1 / 36

Chemical Kinetics

Chemical Kinetics. Kinetics. Kinetics in chemistry is concerned with how quickly a reaction proceeds Factors that affect rate Physical state of the reactants Concentration of the reactants Temperature at which the reaction occurs The presence of a catalyst. Reaction Rates.

colman
Download Presentation

Chemical Kinetics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chemical Kinetics

  2. Kinetics • Kinetics in chemistry is concerned with how quickly a reaction proceeds • Factors that affect rate • Physical state of the reactants • Concentration of the reactants • Temperature at which the reaction occurs • The presence of a catalyst

  3. Reaction Rates • Reaction rates depend on the frequency of collisions between molecules • Reaction rate = speed of a reaction (M/s) • A → B • Δ[B]/Δt = -Δ[A]/Δt • Average Rate

  4. Change of Rates with Time • C4H9Cl(aq) + H2O(l) → C4H9OH(aq) + HCl(aq) • What is happening to the rate as the reaction proceeds? • Graphs of the data allow you to find the instantaneous rate

  5. Reaction Rates and Stoichiometry • Stoichiometry will affect the rates of disappearance and formation • 2HI(g) → H2(g) + I2(g) • -1/2Δ[HI]/Δt = Δ[H2]/Δt = Δ[I2]/Δt • For any general reaction • aA + bB → cC + dD

  6. Practice • How is the rate of disappearance of ozone related to the rate of appearance of oxygen in the following equation: 2O3(g) → 3O2(g)? If the rate of appearance of oxygen is 6.0x10- 5 M/s at a particular instant what is the value of the rate of disappearance of ozone at this same time? • Answer: 4.0x10- 5 M/s

  7. Practice • The decompostion of N2O5 proceeds according to the following equation: 2N2O5(g) → 4NO2(g) + O2(g) If the rate of decomposition of dinitrogen pentoxide at a particular instant in a reaction vessel is 4.2x10- 7 M/s, what is the rate of appearance of NO2 and O2? • Answer: 8.4x10-7 M/s, 2.1x10-7 M/s

  8. Beer's Law • Spectroscopic methods are useful in seeing how concentration changes with time • 2HI(g) → H2(g) + I2(g) • A = abc • A : absorbance • a : molar absorptivity • b : path length • c : concentration

  9. Concentration and Rate • To determine the effect of concentration of rate, you can vary the concentration of reactants and monitor the change in initial rate • NH4+(aq) + NO2-(aq) → N2(g) + 2H2O(l) • What happens to the initial rate when the concentrations are changed?

  10. Rate Law • Rate law shows the rate of a reaction is related to the concentrations of the reactants • Rate = k[NH4+][NO2-] • For a general reaction: aA + bB → cC + dD • Rate = k[A]m[B]n • k = rate constant • The magnitude of k is affected by changes in temp. • If we know the rate law we can calculate k • From exp. 1: r = 5.4x10-7 M/s = k(0.0100M)(0.200M) • k = 2.7x10-4 M-1·s-1

  11. Reaction Orders • Rate = k[A]m[B]n • m and n are reaction orders • Rate = k[NH4+][NO2-] • Each compound is 1s t order but the overall order is 2n d (just add the exponents) • Reaction orders must be determined experimentally • 2N2O5(g) → 4NO2(g) + O2(g) : Rate = k[N2O5] • 2HI(g) → H2(g) + I2(g) : Rate = k[H2][I2]

  12. Units of Rate Constants • Units of the rate constant depend on the overall order of the rate law • What are the overall reaction orders and units of the rate constant for the following reactions? • 2N2O5(g) → 4NO2(g) + O2(g) • Rate = k[N2O5] • 2HI(g) → H2(g) + I2(g) • Rate = k[H2][I2] • CHCl3(g) + Cl2(g) → CCl4(g) + Hcl(g) • Rate = k[CHCl3][Cl]1/2

  13. Using Initial Rates • Observing the effect of changing the initial concentrations of the reactants on the initial rate allows us to determine reaction orders • Exponents will commonly be 0, 1, 2 • What effect will a reactant with a reaction order of 0 have on the reaction? 1? 2? • Remember only the rate depends on concentration

  14. Practice • The initial rate of a reaction A + B → C was measured for several different starting concentrations of A and B. Using this data determine the rate law for the reaction; the magnitude of the rate constant; and the rate when [A] = 0.050M and [B] = 0.100M • Answer: r = k [A]2, 4.0x10- 3M- 1s- 1, r = 1.0x10- 5M/s

  15. Practice • The following data were measured for the reaction of nitric oxide with hydrogen: 2NO(g) + 2H2(g) → N2(g) + 2H2O(g) Determine the rate law for this reaction, the value of the rate constant and the rate when [NO] = 0.050M and [H2] = 0.150M • Answer: r = k[NO]2[H2], k = 1.2 M- 2s- 1, r = 4.5x10- 4 M/s

  16. Changing Concentration with Time • Rate laws tell us how the rate of a reaction changes at a given temperature as concentration changes. • Rate laws can be converted to tell us what the concentration of a substance is at any time during a reaction. • There are two special cases and you must be able to use them.

  17. First Order Reactions • For a first order reaction that proceeds A → products the rate law is • Rate = -Δ[A]/Δt = k[A] • After some math magic (involving integration) • ln[A]t – ln[A]0 = -kt • With some rearrangement we get something similar to y = mx + b • ln[A]t = -kt + ln[A]0 • What would you graph to see if the reaction was first order?

  18. Practice • The first order rate constant for the decomposition of a certain insecticide in water at 12°C is 1.45 yr - 1. A quantity of the insecticide is washed into a lake on June 1, leading to a concentration of 5.0x10- 7g/mL. What is the concentration of insecticide after 1 year? How long will it take for the concentration to drop to 3.0x10- 7g/mL? • Answer: 1.2x10- 7g/mL, 0.35 years

  19. Practice • The decomposition of dimethyl ether, (CH3)2O, at 510°C is a first order process with a rate constant of 6.8x10- 4 s- 1: (CH3)2O(g) → CH4(g) + H2(g) + CO(g) If the initial pressure of dimethyl ether is 135 torr, what is the partial pressure after 1420s? • Answer: 51 torr

  20. Second Order Reactions • For a reaction that proceeds A → products or A+B → products that are second order in just one reactant A: • Rate = - Δ[A]/Δt = k[A]2 • After some calculus magic this becomes: • 1/[A]t = kt + 1/[A]0 • If plotting 1/[A]t creates a straight line the reaction is second order

  21. Practice • The following data were obtained for the gas phase decomposition of nitrogen dioxide at 300°C, NO2(g) → NO(g) + ½ O2(g): Is the reaction first or second order in NO2? What is k? If the initial concentration of NO2 is 0.0500M, what is the concentration after 0.500hr? • Answer: 2nd order r = k[NO2]2, k = 0.543 M- 1s- 1, 1.00x10- 3M

  22. Half Life • Half life (t½) is the time it takes for the concentration of a reactant to drop to one half of its initial value • Using algebra you can find the half life of a 1st order reaction • t½ = 0.693/k • Half life for a second order reaction depends on concetration • t½ = 1/k[A]0

  23. Temperature and Rate • Rate of most chemical reactions increase as temperature increases • Increasing temperature increases the rate constant and thus the rate • Glow stick fun • What effect did the ice and hot water have on the reaction?

  24. Collision Model • Molecule must collide with enough energy to react • What does increasing temperature do? • Collision Theory • Orientation Factor • Activation Energy • Ea

  25. Activation Energy • Higher activation energy the lower the rate • Only a fraction of molecules have energy to generate products upon collision • f = e-Ea/RT

  26. Arrhenius Equation • Rates depend on • Fraction of molecules with an energy of Ea or greater • Collisions per second • Fraction of collision with proper orientation • Arrhenius used these ideas to relate k and Ea • k = Ae- E a / R T • Taking the natural log of both sides • ln k = (-Ea/R)T + ln A • So we can graph ln k versus 1/T to find Ea • We can relate the rate constants of a reaction at different temperatures • ln(k1/k2) = Ea/R (1/T2 - 1/T1)

  27. Practice • The following table shows the rate constants for the rearrangement of methyl isonitrile at various temperatures. From these data, calculate the activation energy for the reaction. What is the value of the rate constant at 430.0K? • Answer: Ea = 160 kJ/mol, k = 1.0x10- 6s - 1

  28. Reaction Mechanisms • Reaction mechanisms show how a reaction occurs • Elementary Steps • NO(g) + O3(g) → NO2(g) + O2(g) • Molecularity • Unimolecular • Bimolecular • Termolecular

  29. Multistep Mechanisms • Many reactions do not happen in just one step • NO2(g) + CO(g) → NO(g) + CO2(g) • NO2(g) + NO2(g) → NO3(g) + NO(g) • NO3(g) + CO(g) → NO2(g) + CO2(g) • Elementary steps must add up to give the overall reaction • Intermediate

  30. Practice • It has been proposed that the conversion of ozone into O2 proceeds via two elementary steps: • O3(g) → O2(g) + O(g) • O3(g) → O(g) + 2O2(g) Describe the molecularity if each step in this mechanism. Write the equation for the overall reaction. Identify any intermediates. • For the reaction Mo(CO)6 + P(CH3)3 → Mo(CO)5P(CH3)3 + CO The proposed mechanism is Mo(CO)6 → Mo(CO)5 + CO Mo(CO)5 + P(CH3)3 → Mo(CO)5P(CH3)3 Is the proposed mechanism consistent with the equation for the overall reaction? Identify any intermediates.

  31. Rate Laws for Elementary Steps • Rate laws cannot normally be predicted from the coefficients of balanced equations. Why? • For elementary steps the equation tells you the rate law. • Rate law is determined by its molecularity

  32. Practice • If the following reaction occurs in a single elementary step, predict the rate law: H2(g) + Br2(g) → 2HBr(g) • Consider the following reaction: 2NO(g) + Br2(g) → 2NOBr(g). Write the rate law for the reaction assuming it involves a single elementary step. Is a single elementary step likely for this reactipon?

  33. Rate Laws for Multistep Mechanisms • In multistep systems the rate laws is set by the slowest step • Rate determining step • Mechanisms with slow first step Step 1: NO2(g) + NO2(g) → NO3(g) + NO(g) (slow) Step 2: NO3(g) + CO(g) → NO2(g) + CO2(g) (fast) Overall: NO2(g) + CO(g) → NO(g) + CO2(g) • Rate = k1[NO2]2

  34. Mechanisms with Initial Fast Step • You need to derive the rate law for a mechanism in which there is an intermediate. 2NO(g) + Br2(g) → 2NOBr(g) Rate = k[NO]2[Br2] • Possible Mechanism • NO(g) + Br2(g) ↔ NOBr2(g) (fast) • NOBr2(g) + NO(g) → 2NOBr(g) (slow) • Algebra fun • When a fast step precedes a slow one we can solve for the concentration of an intermediate by assuming that an equilibrium is established in the fast step.

  35. Practice • Show that the following mechanism for the reaction producing NOBr also produces a rate law consistent with the experimentally observed one: Step 1: NO(g) + NO(g) ↔ N2O2(g) (fast) Step 2: N2O2(g) + Br2(g) → 2NOBr(g) (slow) • The first step of a mechanism involving the reaction of bromine is: Br2(g) ↔ 2Br(g) (fast). What is the expression relating the concentration of Br(g) to that of Br2(g)

  36. Catalysis • What does a catalyst do? • Types of catalysts • Homogeneous – same phase as the reactants • Heterogeneous – different phase from the reactants • Adsorption happens first • Enzymes

More Related