Chap14-Part4

Chap14-Part4 - Second‐Order Processes

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Unformatted text preview: Second‐Order Processes  The decomposi,on of NO2 at 300°C is described by the  equa,on  1  NO2 (g)  NO (g) +      O2 (g)  2  and yields data comparable to this:  Time (s) [NO2], M 0.0 0.01000 50.0 0.00787 100.0 0.00649 200.0 0.00481 300.0 0.00380 2  Second‐Order Processes  •  PloDng ln [NO2] vs. t yields the  graph at the right.  •  The plot is not a straight line,  so the process is not ﬁrst‐ order in [A].  Time (s) [NO2], M ln [NO2] 0.0 0.01000 −4.610 50.0 0.00787 −4.845 100.0 0.00649 −5.038 200.0 0.00481 −5.337 300.0 0.00380 −5.573 3  Second‐Order Processes  •  Time (s) [NO2], M 1/[NO2] 0.0 0.01000 100 50.0 0.00787 127 100.0 0.00649 0.00481 0.00380 263 •  Because this is a  straight line, the  process is second‐ order in [A].  208 300.0 however, gives this  plot.  154 200.0 1  Graphing         vs. t,  [NO2]  4  Sucrose (a disaccharide) reacts in acidic solu,ons to form two  monosaccharides (glucose & fructose).  At 23oC in 0.5 M HCl,   the following data were obtained for the disappearance of sucrose:  ,me (min)    0    39    80    140  210            sucrose (M)   0.316   0.274   0.238   0.190   0.146  Is the reac,on ﬁrst or second order with respect to sucrose?  5  Sucrose (a disaccharide) reacts in acidic solu,ons to form two  monosaccharides (glucose & fructose).  At 23oC in 0.5 M HCl,   the following data were obtained for the disappearance of sucrose:  ,me (min)    0    39    80    140  210            sucrose (M)   ln[sucrose]    1/[sucrose]   0.316  0.274  0.238  0.190  0.146                      ‐1.152    ‐1.295    ‐1.435    ‐1.661    ‐1924              3.16    3.65    4.20   5.26   6.85          Is the reac,on ﬁrst or second order with respect to sucrose? First order  6  Temperature Dependence of Reac;on Rates  Rate constants k increase with temperature.  Example:  A chemiluminescent reac,on.  At the higher temperature the reac,on   proceeds faster.    7  Temperature Dependence of Reac;on Rates  CH3NC  CH3CN  gas‐phase isomeriza,on  8  A ﬁrst order reac,on is examined  for three diﬀerent condi,ons.  (a) Which two lines represent data  obtained at the same temperature?  (b) Which two lines represent data  obtained with the same ini,al condi,on?  (c) Which line represents data taken  at the lower temperature?  9  A ﬁrst order reac,on is examined  for three diﬀerent condi,ons.  (a) Which two lines represent data  obtained at the same temperature?  Line 1 and 2: Same slope  Started at 2 diﬀerent ini,al concentra,ons  (b) Which two lines represent data  obtained with the same ini,al condi,on?  (c) Which line represents data taken  at the lower temperature?  10  A ﬁrst order reac,on is examined  for three diﬀerent condi,ons.  (a) Which two lines represent data  obtained at the same temperature?  Line 1 and 2: Same slope  Started at 2 diﬀerent ini,al concentra,ons  (b) Which two lines represent data  obtained with the same ini,al condi,on?  Line 2 and 3: Same ini,al concentra,on  But 2 diﬀerent temperatures  (c) Which line represents data taken  at the lower temperature?  11  A ﬁrst order reac,on is examined  for three diﬀerent condi,ons.  (a) Which two lines represent data  obtained at the same temperature?  Line 1 and 2: Same slope  Started at 2 diﬀerent ini,al concentra,ons  (b) Which two lines represent data  obtained with the same ini,al condi,on?  Line 2 and 3: Same ini,al concentra,on  But 2 diﬀerent temperatures  (c) Which line represents data taken  at the lower temperature?  Line 3: Smaller slope  12  The Collision Model  For bimolecular or higher order reac,ons, the reactants must  collide for the reac,on to proceed.  Higher temperatures lead to more frequent collisions.  Higher reactant concentra,ons lead to more frequent collisions.  Higher temperatures lead to more forceful collisions.  All three of these factors give higher reac,on rates.  For collisions to be successful, the reactants must be properly  oriented.   Example:  Cl + NOCl → Cl2 + NO  eﬀec,ve   ineﬀec,ve  Properly oriented molecular collisions are important for a reac,on  Ac;va;on Energy  For a reac,on to proceed, the colliding molecules must have a total  kine,c energy greater than an energy called the ac,va,on energy = Ea.  The value of Ea is diﬀerent for diﬀerent reac,ons.  The value of Ea describes the ac,va,on barrier that must be overcome  for the reac,on to proceed.  Example:  Methyl Isonitrile Gas‐Phase Isomeriza,on  CH3NC  CH3CN     “intermediate”  or “ac,vated complex”  or “transi,on state”  Reac,on Coordinate Diagrams   It is helpful to  visualize energy  changes throughout  a process on a  reac,on coordinate  diagram like this one  for the  rearrangement of  methyl isonitrile.  © 2009, Pren,ce‐Hall, Inc.  Reac,on Coordinate Diagrams  •  The diagram shows the  energy of the reactants  and products (and,  therefore, ΔE).  •  The high point on the  diagram is the transi,on  state.  •  The species present at the transi,on state is called  the ac,vated complex.  •  The energy gap between the reactants and the ac,vated  complex is the ac,va,on energy barrier.  © 2009, Pren,ce‐Hall, Inc.  The bond between the H3C group and the N≡C group must be stretched by the  incoming energy, allowing the N≡C group to rotate.  Amerwards, the C‐C bond  begins to form and the energy drops.  Ea is the ac,va,on energy and ΔE is the  energy between the ini,al and ﬁnal states.  The rate is determined by Ea.  Maxwell–Boltzmann Distribu,ons  •  Temperature is  deﬁned as a  measure of the  average kine,c  energy of the  molecules in a  sample.  •  At any temperature there is a wide distribu,on  of kine,c energies.  © 2009, Pren,ce‐Hall, Inc.  Maxwell–Boltzmann Distribu,ons  •  As the temperature  increases, the curve  ﬂaqens and  broadens.  •  Thus at higher  temperatures, a  larger popula,on of  molecules has higher  energy.  © 2009, Pren,ce‐Hall, Inc.  Maxwell–Boltzmann Distribu,ons  •  If the doqed line represents the ac,va,on energy,  then as the temperature increases, so does the  frac,on of molecules that can overcome the  ac,va,on energy barrier.  • As a result, the  reac,on rate  increases.  © 2009, Pren,ce‐Hall, Inc.  Maxwell–Boltzmann Distribu,ons  This frac,on of molecules that has an energy equal to or greater than  Ea is given by,  f = e  ‐Ea  RT  where R is the gas constant and T is the Kelvin temperature.  © 2009, Pren,ce‐Hall, Inc.  The ac,va,on energy for the isomeriza,on of methyl isonitrile  is 160 kJ/mol.  a) Calculate the frac,on of molecules with an energy > Ea at 500 K.  b) Calculate the frac,on of molecules with an energy > Ea at 510 K.  c) Find the ra,o.  The ac,va,on energy for the isomeriza,on of methyl isonitrile  is 160 kJ/mol.  a) Calculate the frac,on of molecules with an energy > Ea at 500 K.  f = e  ‐Ea  RT  Ea = 160 kJ/mol = 1.6X105 J/mol , T=500K R= 8.314J/mol‐K   ‐Ea/RT = 160000/(500X8.314) =‐38.5   f = e‐38.5 =  2X10‐17  b) Calculate the frac,on of molecules with an energy > Ea at 510 K.  Ea = 160 kJ/mol = 1.6X105 J/mol , T=510K R= 8.314J/mol‐K   ‐Ea/RT = 160000/(510X8.314) =‐37.7   f = e‐38.5 =  4X10‐17  c) Find the ra,o.  f at 510 /f at 500 = 2  An Increase of 10K means that twice more molecules having this  energy.  Arrhenius Equa,on   Svante Arrhenius developed a mathema,cal  rela,onship between k and Ea:  k = A e  ‐Ea  RT   where A is the frequency factor, a number that  represents the likelihood that collisions would  occur with the proper orienta,on for reac,on.  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 Ac;va;on 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 ...
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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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