How fast do chemical processes occur? There is an enormous range of time scales. Chapter 14 Chemical Kinetics Kinetics also sheds light on the reaction mechanism (exactly how the reaction occurs). Why is it important? Increase the shelf life of products like food, beverages, pharmaceutical drugs, etc. (How?) slowing down some reactions. Remove hazardous pollutants from the environment (How?) speeding up some reactions (biodegradable materials). 14.1 Factors that affect Reaction Rates 1. Physical state of reactants (s, l, g, aq). Molecules must come in contact with each other in order to react, therefore: The more homogeneous the mixture of reactants, the faster the molecules can react. Gases tend to react faster than liquids, which react faster than solids. For solids, however, surface area is important (powders react faster than chunks). 2. Reactant Concentrations. Generally, the larger the concentration of reactant molecules, the faster the reaction will proceed (Why?). The likelihood that reactant molecules will collide increases. 3. Reaction Temperature. Higher T leads to higher kinetic energy, molecules move faster and collide more often and with greater energy (i.e. more molecules have enough energy to react). chemist s rule of thumb: ΔT = 10 C 2(rate) 4. The presence/concentration of a catalyst. Catalysts speed up reactions by changing the mechanism of the reaction. Catalysts are not consumed during the course of the reaction. 14.2 Reaction Rates ( ) For the hypothetical reaction: A B A will decrease and B will increase (every 1 A that reacts produces 1 B). The rate of reaction is positive (+) by convention; however, the change of reactants and products can be positive (+) or negative ( ). 1
The average rate over a certain time interval depends on the time interval chosen. The instantaneous rate refers to the rate at a particular moment. To determine the instantaneous rate from a graph of concentration vs time: 1. Draw a tangent to the curve at the desired time point. 2. Determine the slope of the tangent. For a chemical reaction, the rate decreases as the time goes on (i.e. the slope gets less steep). Reaction Rate Changes Over Time Because the concentration of the reactants decreases. At some time the reaction stops, either because the reactants run out or because the system has reached equilibrium. Problem 1 Using Figure 14.4, determine the instantaneous rate of disappearance of C 4 H 9 Cl at t = 300 s. Relating rates Rates of products and reactants can be related: N 2 O 5 (g) 2 NO 2 (g) + ½ O 2 (g) As 1 mol N 2 O 5 (g) is consumed, 2 mol NO 2 (g) and ½ mol of O 2 (g) are formed. So [NO 2 ] changes twice as fast as [N 2 O 5 ]. So rate [N 2 O 5 ] = ½ rate [NO 2 ]. Reaction Rates and Stoichiometry To generalize, then, for the reaction: Instantaneous Rate Example Using [H 2 ], the instantaneous rate at 50 s is: Using [HI], the instantaneous rate at 50 s is 2
Problem 2 a. How is the rate at which ozone disappears related to the rate at which oxygen appears in the reaction? 2 O 3 (g) 3 O 2 (g)? b. If the rate at which O 2 appears, [O 2 ]/ t, is 6.0 10 5 M/s at a particular instant, what is the rate of change of O 3? c. What is the overall rate at this instant? Measuring the Reaction Rate To measure the reaction rate we need to measure the concentration of at least one component (reactant or product) in the mixture at many points in time. - Continuous monitoring for reactions that take place in less than an hour. - Sampling of the mixture at various times for reactions that happen in a long time. Possible tests: Measure the absorbance of visible or UV light, ph, electrical conductivity, and pressure of a gas. 14.3 Concentration and Rate Laws We can gain information about the rate of a reaction by seeing how the rate changes with changes in concentration. The rate law of a reaction is the mathematical relationship between the rate of the reaction and the concentrations of the reactants. The rate law must be determined experimentally!! (There is no way to predict it!!) For the reaction: N 2 O 5 (g) 2 NO 2 (g) + ½ O 2 (g) It was determined experimentally that when [N 2 O 5 ] doubles, the reaction rate doubles. If [N 2 O 5 ] triples, the rate triples. If [N 2 O 5 ] is halved, the rate is halved. Therefore: rate α [N 2 O 5 ]. 3
The rate law for this reaction:. N 2 O 5 (g) 2 NO 2 (g) + ½ O 2 (g) Where is the rate constant, which depends on the reaction and on T. (It is determined by experiment.) In general: Again: are called the orders for each reactant (usually 0, 1, 2, but they can be fractions or negative). is called the rate constant. The rate law must be determined experimentally, and it is not related to the balanced equation. For: The rate law is: exponents coefficients (except by coincidence) 2 NO(g) + O 2 (g) 2 NO 2 (g) The rate law for the reaction is: Rate = k[no] 2 [O 2 ] The reaction is: second order with respect to [NO], first order with respect to [O 2 ], and third order overall (order of the reaction) For: First order in First order in Second order in Fourth order overall (add all exponents). For: ( ) ( ) ( ) Zero order in NH 3 Experimental determination of Rate Law: one method is the Method of Initial Rates. Initial rate rate at the very beginning of the reaction ( 1-2% complete. Convenient because: initial are known, avoids complications with products and side reactions). Approach: vary initial [reactants] and see how the rate is affected. 4
( ) ( ) ( ) Exp. Initial [NO] Initial [O 2 ] Initial rate M/sec 1 0.020 0.010 0.028 2 0.020 0.020 0.057 3 0.020 0.040 0.114 4 0.040 0.020 0.227 5 0.010 0.020 0.014 1 st solve by inspection, then by brute force: If we compare reactions 1 and 2, we notice that: Rate α [O 2 ] so first order in O 2. Compare experiments 2 and 4 ([O 2 ] constant). Rate is second order in [NO]. Brute Force Method (Math) Take a ratio of the rate laws (larger/smaller): For reactions 1 and 2, [NO] is constant. k is also constant (it is the same reaction and the same T). So: ( ) ( ) The rate law is first order in O 2, as we previously found by inspection (you can skip the math, if it is simple!). For reactions 2 and 4, [NO] doubles, but [O 2 ] and k are constant. We have: ( ) 5
Once the orders are known, we can use the results of any experiment to determine k. Using experiment 3: (Check the units; they depend on the overall reaction order) Q. For the reaction: 2 NO + 2 H 2 N 2 + 2 H 2 O 1. Orders? 2. If [NO] is doubled, what happens to the rate? 3. If [H 2 ] is halved, what happens to the rate? Q. Determine the rate law and the rate constant: Exp. Initial Initial Initial rate M/sec 1 0.080 0.034 2.2 10-4 2 0.080 0.017 1.1 10-4 3 0.16 0.017 2.2 10-4 14.4 The Change of Concentration with Time From the rate laws, we can obtain equations that relate concentrations of reactants or products to time. - calculate concentrations after a certain amount of time - how long it will take reach a certain concentration level - obtain k First Order Reactions If we rearrange, integrate and evaluate both sides (calculus required here!): Notes: we use the natural logarithm. Also, there is a concentrations ratio, just make sure the units of concentration are the same and you ll be fine (M, mol, m, g, mg, P, etc). 6
Q. Sucrose H + Glucose + Fructose rate = k[sucrose] ; k=0.21 hr -1 (first order) If [Sucrose] initially = 0.010 M, what is [Sucrose] after 5.0 hr? Q. For the same reaction. How long will it take for [Sucrose] to decrease to 20.0% of its initial value? Q. For the same reaction. How long will it take for the reaction to be 95.0% complete? 7
One more thing: ln[a] ln[a] time Second Order Reactions After we rearrange and integrate: Q. 2 HI H 2 + I 2 Calculate [HI] after 10. min if the original [HI] = 0.010 M 8
One more thing on Second Order Reactions: [A] Zero Order Reactions time After we integrate: ( ) [A] [A] A way to determine the rate law Plot [A] vs t, if a line, rxn is zero order in A. Plot ln[a] vs t, if a line, rxn is first order in A. Plot 1/[A] vs t, if a line, rxn is second order in A. time Sample Exercise (p. 572) The decomposition of NO 2 at 300 C is described by the equation NO 2 (g) NO (g) + ½ O 2 (g) and yields data comparable to this table: Time (s) [NO 2 ], M 0.0 0.01000 50.0 0.00787 100.0 0.00649 Is the reaction 1 st or 2 nd order in NO 2? 200.0 0.00481 300.0 0.00380 9
Half life ( ) It refers to the time it takes for the reactant concentration to decrease to half of its initial value. For first order reactions: ( ) ( ) ( ) For first order reactions, t ½ is independent of the initial [A]. The half-life expression for second order and zero order reactions are: Second order Zero order For a first order reaction: Time (sec) [A] t ½ left 0 2.0 100 1.0 1 1/2 left 200 0.50 2 1/4 left 300 0.25 3 1/8 left 400 0.125 4 1/16 left Q. cyclopropane propene is first order.. What is the half life? What fraction of cyclopropane remains after 51.2 hr? What fraction remains after 18.0 hr? 10
14.5 Temperature and Rate Generally, as temperature increases, so does the reaction rate. This is because k is temperature-dependent. Rule of Thumb: rate doubles every ΔT=10 C. The Collision Model In a chemical reaction, bonds are broken and new bonds are formed. Molecules can only react if they collide with each other. Two requirements: molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation. Lots of collisions are not contributing to the reaction due to incorrect orientation. However, increasing the concentration of reactants leads to more frequent collisions and a faster reaction. Activation Energy (E a ) It is the minimum energy required for a reaction to occur (it varies). Higher E a leads to a slower reaction. Reaction Coordinate Diagrams The diagram shows the energy of the reactants and products (and, therefore, E). The energy gap between the reactants and the activated complex is the activation-energy barrier. The transition state contains partial bonds. Maxwell Boltzmann Distributions As the temperature increases, so does the fraction of molecules that can overcome the activation-energy barrier. As a result, the reaction rate increases. The fraction of molecules possessing enough energy can be found through the expression: where R (8.314 J/mol K) is the gas constant and T is the Kelvin temperature. Arrhenius Equation Changing the temperature changes the rate constant of the rate law. Svante Arrhenius investigated this relationship and showed that: where A is the frequency factor, a number that represents the likelihood that collisions would occur with the proper orientation for reaction. 11
The equation can be algebraically solved to give: ( ) Therefore, if k is determined experimentally at several temperatures, E a can be calculated from the slope of a plot of ln k vs. 1/T. Another form of the Equation (p. 579) The two points form: ( ) We can calculate at some if we know, the rate constant and some other temperature. Q. Acetaldehyde CH 3 CHO CH 4 + CO a. Find Ea. b. Find K at 865 K. 12
14.6 Reaction Mechanisms A chemical equation shows all the reactant and product molecules; however, the probability of more than 3 molecules colliding - at the same instant - with the proper orientation - and sufficient energy is negligible. Most reactions occur in a series of small reactions (elementary reactions or elementary processes) involving 1, 2, or at most 3 molecules. The molecularity of a process tells how many molecules are involved in the process. The sequence of steps that describes the actual process by which reactants become products is called the reaction mechanism. An Example of a Reaction Mechanism Overall reaction: H 2(g) + 2 ICl (g) 2 HCl (g) + I 2(g) Mechanism: 1. H 2(g) + ICl (g) HCl (g) + HI (g) 2. HI (g) + ICl (g) HCl (g) + I 2(g) Two bimolecular elementary steps (they cannot be broken down into simpler steps). Intermediates HI is made but then consumed (it does not show up in the overall reaction). Materials that are products in an early mechanism step, but then reactants in a later step, are called intermediates. 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. For elementary reactions ONLY, the rate law can be written from the balanced equation: H 2 (g) + ICl(g) HCl(g) + HI(g) Rate = k 1 [H 2 ][ICl] HI(g) + ICl(g) HCl(g) + I 2 (g) Rate = k 2 [HI][ICl] H 2 (g) + 2 ICl(g) 2 HCl(g) + I 2 (g) Rate =? 13
Q. Write the rate laws for the following elementary steps. Overall reaction? Intermediate? 1. NO 2 (g) + NO 2 (g) NO 3 (g) + NO (g) 2. NO 3 (g) + CO (g) NO 2 (g) + CO 2 (g) Keep in mind the following In general, we cannot predict rate laws from balanced equations because many reactions involve 2 or more steps. The rate law for the overall rxn is some combination of the rate laws for the individual steps. Mechanisms of Reaction are theories (i.e. we think them up. They can be disproven, but cannot be proved.) Validating a Mechanism 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. Rate Determining Step In most mechanisms, one step occurs slower than the other steps (the formation of product cannot occur any faster than the rate of this slowest step). We call the slowest step in the mechanism the rate determining step (it has the largest activation energy) The overall rate of reaction is usually equal to the rate of the slow step. Rate Determining Step 1. NO 2 (g) + NO 2 (g) NO 3 (g) + NO (g) SLOW 2. NO 3 (g) + CO (g) NO 2 (g) + CO 2 (g) FAST NO 2 (g) + CO (g) NO (g) + CO 2 (g) (Overall) Rate obs = k[no 2 ] 2 14
Mechanisms with a Fast Initial Step Proposed mechanism: 1. k 1 N 2 O 5 NO 2 + NO k- 3 1 Fast, equilibrium 2. NO 2 + NO 3 k 2 NO + NO 2 + O 2 Slow 3. NO 3 + NO k 3 2 NO 2 Fast Problems to solve: a. the rate observed for the overall reaction and the slow step don t match each other (wrong mechanism?). b. The rate of the slow step includes [NO 3 ], which is an intermediate (they are usually unstable and have low, unknown concentrations). At equilibrium, the rates of the forward and reverse reactions are equal: Q. Show that the proposed mechanism for the reaction 2 O 3(g) 3 O 2(g) matches the observed rate law: Rate = k[o 3 ] 2 [O 2 ] 1 1. O 3(g) O 2(g) + O (g) Fast 2. O 3(g) + O (g) 2 O 2(g) Slow 14.7 Catalysts Catalysts increase the rate of a reaction by decreasing the activation energy of the reaction. Catalysts change the mechanism by which the process occurs. Note: ΔE doesn t change. The catalyst speeds up both, the forward and the reverse reactions. 15
Catalysts are consumed in an early mechanism step, then made in a later step; therefore, they do not appear in the balanced equation. mechanism without catalyst mechanism with catalyst O 3(g) + O (g) 2 O 2(g) V. Slow Cl (g) + O 3(g) O 2(g) + ClO (g) Fast ClO (g) + O (g) O 2(g) + Cl (g) Slow Heterogeneous catalysts (in a different phase than reactants) hold one reactant molecule in proper orientation for reaction to occur when the collision takes place. Homogeneous catalysts react with one of the reactant molecules to form a more stable activated complex with a lower activation energy. Heterogeneous Catalysts Often composed of metals or metal oxides. Hydrogenation is much faster in the presence of Pd, Pt or Ni. H 2 gets adsorbed on the surface of the metal. Catalytic Converters (p. 592) CO and HC s to CO 2 and H 2 O. And NO x s to N 2. Enzymes Enzyme: A protein or other molecule that acts as a catalyst for a biological reaction. Substrate: Substance upon which the enzyme acts (the reactant). Active site: The site on the enzyme where catalysis occurs. Very specific; its shape matches only a certain type of molecule (Lock-and-Key model). How do enzymes catalyze reactions? Bring substrate(s) and catalytic sites together (proximity effect). Hold substrate(s) at the exact distance and in the exact orientation necessary for reaction (orientation effect). Provide acidic, basic, or other types of groups required for catalysis (catalytic effect). Lower the energy barrier by inducing strain in bonds in the substrate molecule (energy effect). Efficiency of Enzymes Enzymes are much more efficient than other types of catalysts. 16