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15 Chemical Kinetics 1 Chemical kinetics : the study of reaction rate,  a quantity conditions affecting it, the molecular events during a chemical reaction.

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Presentation on theme: "15 Chemical Kinetics 1 Chemical kinetics : the study of reaction rate,  a quantity conditions affecting it, the molecular events during a chemical reaction."— Presentation transcript:

1 15 Chemical Kinetics 1 Chemical kinetics : the study of reaction rate,  a quantity conditions affecting it, the molecular events during a chemical reaction (mechanism), and presence of other components (catalysis). Factors affecting reaction rate : Concentrations of reactants Catalyst Temperature Surface area of solid reactants or catalyst What quantities do we study regarding chemical reactions?

2 15 Chemical Kinetics 2 Reaction Rate Defined Reaction rate : changes in a concentration of a product or a reactant per unit time.  [ ] concentration Reaction rate = ——  t [ ] t  [ ] tt change Define reaction rate and explain Average reaction rate Instantaneous reaction rate (2 tangents shown) Initial reaction rate

3 15 Chemical Kinetics 3 Expressing reaction rates For a chemical reaction, there are many ways to express the reaction rate. The relationships among expressions depend on the equation. Note the expression and reasons for their relations for the reaction 2 NO + O 2 (g) = 2 NO 2 (g)  [O 2 ] 1  [NO] 1  [NO 2 ] Reaction rate = – ——— = – — ———— = — ———  t 2  t 2  t Make sure you can write expressions for any reaction and figure out the relationships. For example, give the reaction rate expressions for 2 N 2 O 5 = 4 NO 2 + O 2 How can the rate expression be unique and universal?

4 15 Chemical Kinetics 4 Calculating reaction rate The concentrations of N 2 O 5 are 1.24e-2 and 0.93e-2 M at 600 and 1200 s after the reactants are mixed at the appropriate temperature. Evaluate the reaction rates for 2 N 2 O 5 = 4 NO 2 + O 2 Solution: (0.93 – 1.24)e-2 – 0.31e-2 M Decomposition rate of N 2 O 5 = – ———————— = – —————— 1200 – s = 5.2e-6 M s -1. Note however, rate of formation of NO 2 = 1.02e-5 M s -1. rate of formation of O 2 = 2.6e-6 M s -1. The reaction rates are expressed in 3 forms Be able to do this type problems

5 15 Chemical Kinetics 5 Determine Reaction Rates To measure reaction rate, we measure the concentration of either a reactant or product at several time intervals. The concentrations are measured using spectroscopic method or pressure (for a gas). For example, the total pressure increases for the reaction: 2 N 2 O 5 (g)  4 NO 2 (g) + O 2 (g) Because 5 moles of gas products are produced from 2 moles of gas reactants. For the reaction CaCO 3 (s)  CaO(s) + CO 2 (g) The increase in gas pressure is entirely due to CO 2 formed. barometer

6 15 Chemical Kinetics 6 Differential Rate Laws Dependence of reaction rate on the concentrations of reactants is called the rate law, which is unique for each reaction. For a general reaction, a A + b B + c C  products the rate law has the general form order wrt A, B, and C, determined experimentally reaction rate = k [ A ] X [ B ] Y [ C ] Z the rate constant For example, the rate law is rate = k [Br - ] [BrO 3 - ] [H + ] for 5 Br- + BrO H +  3Br H 2 O The reaction is 1 st order wrt all three reactants, total order 3. Use differentials to express rates

7 15 Chemical Kinetics 7 Variation of Reaction rates and Order First order, rate = k [A] k = rate, 0 th order [A] rate 2nd order, rate = k [A] 2 The variation of reaction rates as functions of concentration for various order is interesting. Mathematical analysis is an important scientific tool, worth noticing. [A] = ___?

8 15 Chemical Kinetics 8 Differential Rate Law determination Estimate the orders and rate constant k from the results observed for the reaction? What is the rate when [H 2 O 2 ] = [I - ] = [H + ] = 1.0 M? H 2 O I H +  I H 2 O Exprmt[H 2 O 2 ][I - ][H + ]Initial rate M s e e e e-6 Solution next Learn the strategy to determine the rate law from this example. Figure out the answer without writing down anything.

9 15 Chemical Kinetics 9 Differential Rate Law determination - continue Estimate the orders from the results observed for the reaction H 2 O I H +  I H 2 O Exprmt[H 2 O 2 ][I - ][H + ]Initial rate M s e e-6 1 for H 2 O e-6 1 for I e-6 0 for H e-6 = k [H 2 O 2 ] x [I - ] y [H + ] z 1.15e-6 k (0.010) x (0.010) y (0.0050) z  exprmt = = e-6 k (0.020) x (0.010) y (0.0050) z  exprmt 22 2 x = 1 ( ) x

10 15 Chemical Kinetics 10 Differential Rate Law determination - continue Other orders are determined in a similar way as shown before. Now, lets find k and the rate Thurs, rate = 1.15e-6 = k (0.010)(0.010) from exprmt 1 k = 1.15e-6 M s -1 / (0.010)(0.010) M 3 = M -1 s -1 And the rate law is therefore, – d [H 2 O 2 ] k rate =—————= [H 2 O 2 ] [I - ]a differential rate law d t total order 2 The rate when [H 2 O 2 ] = [I - ] = [H + ] = 1.0 M: The rate is the same as the rate constant k, when concentrations of reactants are all unity (exactly 1), doesn’t matter what the orders are.

11 15 Chemical Kinetics 11 Differential Rate Law determination – continue The reaction rate – d [H 2 O 2 ]/ dt = [H 2 O 2 ] [I – ], for H 2 O I H +  I H 2 O What is – d [I – ]/ dt when [H 2 O 2 ] = [I – ] = 0.5? Solution : Please note the stoichiometry of equation and how the rate changes. – d [I – ]/ dt = – 3 d [H 2 O 2 ]/ dt = 3* [H 2 O 2 ] [I – ] = * 0.5 * 0.5 = M s -1 In order to get a unique rate constant k, we evaluate k for the reaction a A + b B  product this way rate = - 1 / a d [A]/ dt = - 1 / b d [B]/ dt = k [A] x [B] y Note the reaction rate expression and the stoichiometry of equation.

12 15 Chemical Kinetics 12 Differential Rate Law determination - continue From the following reaction rates observed in 4 experiments, derive the rate law for the reaction A + B + C  products where reaction rates are measured as soon as the reactants are mixed. Expt [A] o [B] o [C] o rate Solution next This example illustrates the strategy to determine, and a reliable method to solve rate-law experimentally.

13 15 Chemical Kinetics 13 Differential Rate Law determination - continue From the following reaction rates, derive the rate law for the reaction A + B + C  products where reaction rates are measured as soon as the reactants are mixed. Expt order [A] o from expt 1 & 2 [B] o expt 1, 2 & 3 [C] o expt 1 & 4 rate k 0.2 x 0.1 y 0.1 z = k 0.1 x 0.1 y 0.1 z Therefore 8 = 2 x log 8 = x log 2 x = log 8 / log 2 = 3 Assume rate = k [A] x [B] y [C] z

14 15 Chemical Kinetics 14 Integrated Rate Laws concentrations as functions of time One reactant A decomposes in 1 st or 2 nd order rate law. Differential rate lawIntegrated rate law – d[A] / dt = k[A] = [A] o – k t d [A] – —— = k [A][A] = [A] o e – k t or ln [A] = ln [A] o – k t d t d [A] 1 1 [A] conc at t – —— = k [A] 2 —— – —— = k t d t [A] [A] o [A] o conc at t = 0 Describe, derive and apply the integrated rate laws Learn the strategy to determine rate-law

15 15 Chemical Kinetics 15 Concentration and time of 1 st order reaction Describe the features of plot of [A] vs. t and ln[A] vs. t for 1 st order reactions. Apply the technique to evaluate k or [A] at various times. [A] t ln[A] t [A] = [A] o e – k t ln [A] = ln [A] o – k t t½t½

16 15 Chemical Kinetics 16 Half life & k of First Order Decomposition The time required for half of A to decompose is called half life t 1/2. Since[A] = [A] o e – k t or ln [A] = ln [A] o – k t When t = t 1/2, [A] = ½ [A] o Thus ln ½ [A] o = ln [A] o – k t 1/2 – ln 2 = – k t 1/2 k t 1/2 = ln 2 =  relationship between k and t 1/2 Radioactive decay usually follow 1 st order kinetics, and half life of an isotope is used to indicate its stability. Evaluate t ½ from k or k from t ½

17 15 Chemical Kinetics 17 1 st order reaction calculation N 2 O 5 decomposes according to 1 st order kinetics, and 10% of it decomposed in 30 s. Estimate k, t ½ and percent decomposed in 500 s. Solution next If the rate-law is known, what are the key parameters?

18 15 Chemical Kinetics 18 1 st order reaction calculation N 2 O 5 decomposes according to 1 st order kinetics, and 10% of it decomposed in 30 s. Estimate k, t ½ and percent decomposed in 500 s. Solution : Assume [A] o = [N 2 O 5 ] o = 1.0, then [A] = 0.9 at t = 30 s or 0.9 = 1.0 e – k t apply [A]o = [A] e – k t ln 0.9 = ln 1.0 – k 30 s – = 0 – k * 30 k = s – 1 t ½ = / k = 197 s apply k t ½ = ln 2 [A] = 1.0 e – *500 = Percent decomposed: 1.0 – = or 82.7 % After 2 t ½ (2*197=394 s), [A] = (½) 2 =¼, 75% decomposed. After 3 t ½ (3*197=591 s), [A] = (½) 3 =1/8, 87.5% decomposed. Apply integrated rate law to solve problems

19 15 Chemical Kinetics 19 Typical Problem wrt 1 st Order Reaction The decomposition of A is first order, and [A] is monitored. The following data are recorded: t / min0248 [A]/[M] Calculate k (What is the rate constant? k = ) Calculate the half life (What is the half life? Half life = 13.89) Calculate [A] when t = 5 min. ( What is the concentration when t = 5 min? ) Calculate t when [A] = (Estimate the time required for 90% of A to decompose.) Work out all the answers

20 15 Chemical Kinetics 20 A 2 nd Order Example Dimerization of butadiene is second order: 2 C 4 H 6 (g) = C 8 H 12 (g). The rate constant k at some temperature is /min. The initial concentration of butadiene [B] is 2.0 M. Calculate the time required for [B] = 1.0 and 0.5 M Calculate concentration of butadiene when t = 1, 5, 10, and 30. Solution next If the rate-law is known, what are the key parameters? Apply the right model and work out all the parameters.

21 15 Chemical Kinetics 21 A 2 nd Order Example Dimerization of butadiene is second order: 2 C 4 H 6 (g) = C 8 H 12 (g). The rate constant k at some temperature is /min. The initial concentration of butadiene [B] is 2.0 M. Calculate the time t required for [B] = 1.0 and 0.5 M Calculate concentration of butadiene when t = 1, 5, 10, and ——–—— = k t [B][B] o 1 1 ——–—— [B][B] o t = ———————— k [B] o [B] = —————— [B] o k t + 1 t = [B] = Work out the formulas and then evaluate values

22 15 Chemical Kinetics 22 Half life of 2 nd Order Chemical Kinetics t = [B] = /[B] How does half life vary in 2 nd order reactions?

23 15 Chemical Kinetics 23 Plot of [B] vs. t & 1/[B] vs. t for 2 nd Order Reactions t = [B] = [B] 1 — [B] tt 1 1 ——–—— = k t [B][B] o [B] o [B] = —————— [B] o k t + 1 What kind of plot is linear for 1 st and 2 nd reactions?

24 15 Chemical Kinetics 24 Chemical Reaction and Molecular Collision Molecular collisions lead to chemical reactions. Thus, the reaction constant, k is determined by several factors. k = Z f p Z : collision frequency p, the fraction with proper orientation f, fraction of collision having sufficient energy for reaction f is related to the potential energy barrier called activation energy, E a. f  e – E a / RT or exp (– E a / R T ) Thus, k = A e – E a / RT constant EaEa Potential energy reaction constant How does temperature affect reaction rates ? Explain energy aspect in a chemical reaction

25 15 Chemical Kinetics 25 Energy in chemical reactions Potential energy Progress of reaction R + A   P + D  H exothermic E a E a for reverse reaction Endothermic rxn RA-PD activated complex Explain the various terms and energy changes in a chemical reaction

26 15 Chemical Kinetics 26 The Arrhenius Equation 1903 Nobel Prize citation” …in recognition of the extraordinary services he has rendered to the advancement of chemistry by his electrolytic theory of dissociation” The temperature dependence of the rate constant k is best described by the Arrhenius equation: k = A e – E a / R T or ln k = ln A – E a / R T If k 1 and k 2 are the rate constants at T 1 and T 2 respectively, then k 1 E a 1 1 ln —— = – — — – — k 2 R T 1 T 2 How does temperature affect reaction rates ? Derive and apply these relationship to solve problems, and recall the Clausius-Clapeyron equation.

27 15 Chemical Kinetics 27 Application of Arrhenius Equation From k = A e – E a / R T, calculate A, E a, k at a specific temperature and T. The reaction: 2 NO 2 (g) -----> 2NO(g) + O 2 (g) The rate constant k = 1.0e-10 s -1 at 300 K and the activation energy E a = 111 kJ mol -1. What are A, k at 273 K and T when k = 1e-11? Method: derive various versions of the same formula k = A e – E a / R T A = k e E a / R T A / k = e E a / R T ln ( A / k ) = E a / R T Make sure you know how to transform the formula into these forms.

28 15 Chemical Kinetics 28 Application of Arrhenius Equation (continue) The reaction: 2 NO 2 (g) -----> 2NO(g) + O 2 (g) The rate constant k = 1.0e-10 s -1 at 300 K and the activation energy E a = 111 kJ mol -1. What are A, k at 273 K and T when k = 1e-11? Use the formula derived earlier: A = k e Ea / R T = 1e-10 s -1 exp ( J mol -1 / (8.314 J mol -1 K –1 *300 K)) = 2.13e9 s -1 k = 2.13e9 s -1 exp ( – J mol -1 ) / (8.314 J mol -1 K –1 *273 K) = 1.23e-12 s -1 T = E a / [ R* ln ( A / k )] = J mol -1 / (8.314*46.8) J mol -1 K -1 = 285 K

29 15 Chemical Kinetics 29 The Effect of Temperature on Reaction Rates Reaction rate = k [A} x [B] y [C] z (Concentration effect at constant T ) k = A exp ( – E a / RT )(Temperature effect) Use graphic method to discuss the variation of k vs. T variation of k vs 1 / T variation of ln( k ) vs T variation of ln( k ) vs 1 / T See a potential multiple choice question in an exam?

30 15 Chemical Kinetics 30 Elementary Reactions and Mechanism Elementary reactions are steps of molecular events showing how reactions proceed. This type of description is a mechanism. The mechanism for the reaction between CO and NO 2 is proposed to be Step 1NO 2 + NO 2  NO 3 + NO(an elementary reaction) Step 2NO 3 + CO  NO 2 + CO 2 (an elementary reaction) Add these two equations led to the overall reaction NO 2 + CO = NO + CO 2 (overall reaction) A mechanism is a proposal to explain the rate law, and it has to satisfy the rate law. A satisfactory explanation is not a proof. Explain terms in red

31 15 Chemical Kinetics 31 Molecularity of Elementary Reactions The total order of rate law in an elementary reaction is molecularity. The rate law of elementary reaction is derived from the equation. The order is the number of reacting molecules because they must collide to react. A molecule decomposes by itself is a unimolecular reaction (step); two molecules collide and react is a bimolecular reaction (step); & three molecules collide and react is a termolecular reaction (step). O 3  O 2 + O rate = k [O 3 ] NO 2 + NO 2  NO 3 + NO rate = k [NO 2 ] 2 Br + Br + Ar  Br 2 + Ar* rate = k [Br] 2 [Ar] Caution: Derive rate laws this way only for elementary reactions.

32 15 Chemical Kinetics 32 Molecularity of elementary reactions - Example Some elementary reactions for the reaction between CH 4 and Cl 2 are Cl 2  2 Cl 2 Cl  Cl 2 2Cl + CH 4  Cl 2 + CH 4 * Cl + CH 4  HCl + CH 3 CH 3 + Cl  CH 3 Cl CH 3 + CH 3  CH 3 -CH 3 CH 3 Cl + Cl  HCl + CH 2 Cl CH 2 Cl + Cl  CH 2 Cl 2 * * * (more) Write down the rate laws and describe them as uni- bi- or ter-molecular steps yourself, please.

33 15 Chemical Kinetics 33 Elementary Reactions are Molecular Events N 2 O 5  NO 2 + NO 3  NO + O 2 + NO 2  NO 2 + NO 3 Explain differences between elementary and over reaction equations

34 15 Chemical Kinetics 34 Rate Laws and Mechanisms A mechanism is a collection of elementary steps devise to explain the the reaction in view of the observed rate law. You need the skill to derive a rate law from a mechanism, but proposing a mechanism is task after you have learned more chemistry For the reaction, 2 NO 2 (g) + F 2 (g)  2 NO 2 F (g), the rate law is, rate = k [NO 2 ] [F 2 ]. Can the elementary reaction be the same as the overall reaction? If they were the same the rate law would have been rate = k [NO 2 ] 2 [F 2 ], Therefore, they the overall reaction is not an elementary reaction. Its mechanism is proposed next.

35 15 Chemical Kinetics 35 Rate-determining Step in a Mechanism The rate determining step is the slowest elementary step in a mechanism, and the rate law for this step is the rate law for the overall reaction. The ( determined ) rate law is, rate = k [NO 2 ] [F 2 ], for the reaction, 2 NO 2 (g) + F 2 (g)  2 NO 2 F (g), and a two-step mechanism is proposed: i NO 2 (g) + F 2 (g)  NO 2 F (g) + F (g) ii NO 2 (g) + F (g)  NO 2 F (g) Which is the rate determining step? Answer : The rate for step i is rate = k [NO 2 ] [F 2 ], which is the rate law, this suggests that step i is the rate-determining or the s-l-o-w step. Explain rate determining step in a mechanism and use it to derive the rate law.

36 15 Chemical Kinetics 36 Deriving a Rate Law From a Mechanism - 0 The decomposition of H 2 O 2 in the presence of I – follow this mechanism, iH 2 O 2 + I –  k 1  H 2 O + IO – slow ii H 2 O 2 + IO –  k 2  H 2 O + O 2 + I – fast What is the rate law? Energy Solve the problem E ai E aii reaction

37 15 Chemical Kinetics 37 Deriving a rate law from a mechanism - 1 The decomposition of H 2 O 2 in the presence of I – follow this mechanism, iH 2 O 2 + I –  k 1  H 2 O + IO – slow ii H 2 O 2 + IO –  k 2  H 2 O + O 2 + I – fast What is the rate law? Solution The slow step determines the rate, and the rate law is: rate = k 1 [H 2 O 2 ] [I – ] Since both [H 2 O 2 ] and [I – ] are measurable in the system, this is the rate law.

38 15 Chemical Kinetics 38 Deriving a rate law from a mechanism - 2 Derive the rate law for the reaction, H 2 + Br 2 = 2 HBr, from the proposed mechanism: i Br 2  2 Brfast equilibrium ( k 1, k -1 ) iiH 2 + Br  k 2  HBr + H slow iii H + Br  k 3  HBr fast Solution: The fast equilibrium condition simply says that k 1 [Br 2 ] = k -1 [Br] 2 and[Br] = ( k 1 / k -1 [Br 2 ]) ½ The slow step determines the rate law, rate = k 2 [H 2 ] [Br] Br is an intermediate = k 2 [H 2 ] ( k 1 / k -1 [Br 2 ]) ½ = k [H 2 ] [Br 2 ] ½; k = k 2 ( k 1 / k -1 ) ½ M -½ s -1 total order 1.5 explain

39 15 Chemical Kinetics 39 Deriving a rate law from a mechanism - 3 The decomposition of N 2 O 5 follows the mechanism: 1N 2 O 5  NO 2 + NO 3 fast equilibrium 2NO 2 + NO 3 — k 2  NO + O 2 + NO 2 slow 3NO 3 + NO — k 3  NO 2 + NO 2 fast Derive the rate law. Solution : The slow step determines the rate, rate = k 2 [NO 2 ] [NO 3 ] NO 2 & NO 3 are intermediate From 1, we have [NO 2 ] [NO 3 ] —————— = KK, equilibrium constant [N 2 O 5 ] K differ from k Thus, rate = K k 2 [N 2 O 5 ]

40 15 Chemical Kinetics 40 Deriving rate laws from mechanisms – steady-state approximation The steady-state approximation is a general method for deriving rate laws when the relative speed cannot be identified. It is based on the assumption that the concentration of the intermediate is constant. Rate of producing the intermediate, R prod, is the same as its rate of consumption, R cons. R prod = R cons [Intermediate] time R prod < R cons R prod > R cons Be able to apply the steady- state approximation to derive rate laws

41 15 Chemical Kinetics 41 Steady-state approximation - 2 Let’s assume the mechanism for the reaction. H 2 + I 2  2 HI as follows. Step (1)I 2 — k 1  2 I Step (  1)2 I — k -1  I 2 Step (2) H I — k 2  2 HI Derive the rate law. Derivation : rate = k 2 [H 2 ] [I] 2 (‘cause this step gives products) but I is an intermediate, this is not a rate law yet. Since k 1 [I 2 ](= rate of producing I) = k -1 [I] 2 + k 2 [H 2 ] [I] 2 (= rate of consuming I) Thus, k 1 [I 2 ] [I] 2 = —————— k -1 + k 2 [H 2 ]  rate = k 1 k 2 [H 2 ] [I 2 ] / { k -1 + k 2 [H 2 ] } Steady state

42 15 Chemical Kinetics 42 Steady-state approximation - 3 From the previous result: k 1 k 2 [H 2 ] [I 2 ] rate = ——————— { k -1 + k 2 [H 2 ] } Discussion : (i) If k -1 << k 2 [H 2 ] then { k -1 + k 2 [H 2 ]} = k 2 [H 2 ], then rate = k 1 k 2 [H 2 ] [I 2 ] / { k 2 [H 2 ] } = k 1 [I 2 ] (pseudo 1st order wrt I 2 ) using large concentration of H 2 or step 2 is fast (will meet this condition). (ii) If step (2) is slow, then k 2 << k 1, and if [H 2 ] is not large, we have { k -1 + k 2 [H 2 ]} = k -1 and rate = k 1 k 2 [H 2 ] [I 2 ] / k 1 = k 2 [H 2 ] [I 2 ]

43 15 Chemical Kinetics 43 Steady-state approximation - 4 In an alkaline solution, peroxydisulfate oxidizes sulfite to yield sulfate, S 2 O SO OH -  3 SO H 2 O. The following mechanism has been proposed: i S 2 O SO 3 2- — k 1  S 2 O SO 4 2- ii S 2 O H 2 O — k 2  2 SO H + iii H + + OH - — k 3  H 2 O (fast equilibrium to be discussed) Steady-state approximation follows these steps: What is or are the intermediates I? Use which step to give the rate law that may involve [I]? Express the rates of producing and consuming intermediate(s) Express [I] of intermediate(s) in terms of [Reactants] Derive the rate law in terms of [Reactants] Discuss See page 607 PHH Text

44 15 Chemical Kinetics 44 Catalysis A catalyst is a substance that changes the rate of a reaction by lowing the activation energy, E a. It participates a reaction in forming an intermediate, but is regenerated. Enzymes are marvelously selective catalysts. A catalyzed reaction, NO (catalyst) 2 SO 2 (g) + O 2 —  2 SO 3 (g) via the mechanism i2 NO + O 2  2 NO 2 (3 rd order) iiNO 2 + SO 2  SO 3 + NO Uncatalyzed rxn Catalyzed rxn rxn Energy

45 15 Chemical Kinetics 45 Catalyzed decomposition of ozone R.J. Plunkett in DuPont discovered carbon fluorine chlorine compounds. The CFC decomposes in the atmosphere: CFCl 3  CFCl 2 + Cl CF 2 Cl 3  CF 2 Cl + Cl. The Cl catalyzes the reaction via the mechanism: iO 3 + h v  O + O 2, ii ClO + O  Cl + O 2 iii O + O 3  O 2 + O 2. The net result or reaction is 2 O 3  3 O 2 Scientists sound the alarm, and the CFC is banned now.

46 15 Chemical Kinetics 46 Homogenous vs. heterogeneous catalysts A catalyst in the same phase (gases and solutions) as the reactants is a homogeneous catalyst. It effective, but recovery is difficult. When the catalyst is in a different phase than reactants (and products), the process involve heterogeneous catalysis. Chemisorption, absorption, and adsorption cause reactions to take place via different pathways. Platinum is often used to catalyze hydrogenation Catalytic converters reduce CO and NO emission.

47 15 Chemical Kinetics 47 Heterogeneous catalysts Ceryx's vision is to design, produce, and commercialize advanced systems that balance Cost, Performance, Emissions Reduction, and Fuel Penalty to make the economics of pollution control viable. We explore new ways to look at the air quality challenges faced by industry and search for potential solutions by combining proven technologies with state-of-the-art science. Catalyzed reactions: CO + O 2  CO 2 2 NO  N 2 + O 2

48 15 Chemical Kinetics 48

49 15 Chemical Kinetics 49 Enzymes – selective catalysts Enzymes are a long protein molecules that fold into balls. They often have a metal coordinated to the O and N sites. Molecules catalyzed by enzymes are called substrates. They are held by various sites (together called the active site ) of the enzyme molecules and just before and during the reaction. After having reacted, the products P 1 & P 2 are released. E nzyme + S ubstrate  ES (activated complex) ES  P 1 + P 2 + E Enzymes are biological catalysts for biological systems.

50 15 Chemical Kinetics 50 X-ray 3-D structure of fumarate reductase. It reduces fumerate, an important role in the metabolism of anaerobic bacteria, from Max Planck Inst.

51 15 Chemical Kinetics 51 Chemical Kinetics - Summary Explain how the various factors affect reaction rates. Define reaction rates, average rates, initial rates and rate constants. Evaluate rate law from experiments Properly apply 1 st and 2 nd differential rate laws and integrated rate laws. Interpret elementary reactions and mechanisms. Derive rate laws from a given mechanism. Apply the steady-state method to derive the rate law of a given mechanism, and discuss the results. Explain the action of catalysts in terms of chemistry and in terms of energy of activation.


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