Chap14-Part5

# Chap14-Part5 - REVIEW SESSION‐6: SATURDAY  ...

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Unformatted text preview: REVIEW SESSION‐6: SATURDAY   CHAPMAN 211‐ 10.30 am to Noon  CHEMICAL KINETICS   Send me your Ques-ons on Ch‐13 and Ch‐14  Arrhenius Equa-on   Taking the natural  logarithm of both sides,  the equa-on becomes  Ea  1  ln k = ‐         (     ) + ln A  T  R   y   =     m      x     +   b  Therefore, if k is determined experimentally at  several temperatures, Ea can be calculated from the  1  slope of a plot of ln k vs.    T  Sample Exercise 14.10  Assume that all three reac-ons have equivalent frequency factors A.  Rank the reac-ons in order of slowest to fastest.  2 is the slowest and 1 is the fastest      Determining the AcHvaHon Energy  Ea k = A exp(− ) RT ln( xy ) = ln( x) + ln( y ) Ea ⇒ ln(k ) = ln( A) + ln[exp(− )] RT Ea ⇒ ln(k ) = ln( A) − RT plot ln(k) vs. 1/T  intercept = ln(A)    &    A = exp[intercept]  slope = ‐Ea/R    &    Ea = ‐ (R)(slope)  Sample Exercise 14.11  temperature (oC)  rate constant k (sec‐1)  189.7 198.9 230.2 251.2                  2.52 x 10‐5   5.25 x 10‐5   6.30 x 10‐4   3.16 x 10‐3  (a)  Calculate ln(k) and 1/T (in 1/K).  (b)  Determine the ac-va-on energy.  (c)  Determine the frequency factor.  (d) Find the rate constant at 430.0 K.  Determining Rate Constants without Frequency Factors  Ea k1 = A exp(− ) RT1 ⇒ & Ea k 2 = A exp(− ) RT2 Ea A exp(− ) Ea Ea k2 RT2 = = exp( − ) Ea k1 RT1 RT2 A exp(− ) RT1 Ea 1 1 ⇒ k 2 = k1 exp[ ( − )] R T1 T2      ReacHon Mechanisms  A balanced equa-on for a chemical reac-on shows the reactants  and products in the proper stoichiometry.  This equa-on provides no direct informa-on about the mechanism  of the reac-on (e.g., intermediates).  An equa-on showing all of the steps, including intermediates,  indicates the elementary mechanism.  If the elementary mechanism is known, its form can be used to  write the rate laws rather than determining them empirically.       ReacHon Mechanisms  The number of reactants in a given step of an elementary reac-on   deﬁnes the molecularity of the step.   one reactant – unimolecular   two reactants – bimolecular   three reactants – termolecular    etc.  Unimolecular steps do not require collisions.  Bimolecular steps do require collisions.  Termolecular steps are rare, requiring three‐fold collisions.  MulHstep Mechanisms         Reac-on mechanisms can be broken down into elementary reac-ons called   steps in the reac-on.  For example:  Overall:    NO2(g) + CO(g) → NO(g) + CO2(g)  1st Step:    NO2 (g) + NO2(g) → NO3(g) + NO(g)  2nd Step:   NO3(g) + CO(g) → NO2(g) + CO2(g)  Sum:  2 NO2(g) + NO3(g) + CO(g) → NO3(g) + NO(g) + NO2(g) + CO2(g)  Simplify:  2 NO2(g) + NO3(g) + CO(g) → NO3(g) + NO(g) + NO2(g) + CO2(g)  Net:   NO2(g) + CO(g) → NO(g) + CO2(g)  NO3(g) is an intermediate because it is neither a reactant nor a product.  MulHstep Mechanisms  Reac-on mechanisms can be broken down into elementary reac-ons called   steps in the reac-on.  For example:  Overall:    2NO (g) + Br2(g) → 2 NOBr(g)  1st Step:    NO(g) + Br2(g) → NOBr2(g)  2nd Step:   NOBr2(g) + NO(g) → 2NOBr(g)  Sum:     2 NO(g) + Br2(g) + NOBr2(g) → NOBr2(g) + 2NOBr(g)   Net:   2 NO(g) + Br2(g) → 2 NOBr(g)  NOBr2(g) is an intermediate because it is neither a reactant nor a product.  Sample Exercise 14.12  It has been proposed that the conversion of ozone (O3) into molecular oxygen  (O2) proceeds by the following two‐step mechanism:       O3(g) → O2(g) + O(g)   O3(g) + O(g) → 2 O2(g)  (a)  Describe the molecularity of each elementary reac-on.       (b) Write the equa-on for the overall reac-on.  (c) Find the intermediate.      Sample Exercise 14.12  It has been proposed that the conversion of ozone (O3) into molecular oxygen  (O2) proceeds by the following two‐step mechanism:       O3(g) → O2(g) + O(g)   O3(g) + O(g) → 2 O2(g)  (a)  Describe the molecularity of each elementary reac-on.     1 and 2  (b) Write the equa-on for the overall reac-on.   2 O3 (g) 3 O2 (g)  (c) Find the intermediate.  O (Nascent Oxygen, [O])      Determining Rate Laws for Elementary ReacHons  What is the molecularity of the following elementary reac-ons?  Write the rate law for each.     (a) 2 NO(g) → N2O2(g)               (b) SO3(g) → SO2(g) + O(g)       What is the molecularity of the following elementary reac-ons?  Write the rate law for each.     (a) 2 NO(g) → N2O2(g)         Molecularity =2    rate = k[NO]2        (b) SO3(g) → SO2(g) + O(g)         Molecula-ty=1   Rate = k[SO3]     Rate Determining Step  Many chemical reac-ons have mul-ple steps.  Each step has its own ac-va-on energy and rate constant.  If one step is much slower than the others, the overall reac-on  cannot proceed faster than this slow step.  In this case, the slow step is called the rate‐determining step, or  the rate‐limi-ng step.  The rate‐limi-ng step determines the overall reac-on rate.  Rate‐LimiHng Step  The rate of traﬃc ﬂow from place   1 to place 3 is determined by the  rate of movement through plaza A.  The rate of traﬃc ﬂow from place   1 to place 3 is determined by the  rate of movement through plaza B.  Two‐Step Mechanisms with a Slow First Step  Example:  NO2(g) + CO(g) → NO(g) + CO2(g)  1st Step:    NO2 (g) + NO2(g) → NO3(g) + NO(g)  (k1, slow)  2nd Step:    NO3(g) + CO(g) → NO2(g) + CO2(g)  (k2, fast)  The overall reac-on rate is controlled by the ﬁrst step.  Thus,  Rate = k1 [ NO2 ] 2 Two‐Step Mechanisms with a Fast First Step   Example:  2NO (g) + Br2(g) → 2 NOBr(g)  1st Step:    NO(g) + Br2(g) ↔ NOBr2(g)  (k1,k‐1 fast)  2nd Step:    NOBr2(g) + NO(g) → 2NOBr(g)  (k2, slow)  Rate = k 2 [ NOBr2 ][ NO ] This equa-on is not too useful because it depends on the intermediate concentra-on.  Assume the ﬁrst step is in rapid equilibrium.  Then  k1 [ NO ][ Br2 ] = k −1 [ NOBr2 ] ⇒ k1 [ NO ][ Br2 ] = [ NOBr2 ] k −1 k1 [ NO ][ Br2 ] k1 k 2 ⇒ Rate = (k 2 [ NO ])( )=( )[ Br2 ][ NO ] 2 k −1 k −1 Consider the diagram to the  right, represen-ng two steps  inan overall reac-on.  red = O, blue = N, green =  F  (a)  Write the equa-on for the overall reac-on.  (c) Find the rate if the ﬁrst  step is slow.  (b) Iden-fy the intermediate.    Consider the diagram to the  right, represen-ng two steps  in an overall reac-on.  red = O, blue = N, green =  F  (a)  Write the equa-on for the overall reac-on.  NO2 + F2  NO2F+ F  NO2+F  NO2F  (b) Iden-fy the intermediate.    (c) Find the rate if the ﬁrst  step is slow.  Consider the diagram to the  right, represen-ng two steps  in an overall reac-on.  red = O, blue = N, green =  F  (a)  Write the equa-on for the overall reac-on.  NO2 + F2  NO2F+ F  NO2+F  NO2F  (b) Iden-fy the intermediate.  F is the intermediate  (c) Find the rate if the ﬁrst  step is slow.  Consider the diagram to the  right, represen-ng two steps  in an overall reac-on.  red = O, blue = N, green =  F  (a)  Write the equa-on for the overall reac-on.  NO2 + F2  NO2F+ F  NO2+F  NO2F  (c) Find the rate if the ﬁrst  step is slow.  Rate = k[NO2][F]  (b) Iden-fy the intermediate.  F is the intermediate  Catalysis  A catalyst is a substance that changes the rate of a chemical  reac-on without undergoing a chemical change itself.  A homogeneous catalyst is present in the same phase as the  reactants and products.  Example (slow):  2 H2O2(aq) → 2 H2O(l) + O2(g)  With bromine as a catalyst (fast):    step 1:  2 Br‐(aq) + H2O2(aq) + 2 H+(aq) → Br2(aq) + 2H2O(l)   step 2:  Br2(aq) + H2O2(aq) → 2Br‐(aq) + 2H+(aq) + O2(g)   sum:   2 H2O2(aq) → 2 H2O(l) + O2(g)  A catalyst usually speeds up the reac-on rate by providing a diﬀerent  pathway for the reac-on having a lower ac-va-on barrier.  The energies  of the reactants and products are not changed, only the speed of the  reac-on.  The diagram above represents an imaginary two‐step mechanism. The orange,   green and blue spheres represent elements A, B & C, respec-vely.   (a)  Write the equa-on for the net reac-on that is occurring.  (b) Iden-fy the intermediate.    (c) Iden-fy the catalyst.      The diagram above represents an imaginary two‐step mechanism. The orange,   green and blue spheres represent elements A, B & C, respec-vely.   (a)  Write the equa-on for the net reac-on that is occurring.  A2+ AB+ AC BA2 + A + AC  BA2+ A + AC  A2+ BA2+ C  Net: AB+AC BA2+ C  (b) Iden-fy the intermediate.  A  (c) Iden-fy the catalyst. A2 is the catalyst      Heterogenous Catalysis  A heterogeneous catalyst exists in a diﬀerent phase than the  reactants and products.  Usually the catalyst is a solid and the reactants and products   are in a gas or liquid phase.  The ini-al step is usually the adsorp-on of the reactants to the  surface of the catalyst.  Example ‐‐  the reac-on of hydrogen gas with ethylene gas to  form ethane gas     C2H4(g) + H2(g) → C2H6(g)  (a)  (b)  (c)  (d)  The hydrogen and ethylene adsorb to a metal surface.  The H‐H bond is broken to give hydrogen atoms.  The hydrogen atoms move to the adsorbed ethylene and react.  Fully formed ethane molecules desorb and return to the gas phase.  Enzymes  The rates of most biochemical reac-ons are highly regulated  by catalysts called enzymes.  Most enzymes are proteins (polymers of amino acids).  These proteins have large molar masses (e.g., 100,000 gm/mol)  and very well‐deﬁned three‐dimensional structures.  Enzymes are generally highly speciﬁc for the reac-on they catalyze.  The ac-vi-es of enzymes themselves are highly regulated.  An enzyme from the liver called catalase causes hydrogen  peroxide to rapidly decompose into water and oxygen.   The reac-on catalyzed by an enzyme occurs at a small por-on  of the enzyme called the ac-ve site.  The reactants are called substrates.  The speciﬁcity is explained in part by the lock‐and‐key model.  An enzyme called lysozyme.  Lysozyme with its bound substrate.  ...
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## This note was uploaded on 09/13/2011 for the course CHEM 102 taught by Professor Todd during the Spring '08 term at UNC.

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