Chemical Kinetics Chapter 14

Reminders Assignment 2 due today (in class) Assignment 3 up now and will be due Mon., Feb. 05 Assignment 4 (Ch. 15) will not be due before Midterm 1, but Ch. 15 will be on the midterm

Energy diagram for a chemical reaction

SAMPLE EXERCISE Relating Energy Profiles to Activation Energies and Speeds of Reaction Consider a series of reactions having the following energy profiles: Assuming that all three reactions have nearly the same frequency factors, rank the reactions from slowest to fastest. Solution The lower the activation energy, the faster the reaction. The value of  does not affect the rate. Hence the order is (2) < (3) < (1).

Reaction Mechanisms The sequence of events that describes the actual process by which reactants become products is called the reaction mechanism.

Reaction Mechanisms Remember that rate laws for reactions must be determined experimentally; they cannot be predicted from the coefficients of the balanced chemical equations. Reactions may occur all at once or through several discrete steps. Each of these processes is known as an elementary reaction or elementary process. Rate laws and relative speeds of these steps will dictate the overall rate law. The next challenge in kinetics is to arrive at reaction mechanisms that lead to rate laws that are consistent with those observed experimentally.

Reaction Mechanisms The molecularity of a process tells how many molecules are involved in the process. The rate law of any elementary reaction is based directly on its molecularity. Remember that we cannot tell by merely looking at a balanced chemical equation whether the reaction involves one or several elementary steps.

SAMPLE EXERCISE Predicting the Rate Law for an Elementary Reaction If the following reaction occurs in a single elementary reaction, predict the rate law: Comment: Experimental studies of this reaction show that the reaction actually has a very different rate law: Rate = k[H 2 ][Br 2 ] 1/2 Because the experimental rate law differs from the one obtained by assuming a single elementary reaction, we can conclude that the mechanism must involve two or more elementary steps. Solution The reaction is bimolecular, involving one molecule of H 2 with one molecule of Br 2. Thus, the rate law is first order in each reactant and second order overall: Rate = k[H 2 ][Br 2 ]

(a) Write the equation for the overall reaction. (b) Write the rate law for the overall reaction. (b) The rate law for the overall reaction is just the rate law for the slow, rate- determining elementary reaction. Because that slow step is a unimolecular elementary reaction, the rate law is first order: Rate = k[N 2 O] SAMPLE EXERCISE Determining the Rate Law for a Multistep Mechanism The decomposition of nitrous oxide, N 2 O, is believed to occur by a two-step mechanism: Solution Solve: (a) Adding the two elementary reactions gives Omitting the intermediate, O(g), which occurs on both sides of the equation, gives the overall reaction:

Answer: Because the rate law conforms to the molecularity of the first step, that must be the rate-determining step. The second step must be much faster than the first one. The experimental rate law is rate = k[O 3 ][NO 2 ]. What can you say about the relative rates of the two steps of the mechanism? PRACTICE EXERCISE Predicting Relative Rates for a Two-step Mechanism Ozone reacts with nitrogen dioxide to produce dinitrogen pentoxide and oxygen: The reaction is believed to occur in two steps

Multistep Mechanisms 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.

Slow Initial Step 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. NO 2 (g) + CO (g)  NO (g) + CO 2 (g)

Slow Initial Step A proposed mechanism for this reaction is Step 1: NO 2 + NO 2  NO 3 + NO (slow) Step 2: NO 3 + CO  NO 2 + CO 2 (fast) The NO 3 intermediate is consumed in the second step. As CO is not involved in the slow, rate-determining step, it does not appear in the rate law.

Fast Initial Step The rate law for this reaction is found to be Rate = k [NO] 2 [Br 2 ] Because termolecular processes are rare, this rate law suggests a two-step mechanism. 2 NO (g) + Br 2 (g)  2 NOBr (g)

Fast Initial Step A proposed mechanism is Step 2: NOBr 2 + NO  2 NOBr (slow) Step 1 includes the forward and reverse reactions. Step 1: NO + Br 2 NOBr 2 (fast)

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 ]? Step 2: NOBr 2 + NO  2 NOBr (slow) Step 1: NO + Br 2 NOBr 2 (fast)

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. Therefore, Rate f = Rate r Step 2: NOBr 2 + NO  2 NOBr (slow) Step 1: NO + Br 2 NOBr 2 (fast)

Fast Initial Step Because Rate f = Rate r, k 1 [NO] [Br 2 ] = k −1 [NOBr 2 ] Solving for [NOBr 2 ] gives us k1k−1k1k−1 [NO] [Br 2 ] = [NOBr 2 ] Step 2: NOBr 2 + NO  2 NOBr (slow) Step 1: NO + Br 2 NOBr 2 (fast)

Fast Initial Step Substituting this expression for [NOBr 2 ] in the rate law for the rate-determining step gives k2k1k−1k2k1k−1 Rate =[NO] [Br 2 ] [NO] = k [NO] 2 [Br 2 ] Step 2: NOBr 2 + NO  2 NOBr (slow) Step 1: NO + Br 2 NOBr 2 (fast)

A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed. k = A exp( -E a /RT )EaEa k uncatalyzedcatalyzed rate catalyzed > rate uncatalyzed E a < E a ‘ 14.6

In heterogeneous catalysis, the reactants and the catalysts are in different phases. In homogeneous catalysis, the reactants and the catalysts are dispersed in a single phase, usually liquid. Haber synthesis of ammonia Ostwald process for the production of nitric acid Catalytic converters Acid catalysis Base catalysis 14.6

N 2 (g) + 3H 2 (g) 2NH 3 (g) Fe/Al 2 O 3 /K 2 O catalyst Haber Process 14.6

Ostwald Process Hot Pt wire over NH 3 solution Pt-Rh catalysts used in Ostwald process 4NH 3 (g) + 5O 2 (g) 4NO (g) + 6H 2 O (g) Pt catalyst 2NO (g) + O 2 (g) 2NO 2 (g) 2NO 2 (g) + H 2 O (l) HNO 2 (aq) + HNO 3 (aq) 14.6

Catalytic Converters 14.6 CO + Unburned Hydrocarbons + O 2 CO 2 + H 2 O catalytic converter 2NO + 2NO 2 2N 2 + 3O 2 catalytic converter

Enzyme Catalysis 14.6

uncatalyzed enzyme catalyzed 14.6 rate =  [P] tt rate = k [ES]