CHE 116: General Chemistry CHAPTER FOURTEEN Copyright © Tyna L. Heise 2002 All Rights Reserved

Chemical Kinetics The area of chemistry that is concerned with the speeds, or rates, at which reactions occur. The rates of chemical reactions are affected by several factors, most notably: Concentration of reactants Temperature at which the reaction occurs The presence of a catalyst The surface area of solid or liquid reactants or catalysts Chem. 14.1

Chemical Kinetics The speed of an event is defined as the change that occurs in a given interval of time: - commonly referred to as the reaction rate - reaction rate is a measure of how quickly A is consumed or how quickly B is produced Average rate = change in # of moles of B change in time Chem. 14.1

Chemical Kinetics Average rate = D(moles of B) Dt Chem. 14.1

Chemical Kinetics The rate expression focuses on the number of moles of b that are produced during the reaction; it is the rate of appearance of B. Chem. 14.1

Chemical Kinetics The rate expression could just as easily been described in the terms of the change in moles of A, or the rate of disappearance of A. -D(moles of A) = D(moles of B) Chem. 14.1

Chemical Kinetics For primarily all reactions, the reaction rates will be determined by following changes in concentration. Units for reaction rates will be M/s. - brackets around a chemical substance indicate the concentration of the substance. - the minus sign always indicates the disappearance of moles of reactants Chem. 14.1

Chemical Kinetics Using a graph such as this The instantaneous rate is obtained using the tangent of the curve Chem. 14.1

Chemical Kinetics Sample exercise: Using the graph below, estimate the instantaneous rate of disappearance of C4H9Cl at t = 300 s. Chem. 14.1

Chemical Kinetics Sample exercise: Using the graph below, estimate the instantaneous rate of disappearance of C4H9Cl at t = 300 s. 1. Draw line at 300 sec that is tangent to curve, make a right triangle Chem. 14.1

Chemical Kinetics Sample exercise: Using the graph below, estimate the instantaneous rate of disappearance of C4H9Cl at t = 300 s. 2. DC4H9Cl = Dt Chem. 14.1

Chemical Kinetics Previous examples were all one to one relationships, but if the relationship is not one to one, the coefficients need to be included. aA + bB --> cC + dD Rate = -1 D[A] = -1 D[B] = 1 D[C] = 1 D[D] a Dt b Dt c Dt d Dt Chem. 14.1

Chemical Kinetics Sample exercise: The decomposition of N2O5 proceeds according to the equation 2N2O5(g) --> 4NO2(g) + O2(g) If the rate of decomposition of N2O5 at a particular instant in a reaction vessel is 4.2 x 10-7 M/s, what is the rate of appearance of (a) NO2; (b) O2 Chem. 14.1

Chemical Kinetics 2N2O5(g) --> 4NO2(g) + O2(g) If the rate of decomposition of N2O5 at a particular instant in a reaction vessel is 4.2 x 10-7 M/s, what is the rate of appearance of (a) NO2; (b) O2 1. Rate = -1 N2O5 = 1 DNO2 2 Dt 4 Dt Chem. 14.1

Chemical Kinetics 2N2O5(g) --> 4NO2(g) + O2(g) If the rate of decomposition of N2O5 at a particular instant in a reaction vessel is 4.2 x 10-7 M/s, what is the rate of appearance of (a) NO2; (b) O2 2. Rate = -4 Dt DN2O5 = 2 Dt DNO2 Chem. 14.1

Chemical Kinetics 2N2O5(g) --> 4NO2(g) + O2(g) If the rate of decomposition of N2O5 at a particular instant in a reaction vessel is 4.2 x 10-7 M/s, what is the rate of appearance of (a) NO2; (b) O2 3. Rate = -4 Dt DN2O5 = DNO2 2 Dt Chem. 14.1

Chemical Kinetics 2N2O5(g) --> 4NO2(g) + O2(g) If the rate of decomposition of N2O5 at a particular instant in a reaction vessel is 4.2 x 10-7 M/s, what is the rate of appearance of (a) NO2; (b) O2 4. Rate = -4 DN2O5 = DNO2 2 Chem. 14.1

Chemical Kinetics 2N2O5(g) --> 4NO2(g) + O2(g) If the rate of decomposition of N2O5 at a particular instant in a reaction vessel is 4.2 x 10-7 M/s, what is the rate of appearance of (a) NO2; (b) O2 5. Rate = -2 DN2O5 = DNO2 Chem. 14.1

Chemical Kinetics 2N2O5(g) --> 4NO2(g) + O2(g) If the rate of decomposition of N2O5 at a particular instant in a reaction vessel is 4.2 x 10-7 M/s, what is the rate of appearance of (a) NO2; (b) O2 5. Rate = -2 DN2O5 = DNO2 = 2(4.2 x 10-7 M/s) Chem. 14.1

Chemical Kinetics 2N2O5(g) --> 4NO2(g) + O2(g) If the rate of decomposition of N2O5 at a particular instant in a reaction vessel is 4.2 x 10-7 M/s, what is the rate of appearance of (a) NO2; (b) O2 5. Rate = -2 DN2O5 = DNO2 = 2(4.2 x 10-7 M/s) = 8.4 x 10-7 M/s Chem. 14.1

Chemical Kinetics 2N2O5(g) --> 4NO2(g) + O2(g) If the rate of decomposition of N2O5 at a particular instant in a reaction vessel is 4.2 x 10-7 M/s, what is the rate of appearance of (a) NO2; (b) O2 1. Rate = -1 DN2O5 = 1 DO2 2 Dt 1 Dt Chem. 14.1

Chemical Kinetics 2N2O5(g) --> 4NO2(g) + O2(g) If the rate of decomposition of N2O5 at a particular instant in a reaction vessel is 4.2 x 10-7 M/s, what is the rate of appearance of (a) NO2; (b) O2 2. Rate = -1 Dt DN2O5 = 2 Dt DO2 Chem. 14.1

Chemical Kinetics 2N2O5(g) --> 4NO2(g) + O2(g) If the rate of decomposition of N2O5 at a particular instant in a reaction vessel is 4.2 x 10-7 M/s, what is the rate of appearance of (a) NO2; (b) O2 3. Rate = -1 Dt DN2O5 = DO2 2 Dt Chem. 14.1

Chemical Kinetics 2N2O5(g) --> 4NO2(g) + O2(g) If the rate of decomposition of N2O5 at a particular instant in a reaction vessel is 4.2 x 10-7 M/s, what is the rate of appearance of (a) NO2; (b) O2 4. Rate = -1 DN2O5 = DO2 2 Chem. 14.1

Chemical Kinetics 2N2O5(g) --> 4NO2(g) + O2(g) If the rate of decomposition of N2O5 at a particular instant in a reaction vessel is 4.2 x 10-7 M/s, what is the rate of appearance of (a) NO2; (b) O2 5. Rate = -1/2 DN2O5 = DO2 = 1/2(4.2 x 10-7 M/s) Chem. 14.1

Chemical Kinetics 2N2O5(g) --> 4NO2(g) + O2(g) If the rate of decomposition of N2O5 at a particular instant in a reaction vessel is 4.2 x 10-7 M/s, what is the rate of appearance of (a) NO2; (b) O2 5. Rate = -1/2 DN2O5 = DO2 = 1/2(4.2 x 10-7 M/s) = 2.1 x 10-7 M/s Chem. 14.1

Dependence of Rate on Conc. Reaction rates diminish as the concentrations of reactions diminish. Reaction rates generally increase when reactant concentrations are increased. An expression, which shows how the rate depends on the concentrations of reactants, is called a rate law. Chem. 14.2

Dependence of Rate on Conc. An expression, which shows how the rate depends on the concentrations of reactants, is called a rate law. NH4+(aq) + NO2(aq) --> N2(g) + 2H2O(l) Chem. 14.2

Dependence of Rate on Conc. An expression, which shows how the rate depends on the concentrations of reactants, is called a rate law. NH4+(aq) + NO2(aq) --> N2(g) + 2H2O(l) Rate = k[NH4+][NO2-] * the constant k in the rate law is called the rate constant. Chem. 14.2

Dependence of Rate on Conc. The rate laws for most reactions have the general form rate = k[reactant 1]m[reactant 2]n… The exponents m and n are called reaction orders, and their sum is the overall reaction order. Notice that the reaction orders do not necessarily correspond to the coefficients in the balanced chemical equation. Chem. 14.2

Dependence of Rate on Conc. Sample exercise: What is the reaction order of the reactant H2 in the following equation: H2(g) + I2(g) --> 2HI(g) Chem. 14.2

Dependence of Rate on Conc. Sample exercise: What is the reaction order of the reactant H2 in the following equation: H2(g) + I2(g) --> 2HI(g) 1 Chem. 14.2

Dependence of Rate on Conc. Sample exercise: What are the units of the rate constant for: CHCl3(g) + Cl2(g) --> CCl4(g) + HCl(g) Chem. 14.2

Dependence of Rate on Conc. Sample exercise: What are the units of the rate constant for: CHCl3(g) + Cl2(g) --> CCl4(g) + HCl(g) units of rate units of concentration Chem. 14.2

Dependence of Rate on Conc. Sample exercise: What are the units of the rate constant for: CHCl3(g) + Cl2(g) --> CCl4(g) + HCl(g) (M/s) M3/2 Chem. 14.2

Dependence of Rate on Conc. Sample exercise: What are the units of the rate constant for: CHCl3(g) + Cl2(g) --> CCl4(g) + HCl(g) M-1/2 s-1 Chem. 14.2

Dependence of Rate on Conc. Using Initial Rates to Determine Rate Laws: The rate law for any chemical reaction must be determined experimentally. - Rate laws of 0, changing concentrations will have no effect on rate - Rate laws of 1, changing concentrations will produce proportional changes - Rate laws of 2, changing concentrations will produce exponential changes ** rate depends on conc. NOT rate constants Chem. 14.2

Dependence of Rate on Conc. Sample exercise: A particular reaction was found to depend on the concentration of the hydrogen ion, [H+]. The initial rates varied as a function of [H+] as follows: [H+] (M) 0.0500 0.100 0.200 Initial rate (M/s) 6.4 x 10-7 3.2 x 10-7 1.6 x 10-7 (a) What is the order of the reaction in [H+]? Chem. 14.2

Dependence of Rate on Conc. Sample exercise: A particular reaction was found to depend on the concentration of the hydrogen ion, [H+]. The initial rates varied as a function of [H+] as follows: [H+] (M) 0.0500 0.100 0.200 Initial rate (M/s) 6.4 x 10-7 3.2 x 10-7 1.6 x 10-7 (a) What is the order of the reaction in [H+]? -1 , rate is inversely proportional Chem. 14.2

Dependence of Rate on Conc. Sample exercise: A particular reaction was found to depend on the concentration of the hydrogen ion, [H+]. The initial rates varied as a function of [H+] as follows: [H+] (M) 0.0500 0.100 0.200 Initial rate (M/s) 6.4 x 10-7 3.2 x 10-7 1.6 x 10-7 (b) Predict the initial reaction rate when [H+] = 0.400M. Chem. 14.2

Dependence of Rate on Conc. Sample exercise: A particular reaction was found to depend on the concentration of the hydrogen ion, [H+]. The initial rates varied as a function of [H+] as follows: [H+] (M) 0.0500 0.100 0.200 Initial rate (M/s) 6.4 x 10-7 3.2 x 10-7 1.6 x 10-7 (b) Predict the initial reaction rate when [H+] = 0.400M. -should divide in half again Chem. 14.2

Dependence of Rate on Conc. Sample exercise: A particular reaction was found to depend on the concentration of the hydrogen ion, [H+]. The initial rates varied as a function of [H+] as follows: [H+] (M) 0.0500 0.100 0.200 Initial rate (M/s) 6.4 x 10-7 3.2 x 10-7 1.6 x 10-7 (b) Predict the initial reaction rate when [H+] = 0.400M. 8.0 x 10-8 Chem. 14.2

Change of Conc. With Time Rate laws can be converted into equations that tell us what the concentrations of the reactants or products are at any time during the course of a reaction. Two types: first order overall second order overall Chem. 14.3

Change of Conc. With Time First Order - rate depends on the concentration of a single reactant raised to the first power A --> products rate = - D[A] = k[A] Dt Chem. 14.3

Change of Conc. With Time First Order - transform previous equation into an equation that relates the concentration of A at the start of the reaction, [A]0, to its concentration at any other time t, [A]t ln [A]t - ln [A]0 = -kt ln [A]t = -kt [A]0 Chem. 14.3

Change of Conc. With Time First Order - transform previous equation into an equation that relates the concentration of A at the start of the reaction, [A]0, to its concentration at any other time t, [A]t ln [A]t - ln [A]0 = -kt ln [A]t = -kt + ln[A]0 y = mx + b * notice the slope of the line is -k, the rate constant Chem. 14.3

Change of Conc. With Time Equation: ln [A]t = -kt + ln[A]0 can be used to determine 1) the concentration of a reactant remaining at any time after the reaction has begun 2) the time required for a given fraction of a sample to react 3) the time required for a reactant concentration to reach a certain level Chem. 14.3

Change of Conc. With Time Sample exercise: The decomposition of dimethyl ether, (CH3)2O, at 510°C is a first-order process with a rate constant of 6.8 x 10-4 s-1. (CH3)2O(g) --> CH4(g) + H2(g) + CO(g) If the initial pressure of (CH3)2O is 135 torr, what is its partial pressure after 1420 s? Chem. 14.3

Change of Conc. With Time Sample exercise: The decomposition of dimethyl ether, (CH3)2O, at 510°C is a first-order process with a rate constant of 6.8 x 10-4 s-1. (CH3)2O(g) --> CH4(g) + H2(g) + CO(g) If the initial pressure of (CH3)2O is 135 torr, what is its partial pressure after 1420 s? 1. ln [A]t = -kt + ln[A]0 Chem. 14.3

Change of Conc. With Time Sample exercise: The decomposition of dimethyl ether, (CH3)2O, at 510°C is a first-order process with a rate constant of 6.8 x 10-4 s-1. (CH3)2O(g) --> CH4(g) + H2(g) + CO(g) If the initial pressure of (CH3)2O is 135 torr, what is its partial pressure after 1420 s? 1. ln [A]t = -kt + ln[A]0 ln [A]1420 = -(6.8x 10-4 s-1)(1420) + ln[135]0 Chem. 14.3

Change of Conc. With Time Sample exercise: The decomposition of dimethyl ether, (CH3)2O, at 510°C is a first-order process with a rate constant of 6.8 x 10-4 s-1. (CH3)2O(g) --> CH4(g) + H2(g) + CO(g) If the initial pressure of (CH3)2O is 135 torr, what is its partial pressure after 1420 s? 1. ln [A]t = -kt + ln[A]0 ln [A]1420 = -(6.8x 10-4 s-1)(1420) + ln[135]0 ln [A]1420 = -0.9656 + 4.905 Chem. 14.3

Change of Conc. With Time Sample exercise: The decomposition of dimethyl ether, (CH3)2O, at 510°C is a first-order process with a rate constant of 6.8 x 10-4 s-1. (CH3)2O(g) --> CH4(g) + H2(g) + CO(g) If the initial pressure of (CH3)2O is 135 torr, what is its partial pressure after 1420 s? 1. ln [A]1420 = -0.9656 + 4.905 [A]1420 = e3.940 [A]1420 = 51 torr Chem. 14.3

Change of Conc. With Time Half Lives are a special case of first order reactions. ln 1/2[A]0 = -kt1/2 [A]0 * initial concentration falls out of equation!! ln 1/2 = -kt1/2 t1/2 = 0.693 k Chem. 14.3

Change of Conc. With Time Sample exercise: Calculate t1/2 for the following reaction (CH3)2O(g) --> CH4(g) + H2(g) + CO(g) Chem. 14.3

Change of Conc. With Time Sample exercise: Calculate t1/2 for the following reaction (CH3)2O(g) --> CH4(g) + H2(g) + CO(g) 1. t1/2 = 0.693 k Chem. 14.3

Change of Conc. With Time Sample exercise: Calculate t1/2 for the following reaction (CH3)2O(g) --> CH4(g) + H2(g) + CO(g) 1. t1/2 = 0.693 = 0.693 k 6.8 x 10-4 s-1 Chem. 14.3

Change of Conc. With Time Sample exercise: Calculate t1/2 for the following reaction (CH3)2O(g) --> CH4(g) + H2(g) + CO(g) 1. t1/2 = 0.693 = 0.693 = 1.02 x 103 s k 6.8 x 10-4 s-1 Chem. 14.3

Change of Conc. With Time Second Order Reactions: rate depends on the reactant concentration raised to the second power or on the concentrations of two different reactants, each raised to the first power. Rate = - D[A] = k[A]2 Dt 1 = kt + 1 [A]t [A]0 y = mx + b Chem. 14.3

Change of Conc. With Time Second Order Reactions: a plot of 1/[A]t versus t will yield a straight line where the slope equals k ** one way to distinguish between first and second order reactions is to look at the graphs. A downward straight line after plotting ln[A] vs t will always be first order, slope equals -k An upward straight line after plotting 1/[A] vs t will always be second order, slope equals k Chem. 14.3

Change of Conc. With Time Second Order Reactions: half lives of elements or compounds that decay at a second order rate will be dependent on the initial concentration: t1/2 = 1 k[A]0 *practice exercise page 524 Chem. 14.3

Temperature and Rate The rates of most chemical reactions increase as the temperature rises - due to an increase in the rate constant with increasing temperature - explained by the collision model theory molecules must collide to react the greater number of collisions occuring per second, the greater the rate as concentration increases, the number of collisions increases Chem. 14.4

Temperature and Rate The rates of most chemical reactions increase as the temperature rises - due to an increase in the rate constant with increasing temperature - explained by the collision model theory molecules move faster, they collide more forcefully and more frequently, increasing reaction rates for a reaction to occur, though, more is required than simply a collision Chem. 14.4

Temperature and Rate Collision theory continued- molecules must possess a certain minimum amount of energy in order to react kinetic energy of the molecules can be used to stretch, bend, and ultimately break bonds, leading to chemical reactions if molecules are moving too slowly, with too little kinetic energy, they merely bounce off one another without changing minimum energy required to initiate a chemical reaction is called the activation energy, Ea Chem. 14.4

Temperature and Rate Activation Energy Chem. 14.4

Temperature and Rate Potential Energy Diagram Chem. 14.4

Temperature and Rate Collision Theory continued… not every collision in which reactants have an energy Ea or greater results in reaction they must be oriented in a certain way for the collision to lead to a reaction Arrhenius Equation: reaction rate data depends on 1) fraction of molecules possessing Ea 2) number of collisions per second 3) fraction of collisions with correct orientation Chem. 14.4

k = Ae-Ea/RT ln k1 = Ea 1 - 1 k2 R T2 T1 Temperature and Rate Arrhenius Equation: reaction rate data depends on 1) fraction of molecules possessing Ea 2) number of collisions per second 3) fraction of collisions with correct orientation k = Ae-Ea/RT ln k1 = Ea 1 - 1 k2 R T2 T1 Chem. 14.4

Reaction Mechanisms The process by which a reaction occurs is called the reaction mechanism. - describes in great detail the order in which bonds are broken and formed and the changes in relative positions of the atoms in the course of the reaction Chem. 14.5

Elementary steps - a process that occurs in a single step Reaction Mechanisms Elementary steps - a process that occurs in a single step if a single molecule was involved, it is unimolecular if two molecules were involved, it is bimolecular if three molecules were involved, it is termolecular Chem. 14.5

Multistep Mechanisms - a process that occurs in a single step Reaction Mechanisms Multistep Mechanisms - a process that occurs in a single step * the elementary steps in a multistep mechanism must always add to give the chemical equation of the overall process - a molecule that is produced in one step and consumed in the very next step is considered to be an intermediate intermediates are not activated complexes Chem. 14.5

Reaction Mechanisms Sample exercise: For the following reaction of Mo(CO)6 Mo(CO)6 + P(CH3)3 --> Mo(CO)5P(CH3)3 + CO the proposed mechanism Mo(CO)6 --> Mo(CO)5 + CO Mo(CO)5 + P(CH3)3 --> Mo(CO)5P(CH3)3 (a) Is the proposed mechanism consistent with the equation for the overall reaction? Chem. 14.5

Reaction Mechanisms Sample exercise: For the following reaction of Mo(CO)6 Mo(CO)6 + P(CH3)3 --> Mo(CO)5P(CH3)3 + CO the proposed mechanism Mo(CO)6 --> Mo(CO)5 + CO Mo(CO)5 + P(CH3)3 --> Mo(CO)5P(CH3)3 (a) Is the proposed mechanism consistent with the equation for the overall reaction? Yes, the two equations add up correctly Chem. 14.5

Reaction Mechanisms Sample exercise: For the following reaction of Mo(CO)6 Mo(CO)6 + P(CH3)3 --> Mo(CO)5P(CH3)3 + CO the proposed mechanism Mo(CO)6 --> Mo(CO)5 + CO Mo(CO)5 + P(CH3)3 --> Mo(CO)5P(CH3)3 (b) Identify the intermediate or intermediates Chem. 14.5

Reaction Mechanisms Sample exercise: For the following reaction of Mo(CO)6 Mo(CO)6 + P(CH3)3 --> Mo(CO)5P(CH3)3 + CO the proposed mechanism Mo(CO)6 --> Mo(CO)5 + CO Mo(CO)5 + P(CH3)3 --> Mo(CO)5P(CH3)3 (b) Identify the intermediate or intermediates Mo(CO)5 Chem. 14.5

Reaction Mechanisms The rate law for a reaction can be determined from its mechanisms, so our next step is to arrive at reaction mechanisms that lead to rate laws that are consistent with those observed experimentally. If a reaction is an elementary step, we know the rate law automatically, and it is first order A --> products rate = k[A] If a reaction is a bimolecular elementary step, the rate law is second order A + B --> products rate = k[A][B] Chem. 14.5

Reaction Mechanisms , Chem. 14.5

Reaction Mechanisms Sample exercise: Consider the following reaction: 2NO(g) + Br2(g) --> 2NOBr(g) (a) Write the rate law for the reaction, assuming it involves a single elementary step. Chem. 14.5

Reaction Mechanisms Sample exercise: Consider the following reaction: 2NO(g) + Br2(g) --> 2NOBr(g) (a) Write the rate law for the reaction, assuming it involves a single elementary step. Rate =k[NO]2[Br2] Chem. 14.5

Reaction Mechanisms Sample exercise: Consider the following reaction: 2NO(g) + Br2(g) --> 2NOBr(g) (b) Is a single-step mechanism likely for this reaction? Chem. 14.5

Reaction Mechanisms Sample exercise: Consider the following reaction: 2NO(g) + Br2(g) --> 2NOBr(g) (b) Is a single-step mechanism likely for this reaction? No, because termolecular reactions are very rare Chem. 14.5

Reaction Mechanisms Often one of the steps is much slower than the others. The overall rate of a reaction cannot exceed the rate of the slowest elementary step of its mechanisms. Slowest Step is the rate-determining step! - the rate determining step governs the rate law for the overall reaction -mechanisms with an initial fast step, can be solved for the concentration of an intermediate be assuming that an equilibrium is established in the fast step Chem. 14.5

Reaction Mechanisms Practice exercise: The first step of a mechanism involving the reaction of bromine is Br2(g) <----> 2Br(g) (fast equilibrium) What is the expression relating the concentration of Br(g) to that of Br2(g)? Chem. 14.5

Reaction Mechanisms Practice exercise: The first step of a mechanism involving the reaction of bromine is Br2(g) <----> 2Br(g) (fast equilibrium) What is the expression relating the concentration of Br(g) to that of Br2(g)? Rate = k1 [Br2] Chem. 14.5

Reaction Mechanisms Practice exercise: The first step of a mechanism involving the reaction of bromine is Br2(g) <----> 2Br(g) (fast equilibrium) What is the expression relating the concentration of Br(g) to that of Br2(g)? Rate = k1 [Br2] Rate = k-1[Br]2 Chem. 14.5

Reaction Mechanisms Practice exercise: The first step of a mechanism involving the reaction of bromine is Br2(g) <----> 2Br(g) (fast equilibrium) What is the expression relating the concentration of Br(g) to that of Br2(g)? Rate = k1 [Br2] Rate = k-1[Br]2 *rates are equal so ... Chem. 14.5

Reaction Mechanisms Practice exercise: The first step of a mechanism involving the reaction of bromine is Br2(g) <----> 2Br(g) (fast equilibrium) What is the expression relating the concentration of Br(g) to that of Br2(g)? k1 [Br2] = k-1[Br]2 Chem. 14.5

Reaction Mechanisms Practice exercise: The first step of a mechanism involving the reaction of bromine is Br2(g) <----> 2Br(g) (fast equilibrium) What is the expression relating the concentration of Br(g) to that of Br2(g)? k1 [Br2] = [Br]2 k-1 Chem. 14.5

Reaction Mechanisms Practice exercise: The first step of a mechanism involving the reaction of bromine is Br2(g) <----> 2Br(g) (fast equilibrium) What is the expression relating the concentration of Br(g) to that of Br2(g)? k1 [Br2] = [Br] k-1 Chem. 14.5

Catalysis A catalyst is a substance that changes the speed of a chemical reaction without undergoing a permanent chemical change itself in the process. Catalysts are very common reactions in the body in the atmosphere in the oceans in industrial chemistry Chem. 14.6

Catalysis A catalyst that is present in the same phase as the reacting molecules is a homogeneous catalyst - the rate constant k is determined by the activation energy, Ea, and the frequency factor, A. - a catalyst may affect the rate of a reaction by altering the value for either Ea or A. - the most dramatic catalytic effects come from lowering Ea - lowers Ea by providing a completely different mechanism for the reaction Chem. 14.6

Catalysis Homogeneous catalyst Chem. 14.6

Catalysis A catalyst that is different in phase as the reacting molecules is a heterogeneous catalyst - usually as a solid in contact with either gaseous reactants or with reactants in the liquid solution - often composed of metals or metal oxides - catalyzed reactions occurs on the surface, special methods are often used to prepare catalysts so that they have very large surface areas - the initial step is ADSORPTION Chem. 14.6

Catalysis Enzymes - many of the most interesting and important examples of catalysis involve reactions within living systems. - A large number of marvelously efficient biological catalysts known as enzymes are necessary for many of these reaction to occur at suitable rates - most enzymes are large protein molecules with molecular weights ranging from 10,000 to about 1 million amu Chem. 14.6