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Chemistry Exam Review: Equations, Reactions, and Electrochemistry - Prof. Jason D. Kahn, Exams of Chemistry

A review for a final exam in a university-level chemistry course, focusing on topics such as thermodynamics, heat capacity, photosynthesis, kinetics, and electrochemistry. It includes various equations, reactions, and electrode potentials. Students are expected to understand concepts related to the first law of thermodynamics, heat capacity, and the role of photosystems in plants, as well as the kinetics of reactions and the steady state approximation.

Typology: Exams

2010/2011

Uploaded on 06/06/2011

koofers-user-gek
koofers-user-gek 🇺🇸

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Download Chemistry Exam Review: Equations, Reactions, and Electrochemistry - Prof. Jason D. Kahn and more Exams Chemistry in PDF only on Docsity!   Chemistry  271,  Section  23xx     University  of  Maryland,  College  Park   General  Chemistry  and  Energetics   Prof.  Jason  Kahn   Final  Exam  (200  points  total)   December  14,  2010   You  have  120  minutes  for  this  exam.     Exams  written  in  pencil  or  erasable  ink  will  not  be  re-­‐graded  under  any  circumstances.   Explanations  should  be  concise  and  clear.  I  have  given  you  more  space  than  you  should  need.  There  is   extra  space  on  the  last  page  if  you  need  it.   You  will  need  a  calculator  for  this  exam.  No  other  study  aids  or  materials  are  permitted.   Partial  credit  will  be  given,  i.e.,  if  you  don’t  know,  guess.   Honor  Pledge:  At  the  end  of  the  examination  time  ,  please  write  out  the  following  sentence  and  sign  it,   or  talk  to  me  about  it:   “I  pledge  on  my  honor  that  I  have  not  given  or  received  any  unauthorized  assistance  on  this   examination.”               Your  Name:       Your  SID  #:     Viewing:  Friday,  December  17,  11-­2  Chemistry  1110-­1111   Useful  Equations:   ∆S  –  q/T  ≥  0   pH  =  –  log([H+])   S  =  klnW   ∆G°  =  –  RTlnKeq   ∆G  =  ∆H  –  T∆S   PV  =  nRT   Ka  =  [H+][A–]/[HA]   ni/n0  =  exp[–(ε  i-­‐  ε  0)/kT]   ∆G  =  –nℱE   °C  =  °K  –  273.15   R  =  8.314  J/mole  K   E  =  E°  –  2.303(RT/nℱ)log10Q   2.303RT/ℱ  =  0.0592  Volts  at  25  °C   ℱ  =  96500  C(oulomb)/mole   Standard  hydrogen  electrode:   2  H+  (aq,  1  M)  +  2  e–  →  H2  (g)   E°  =  0.000  V   lnKeq  =  –∆H°/RT  +  ∆S°/R   ln  k  =  (–Ea/RT)  +  ln  A   1  Volt  =  1  Joule/Coulomb   [A]  =  [A]0  –  kt   ln[A]  =  ln[A]0  –  kt   1/[A]  =  1/[A]0  +  2kt   P(v)dv  =  Cv2exp(–mv2/2kT)   dS  =  (cp/T)  dT   cp  =  qp/dT  =  dH/dT   Chemistry  271,  section  23xx  Final  Exam,  12/14/10   2/10   Score  for  the  page     1.   Multiple  Choice  (24  pts)   (i,  4  pts)  The  Third  Law  of  Thermodynamics  gives  us  a  reference  state  of  zero  entropy.  What  else  is   necessary  for  measuring  absolute  entropies  of  pure  substances  at  non-­‐zero  temperature?   (a)  The  First  Law  of  Thermodynamics.   (b)  Measurements  of  heat  capacity  as  a  function  of  temperature.   (c)  Calculation  of  the  free  energy  of  formation  from  tabulated  data.   (d)  Enumeration  of  microstates.   (e)  We  cannot  measure  absolute  entropy,  only  changes  in  entropy.   (ii;  4  pts)  A  second-­‐order  rate  constant     (a) has  units  of  time–1.   (b) changes  during  the  course  of  the  reaction  as  the  reactant  is  depleted.   (c) can  almost  never  be  faster  than  diffusion-­‐controlled.   (d) has  units  of  M–1s–1.   (e) (c)  and  (d).   (iii,  4  pts)  The  Steady  State  Approximation     (a) is  not  useful  in  enzyme  kinetics  because  substrate  concentration  is  constantly  changing.   (b) can  be  applied  only  to  reactions  of  the  type  A  →  B  →  C.   (c) holds  only  at  equilibrium,  when  the  state  of  the  system  is  steady.   (d) includes  rapid  pre-­‐equilibrium  as  a  special  case.   (e) is  applicable  to  the  product  of  any  rapid  reaction.   (iv;  4  pts)  The  half  life  t1/2  of  a  second  order  reaction  A+A  →  B   (a) =  (ln2)/k   (b) =  1/{k([A]0)  2}   (c) =  [A]0/k   (d) is  always  shorter  than  the  half  life  of  a  first-­‐order  reaction.   (e) none  of  the  above   (v,  4  pts)  The  function  of  Photosystem  I  in  plants  is  to     (a) use  light  energy  to  reduce  NADP+  to  NADPH.   (b) collect  light  energy  and  deliver  it  to  Photosystem  II.   (c) oxidize  water  to  provide  light.   (d) re-­‐reduce  the  reaction  center  of  Photosystem  II.   (e) reduce  water  to  maintain  a  basic  pH.   (vi,  4  pts)  Assimilation  is  defined  as   (a) using  a  substance  as  a  terminal  electron  acceptor  and  making  no  further  use  of  the  product.   (b) converting  (typically  through  reduction)  inorganic  material  into  biomass.   (c) nitrogen  fixation  and  all  other  processes  that  split  diatomic  molecules.   (d) absorption  of  enemies  by  an  oppressive  regime.   (e) none  of  the  above.   Chemistry  271,  section  23xx  Final  Exam,  12/14/10   5/10   Score  for  the  page     (c;  4  pts)  Two  possible  failure  pathways  for  photosynthesis  are  _____________________________________,  if  the   electron  is  not  transferred  away  from  the  excited  state  fast  enough,  or  __________________________________   ___________________________________,  if  the  electron  is  not  removed  from  the  immediate  vicinity  fast  enough.   (d;  9  pts)  How  does  increased  atmospheric  CO2  cause  global  warming?                     4.  Kinetics  and  Thermodynamics  (45  pts)   (a;  7  pts)  The  Maxwell-­‐Boltzmann  distribution  and  Arrhenius  equation  state  that  the  rate  of  an   elementary  reaction  always  increases  with  increasing  temperature.  So  how  is  it  possible  for  the   equilibrium  constant  of  a  reaction  A  ⇌  B  to  decrease  as  temperature  increases?  Would  the   reaction  be  (circle  one)        exothermic          or          endothermic?       Chemistry  271,  section  23xx  Final  Exam,  12/14/10   6/10   Score  for  the  page     (b;  24  pts)  Fill  in  the  blanks.    A  catalyst  can  change  the  ________________________  and/or  the   _____________________________of  a  reaction  but  not  its  ___________________________  constant.   The  underlined  “v2”  factor  in  the  Maxwell-­‐Boltzmann  distribution  Cv2(exp(–mv2/2kT))  arises  because   of  the  ω(ε)  factor  in  the  Boltzmann  distribution,  the  __________________________________________,  which  is  the   number  of  ways  in  which  a  particle  can  have  _______________________________________.  The  underlined  v2   comes  from  the  _____________________________________  of  a  sphere  of  radius  v.     A  rate  law  can  be  determined  by  inspection  only  for  a(n)  ________________________________________reaction.   The  “2”  in  SN2  comes  from  the  fact  that_______________________________________________________________________.   (c;  8  pts)  Sketch  the  Arrhenius  plot  used  to  measure  activation  energy  Ea  and  the  preexponential   factor  A.  Label  the  axes.  What  is  the  molecular  meaning  of  Ea  according  to  collision  theory  for  gas   phase  reactions?                         (d;  6  pts)  In  discussing  gas-­‐phase  reactions  we  frequently  discussed  the  need  for  a  collision  gas  M  to   catalyze  either  dissociation  or  recombination  reactions.  It  is  easy  to  understand  that  the  collision  gas   can  deliver  energy  to  break  apart  molecules  (think  of  any  parking  lot).  Why  is  the  collision  gas  needed   to  allow  atoms/molecules  to  recombine?   Chemistry  271,  section  23xx  Final  Exam,  12/14/10   7/10   Score  for  the  page     5.  Kinetics  and  the  Steady  State  Approximation  (35  pts)     Leucine  zipper  proteins  (Z)  are  monomeric  in  solution  by  themselves  but  they  bind  to  DNA  (D)  as   dimers,  with  the  overall  reaction  being   Z + Z + D Kbind   Z2D   There  has  been  debate  about  the  reaction  mechanism.  One  possibility  is  that  a  monomer  Z  must   dimerize  to  form  an  unstable  intermediate  Z2  that  either  readily  falls  apart  or  else  binds  to  DNA.   The  proposed  mechanism  is  as  follows,  where  we  ignore  the  reverse  of  the  second  step  (this  is   reasonable  if  the  complex  is  very  stable):   Z + Z k1 k−1    Z2 Z2 + D k2⎯ →⎯ Z2D   (a;  5  pts)  Write  down  the  differential  rate  law  for  the  appearance  of  Z2D.  It  includes  the  unknown   [Z2].             (b;  12  pts)  Apply  the  Steady  State  Approximation  to  the  unstable  Z2  intermediate  to  determine  its   steady  state  concentration  during  the  binding  reaction.  
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