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Home , Collision theory, Elementary reaction, Molecularity, Reaction intermediate, 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 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 of the reaction and the 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.

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 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 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) () 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 that must collide for the elementary reaction to occur. •A unimolecular reaction involves only one reactant . •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. . • 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 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 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 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 • 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 . 2 for a collision between two BrNO molecules. NO (g) + Cl2 (g) º NOCl (g) + Cl (g) Orientations (a) and (b) can lead to a reaction, but orientation (c) cannot.

2 NOBr (g) º 2 NO (g) + Br2 (g)

Collision Model Collision Theory and the •Key Idea: Molecules must collide to react.  Collisions must have enough energy to produce the reaction (must equal or exceed •However, only a small fraction of the activation energy). collisions produces a reaction. Why?  Orientation of reactants must allow •Arrhenius: An activation energy must be formation of new bonds. overcome.

Figure 12.11: (a) The change in potential energy as a function of reaction progress for the reaction 2BrNO

2NO + Br2. The activation energy Ea represents the energy Collision Theory needed to disrupt the BrNO molecules so that they can form products. The quantity DE represents the net change in • Rate constants vary with . energy in going from reactant to products. (b) A molecular Consequently, the actual rate of a reaction is representation of the reaction. very temperature dependent. • Why the rate depends on temperature can by explained by collision theory.

7 Collision Theory and the Figure 12.12: Plot showing the number of collisions Arrhenius Equation with a particular energy at T1 and T2, where T2 > T1. • Z is only slightly temperature dependent.

• This alone does not account for the observed increases in rates with only small increases in temperature. • From kinetic theory, it can be shown that a 10 oC rise in temperature will produce only a 2% rise in collision frequency.

Collision Theory and the Collision Theory and the Arrhenius Equation Arrhenius Equation • Collision theory maintains that the rate constant • On the other hand, f, the fraction of molecules for a reaction is the product of three factors. with sufficient activation energy, turns out to be 1. Z, the collision frequency very temperature dependent.

2. f, the fraction of collisions with sufficient • It can be shown that f is related to Ea by the energy to react following expression. 3. p, the fraction of collisions with the proper -Ea RT orientation to react f = e • Here e = 2.718… , and R is the ideal gas k = Zpf constant, 8.31 J/(mol.K).

Collision Theory and the Collision Theory and the Arrhenius Equation Arrhenius Equation • Z is only slightly temperature dependent. • On the other hand, f, the fraction of molecules with sufficient activation energy turns out to be • This is illustrated using the kinetic theory of very temperature dependent. gases, which shows the relationship between • From this relationship, as temperature the velocity of gas molecules and their increases, f increases. absolute temperature. -Ea RT 3RT f = e velocity = abs or velocity ∝ T abs • Also, a decrease in the activation energy, Ea, Mm increases the value of f.

8 Arrhenius Equation Collision Theory and the (continued) Arrhenius Equation • On the other hand, f, the fraction of molecules with sufficient activation energy turns out to be −E / RT very temperature dependent. kAe= a • This is the primary factor relating temperature • k = rate constant increases to observed rate increases. • A = frequency factor -Ea RT • E = activation energy f = e a • T = temperature • R =

Collision Theory and the Figure 12.10: Arrhenius Equation A plot showing • The reaction rate also depends on p, the fraction the exponential of collisions with the proper orientation. dependence of • This factor is independent of temperature the rate constant changes. on absolute • So, with changes in temperature, Z and p temperature. remain fairly constant. • We can use that fact to derive a mathematical relationship between the rate constant, k, and the absolute temperature.

The Arrhenius Equation • If we were to combine the relatively constant terms, Z and p, into one constant, let’s call it A. Figure 14.15: Plot of We obtain the Arrhenius equation: ln k versus 1/T.

-Ea k = A e RT • The Arrhenius equation expresses the dependence of the rate constant on absolute temperature and activation energy.

9 The Arrhenius Equation • It is useful to recast the Arrhenius equation in logarithmic form. • We can relate this equation to the (somewhat rearranged) general formula for a straight line.

Ea 1 ln k = ln A - R (T) y = b + m x • A plot of ln k versus (1/T) should yield a straight

line with a slope of (-Ea/R) and an intercept of ln A. (see Figure 14.15)

Figure 12.14: Plot The Arrhenius Equation of ln(k) versus 1/T for the reaction • A more useful form of the equation emerges if 2N2O5(g) we look at two points on the line this equation 4NO2(g) + O2(g). describes that is, (k , (1/T )) and (k , (1/T )). The value of the 1 1 2 2 • The two equations describing the relationship activation energy at each coordinate would be for this reaction E can be obtained a 1 ln k1 = ln A - R (T ) from the slope of 1 the line, which and Ea 1 equals -Ea/R. ln k = ln A - ( ) 2 R T2

The Arrhenius Equation The Arrhenius Equation • It is useful to recast the Arrhenius equation in • A more useful form of the equation emerges if logarithmic form. we look at two points on the line this equation describes that is, (k , (1/T )) and (k , (1/T )). • Taking the natural logarithm of both sides of 1 1 2 2 the equation, we get: • We can eliminate ln A by subtracting the two equations to obtain

Ea ln k = ln A - RT k E ln 2 = a ( 1 − 1 ) k1 R T1 T2

10 The Arrhenius Equation A Problem to Consider • A more useful form of the equation emerges if • The rate constant for the formation of hydrogen we look at two points on the line this equation iodide from its elements

describes that is, (k1, (1/T1)) and (k2, (1/T2)). H2 (g) + I2 (g) → 2HI(g) • With this form of the equation, given the is 2.7 x 10-4 L/(mol.s) at 600 K and 3.5 x 10-3 L/(mol.s)

activation energy and the rate constant k1 at a at 650 K. Find the activation energy, Ea. given temperature T , we can find the rate 1 • Solving for Ea: constant k2 at any other temperature, T2. 1.11× 8.31J / mol 5 k 2 Ea 1 1 E = = 1.66×10 J ln = ( − ) a 1.28×10−4 k1 R T1 T2

A Problem to Consider Transition-State Theory • The rate constant for the formation of hydrogen • Transition-state theory explains the reaction iodide from its elements resulting from the collision of two molecules H2 (g) + I2 (g) → 2HI(g) in terms of an . is 2.7 x 10-4 L/(mol.s) at 600 K and 3.5 x 10-3 L/(mol.s) •An activated complex () is an at 650 K. Find the activation energy, Ea. unstable grouping of that can break up • Substitute the given data into the Arrhenius to form products. equation. • A simple analogy would be the collision of -3 3.5×10 Ea 1 1 three billiard balls on a billiard table. ln −4 = ( − ) 2.7×10 8.31 J/(mol ⋅K) 600K 650K

A Problem to Consider Transition-State Theory • The rate constant for the formation of hydrogen • Transition-state theory explains the reaction iodide from its elements resulting from the collision of two molecules H2 (g) + I2 (g) → 2HI(g) in terms of an activated complex. is 2.7 x 10-4 L/(mol.s) at 600 K and 3.5 x 10-3 L/(mol.s) • Suppose two balls are coated with a slightly at 650 K. Find the activation energy, Ea. stick adhesive. • Simplifying, we get: • We’ll take a third ball covered with an extremely sticky adhesive and collide it with 1 E −4 ln (1.30×10 ) = 1.11 = a ×(1.28×10 ) our joined pair. 8.31 J/(mol)

11 Figure 14.13 Potential-energy curve for the endothermic reaction of nitric oxide and chlorine. Transition-State Theory • Transition-state theory explains the reaction resulting from the collision of two molecules in terms of an activated complex. • The “incoming” billiard ball would likely stick to one of the joined spheres and provide sufficient energy to dislodge the other, resulting in a new “pairing.” • At the instant of impact, when all three spheres are joined, we have an unstable transition- state complex.

Potential-Energy Diagrams for Transition-State Theory Reactions • Transition-state theory explains the reaction • The potential-energy diagram for an resulting from the collision of two molecules exothermic reaction shows that the products in terms of an activated complex. are more stable than the reactants. • If we repeated this scenario several times, • Figure 14.14 illustrates the potential-energy some collisions would be successful and others diagram for an exothermic reaction. (because of either insufficient energy or • We see again that the forward activation energy improper orientation) would not be successful. is required to form the transition-state complex. • We could compare the energy we provided to • In both of these graphs, the reverse reaction

the billiard balls to the activation energy, Ea. must still supply enough activation energy to form the activated complex.

Figure 14.14 Potential-energy curve for an exothermic reaction. Potential-Energy Diagrams for Reactions • To illustrate graphically the formation of a transition state, we can plot the potential energy of a reaction versus time. • Figure 14.13 illustrates the endothermic reaction of nitric oxide and chlorine gas. • Note that the forward activation energy is the energy necessary to form the activated complex. • The ∆H of the reaction is the net change in energy between reactants and products.

12 Catalysis Figure 12.16: Effect of a catalyst on the number of reaction-producing •Catalyst: A substance that speeds up a collisions. reaction without being consumed

•Enzyme: A large molecule (usually a protein) that catalyzes biological reactions.

•Homogeneous catalyst: Present in the same phase as the reacting molecules.

•Heterogeneous catalyst: Present in a different phase than the reacting molecules.

Catalysts Catalysts •A catalyst is a substance that provides a good •A catalyst is a substance that provides a good “environment” for a reaction to occur, thereby “environment” for a reaction to occur, thereby increasing the reaction rate without being increasing the reaction rate without being consumed by the reaction. consumed by the reaction. • Its presence increases the rate of reaction by • To avoid being consumed, the catalyst must either increasing the frequency factor, A (from participate in at least one step of the reaction and the Arrhenius equation) or lowering the then be regenerated in a later step.

activation energy, Ea.

Figure 12.15: Energy plots for a catalyzed and an uncatalyzed Catalysts pathway for a given reaction. • Homogeneous catalysis is the use of a catalyst in the same phase as the reacting species. • The oxidation of sulfur dioxide using nitric oxide as a catalyst is an example where all species are in the gas phase.

NO(g) 2SO2 (g) + O2 (g) → 2SO3 (g)

13 Figure 14.17: Comparison of activation energies in the uncatalyzed and catalyzed decompositions of Enzyme Catalysis ozone. • have enormous catalytic activity. • The substance whose reaction the enzyme catalyzes is called the substrate. (see Figure 14.21) • Figure 14.22 illustrates the reduction in acivation energy resulting from the formation of an enzyme-substrate complex.

Figure 12.21: Protein-substrate interaction. The substrate is shown in black and red, with the red Catalysts representing the terminal amino acid. Blue indicates side chains from the enzyme that help bind the • Heterogeneous catalysis is the use of a substrate. catalyst that exists in a different phase from the reacting species, usually a solid catalyst in contact with a liquid or gaseous solution of reactants. • Such surface catalysis is thought to occur by chemical adsorbtion of the reactants onto the surface of the catalyst. • Adsorbtion is the attraction of molecules to a surface. (see Figure 14.18)

Heterogeneous Catalysis Figure 14.22: Potential-energy curves for the reaction of substrate, S, to products, P. Steps: • 1. Adsorption and activation of the reactants. • 2. Migration of the adsorbed reactants on the surface. • 3. Reaction of the adsorbed substances. • 4. Escape, or desorption, of the products.

14 Figure 12.18: The exhaust gases from an automobile engine are passed through a catalytic converter to minimize environmental damage.

Figure 14.19: Proposed mechanism of a catalytic converter.

Figure 12.17: Heterogeneous catalysis of the hydrogenation of .

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