CHAPTER 12 CHEMICAL KINETICS

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1 5/9/202 CHAPTER 2 CHEMICAL KINETICS CHM52 GCC Kinetics Some chemical reactions occur almost instantaneously, while others are very slow. Chemical Kinetics - study of factors that affect how fast a reaction occurs and the step-bystep processes involved in chemical reactions. Factors That Affect Reaction Rate A. Concentration - higher concentration of reactants increases rate B. Temperature - higher T increases rate C. Catalysts - accelerate reaction rate D. Surface area of solid - smaller particles increase rate Rate of Reaction - the change in the amount of a reactant or product per unit time - Rate = concentration [ A ] time time [ ] means concentration Typical Rate Units: M/time or mol/(ltime) Recall: M = moles solute/l of solution Rates for products or reactants For the reaction A B Rate of formation of products B B B 2 Rate t time2 time A A A 2 or Rate t time2time Rate of disappearance of reactants Rate Characteristics Rate is a measure of the speed of a reaction. Rate can be expressed as rate of formation of products or rate of disappearance of reactants. Reaction rate is always positive: (-) sign used for reactants since they decrease with time (this gives rate a + sign.) Reaction rates decrease with time as reactants are used up.

2 5/9/202 Reaction Rates 2N 2 O 5 (g) 4NO 2 (g) + O 2 (g) Colorless brown colorless The reaction rate can be defined in terms of the increase of [NO 2 ] or [O 2 ] or the decrease of [N 2 O 5 ]. Experimental data for this reaction is given in table 2. and figure 2. Figure 2. Different ways to measure rate Initial rate is measured by the slope of the tangent line when initial reactant concentrations are measured (find the slope of the tangent line at time (t) = 0). Instantaneous rate is measured at a specific point in time (find the slope of a tangent line at a specified time). Average rate is measured as the average between two times; similar to calculating slope: [(y 2 - y ) / (x 2 - x )]. Figure 2.2 Different Rates Initial at t = 0 s Instantaneous at t = 350 s Average triangles tri Example Average Rate Calculation Use the data in Table 2. to calculate the average rate of NO 2 formation for t = 300 s to t = 400 s. 2

3 5/9/202 Rate Expressions E.g. For 3H 2 + N 2 2NH 3, Reaction: 2N 2 O 5 4NO 2 + O 2 The rate of disappearance of N 2 O 5 is twice the rate of formation of O 2 To make rates equal, divide rates by their stoichiometric coefficients: Write the rate expressions in terms of the disappearance of the reactants and the appearance of the products: Rate 2 N O NO O t 4 t t Example. For the reaction, 2N 2 O 5 4NO 2 + O 2, if the rate of decomposition of N 2 O 5 is 4.2x0-7 mol/(ls), what is the rate of appearance of (a) NO 2 ; (b) O 2? RATE LAW Rate law: relationship between the reaction rate and the concentration of each reactant. For a reaction aa + bb Products Rate = k[a] m [B] n k = Rate Constant: it is a numerical constant for a reaction at a given temperature. [A], [B] are concentrations of reactants Rate Law continued Rate = k[a] m [B] n m & n = order of reactants = exponent of reactant in rate law Overall Reaction Order = sum of all exponents in rate law Exponents cannot be obtained by looking at the equation; they are experimentally determined values. exponents are usually 0, or 2 Reaction Order Example E.g. 2ClO 2 + F 2 2FClO 2 Rate = k[clo 2 ][F 2 ] order of ClO 2 = ; order of F 2 = Overall order = 3

4 5/9/202 Determination of Rate Law from Initial Rates Data To determine the rate law, we observe the effect of changing initial concentrations of reactants on the initial rate of reaction. Exp. Data: Initial rates ([Products]/t after - 2% of limiting reactant has been consumed) are usually given; there is less chance of error from competing side reactions & reversible reactions. Inspection Method A reactant is st order if doubling [A] causes rate to double; rate is directly proportional to [A]. A reactant is 2 nd Order if doubling [A] causes rate to quadruple (or tripling [A] causes 3 2 increase in rate). A reactant is Zero order if changing [A] does not affect the rate. Reaction Orders may also be found mathematically To find the order for a specific reactant, examine what happens to the rate as the concentration of only that reactant changes (note: the rate constant k and other concentration terms cancel out, since they don t change!): concentration 2 concentration m Rate2 Rate Repeat this process for each reactant. e.g. Finding Rate Law using Initial Rates Experiment [A] M [B] M Rate M/s Order A =? Order B =? Rate =? k =? Find the orders for H 2 O 2 and I -, then write the rate law and find k for H 2 O 2 + 3I - + 2H + I H 2 O (note: the order for H + is zero so its not part of the rate law). [H 2 O 2 ] M [I - ] M Rate (M/s) x x x x Integrated Rate Laws st order For a reaction, A Products [A] Rate k[ A] t Integrate to get st order IRL (calculus ) [A] st t order IRL: ln kt [ A] 0 rearrange ln[a] ln[ o A ] t kt 4

5 5/9/202 IRL variables t = time k = rate constant [A] 0 = initial concentration [A] t = concentration at time t Units for A can be g, moles, M, torr, etc A t is always less than A o st order problems. The rate law for the decomposition of N 2 O 5 is Rate = k[n 2 O 5 ], where k = 5.0 x 0-4 s -. What is the concentration of N 2 O 5 after 900 s, if the initial concentration is 0.56 M? 2. The first order reaction, SO 2 Cl 2 SO 2 + Cl 2, has a rate constant of 0.7 h -. If the initial concentration of SO 2 Cl 2 is.25 x 0-3 M, how many seconds does it take for the concentration to drop to 0.3 x 0-3 M? st Order IRL linear format Can modify to get linear form: (refer to Appendix A, p. A5-6 for log functions) Reaction: A products st order reaction [A] vs t is not linear ln [A] vs t is linear ln [A] t = -kt + ln [A] o y = ln [A] t m = -k ; thus k = - slope x = t b = ln [A] o = y intercept Only st order reactions will give you a straight line when plotting ln [A] t vs. time! Half Life of First Order Reactions Half-life (t /2 ): time it takes for half (50%) of a reactant to be consumed. 50% of the reactant also remains unreacted. Half-Life Concept How much of a sample remains after 3 halflives? 5

6 5/9/202 Derivation of st order Half life At t = t /2 Plug into IRL: [A] ln 2 [A] 2 A t A kt ln 0.5 = -kt /2 kt /2 = /2 t /2 does not depend on [A], only for st order reactions! st order half life problems. Cobalt-60 is a radioisotope that decays by first-order kinetics and has a half-life of 5.26 years. The Co-60 in a radiotherapy unit must be replaced when the concentration of Co decreases to 75% of its initial value. When does this occur? 2. The first order reaction, CH 3 NC CH 3 CN, has a rate constant of 6.3x0-4 s - at 230 C. a) What is the half-life of the reaction? b) How much of a 0.0 g sample of CH 3 NC will remain after 5 half-lives? c) How many seconds would be required for 75% of a CH 3 NC sample to decompose? Comparison Table for Typical Reaction Orders Order Rate Law Integrated Law Linear graph slope Half-life 0 Rate = k [A]t = -kt +[A]0 [A] vs. t -k Rate = k[a] 2 Rate = k[a] 2 [A] t ln = - kt [A] 0 = kt + [A] t [A] 0 ln [A] vs. t -k [A] t vs. t +k [A] t/2 = 0 2k t/2 = k t/2 = k[a] 0 Second Order Plot slope = +k [A] 0 [A] Rate k[ A] t 2 nd order IRL: time 2 kt [A] t [A] 0 Only 2 nd order reactions will give you a straight line when plotting [A] t vs. time. Zero Order Plot Δ[A] Rate k[a] Δt 0 0 order IRL: [A] t = -kt + [A] 0 k Only 0 order reactions will give you a straight line when plotting [A] vs. time. Graphical Method of Determining Rate Law ) Make 3 plots: [A] vs time; ln [A] vs. time; and /[A] vs. time. 2) The most linear plot gives the correct order for A; the other 2 graphs should be curves. 6

7 5/9/202 What order is this reaction? Reaction Mechanisms Many chemical reactions occur by a sequence of 2 or more steps. Each individual event in the overall reaction is called an elementary step. Molecularity: the number of molecules that react in an elementary step. Molecularity Unimolecular: molecule A products Bimolecular: 2 molecules 2A products or A + B products Termolecular (uncommon): 3A products or 2A + B Products or A + B + C Products Mechanism example E.g. A 2 step Overall rxn: Br 2 + 2NO 2BrNO Step : Br 2 + NO Br 2 NO (Bimolecular step) Step 2: Br 2 NO + NO 2BrNO (Bimolecular step) Intermediates Intermediates are short lived species that are formed during the reaction, then are subsequently consumed Its st a product, then a reactant Br 2 NO in last example is intermediate Rate Law for Elementary Step For an elementary step, the rate law can be written using stoichiometric coefficients of the reactants. (molecularity = order). E.g. Step : Rate = k[br 2 ][NO] 7

8 5/9/202 Rate Determining Step The slowest step in the reaction is the rate determining step; this step limits how fast products can form. Analogy: freeway during rush hour The rate law for the overall rxn is determined by the rate of this slow step. 2-Step Reaction Mechanism Step : NO 2 (g) + NO 2 (g) NO(g) + NO 3 (g) Step 2: NO 3 (g) + CO(g) NO 2 (g) + CO 2 (g) 2-Step Reaction Mechanism Step : NO 2 (g) + NO 2 (g) NO(g) + NO 3 (g) Step 2: NO 3 (g) + CO(g) NO 2 (g) + CO 2 (g) What is the overall equation? (Hint: Think Hess s Law - adding equations) 2 Step Reaction Mechanisms Rate laws for these elementary steps: Step : NO 2 (g) + NO 2 (g) NO(g) + NO 3 (g) Step : rate = k[no 2 ][NO 2 ] (bimolecular) Step 2: NO 3 (g) + CO(g) NO 2 (g) + CO 2 (g) Step 2: rate = k[no 3 ][CO] (bimolecular) Intermediates: product in one step and consumed in a later step Catalyst: reactant in one step and product in later step Note: Intermediates and catalyst cannot be part of overall reaction! Rate Laws/Reaction Mechanisms Step (slow):no 2 (g)+no 2 (g) NO(g) + NO 3 (g) Step 2 (fast):no 3 (g)+co(g) NO 2 (g) +CO 2 (g) a) What is the rate law for this 2-step reaction? b) What is the intermediate? Catalyst? Collision Theory Collision Frequency affects Reaction Rate concentrations of reactants result in more collisions, thus the reaction rate. Temperatures cause molecules to move faster and collide more often, increasing the rate. 8

9 5/9/202 Most collisions don t cause a reaction because ) molecules must have enough Kinetic Energy to: overcome electron cloud repulsions between atoms/molecules weaken/break reactant bonds Kinetic Energy Distribution 2) molecules must have the proper orientation to have an effective collision Activation Energy, E a energy barrier that molecules have to surmount in order to react. Energy is needed to break reactant bonds (endothermic process). analogies: pushing a boulder up a hill getting started on an unpleasant task Comments on E a Only a small fraction of molecules have enough KE to initiate reaction. E a is different for each reaction. Reactions with low E a are faster & k is larger because more molecules can overcome E a. At higher T, a larger fraction of molecules have enough KE to overcome E a. (This is the main reason why the rxn rate increases dramatically as T ) Molecules must be properly oriented to react. Reaction: A + BC AB + C P.E. Curve for Exothermic Reaction (-E) Effective collision Ineffective collision 9

10 5/9/202 Potential Energy curves transition state a highly unstable species formed by the collision of the reactant molecules; arrangement of atoms at the top of the energy barrier. E = Heat of reaction E = E(products) E(reactants) E a shown for curve is the activation energy for the forward reaction, E a (forward). This is the difference in energy between the transition state and the reactants. P.E. Curve for Endothermic Reaction (+E) Transition state O=N-- Cl -- Cl P.E. O=N- Cl + Cl E a E O=N + Cl -Cl Reaction progress Arrhenius Equation gives the relationship between k and T k = Ae -E a /RT Arrhenius plot E a can be found graphically: R = gas constant = 8.34 J/K-mol T = temperature in K e -E a /RT = fraction of molecules that have enough KE to react A = frequency factor (depends on # of collisions that are properly oriented) Linear form: ln k = -E a /RT + ln A plot of ln k (y axis) vs /T (x axis) yields a straight line. slope= -E a /R; y intercept = ln A Thus E a = -R slope 2.2 Arrhenius If you have 2 sets of conditions, solve for k, k 2, T, T 2 or E a using: k ln k 2 E a R T2 T 2.2 Arrhenius Example (p 470): 2HI (g) H 2(g) + I 2(g) Temperature ( C) a) Find E a using all five data points k(m - s - ) b) Given the rate constant at 283 C and E a = 87 kj/mol, what is the rate constant at 293 C? 0

11 ln k 5/9/202 Catalysts 0 /T (/K) y = x E a = -slope R J = (22400 K)x(8.34 K mol ) =.86x0 5 J or 86 kj A Catalyst is a substance that increases the rate of a reaction, but is not itself used up. A catalysts works by lowering E a - usually they help weaken or break reactant bonds. A catalyst alters the reaction mechanism, but does not change the overall reaction. Catalyst may show up in experimental rate law a reaction may have more than one rate law. Enzymes are large protein molecules with one or more active sites that serve as biological catalysts in living organisms Homogeneous Catalyst A homogeneous catalyst is one that is in the same phase as the reactants. Example: 2H 2 O 2 (aq) 2H 2 O(l) + O 2 (g) Effect of Catalyst on E A Step Step 2 Step H 2 O 2 (aq) + I - (aq) H 2 O(l) + IO - (aq) Slow Step 2 H 2 O 2 (aq) + IO - (aq) O 2 (g) + H 2 O(l) + I - (aq) Fast Net rxn: 2H 2 O 2 (aq) 2H 2 O(l) + O 2 (g) I - (aq) is the catalyst Note: E a is lower but E is same. Heterogenous Catalyst A heterogenous catalyst is in a different phase than the reactants: C 2 H 4 (g) + H 2 (g) catalyst C 2 H 6 (g) The catalyst is usually an inactive metal (Ni, Pd, or Pt). Figure 2.9 Catalytic Converters Convert pollutants (hydrocarbons, carbon monoxide, and nitric oxide) into CO 2, H 2 O, N 2, and O 2 using a heterogeneous catalyst (Pt, Pd, Rh).