Half lives Half life: t 1/2 t 1/2 is the time it takes for the concentration of a reactant to drop to half of its initial value. For the reaction A! products Half Life of a First Order Reaction! Pressure of methyl isonitrile as a function of time! What is the half life according to the graph below?! 1.! 75 s! 2.! 13600 s! 3.! 27200 s! t 1/2 is where [A] = [A] o 2 For 1 st order reactions: What is t 1/2?!n[A] =!n[a] o " kt First order t 1/2 doesn t depend on concentration CH112 LRSVDS Kinetics 4-6 1 CH112 LRSVDS Kinetics 4-6 2
HALF LIFE OF 2 ND ORDER REACTIONS 1 A [ ] = k t + 1 A [ ] 0 To get t 1/2, we let [A] = 1/2 [A] o HALF LIFE OF A SECOND ORDER REACTION NO 2 (g)! NO(g) + 1/2 O 2 (g) 0.01 0.009 0.008 [NO 2 ] M 0.007 0.006 0.005 0.004 0.003 0.002 0.001 t 1/2 = 0 0 100 200 300 400 500 600 Time (sec) t 1/2 of a Second Order Reaction depends on the initial concentration: [A] 0 CH112 LRSVDS Kinetics 4-6 3 CH112 LRSVDS Kinetics 4-6 4
Half Life Problem It requires 100 years for a first-order reaction to go 30.0% of the way to completion. What is the halflife of the reaction? Temperature Dependence of Reaction Rates CH 3 "N#C:! CH 3 " C # N: Rate = k [CH 3 "N#C:] Measure rate at different temperatures: Rates remain first order in [CH 3 "N#C:] What variable must change? CH112 LRSVDS Kinetics 4-6 5 CH112 LRSVDS Kinetics 4-6 6
Temperature Dependence of Reaction Rates For a reaction to occur; 1.!A molecular collision must occur 2.! The orientation of the molecules may determine the effectiveness of the collision. Temperature dependence of Reaction Rates As T increases what happens to reaction rate? Why? What about unimolecular reactions? Look at energy profile for a typical reaction: isonitrile isomerizes to form acetonitrile Cl + NOCl!! Cl 2 + NO! Cl + NOCl!! Cl + NOCl! CH112 LRSVDS Kinetics 4-6 7 CH112 LRSVDS Kinetics 4-6 8
Activation Energy %E= Temperature Dependence of Kinetic Energy Energy required to go from reactants to a transition state is Activation Energy Transition State or Activated Complex: How do molecules get the energy needed to get over the barrier? The fraction of molecules with a certain value of energy is f f $ e -Ea/RT What does increasing T do to f? CH112 LRSVDS Kinetics 4-6 9 CH112 LRSVDS Kinetics 4-6 10
Reaction Profiles For the reaction profiles below: 1.! What is %E? 2.! What is the activation energy? 3.! Which reaction will be fastest? Molecular level picture:!!reaction rate depends on frequency of collisions (& to concentration of reactants)!!reactants must come together with right orientation and enough KE to overcome E a How do orientation and activation energy factor into the rate laws? CH112 LRSVDS Kinetics 4-6 11 CH112 LRSVDS Kinetics 4-6 12
Arrhenius Equation k = A e E a RT ln k = ln A " E a RT! ARRHENIUS PLOT! ln k = ln A E a /RT How does this reaction happen? Does the C-N bond break? Find E a, compare to C-N Bond Energy of 293 kj/mole k = rate constant (temperature dependent) H 3 C N C! H 3 C C N A = frequency factor (Related to collision frequency and orientation) E a = Activation energy R = gas constant (usually 8.314 J/mol-K) T = temperature in K Slope? Y Intercept? CH112 LRSVDS Kinetics 4-6 13 CH112 LRSVDS Kinetics 4-6 14
Compare k at different temperatures: eliminates A from the equation Understanding the high-temperature behavior of nitrogen oxides is essential for controlling pollution generated in automobile engines. 2NO(g)! N 2 (g)+ O 2 (g) The decomposition of nitric oxide (NO) to N 2 and O 2 is second order with a rate constant of 0.0796 M -1 s -1 at 737 C and 0.0815 M -1 s -1 at 947 C. Calculate the activation energy for the reaction. k ln k 2 = E R & 1 $ % T2 ' 1 a 1 T 1 #! " FYI Where does this equation come from? k ln k 2 = E R & 1 $ % T2 We want to eliminate A from the Arrhenius equation. This can be done if we know the reaction rate at two different temperatures: ln k 1 = " E a RT 1 ' 1 a 1 " ln A T 1 #! " ln k 2 = " E a RT 2 " ln A Subtracting eq 2 from equation one eliminates A ln k 1 " ln k 2 = " E a RT 1 ln k 1 k 2 = E a R # 1 " 1 & % ( $ T 2 T 1 ' # " "E & a % ( $ RT 2 ' R = 8.314 J/mol-K CH112 LRSVDS Kinetics 4-6 15 CH112 LRSVDS Kinetics 4-6 16
How much does temperature affect rates? Most reactions have E a = 20 200 kj/mol A typical E a might be 50 kj/mol How does rate vary over a 10 temperature range? (e.g. from 300 to 310 K) k ln k 2 = E & 1 R $ % T2 ' T 1 a 1 1 #! " REACTION MECHANISMS Reaction Mechanism: Process by which a reaction occurs; could be a multi-step reaction Start with elementary reactions: Process occurs in a single step. (Reaction proceeds as written with no other steps in between.) NO + O 3! NO 2 +O 2 k 2 /k 1 = e 0.65 = 1.9 (about twice as fast at 310 K compared to 300 K) For an elementary process Rate $ [reactants] Rate = k [NO][O 3 ] Rule of thumb: reaction rates double for every 10 rise in temperature (assumes E a ' 50 kj/mol) CH112 LRSVDS Kinetics 4-6 17 CH112 LRSVDS Kinetics 4-6 18
Reaction Mechanisms Most balanced reactions do not occur as a single elementary step. They occur as the net result of several elementary steps which could include such events as: "!Collisions "!Bonds Breaking "!Bonds Forming "!Groups of atoms changing orientation or relative position Elementary Reaction Molecularity: Unimolecular Bimolecular Termolecular This series of steps is called the REACTION MECHANISM CH112 LRSVDS Kinetics 4-6 19 CH112 LRSVDS Kinetics 4-6 20
Rates of Elementary Steps Reaction Rate A! B + C A + B! C + D For this reaction: NO 2 +CO! NO + CO 2 If rate obs = k[no 2 ][CO], is it an elementary process?! If rate obs " k[no 2 ][CO], is it an elementary process?! ( Propose a multistep mechanism: NO 2 + NO 2! NO 3 + NO step 1 NO 3 + CO! NO 2 + CO 2 step 2 NO 2 + CO! NO + CO 2 2A! C + D A + B + C! D + E CH112 LRSVDS Kinetics 4-6 21 Possible Rate Laws for this multi-step mechanism: or [ ] rate = "# NO 2 #t [ ] #t "# CO rate = = k[ NO 2 ] 2 = k[ NO 3 ][ CO] Which is the rate law? Compare to experimental data to see which rate law matches. CH112 LRSVDS Kinetics 4-6 22
Experimental Data indicates: NO 2 +CO! NO + CO 2 Rate $ [NO 2 ] 2 It is possible to form Intermediates in a Mechanism NO 2 + NO 2! NO 3 + NO step 1 NO 3 + CO! NO 2 + CO 2 step 2 Which step of the mechanism is consistent with this data? NO 2 + CO! NO + CO 2 overall rxn Which step determines the rate of the entire reaction? The rate can t proceed any faster then the slowest step. SO: Rate Determining Step = Slow Step NO 3 is produced in step 1 and consumed in step 2. Note: an intermediate is NOT the same as an activated complex.!!intermediates do not appear in the rate law for the overall reaction.!!elementary steps must add up to eliminate intermediates and give the balanced equation for the overall process. CH112 LRSVDS Kinetics 4-6 23 CH112 LRSVDS Kinetics 4-6 24
To determine a mechanism for a reaction: 1.! Find the experimental rate law 2.! (Postulate elementary steps: this step will be done for you in Chem 112) 3.! Find the rate law predicted by the mechanism and compare it to the experimental data. "! No rate can be written in terms of intermediates If the first step is not rate determining, intermediates become involved in the rate law. Cl 2 + CHCl 3! HCl + CCl 4 Observed Rate = k obs [Cl 2 ] 1/2 [CHCl 3 ] Q: Is the following mechanism consistent with the experimental rate law? 1)! Cl 2 2Cl fast 2)! Cl + CHCl 3! HCl + CCl 3 slow 3)! Cl + CCl 3! CCl 4 fast CH112 LRSVDS Kinetics 4-6 25 CH112 LRSVDS Kinetics 4-6 26
Strategy for Determining Consistency of Proposed Mechanism Strategy: a) Overall reaction? b) Rate determining step? c)! Rate Law? d)! Define intermediate concentration in terms of reactants only In General: If a fast step comes before a slow step, solve for the concentration of an intermediate by assuming that the first step is an equilibrium. g)!substitute for intermediate in rate law h)!is it consistent? CH112 LRSVDS Kinetics 4-6 27 CH112 LRSVDS Kinetics 4-6 28
CATALYSIS Catalysis and Enzymes Read the supplemental reading on the class website: http://courses.chem.psu.edu/chem112/ fall/handouts/supp_reading2.htm Definition of a Catalyst: How does it do this? 1. 2. Thermodynamic state functions are unaffected by catalysis (%E, % H, % G, % S ) (Why??) A Catalyst:!!!!!!!!!! CH112 LRSVDS Kinetics 4-6 29 CH112 LRSVDS Kinetics 4-6 30
Homogeneous catalysis: Catalyst is in the same phase as the reactants. Example: 2H 2 O 2 (l)! 2H 2 O(g) + O 2 (g) H H O O O O H H 2! H O O +! O H Heterogeneous catalysis: Catalyst is in a different phase than the reactants. H 2 + 1/2 O 2! H 2 O This reaction requires breaking strong H"H and O=O bonds ) ) 435 kj 498 kj (! Rate is negligible without a catalyst (!The stronger the bond, the more need for a catalyst to get a useful rate gas molecules H-H H-H O=O Pt surface solid adsorbed atoms H H O O diffusion on surface = = H O H H H O = CH112 LRSVDS Kinetics 4-6 31 CH112 LRSVDS Kinetics 4-6 32
Examples of Heterogeneous Catalysis Heterogeneous Catalysis in Nitrogen Fixation N 2 + 3H 2! 2NH 3 %G = -33 kj/mol @ 298 K N#N triple bond (D = 946 kj) 1. Reaction of ethylene with hydrogen gas 2. Catalysis of Hydrogen Peroxide Decomposition by Manganese! Similar Mechanism! 3. Catalytic Converters! O 2 1) CO, C x H 2x+2! CO 2 (g) + H 2 O 2) NO, NO 2! N 2 (g) Catalysts: CuO, Cr 2 O 3, Pt, Pd, Rh CH112 LRSVDS Kinetics 4-6 33 *!Microorganisms convert waste into N 2 (g), which must be turned into a form plants can use *!Plants accomplish this with the Enzyme. *!The best we can do is the Haber Process for Nitrogen Fixation CH112 LRSVDS Kinetics 4-6 34
Enzymes!! Biological catalysts.! Accelerate and control reaction rates.! Enormous protein molecules or combinations of proteins with other molecules.! Reactant is called a.! The region where substrates bind is called the. How Does Lowering E a AFFECT RATES? Uncatalyzed H 2 O 2! H 2 O + 1/2 O 2 Catalase (enzyme in liver) k cat k uncat = A cat e -E E a = 72 kj E a = 28 kj Compare k cat /k uncat at 37 C (body temp.) Assume A cat = A uncat (Is this a good assumption?) /RT a,cat A e -E a,uncat /RT uncat Catalase enzyme speeds up the decomposition rate by a factor of! Lysozyme CH112 LRSVDS Kinetics 4-6 35 Peptidase enzymes break up proteins into amino acids (in your stomach). Similar effect on E a. CH112 LRSVDS Kinetics 4-6 36
Enzymes are superior to man-made catalysts 1.! More efficient ENZYME CATALYSIS Nature s catalysts how do they increase rates by > 10 10? Lock and Key Model: enzyme 2.! Absolute specificity binding sites k = A e -E /RT a reactant molecules enzyme-substrate complex 3.! Regulated 1.! Active site lowers E a how? products 2.! Enzyme-substrate complex stabilizes the transition state, activates the substrate 3.! Holds reactants in place ( high effective concentration (increases A) Each factor enhances rate by! 10 5 CH112 LRSVDS Kinetics 4-6 37 CH112 LRSVDS Kinetics 4-6 38
Enzyme Activity Depends on:! Temperature! ph! Concentration of substrate These conditions affect the active site by affecting the enzyme conformation ( ). (Left) Representation of an active site in an enzyme. Carbonic Anhydrase Enzyme Metal ions are often part of the active site and serve as the reaction center of the enzyme.! Zn 2+ ion at the active site! bound H 2 O displaced by CO 2! CO 2 activated & converted to HCO 3 - CO 2 + H 2 O! HCO 3 - + H + (Right) Denatured enzyme parts of the active site are no longer in close proximity. CH112 LRSVDS Kinetics 4-6 39 CH112 LRSVDS Kinetics 4-6 40
VITAMINS!! Non-protein parts of enzymes, called co-enzymes.!! Combined with the protein they make an enzyme.!! Enzymes derived from vitamins play critical roles in redox chemistry in the body which is the source of heat and energy. Enzyme activity can be inhibited 1.! Denature Change shape via Temp. or ph change 2.! Competitive inhibition Bind a molecule to its active site, blocking any catalytic activity Many drugs and toxins work by this mechanism. Competitive inhibitors can be overcome by increasing the concentration of substrate. 3.! Non-competitive inhibition Bind a molecule to another site on enzyme: this results in a modification of the active site. 4. Irreversible Inhibitors Form strong covalent bonds to enzyme Increasing concentration of substrate has no effect. CH112 LRSVDS Kinetics 4-6 41 CH112 LRSVDS Kinetics 4-6 42
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) CH112 LRSVDS Kinetics 4-6 43