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Chemical Kinetics: Rate Laws ORDER OF REACTION

Chemical Kinetics: Rate Laws ORDER OF REACTION. rate (=  d [A ] / d t ) = k[A] x [B] y Overall order of reaction = x + y Example: rate = k[A] 2 [B] The reaction is second order in A first order in B overall reaction order = 1+ 2= 3 (sum of the exponents).

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Chemical Kinetics: Rate Laws ORDER OF REACTION

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  1. Chemical Kinetics: Rate LawsORDER OF REACTION • rate (=d[A] /dt) = k[A]x[B]y • Overall order of reaction = x + y • Example: rate = k[A]2[B] • The reaction is • second order in A • first order in B • overall reaction order = 1+ 2= 3 • (sum of the exponents)

  2. DETERMINING THE RATE LAW Method of initial rates: Initially, we know [A] and [B] (and [C] = [D] = 0) initial rate = k1[A]ox[B]oy “o”  “initial” (t = 0) Vary [A]0 and [B]0, measure initial rates

  3. Sample problem Given the following data for the reaction A + B  Z What is the overall reaction order? [A]0[B]0initial rate 1) 1.0M 1.0M 0.8x102 M/s • 2.0 1.0 1.6x102 • 2.0 2.0 6.4x102 • 1.0 2.0 3.2x102 Rate = k [A]a [B]b a = overall order = a + b b= • 0 • 1 • 2 • 3 • 4

  4. 2N2O5 4NO2 +O2 [N2O5]i Initial rate 0.01 M 0.018 0.02 M 0.036 0.04 M 0.072 What is the rate law? Rate = k [N2O5]x x = ???

  5. INTEGRATED RATE LAW So far we have used differential rate laws e.g., for aA  products rate = -[A] = k[A]x (a differential eqn) t Integration gives [A] as a function of t (more useful) 1st order reaction (x = 1): Integrate to get: [A] = [A]0 e-kt or ln[A] = ln[A]0 k t (2 forms of same equation)

  6. GRAPHICAL ANALYSIS • 1st order: • ln [A] =ln [A]0 k t • y = b + mx • Plot of ln [A] vs. t gives a straight line • slope = k • intercept = ln [A]0

  7. CH3NC: CH3 C N: (1st Order Reaction) [A] = [A]0 e-kt ln[A] = ln[A]0 kt

  8. INTEGRATED RATE LAW 2nd order reaction 2nd order reaction (x = 2): differential rate law Integrate to get: y = m x + b

  9. Reaction: 2NO2(g)  2NO + O2 Rate = k[NO2]2 Rate Law:

  10. Half lives • Half life: t1/2 • time it takes for the concentration of a reactant to drop to half of its initial value. • e.g. A  products • t1/2 is where [A] = 1/2[A]0 • For 1st order reactions: • t1/2doesn’t depend on concentration • E.g. nuclear decays: 14C 14N + e • t1/2 = 5730 years • (carbon dating)

  11. Half Life Problem He nucleus 239Pu 235U +   Plutonium – very toxic (lethal dose ≈ 5x10-5g) If we bury 1 lb. spent nuclear fuel containing 1g 239Pu, how long until it’s safe to dig it up? “safe”   5x10-5g left t1/2 = 24,400 yr

  12. Temperature dependence of reaction rates As T increases, reaction rates increase Why? Look at energy profile for a typical reaction

  13. Fraction of molecules with enough energy e-Ea/RT

  14. Arrhenius Equation or k = rate constant A = frequency factor Related to collision frequency and orientation Ea = Activation energy R = gas constant (usually 8.314 J/mol-K) T = temperature in K

  15. ARRHENIUS PLOT • ln k = ln A – Ea/RT • plot of ln k vs 1/T is a straight line • slope = -Ea/R • intercept = ln A • Arrhenius plot • The bigger Ea is, the more the rate varies with T

  16. RATE VS TEMPERATURE Most reactions have Ea = 20 – 200 kJ/mol A “typical” Ea might be 50 kJ/mol How does rate vary over a 10° temperature range? (e.g. from 300 to 310 K) • k2/k1 = e0.65 = 1.9 • (about twice as fast at 310 K compared to 300 K) • rule of thumb: reaction rates double for every 10° rise in temperature • (assumes Ea 50 kJ/mol)

  17. Rxn: CH3NC  CH3CN Using the plot, find Ea for the reaction.

  18. SAMPLE PROBLEM For the same reaction, CH3NC  CH3CN if k = 2.52x10-5 s-1 at 189.7C, what is the rate constant at 430K?

  19. REACTION MECHANISMS • Reaction Mechanism: • process by which a reaction occurs • For elementary reaction steps • Reaction proceeds as written • NO + O3 NO2 +O2 • Rate = k [NO][O3] • Most reactions do not occur as a single elementary step. They occur as the result of several elementary steps.

  20. Example:NO2 +CO  NO + CO2 Probable mechanism: NO2 + NO2  NO3 + NO 1 NO3 + CO  NO2 + CO2 2 NO2 + CO  NO + CO2 Rate Law for multi-step mechanism: If step 1 is slow or If step 2 is slow Compare with experiment to determine which is correct.

  21. Example:NO2 +CO  NO + CO2 According to experiment: Rate  [NO2]2 Consistent with step 1 as the RATE DETERMINING (or slowest) step Rate cannot proceed any faster then the slowest step Rate determining step = slow step

  22. NO2 +CO  NO + CO2 • Elementary steps in a mechanism must add up to give the balanced overall reaction. • NO3 is produced in step 1 and consumed in step 2 Intermediate: a stable molecule Note: it is NOT the same as the activated complex. • Intermediates do not (should not) appear in the rate law

  23. To find mechanisms • Find the experimental rate law • Postulate elementary steps • Find the rate law predicted by the mechanism and compare to experiment. No rate can be written in terms of intermediates

  24. Example Cl2 + CHCl3 HCl + CCl4 Observed rate = kobs [Cl2]1/2 [CHCl3] Postulate the following mechanism - is it consistent with the experimental rate law?? Cl2 2Cl fast Cl + CHCl3 HCl + CCl3 slow Cl + CCl3 CCl4 fast

  25. REACTION MECHANISMS • Another example: • 2NO(g) + Br2(g)  2ONBr(g) observed rate = k[NO]2[Br2] • Does this mean mechanism is: • No! 3 body collisions are very rare (unlikely) compared to 2-body (bimolecular) • In general: All elementary reaction steps • involve only unimolecular or bimolecular processes • (exception – solute reactions with solventmolecules)

  26. Explaining the rate law for 2NO(g) + Br2(g)  2ONBr(g) k1 NO + Br2 ONBr2 (fast) ONBr2 + NO 2 ONBr (slow) 2 NO(g) + Br2(g) 2 ONBr(g) (overall) k-1 k2 • step (2) is slow (rate determining) •  rate = k2[ONBr2][NO] •  • intermediate • need to express [ONBr2] in terms of [reactants]

  27. step (1): fast equilibrium  forward rate = back rate k1[NO][Br2] = k-1[ONBr2] [ONBr2] = k1/k-1 [NO][Br2] Now plug this into the rate law: Conclusion: more than one mechanism may fit the rate law

  28. Catalysis Catalyst:substance that speeds up a reaction without undergoing permanent change. How:changes the mechanism lowers the activation energy Example: 2H2O2 2H2O + O2 Catalyst: Br2, MnO2, or catalase (enzyme) Homogeneous catalysis: catalyst is in the same phase as the reactant.(Br2, catalase) Heterogeneous catalysis: catalyst is in a different phase from the reactants (MnO2) • usually a solid catalyst and gas or solution reactants • Reaction happens on the surface of the catalyst

  29. Energy profiles for catalyzed and uncatalyzed H2O2 decomposition

  30. CATALYSIS Thermodynamic state functions (E, H, G, S…) are unaffected by catalysis (Changes are path-independent)

  31. HETEROGENOUS CATALYSIS • EXAMPLE: H2 + 1/2 O2 H2O • reaction requires breaking • strong H-H and O=O bonds •  • 435 kJ 498 kJ • negligible rate without catalyst • Usually, the stronger the bonds in reactants, the more we need a catalyst. • e.g. 3H2 + N2 2NH3G° = 33 kJ/mol at 298 K • NN triple bond (D = 946 kJ)

  32. NN triple bond (D = 946 kJ) Nitrogen fixation • must break NN triple bond (difficult) • Important in biological systems (proteins, nucleic acids) & industrially (fertilizer, polymers, explosives, …) • Beans, bacteria, etc: nitrogenase • enzyme reduces N2 to NH3 at room temp, 1 atm pressure • Haber process: uses Fe/Al2O3 catalyst at • 400-500°C, 300 atm. • N2 + 3H2 2 NH3

  33. CATALYTIC CONVERTER CATALYTIC CONVERTER O2 1) CO  CO2(g) + H2O Hydrocarbons 2) NO, NO2 N2(g) Catalysts: CuO, Cr2O3, Pt, Rh

  34. HOW DOES LOWERING Ea AFFECT RATES? EXAMPLE. H2O2 H2O + 1/2 O2 hydrogen peroxide (toxic) Uncatalyzed reaction has Ea = 72 kJ Catalase (enzyme in liver) lowers Ea to 28 kJ What is the ratio of kcat/kuncat at 37°C? (body temperature) Assume Acat = Auncat (Q: is this a good assumption?)

  35. CATALYZED VS UNCATALYZED kcat -(Ea,cat - Ea,uncat)/RT = e kuncat • speeds up by a factor of 30 million! • Peptidase enzymes – break up proteins into amino acids (in your stomach) • similar effect on Ea • Without these it would take ~ 300 years to digest a steak!

  36. ENZYMES Enzymes are biological catalysts. Enzymes are produced by organisms to accelerate and to control reaction rates. Enzymes are typically large protein molecules or combinations of proteins with other molecules. The region where the substrate/s (reactant/s) bind is called the active site. Enzymes differ from man-made catalysts: More efficient. More specific. Rate can be controlled by changing enzyme activity.

  37. ENZYME CATALYSIS Nature’s catalysts – big organic molecules specifically designed for certain reactions. Rate acceleration by > 1010 (how?) Enzyme active sites are ideally suited for transition state binding (lowers Ea) Juxtaposition of reactants  high effective concentration (increases A) Each factor enhances rate by ≥ 105

  38. CONTROL OF ENZYMES • Some enzymes wait in the “off” state, such as blood-clotting and digestive proteins • They are activated (reacted to make the active form) when needed. • The active site depends on the enzyme conformation (shape): • Metal ions are held in place by different sections of the protein sitting in close proximity. • If this shape is altered, the active site no longer functions and the enzyme is “turned off.” • Molecular shape depends on pH, temperature, and reactions of the enzyme.

  39. ENZYME ACTIVE SITES • Denatured enzyme – parts of the active site are no longer in close proximity. • Representation of an active site in an enzyme. Competitive Inhibition: Another way to inhibit an enzyme is to bind a molecule to its active site, blocking any catalytic activity. Many drugs and poisons work by this mechanism.

  40. DRUGS Penicillin (antibiotic) blocks an enzyme that bacteria use to build cell walls. People do not have this enzyme Bacterial cells only are poisoned. HIV-protease inhibitors bind to the active site of an enzyme that releases the viral coat proteins, preventing the production of the HIV virus. HIV protease Active site Ritonavir (inhibitor)

  41. ENZYMES Metal ions are often bound at the active site and serve as the reaction center of the enzyme. The enzyme carbonic anhydrase uses a Zn2+ ion at its active site to accelerate the reaction: CO2 + H2O  H2CO3 In red blood cells, CO2 is converted to H2CO3 which deprotonates to form HCO3-. HCO3- leaves the cell and serves as a buffer for blood plasma. In the lungs, HCO3- is re-protonated to form H2CO3. Carbonic anhydrase converts H2CO3 back to CO2(g) and H2O. Exhale!

  42. VITAMINS Vitamins are non-protein parts of enzymes, called co-enzymes. When combined with the protein part they make enzymes. Enzymes derived from vitamins play critical roles in redox chemistry in the body, which is the source of heat and energy.

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