Chapter 12. Kinetics. Factors That Affect Reaction Rates. Factors That Affect Reaction Rates. Chemical. Kinetics

Transcription

1 PowerPoint to accompany Kinetics Chapter 12 Chemical Kinetics Studies the rate at which a chemical process occurs. Besides information about the speed at which reactions occur, kinetics also sheds light on the reaction mechanism (exactly how the reaction occurs). Factors That Affect Reaction Rates Factors That Affect Reaction Rates Physical State of the Reactants In order to react, molecules must come in contact with each other. The more homogeneous the mixture of reactants, the faster the molecules can react. Concentration of Reactants As the concentration of reactants increases, so does the likelihood that reactant molecules will collide. Figure 12.2

2 Factors That Affect Reaction Rates Factors That Affect Reaction Rates Temperature At higher temperatures, reactant molecules have more kinetic energy, move faster, and collide more often and with greater energy. Presence of a Catalyst Catalysts speed up reactions by changing the mechanism of the reaction. Catalysts are not consumed during the course of the reaction. Reaction Rates Reaction Rates A B Figure 12.3 Rates of reactions can be determined by monitoring the change in concentration of either reactants or products as a function of time. The average rate of the reaction over each interval is the change in concentration divided by the change in time: Average rate of appearance of B = change in concentration of B change in time

3 Reaction Rates Reaction Rates C 4 H 9 Cl(aq) + H 2 O(l) C 4 H 9 OH(aq) + HCl(aq) Note that the average rate decreases as the reaction proceeds. A plot of concentration versus time for this reaction yields a curve like this. Table 12.1 This is because as the reaction goes forward, there are fewer collisions between reactant molecules. The slope of a line tangent to the curve at any point is the instantaneous rate at that time. Figure 12.4 Reaction Rates Reaction Rates and Stoichiometry All reactions slow down over time. Therefore, the best indicator of the rate of a reaction is the instantaneous rate near the beginning. C 4 H 9 Cl(aq) + H 2 O(l) C 4 H 9 OH(aq) + HCl(aq) In this reaction, the ratio of C 4 H 9 Cl to C 4 H 9 OH is 1:1. Thus, the rate of disappearance of C 4 H 9 Cl is the same as the rate of appearance of C 4 H 9 OH. Figure 12.4 Rate = -[C 4H 9 Cl] t = [C 4H 9 OH] t Figure 12.4

4 Reaction Rates and Stoichiometry What if the ratio is not 1:1? 2 HI(g) H 2 (g) + I 2 (g) Rate = 1 2 [HI] t Therefore, = [I 2] t Reaction Rates and Stoichiometry To generalise, then, for the reaction aa + bb dd + ee Rate = 1 a [A] t = 1 b [B] t = 1 d [D] t = 1 e [E] t Concentration and Rate Concentration and Rate One can gain information about the rate of a reaction by seeing how the rate changes with changes in concentration. Table 12.2 NH 4 + (aq) + NO 2 (aq) N2 (g) + 2 H 2 O(l) Comparing Experiments 1 and 2, when [NH 4+ ] doubles, the initial rate doubles.

5 Concentration and Rate Concentration and Rate NH 4 + (aq) + NO 2 (aq) Table 12.2 N2 (g) + 2 H 2 O(l) This means: Rate [NH 4+ ] Rate [NO 2 ] Rate [NH + ] [NO 2 ] or Rate = k [NH 4+ ] [NO 2 ] Likewise, comparing Experiments 5 and 6, when [NO 2 ] doubles, the initial rate doubles. This equation is called the rate law, and k is the rate constant. Rate Laws For the general reaction aa + bb dd + ee the rate law has the form Rate = k[a]m[b]n Exponents in the Rate Law Generally, Rate = k[reactant 1]m[reactant 2]n m and n are called the reaction orders A rate law shows the relationship between the reaction rate and the concentrations of reactants. The exponents m and n tell the order of the reaction with respect to each reactant. This reaction is First-order in [NH 4+ ] First-order in [NO 2 ] The overall reaction order can be found by adding the exponents on the reactants in the rate law. This reaction is second-order overall.

6 The Change of Concentration with Time (Integrated Rate Equations) Using calculus to integrate the rate law for a first-order process gives us ln[a] t - ln[a] 0 = -kt or ln [A] t [A] 0 = -kt [A] 0 is the initial concentration of A. [A] t is the concentration of A at some time, t, during the course of the reaction. First-Order Processes Consider the process in which methyl isonitrile is converted to acetonitrile. or which is in the form ln[a]t = -kt - ln[a] 0 y = mx + b CH 3 NC CH 3 CN therefore, if a reaction is first-order, a plot of ln [A] versus t will yield a straight line, and the slope of the line will be -k. Figure 12.5 First-Order Processes First-Order Processes CH 3 NC CH 3 CN This data was collected for this reaction at C. Figure 12.6 (a) Figure 12.6 (a,b) When ln P is plotted as a function of time, a straight line results. Therefore, The process is first-order. k is the negative slope: s 1.

7 Second-Order Processes Similarly, integrating the rate law for a process that is second-order in reactant A, we get 1 [A] = kt + 1 t [A] 0 which, like the first order process, is also in the form: y = mx + b Second-Order Processes 1 [A] = kt + 1 t [A] 0 So if a process is second-order in A, a plot of 1/[A] versus t will yield a straight line, and the slope of that line is k. To distinguish between first and second order rate laws, graph both ln([a]t) and 1/[A]t versus t. If the ln([a]t) plot is linear then the reaction is first order. If the 1/[A]t plot is linear then the reaction is second order. Second-Order Processes Second-Order Processes The decomposition of NO 2 at 300 C is described by the equation NO 2 (g) NO (g) + 1/2 O 2 (g) Graphing ln([no 2 ]) verus t yields: Graphing ln 1/[NO 2 ] versus t, however, gives this plot. and yields data comparable to this: The plot is not a straight line, so the process is not first-order in [A]. Figure 12.6 (a,b) This is a straight line, so the process is secondorder in [A].

8 Half-Life Half-Life Half-life is defined as the time required for onehalf of a reactant to react. For a first-order process, this becomes 0.5 [A] ln 0 [A] 0 = kt 1/2 ln 0.5 = kt 1/2 Figure 12.8 Because [A] at t 1/2 is one-half of the original [A], [A] t = 0.5 [A] = kt 1/ k = t 1/2 NOTE: For a first-order process, the half-life does not depend on [A] 0. Half-Life Temperature and Rate For a second-order process, 1 k[a] 0 = t 1/2 Generally, as temperature increases, so does the reaction rate. This is because k is temperature dependent. i.e. the half-life for second order (and other) reactions depends on reactant concentrations and therefore changes as the reaction progresses. Figure Figure 12.10

9 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. The Collision Model Furthermore, molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation. Figure Activation Energy In other words, there is a minimum amount of energy required for reaction: the activation energy, E a. Just as a ball cannot get over a hill if it does not roll up the hill with enough energy, a reaction cannot occur unless the molecules possess sufficient energy to get over the activation energy barrier. Figure Reaction Coordinate Diagrams It is helpful to visualise energy changes throughout a process on a reaction coordinate diagram like this one for the rearrangement of methyl isonitrile. Figure 12.14

10 Reaction Coordinate Diagrams It shows the energy of the reactants and products (and, therefore, E). The arrangement of atoms at the top of the barrier is called the activated complex or transition state. The energy gap between the reactants and the activated complex is the activation energy barrier. Figure Arrhenius Equation Svante Arrhenius developed a mathematical relationship between k and E a : k = A e E a/rt where A is the frequency factor, a number that represents the likelihood that collisions would occur with the proper orientation for reaction. Arrhenius Equation Reaction Mechanisms Figure Taking the natural logarithm of both sides, the equation becomes: ln k = -E a / RT + ln A Therefore, if k is determined experimentally at several temperatures, E a can be calculated from the slope of a plot of ln k versus 1/T. The sequence of events that describes the actual process by which reactants become products is called the reaction mechanism. Reactions may occur all at once or through several discrete steps. Each of these processes is known as an elementary reaction or elementary process. The molecularity of a process tells how many molecules are involved in the process, e.g. CH 3 NC CH 3 CN is unimolecular

11 In a multistep process, one of the steps will be slower than all others. The overall reaction cannot occur faster than this slowest, rate-determining step. NO 2 (g) + CO (g) NO (g) + CO 2 (g) The rate law for this reaction is found experimentally to be: Rate = k [NO 2 ] 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. (Slow Initial Step) A proposed mechanism for this reaction is Step 1: NO 2 + NO 2 NO 3 + NO Step 2: NO 3 + CO NO 2 + CO 2 It follows then that: Step 1, NO 2 + NO 2 NO 3 + NO, is slow, and The NO 3 intermediate is consumed in the second step. Step 2, NO 3 + CO NO 2 + CO 2, is fast or, k 2 >> k 1 As CO does not appear in the rate law, it is not involved in the slow step. In other words, Step 1 is slow and therefore is the rate determining step. Because NO 3 is neither a reactant nor a product in the overall reaction, it is called an intermediate. Step 1 is a bimolecular process that has the rate law, Rate = k 1 [NO 2 ] 2

12 Some molecularity and Rate Law interpretations are tabulated below: (Fast Initial Step) 2 NO(g) + Br 2 (g) 2NOBr(g) The rate law for this reaction is found to be: Rate = k [NO] 2 [Br 2 ] Table 12.3 Note that termolecular reactions are rare. Because termolecular processes are rare, this rate law suggests a two-step mechanism. (Fast Initial Step) A proposed mechanism is: Step 1: NO + Br 2 NOBr 2 (fast) Step 2: NOBr 2 + NO 2 NOBr (Step 1 includes the forward and reverse reactions.) (slow) (Fast Initial Step) The rate of the overall reaction depends upon the rate of the slow step. The rate law for that step would be: Rate = k 2 [NOBr 2 ] [NO] But how can we find [NOBr 2 ]?

13 (Fast Initial Step) NOBr 2 can react two ways: With NO to form NOBr By decomposition to reform NO and Br 2 The reactants and products of the first step are in equilibrium with each other. (Fast Initial Step) Because Rate f = Rate r, k 1 [NO] [Br 2 ] = k 1 [NOBr 2 ] Solving for [NOBr 2 ] gives us Therefore: Rate f = Rate r k 1 k 1 [NO] [Br 2 ] = [NOBr 2 ] (Fast Initial Step) Substituting this expression for [NOBr 2 ] in the rate law for the rate-determining step gives Rate = k 2 k 1 k 1 [NO] [Br 2 ] [NO] 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 increase the rate of a reaction by decreasing the activation energy of the reaction. Catalysts change the mechanism by which the process occurs. = k [NO] 2 [Br 2 ] Figure 12.18

14 Catalysts Homogeneous Catalyst One that is present in the same phase as the reacting molecules, e.g. Br - assisted decomposition of acidified H 2 O 2 solution. Heterogeneous Catalyst One that exists in a different phase from the reacting molecules. These are often comprised of metals or metal oxides, e.g. Pd or Pt.