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Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions

Concentration-Time Equation. Order. Rate Law. Half-Life. 1. 1. =. + kt. [A]. [A] 0. =. [A] 0. t ½ =. t ½. t ½ =. ln2. 2 k. k. 1. k [A] 0. Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions. [A] = [A] 0 - kt. rate = k. 0.

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Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions

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  1. Concentration-Time Equation Order Rate Law Half-Life 1 1 = + kt [A] [A]0 = [A]0 t½ = t½ t½ = ln2 2k k 1 k[A]0 Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions [A] = [A]0 - kt rate = k 0 ln[A] = ln[A]0 - kt 1 rate = k [A] 2 rate = k [A]2 13.3

  2. The decomposition of a certain antiobiotic in water has a rate constant at 20°C of 1.29 yr-1. A. Determine the molarity of an antiobiotic solution that sits for 1.00 month, if its original concentration was 5.00 x 10-3M? What is the order of the reaction? Units on rate constant tell us it is first order ln[A] = ln[A]0 - kt [A]0 = 5.00 x 10-3M t = (1.00 month)(1 yr/12 months) = 0.0833 yr k = 1.29 yr-1 ln[A] = ln (5.00 x 10-3M) - (1.29 yr-1)(0.0833 yr) ln[A] = -5.397 take antilog [A] = 4.53 x 10-3M 13.3

  3. The decomposition of a certain antiobiotic in water has a rate constant at 20°C of 1.29 yr-1. B. If the antiobiotic solution is considered to be “no longer effective” at concentrations lower than 5.00 x 10-4M, how long will the solution be effective? first order reaction ln [A]0 = kt [A] t [A]0 = 5.00 x 10-3M [A]t = 5.00 x 10-4M k = 1.29 yr-1 ln (5.00 x 10-3M/ 5.00 x 10-4M) = (1.29 yr-1) t t = ln(10)/ (1.29 yr-1 ) = 1.78 yr 13.3

  4. The decomposition of a certain antiobiotic in water has a rate constant at 20°C of 1.29 yr-1. C. A 70. % depletion of the initial concentration of the antiobiotic solution results in a significantly lower effectivness. How long before this lower effectiveness will be noticed? first order reaction ln [A]0 = kt [A] t [A]t = 0.30 [A]0 k = 1.29 yr-1 ln [A]0 = (1.29 yr-1) t 0.30[A]0 t = ln(3.3)/ (1.29 yr-1 ) = 0.933 yr 13.3

  5. The decomposition of a certain antiobiotic in water has a rate constant at 20°C of 1.29 yr-1. D. Determine the half-life of the antiobiotic. first order reaction k = 1.29 yr-1 t 1/2 = ln 2 / k t 1/2 = ln(2)/ (1.29 yr-1 ) = 0.537 yr 13.3

  6. A second order decomposition reaction takes 2.54 x 103 seconds for the initial concenteration to fall to one half of its origianal value. What is the value of the rate constant if the initial concentration was 6.23 x 10-2M?. Second order reaction t 1/2 = 2.54 x 103 seconds [A]0 = 6.23 x 10-2M t 1/2 = 1 / k [A]0 k = 1 / [A]0 t 1/2 k = 1 / (6.23 x 10-2M)(2.54 x 103 seconds) = 6.32 x 10-3 /Ms 13.3

  7. A + B C + D Endothermic Reaction Exothermic Reaction The activation energy (Ea) is the minimum amount of energy required to initiate a chemical reaction. 13.4

  8. lnk = - + lnA Ea 1 T R Temperature Dependence of the Rate Constant k = A •exp( -Ea/RT ) (Arrhenius equation) Eais the activation energy (J/mol) R is the gas constant (8.314 J/K•mol) T is the absolute temperature A is the frequency factor 13.4

  9. lnk = - + lnA Ea 1 T R 13.4

  10. 2NO (g) + O2 (g) 2NO2 (g) Elementary step: NO + NO N2O2 + Elementary step: N2O2 + O2 2NO2 Overall reaction: 2NO + O2 2NO2 Reaction Mechanisms The overall progress of a chemical reaction can be represented at the molecular level by a series of simple elementary steps or elementary reactions. The sequence of elementary steps that leads to product formation is the reaction mechanism. N2O2 is detected during the reaction! 13.5

  11. Elementary step: NO + NO N2O2 + Elementary step: N2O2 + O2 2NO2 Overall reaction: 2NO + O2 2NO2 Intermediates are species that appear in a reaction mechanism but not in the overall balanced equation. An intermediate is always formed in an early elementary step and consumed in a later elementary step. • The molecularity of a reaction is the number of molecules reacting in an elementary step. • Unimolecular reaction – elementary step with 1 molecule • Bimolecular reaction – elementary step with 2 molecules • Termolecular reaction – elementary step with 3 molecules 13.5

  12. Unimolecular reaction Bimolecular reaction Bimolecular reaction A + B products A + A products A products The rate-determining step is the sloweststep in the sequence of steps leading to product formation. Rate Laws and Elementary Steps rate = k [A] rate = k [A][B] rate = k [A]2 • Writing plausible reaction mechanisms: • The sum of the elementary steps must give the overall balanced equation for the reaction. • The rate-determining step should predict the same rate law that is determined experimentally. 13.5

  13. Step 1: Step 2: NO2 + NO2 NO + NO3 The experimental rate law for the reaction between NO2 and CO to produce NO and CO2 is rate = k[NO2]2. The reaction is believed to occur via two steps: NO2+ CO NO + CO2 NO3 + CO NO2 + CO2 What is the equation for the overall reaction? What is the intermediate? NO3 What can you say about the relative rates of steps 1 and 2? rate = k[NO2]2 is the rate law for step 1 so step 1 must be slower than step 2 13.5

  14. But what happens if the first step is NOT the rate determining step? 1: NO + O2  NO 3 fast 2: NO3 NO + O2 fast 3: NO3 +NO 2 NO2 slow rate = k3 [NO3][NO], but NO3 is an intermediate Since the first two steps are fast, the rate for creating NO3 = rate for destroying NO3 k1 [NO] [O2] = k2 [NO3] + k3 [NO3][NO] negligibly small rate => very small number solve for [NO3]: [NO3] = k1 [NO] [O2] k2 substitute into rate expression: rate = k1k3[NO]2 [O2] k2 => observed rate = kobs [NO]2 [O2]

  15. Ea Potential Energy A + B C + D Reaction Progress What does the potential energy diagram look like for a reaction with multiple steps ? Ea Potential Energy A + B C + D Reaction Progress Reaction with first step as slow step Reaction with last step as slow step Each hill represents a reaction Any dip or valley represents an intermediate

  16. Consider the following mechanism for a gas phase reaction: 1: Cl2 Cl .fast 2: Cl . + CHCl3 HCl + .CCl3 slow 3: Cl .+ .CCl3  CCl4 fast A. What is the overall reaction? Cl2 + CHCl3 HCl + CCl4 B. What are the intermediates in the reaction? Cl . & .CCl3 C. What is the molecularity of each step in the mechanism? Step 1 = unimolecular; Steps 2 & 3 = bimolecular D. What is the rate determining step? Step 2 = slow step = rate determing step

  17. E. What is the rate law predicted by this mechanism? Step 2 = slow step = rate determing step rate = k2 [Cl .][CHCl3 ] But Cl . is an intermediate formed in Step 1 and , thus, cannot appear in the rate law. Cl . is in equilibrium with Cl2 By definition, this means that the forward and reverse rates of this reaction are equal. k1 [Cl2 ] = k-1[Cl .]2 Solving for [Cl .] in terms of[Cl2], [Cl .]2 =k1/ k-1 [Cl2 ] => [Cl .]= {k1/ k-1 [Cl2 ]} 1/2 Substituting into the overall rate law: rate = k2{k1/ k-1 [Cl2 ]} 1/2 [CHCl3 ] rate = kobs [Cl2 ]1/2 [CHCl3 ] overall reaction order is 3/2

  18. Chain Reactions: a series of reactions in which one reaction initiates the next Initiation: I2  2 I . slow - produces reactive intermediate Propagation: I . + Br2 IBr +Br . fast Br . + I2 IBr +I . fast Termination: Br . +Br .  Br2 rare reactions since I . +I .  I2 concentrations of species Br . +I . IBr2 are small

  19. uncatalyzed catalyzed Ea k ‘ Ea< Ea A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed. k = A •exp( -Ea/RT ) ratecatalyzed > rateuncatalyzed 13.6

  20. In heterogeneous catalysis, the reactants and the catalysts are in different phases. • Haber synthesis of ammonia • Ostwald process for the production of nitric acid • Catalytic converters In homogeneous catalysis, the reactants and the catalysts are dispersed in a single phase, usually liquid. • Acid catalyses • Base catalyses A catalyst can change the mechanism of a reaction, but it will not cause a reaction to proceed that is not favorable. 13.6

  21. Fe/Al2O3/K2O N2 (g) + 3H2 (g) 2NH3 (g) catalyst Haber Process 13.6

  22. 4NH3(g) + 5O2(g) 4NO (g) + 6H2O (g) 2NO (g) + O2(g) 2NO2(g) 2NO2(g) + H2O (l) HNO2(aq) + HNO3(aq) Pt-Rh catalysts used in Ostwald process Hot Pt wire over NH3 solution Ostwald Process Pt catalyst 13.6

  23. catalytic CO + Unburned Hydrocarbons + O2 CO2 + H2O converter catalytic NO + NO2 N2 + O2 converter Catalytic Converters 13.6

  24. Enzyme Catalysis 13.6

  25. enzyme catalyzed uncatalyzed 13.6

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