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CHEMICAL KINETICS Pt 2 Reaction Mechanisms Reaction Mechanism

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CHEMICAL KINETICS Pt 2 Reaction Mechanisms Reaction Mechanism Reaction Mechanism (continued) CHEMICAL The reaction KINETICS 2 C H O +→5O + 6CO 4H O Pt 2 3 4 3 2 2 2 • has many steps in the reaction mechanism. Objectives ! Be able to describe the collision and Reaction Mechanisms transition-state theories • Even though a balanced chemical equation ! Be able to use the Arrhenius theory to may give the ultimate result of a reaction, determine the activation energy for a reaction and to predict rate constants what actually happens in the reaction may take place in several steps. ! Be able to relate the molecularity of the reaction and the reaction rate and • This “pathway” the reaction takes is referred to describle the concept of the “rate- as the reaction mechanism. determining” step • The individual steps in the larger overall reaction are referred to as elementary ! Be able to describe the role of a catalyst and homogeneous, heterogeneous and reactions. enzyme catalysis Reaction Mechanisms Often Used Terms •Intermediate: formed in one step and used up in a subsequent step and so is never seen as a product. The series of steps by which a chemical reaction occurs. •Molecularity: the number of species that must collide to produce the reaction indicated by that A chemical equation does not tell us how step. reactants become products - it is a summary of the overall process. •Elementary Step: A reaction whose rate law can be written from its molecularity. •uni, bi and termolecular 1 Elementary Reactions Elementary Reactions • Consider the reaction of nitrogen dioxide with • Each step is a singular molecular event carbon monoxide. resulting in the formation of products. NO (g) + CO(g) → NO(g) + CO (g) • The overall chemical equation is obtained by 2 2 adding the two steps together and canceling • This reaction is believed to take place in two any species common to both sides. steps. NO2 (g) + NO2 (g) → NO3 (g) + NO(g) NO2 (g) + NO2 (g) → NO3 (g) + NO(g) (elementary reaction) NO3 (g) + CO(g) → NO2 (g) + CO2 (g) NO (g) + NO (g) + NO (g) + CO(g) → NO (g) + NO(g) + NO (g) + CO (g) NO3 (g) + CO(g) → NO2(g) + CO2 (g) (elementary reaction) 2 2 3 3 2 2 Figure 12.9: A molecular representation of the elementary Molecularity steps in the reaction of NO2 and • We can classify reactions according to their CO. molecularity, that is, the number of molecules that must collide for the elementary reaction to occur. •A unimolecular reaction involves only one reactant molecule. •A bimolecular reaction involves the collision of two reactant molecules. •A termolecular reaction requires the collision NO2 (g) + CO (g) º NO (g) + CO2 (g) of three reactant molecules. Elementary Reactions Molecularity • Each step is a singular molecular event • We can classify reactions according to their resulting in the formation of products. molecularity, that is, the number of molecules that must collide for the elementary • Note that NO3 does not appear in the overall equation, but is formed as a temporary reaction to occur. reaction intermediate. • Higher molecularities are rare because of the small statistical probability that four or more molecules would all collide at the same instant. 2 Rate Equations for Elementary Rate Equations for Elementary Reactions Reactions • Since a chemical reaction may occur in • For a termolecular reaction several steps, there is no easily stated A + B + C → products relationship between its overall reaction and its rate law. the rate is dependent on the populations of all three participants. • For elementary reactions, the rate is proportional to the concentrations of all Rate = k[A][B][C] reactant molecules involved. Rate Equations for Elementary Reactions • For example, consider the generic equation below. A → products The rate is dependent only on the concentration of A; that is, Rate = k[A] Rate Equations for Elementary Rate Equations for Elementary Reactions Reactions • However, for the reaction • Note that if two molecules of a given reactant are required, it appears twice in the rate law. A + B → products For example, the reaction the rate is dependent on the concentrations 2A + B → products of both A and B. would have the rate law: Rate = k[A][B] Rate = k[A][A][B] or Rate = k[A]2[B] 3 Rate Equations for Elementary Rate Laws and Mechanisms Reactions • So, in essence, for an elementary reaction, • Consider the reaction below. the coefficient of each reactant becomes the 2 NO (g) + F (g) → 2 NO F(g) power to which it is raised in the rate law for 2 2 2 that reaction. • Experiments performed with this reaction show that the rate law is • Note that many chemical reactions occur in multiple steps and it is, therefore, impossible to Rate = k[NO2 ][F2 ] predict the rate law based solely on the overall reaction. • This implies that the reaction above is not an elementary reaction but rather the result of multiple steps. Rate-Determining Step Rate-Determining Step • In multiple-step reactions, one of the •In a multistep reaction, it is the elementary reactions in the sequence is often slower than the rest. slowest step. It therefore • The overall reaction cannot proceed any faster determines the rate of reaction. than this slowest rate-determining step. Rate Laws and Mechanisms Rate-Determining Step • Consider the reaction below. • In multiple-step reactions, one of the elementary reactions in the sequence is often 2 NO (g) + F (g) → 2 NO F(g) 2 2 2 slower than the rest. • Experiments performed with this reaction show • Our previous example occurs in two elementary that the rate law is steps where the first step is much slower. Rate = k[NO2 ][F2 ] k1 (slow) NO2 (g) + F2 (g) → NO2F(g) + F(g) k2 (fast) • The reaction is first order with respect to each NO2 (g) + F(g)→ NO2F(g) reactant, even though the coefficient for NO in 2 2 NO (g) + F (g) → 2 NO F(g) the overall reaction is. 2 2 2 4 Rate-Determining Step Figure 14.16: • In multiple-step reactions, one of the Representation of elementary reactions in the sequence is often the mechanism of slower than the rest. decomposition of • Since the overall rate of this reaction is N2O5 using determined by the slow step, it seems logical that molecular models. the observed rate law is Rate = k1[NO2][F2]. Rate = k3[NO2][NO3] k1 and from step 1 NO2 (g) + F2 (g) → NO2F(g) + F(g) (slow) k1[N2O5] = k-1 [NO2][NO3] so that Rate = (k3k1/k-1 ) [N2O5] Rate-Determining Step Rate-Determining Step • In a mechanism where the first elementary • Consider the reduction of nitric oxide with H2. step is the rate-determining step, the overall 2NO(g) + 2H2 (g) → N2(g) + 2H2O(g) rate law is simply expressed as the • A proposed mechanism is: k elementary rate law for that slow step. 1 (fast, equilibrium) 2NO N2O2 k • A more complicated scenario occurs when the -1 k 2 (slow) rate-determining step contains a reaction N2O2 + H2 → N2O + H2O intermediate, as you’ll see in the next section. k 3 (fast) N2O + H2 → N2 + H2O • It has been experimentally determined that the 2 rate law is Rate = k [NO] [H2] Rate-Determining Step Rate-Determining Step • Mechanisms with an Initial Fast Step • The rate-determining step (step 2 in this case) • There are cases where the rate-determining step generally outlines the rate law for the overall of a mechanism contains a reaction reaction. intermediate that does not appear in the Rate = k 2[N2O2 ][H2 ] overall reaction. (Rate law for the rate-determining step) • The experimental rate law, however, can be • As mentioned earlier, the overall rate law can be expressed only in terms of substances that expressed only in terms of substances appear in the overall reaction. represented in the overall reaction and cannot contain reaction intermediates. 5 Rate-Determining Step Rate-Determining Step • The rate-determining step (step 2 in this case) • The rate-determining step (step 2 in this case) generally outlines the rate law for the overall generally outlines the rate law for the overall reaction. reaction. Rate = k 2[N2O2 ][H2 ] Rate = k 2[N2O2 ][H2 ] (Rate law for the rate determining step) (Rate law for the rate-determining step) • It is necessary to re-express this proposed rate • Therefore, law after eliminating [N O ]. 2 2 2 [N2O2 ] = (k1 / k −1 )[NO] • If we substitute this into our proposed rate law we obtain: Rate-Determining Step Rate-Determining Step • The rate-determining step (step 2 in this case) • The rate-determining step (step 2 in this case) generally outlines the rate law for the overall generally outlines the rate law for the overall reaction. reaction. Rate = k 2[N2O2 ][H2 ] Rate = k 2[N2O2 ][H2 ] (Rate law for the rate determining step) (Rate law for the rate determining step) k 2k1 2 • We can do this by looking at the first step, which Rate = [NO] [H2 ] is fast and establishes equilibrium. k −1 • If we replace the constants (k2k1/k-1) with k, we obtain the observed rate law: Rate = 2 k[NO] [H2]. Rate-Determining Step Collision Theory • The rate-determining step (step 2 in this case) • Collision theory assumes that for a reaction generally outlines the rate law for the overall to occur, reactant molecules must collide reaction. with sufficient energy and the proper Rate = k 2[N2O2 ][H2 ] orientation. (Rate law for the rate-determining step) • The minimum energy of collision required • At equilibrium, the forward rate and the reverse for two molecules to react is called the rate are equal. activation energy, Ea. 2 k1[NO] = k −1[N2O2 ] 6 Figure 14.12: Importance of molecular orientation in the reaction of Figure 12.13: Several possible orientations NO and Cl .
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