CHAPTER 13 (MOORE) CHEMICAL KINETICS: RATES AND MECHANISMS OF CHEMICAL REACTIONS This chapter deals with reaction rates, or how fast chemical reactions occur. Reaction rates vary greatly some are very fast and some are very slow. Chemical kinetics is the study of the rates of chemical reactions, as well as the factors that affect these rates, and the reaction mechanisms by which reactions occur. Factors That Affect Reaction Rate 1. nature of the reactants (relative reactivity) 2. concentration(s) of the reactants 3. temperature 4. presence of catalyst(s) We will begin by considering the meaning of the term, reaction rate. We can think of reaction rate as being similar to speed. A car travels at 65 miles per hour. Note the units: distance/time. What about reaction rates? Since they relate to chemical equations, we can express their rates as changes in reactant (or product) concentrations per unit time: where Δ = change in. rate (disappearance) = Δ[reactant]/Δt here, [ ] means concentration in mol/l Thus, the rate of a reaction is the change in concentration of a species per unit of time. We can graph changes in reactant (or product) concentrations with time Expressing Reaction Rates (see also Table 13.1, Moore, p. 610) The rate of reaction can be expressed as moles per liter per unit time (mol L 1 s 1 or M s 1 ) General reaction rates are calculated by dividing rate expressions by stoichiometric coefficients. Consider: 2 H 2 O 2 2 H 2 O + O 2, where we can write H 2O2 ] Δ[ O2 ] Rate = = 2 Δ t Δ t In general, for a reaction where aa + bb cc + dd, A] B] 1Δ[ C] D] Rate= = = = a Δt b Δt c Δt d Δt Rate is also viewed as the negative of the change in concentration of a reactant per unit of time (rate of disappearance of reactant). Thus, rate = - ½ Δ[H 2 O 2 ]/Δt for 2 H 2 O 2 2 H 2 O + O 2 Problem-Solving Practice 13.2 Consider the reaction 4 NO 2 (g) + O 2 (g) 2 N 2 O 5 (g) (a) express the rate of formation of N 2 O 5 in terms of the rate of disappearance of O 2 (b) if the rate of disappearance of O 2 (g) is 0.0037 mol L 1 s 1, what is the rate of disappearance of NO 2?
Solution. (a) Average and Instantaneous Rates (see Figure 13.3, Moore, p. 613) The Rate Law of A Chemical Reaction - Reaction Rates and Rate "Laws" The rate law for a chemical reaction relates the rate of reaction to the concentrations of reactants. Consider the reaction in which aa + bb + cc Products rate = k[a] n [B] m [C] p for n, m and p = 0, 1, 2 or 3 Important! Note that the exponents (m, n, p ) are determined by experiment. Rate Laws and Rate Constants The rate constant remains constant throughout a reaction, regardless of the initial concentrations of the reactants. The rate of a reaction is the change in concentration with time, whereas the rate constant is the proportionality constant relating reaction rate to the concentrations of reactants, e.g. Rate [A] x [B] y Exponents are not derived from the coefficients in the balanced chemical equation, though in some instances the exponents and the coefficients may be the same. The value of an exponent in a rate law is the order of the reaction with respect to the reactant in question. rate = k[a] n [B] m [C] p The proportionality constant, k, is the rate constant. Reaction "Order" If Rate = k[a] 1 = k[a] If Rate = k[a] 2 If Rate = k[a]3 reaction is first order in A reaction is second order in A Reaction is third order in A If we triple the concentration of A in a second-order reaction, the rate increases by a factor of. Answer: a factor of 9 (or, 9X) Determining Rate Laws From Data - The Method of Initial Rates - a method of establishing the rate law for a reaction finding the values of the exponents in the rate law, and the value of k A series of experiments is performed in which the initial concentration of one reactant is varied. Concentrations of the other reactants are held constant.
When we double the concentration of a reactant, A, - if there is no effect on the rate, the reaction is zero-order in A - if the rate doubles, the reaction is first-order in A - if the rate quadruples, the reaction is second-order in A - if the rate increases eight times, the reaction is third-order in A See Table 13.2 Initial Rates of Reaction for 2 NO(g) + Cl 2 (g) 2 NOCl(g) General rate equation: rate = k [NO] x [Cl 2 ] y Which two experiments are used to find the order of the reaction in NO? Why?? Which two experiments are used to find the order of the reaction in Cl 2? How do we find the value of k after obtaining the order of the reaction in NO and in Cl 2? See Example 13.3 For the reaction 2 NO(g) + Cl 2 (g) 2 NOCl(g) described in the text and in Table 13.2, (a) what is the initial rate for a hypothetical Experiment 4, which has [NO] = 0.0500 M and [Cl 2 ] = 0.0255 M? (b) What is the value of k for the reaction? (a) what is the initial rate for a hypothetical Experiment 4, which has [NO] = 0.0500 M and [Cl 2 ] = 0.0255 M? We have to see how values of concentrations change from Exp. 3 to Exp. 4 [NO] from 0.0250 M to 0.0500 M means doubling and [Cl 2 ] from 0.0255 to 0.0255 M stays constant we see that rate = = k [NO] 2 [Cl 2 ] 1 and that doubling [NO] quadruples rate (4X) while Cl 2 stays the same and has no effect. b) What is the value of k for the reaction? We can use values from any of the experiments to solve for k using rate = k [NO] 2 [Cl 2 ] 1. Let s use data from Exp. 2: [NO] = 0.0125 M and [Cl 2 ] = 0.0510 M so 4.55 x 10-5 M s -1 = k(0.0125 M) 2 (0.0510 M) 1 k = (4.55 x 10-5 M s -1 )/{(0.0125 M) 2 (0.0510) 1 M } k = 5.71 M -2 s -1 FIRST-ORDER RATE In a first-order reaction, the exponent in the rate law is 1. Rate = k[a]1 = k[a] Integrated rate law: ln {[A] 0 /[A] t } = kt ln [A] t ln [A] 0 = kt ln [A] t = kt + ln [A] 0 A straight line!! See Table 13.3, Hill, p. 539. Decomposition of H 2 O 2 Example 13.4, Hill, p. 540 For the first-order decomposition of H 2 O 2 (aq), given k = 3.66 x 10 3 s 1 and [H 2 O 2 ] 0 = 0.882 M, determine (a) the time at which [H 2 O 2 ] = 0.600 M and (b) [H 2 O 2 ] after 225 s.
Half-Life of a Reaction The half-life (t ½ ) of a reaction is the time required for one-half of the reactant originally present to be consumed. At t½, [A]t = ½[A] 0, and for a first order reaction: ln { [A] 0 /(½ [A] 0 )} = kt ½ ln 2 = 0.693 = kt 1/2 t 1/2 = 0.693/k Thus, the half-life is a constant and depends only on the rate constant, k, and not on the concentration of reactant. Decomposition of N 2 O 5 at 67 C. Molecules remaining after 1 half-life; after 2 half-lives If k is known, t ½ can be calculated, and if t ½ is known, k can be calculated. Common applications include half-life of radioactive isotopes used in medicine, nuclear power generation and nuclear waste calculations. Zero-Order Reactions Rate has the same value at all points, and is independent of initial concentration. Second-Order Reactions A second-order reaction has a rate law with a sum of the exponents equal to 2 Rate = k[a][b] m + n = 2 Rate = k[a] 2 m = 2 The integrated rate law which expresses [A] as a function of time has the following form: 1/[A]t = kt + 1/[A] o Second-order half life is t½ = 1/k[A] o Second Order Reaction Plot Plot of 1/[A] vs. time is linear with a positive slope. Example 13.8 A Conceptual Example Shown here are graphs of [A] versus time for two different experiments dealing with the reaction A products. What is the order of this reaction? Collision Theory Before atoms, molecules, or ions can react, they must first come together, or collide. An effective collision between two molecules puts enough energy into key bonds to break them. Activation Energy The activation energy (E a ) is the minimum energy that must be supplied by collisions for a reaction to occur. The spatial orientations of the colliding species also affect the reaction rate. Transition State Theory The configuration of the atoms at the time of the collision is called the transition state. This transitory species is the activated complex.
Effect of Temperature on Rates Arrhenius proposed the following mathematical expression for the effect of temperature on the rate constant, k: k = A exp{ E a /(RT)} or, ln k = Ea/(RT) + ln A Think: y = m x + b where y = ln k and x = Ea/R Plot ln k vs. 1/T then, ln k = (Ea/R)(1/T) + ln A The Arrhenius Equation The constant A, called the frequency factor, is the product of the collision frequency and a probability factor that takes into account the orientation required for effective molecular collisions. exp(-e a /RT) represents the fraction of molecular collisions sufficiently energetic to produce a product molecule Reaction Mechanisms reaction mechanism: a series of simple steps that lead from the initial reactants to the final products in a reaction. An elementary reaction (or process) represents a single step in the progress of the overall reaction. The mechanism must be consistent with the experimentally determined rate law! Elementary Reactions The molecularity of an elementary reaction refers to the number of free reactant atoms, ions, or molecules that enter into the reaction. The rate-determining step is the slowest step in establishing the rate of the overall reaction. Effect of Catalysts on a Reaction A catalyst speeds up reaction rate by reducing the activation energy. Homogeneous Catalysis Reaction profile for the uncatalyzed and catalyzed decomposition of ozone. Heterogeneous Catalysis Many reactions are catalyzed by the surfaces of appropriate solids. Enzyme Catalysis Enzymes are high-molecular-mass proteins that usually catalyze one specific reaction or a set of quite similar reactions - but no others. Summary of Concepts Rates of reactions are based on the rate of disappearance of a reactant or formation of a product An integrated rate law relates concentration and time The half-life of a reaction is the time in which one-half of the reactant initially present is consumed Chemical reactions occur when sufficiently energetic molecules collide in the proper orientation Reactions generally go faster at higher temperatures or in the presence of a catalyst Reaction mechanisms provide a plausible explanation of how a reaction proceeds