Reactions

Second-Order Rate Law For aA  products in a second-order reaction,
Integrated rate law is

Second-Order Reactions For a second order reaction, a plot of ln[A] vs. t is not linear. A plot of 1/[A] versus t is a straight line with slope k and intercept 1/[A]0 For a second order reaction with just one reactant the best-fit line equation is

Second-Order Reactions y = mx + b Note: k = slope of the best-fit line A reaction can have a rate law expression of the form rate = k[A][B], which has first order dependence on A and B, but is second order overall.

Second-Order Rate Law –
Butadiene reacts to form its dimer 2C4H6(g)  C8H12(g) Data: Reaction order? Value of k? Half-life? [C4H6] Time 1000 1800 2800 3600 4400 5200 6200 1/[C4H6] ln[C4H6] 100 -4.605 160 -5.075 210 -5.348 270 -5.599 320 -5.767 370 -5.915 415 -6.028 481 -6.175

Butadiene reacts to form its dimer 2C4H6(g)  C8H12(g) Data: Reaction order? Rate = k[C4H6]2 [C4H6] Time 1000 1800 2800 3600 4400 5200 6200 1/[C4H6] ln[C4H6] 100 -4.605 160 -5.075 210 -5.348 270 -5.599 320 -5.767 370 -5.915 415 -6.028 481 -6.175

Butadiene reacts to form its dimer 2C4H6(g)  C8H12(g) Data: b) Value of k? k = slope [C4H6] Time 1000 1800 2800 3600 4400 5200 6200 1/[C4H6] ln[C4H6] 100 -4.605 160 -5.075 210 -5.348 270 -5.599 320 -5.767 370 -5.915 415 -6.028 481 -6.175

Half-Life of a 2nd-Order Reaction
t1/2 = half-life of the reaction k = rate constant Ao = initial concentration of A The half-life is dependent upon the initial concentration.

Butadiene reacts to form its dimer 2C4H6(g)  C8H12(g) k = 6.14x10-2 L/mol*s [A]0= 1.000x10-2 mol/L c) Half-life?

Practice Homework #34 Because the 1/[A] vs. time plot was linear, the reaction is second order in A. The slope of the 1/[A] vs. time plot equals the rate constant k. Therefore, the rate law, the integrated rate law and the rate constant value are: Rate = k[A]2; ; k = 3.60 × L/mol . s

Practice Homework #34 The half-life expression for a second-order reaction is: t1/2 = For this reaction: t1/2 = = 9.92 × 103 s Note: We could have used the integrated rate law to solve for t1/2 where [A] = (2.80 × /2) mol/L.

Practice Homework #34 Since the half-life for a second-order reaction depends on concentration, we will use the integrated rate law to solve. (1.43 × 10 3)– (357) = 3.60 ×10-2 t, t = 2.98 × 104 s

Zero-Order Rate Law Zero-order reaction the rate is constant.
Rate does not change with respect to concentration Rate = k[A]o = k(1) = k Integrated rate law [A] = -kt + [A]o Half Life = t1/2 = [A]o/2k Line of best fit is [A] vs. t straight line

Order zero 1st 2nd Rate law rate = k rate = k[A] rate = k[A]2
Integrated rate law [A]t=−kt+[A]0 ln[A]t=−kt+ln[A]0 1/[A]t=kt+1/[A]0 Straight-line plot [A] vs. t ln[A] vs. t 1/[A] vs. t Slope −k k Half-life (t1/2) [A]o/2k 0.693/k 1/k[A]0

A Summary 1. Simplification: Conditions are set such that only forward reaction is important. 2. Two types: differential rate law integrated rate law 3. Which type? Depends on the type of data collected - differential and integrated forms can be interconverted. 4. Most common: method of initial rates. 5. Concentration v. time: used to determine integrated rate law, often graphically. 6. For several reactants: choose conditions under which only one reactant varies significantly (pseudo first-order conditions).

12.7 Collision Model Temperature and Rates

As temperature increases, the rate increases.
Temperature and Rate Most reactions speed up as temperature increases (EX. food spoils faster when not refrigerated.) As temperature increases, the rate increases. Typically the rate doubles for every 10ºC increase in temperature.

Why? Temperature and Rate
Since the rate law has no temperature term in it, the rate constant must depend on temperature. Why? The temperature effect is quite dramatic.

Temperature and Rate The Collision Model Observations: rates of reactions are affected by concentration and temperature. In order for a reaction to occur the reactant molecules must collide in the correct orientation and with enough energy to form products.

Temperature and Rate The Collision Model The more molecules present, the greater the probability of effective collisions and the faster the rate. The higher the temperature, the more energy available to the molecules and the faster the rate.

Temperature and Rate The Collision Model Complication: Not all collisions lead to products. In fact, only a small fraction of collisions lead to product.

Temperature and Rate Activation Energy

Temperature and Rate Activation Energy Arrhenius: molecules must possess a minimum amount of energy to react. In order to form products, bonds must be broken in the reactants. Bond breakage requires energy. Activation energy, Ea, is the minimum energy required to initiate a chemical reaction.

Temperature and Rate Activation Energy
Consider the rearrangement of acetonitrile: Activated In H3C-NC, the C-NC bond bends until the C-N bond breaks and the NC portion is perpendicular to the H3C portion. This structure is called the activated complex or transition state. The energy required for the twisting and breaking of the reactant bonds is the activation energy, Ea. Once the C-N bond is broken, the NC portion can continue to rotate forming a C-CN bond.

Temperature and Rate Key Points for Activation Energy
The change in energy, ∆E, for the reaction is the difference in energy between reactants (CH3NC) and products (CH3CN). The activation energy is the difference in energy between the reactants and the transition state. The reaction rate depends on Ea. The lower the Ea, the faster the reaction. Notice that if a forward reaction is exothermic (CH3NC  CH3CN), then the reverse reaction is endothermic (CH3CN  CH3NC).

Temperature and Rate Activation Energy
Consider the reaction between Cl and NOCl: If the Cl effectively collides with the Cl end of NOCl, then the products are Cl2 and NO. If the Cl collided with the O of NOCl, then no products are formed. We need to quantify this effect.

Temperature and Rate As the temperature increases, more molecules have sufficient energy to have effective collisions and react. Lower temp Minimum energy needed for reaction, Ea Higher temp Fraction of molecules Kinetic energy

More About Activation Energy
Arrhenius equation — Temp (K) Rate constant Activation energy 8.31 x 10-3 kJ/K•mol Frequency factor Ae is called the frequency factor. It is a measure of the probability of a favorable collision. Both Ae and Ea are specific to a given reaction. Plot ln k vs. 1/T straight line. slope = -Ea/R

Temperature and Rate The Arrhenius Equation
If you have a lot of data, you can determine Ea and A graphically by rearranging the Arrhenius equation: Ln k = y Ea/R= m(slope) 1/T= x ln A=b If you have only two data points, then you can use

Arrhenius Equation Sample
RXN: 2N2O5(g)  4NO2(g) + O2(g) Calculate the value of Ea for this reaction. K and T given. T(oC) T(K) k(s-1) 20 293 2.0x10-5 30 303 7.3x10-5 40 313 2.7x10-4 50 323 9.1x10-4 60 333 2.9x10-4 1/T(K) Ln(K) 3.41x10-3 -10.82 3.30x10-3 -9.53 3.19x10-3 -8.22 3.10x10-3 -7.00 3.00x10-3 -5.84 Slope = -Ea/R  Ea = -R(slope) = -(8.3145J/K mol)(-1.2x104 K) = 1.0x105 J/mol

Arrhenius Equation Homework Example 58
(CH3)3CBr + OH-  (CH3)3COH + Br- This reaction in a certain solvent is first order with respect to (CH3)3CBr and zero order with respect to OH-. In several experiments, the rate constant was determined at different temperatures. A plot of ln(k) vs. 1/T was constructed resulting in a straight line with a slope value of x 104 K and y-intercept of Assume k has units of s-1. Determine Ea Determine factor A Calculate the value of K at 25oC.

Arrhenius Equation Example 58 Determine Ea Determine factor A
Calculate the value of k at 25oC. Given slope = x 104 K and y-intercept of 33.5. Slope = -Ea/R Ea = x 104 K x (J/K mol) = 9.15 x 104 (J/mol) b) y-intercept = 33.5 = ln A A = e33.5 = 3.54 x 1014 s-1 c) k = Ae-Ea/RT K = 3.24 x 10-2s-1

Slow and Fast Steps Reaction Mechanisms

Reaction Mechanisms The balanced chemical equation provides information about the beginning and end of the reaction. The reaction mechanism gives the path of the reaction. Mechanisms provide a very detailed picture of which bonds are broken and formed during the course of a reaction. A chemical equation does not tell us how reactants become products - it is a summary of the overall process.

Reaction Mechanisms Reactions can be thought of as being a series of elementary steps, which is any process that occurs in a single step. Molecularity is the number of molecules present in an elementary step. Unimolecular: 1 molecule in the elementary step, Bimolecular: 2 molecules in the elementary step, Termolecular: 3 molecules in the elementary step. It is not common to see termolecular processes (statistically improbable).

Reaction Mechanisms Rate Laws of Elementary Steps The rate law of an elementary step is determined by its molecularity: Unimolecular processes are first order, A products rate=k [A] Bimolecular processes are second order, A + A  Products or A+B products Rate=k[A] rate=[A][B] Termolecular processes are third order.

Elementary steps must add to give the balanced chemical equation.
Reaction Mechanisms Elementary Steps Elementary steps must add to give the balanced chemical equation. Often Rate laws are assumed to be the coefficients of the elementary step. Ex. A + 2B → AB2 Rate= [A][B]2 (must be confirmed by experiment) Intermediates can form, which are species that appear in an elementary step, but are not an original reactant or product. They are consumed during the reaction.

Reaction Mechanism We can define a reaction mechanism.
It is a series of elementary steps that must satisfy two requirements: The sum of the elementary steps must give the overall balanced equation for the reactions. The mechanism must agree with the experimentally determined rate law.

Reaction Mechanisms 2NO(g) + Br2(g)  2NOBr(g)
Rate Laws of Multistep Mechanisms The rate-determining step is the slowest of any of the elementary steps. Therefore, the rate-determining step governs the overall rate law for the reaction. Rate law can be written using the coefficients of the slowest step in a reaction. Consider 2NO(g) + Br2(g)  2NOBr(g)

Reaction Mechanisms Mechanisms with an Initial SLOW Step slow fast
2NO(g) + Br2(g)  2NOBr(g) Consider the following mechanism Based on the slow step, the rate law would be Rate = k [NO] [Br2] slow fast

Reaction Mechanisms Mechanisms with an Initial FAST Step
2NO(g) + Br2(g)  2NOBr(g) Consider the following mechanism for which the rate law is (based on Step 2):

Reaction Mechanisms Mechanisms with an Initial Fast Step
The rate law is (based on Step 2): Rate = k2[NOBr2][NO] The rate law should not depend on the concentration of an intermediate because intermediates are usually unstable. Assume NOBr2 is unstable, so we express the concentration of NOBr2 in terms of NOBr and Br2 assuming there is an equilibrium in step 1 we have

Reaction Mechanisms Mechanisms with an Initial Fast Step
By definition of equilibrium: k1[NO][Br2] = k-1[NOBr2] Therefore, the overall rate law becomes The experimentally determined rate law is Rate = k [NO]2[Br2] Note the final rate law is consistent with the experimentally observed rate law.

Problem #51 A mechanism consists of a series of elementary reactions where the rate law for each step can be determined using the coefficients in the balanced equations. For a plausible mechanism, the rate law derived from a mechanism must agree with the rate law determined from experiment. To derive the rate law from the mechanism, the rate of the reaction is assumed to equal the rate of the slowest step in the mechanism. Because step 1 is the rate-determining step, the rate law for this mechanism is: Rate = [C4H9Br].

Problem #51 To get the overall reaction, we sum all the individual steps of the mechanism. Summing all steps gives: C4H9Br → C4H9+ + Br C4H9+ + H2O → C4H9OH2+ C4H9OH2+ + H2O → C4H9OH + H3O+ ____________________________________ C4H9Br + 2 H2O → C4H9OH + Br + H3O+ The intermediates for this mechanism are C4H9+ and C4H9OH2+.

Catalysts

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.

Catalysis Increases the number of effective collisions by providing a reaction pathway with a lower activation energy

Heterogeneous Catalysis
Heterogeneous catalysis most often involves gaseous reactant being absorbed on the surface of a solid surface. Hydrogenation is an example: Changes C=C into saturated H-C-C-H Adsorption to addition of a substance to the surface of another. Migration of absorbed reactants on the surface Reaction of absorbed substances Desorption of products

Homogeneous Catalysis
Reactants and Catalysis are in the same phase. Gas-Gas or Liquid-Liquid N2(g) + O2(g)  2NO(g) Product of high-temperature combustion when N2 is present. However catalytic in production of ozone 2NO(g) + O2(g)  2NO2(g) NO2(g)  NO(g) + O(g) (light) O2(g) + O(g)  O3(g) 3/2O2(g)  O3(g)

Homogeneous Catalysis
In the upper atmosphere, NO has opposite effect. 2NO(g) + O3(g)  NO2(g) + O2(g) O + NO2(g)  NO(g) + O2(g) O3(g) + O(g)  2O2(g) Nitric Oxide is catalytic in production of O2. O3 required in upper atmosphere to block uv radiation.

Chapter 12 – Chemical Kinetics

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