Final exam

1
Final exam

• Format Much like hour exams
• You may bring a 3x5 index card with any
  HANDWRITTEN information on it that you like. I
  would suggest formulas, constants, etc.
  (tabulated values like activity series,
  enthalpies, bond energies, and the like will be
  provided).
• During the exam you may buy formulas for a 2
  point penalty.
• 80 old material
• Read notes carefully and try taking old exams.
  Anything you get wrong---write out the concept
  that is behind the correct answer
• 20 new material (Chapter 14)
• Read through notes carefully and do practice
  problems

2
Chapter 14 problems

• 14.49
• The rate of the reaction
• CH3COOC2H5 (aq) OH (aq) ? CH3COO (aq)
  C2H5OH (aq)
• was measured at several temperatures, and the
  following data were collected. Using these data,
  graph ln k versus 1/T. Using your graph,
  determine the value of Ea.
• T (ºC) k(M-1s-1) 1/T ln k
• 15 0.0521 0.00347 -2.955
• 25 0.101 0.00336 -2.293
• 35 0.184 0.00325 -1.693
• 45 0.332 0.00314 -1.103

3
The slope is -5.71 x 103 Recall ln k
-Ea/RT ln A so the slope is -Ea /R
4
Problem 14.58 Based on the following reaction
profile a) how many intermediates are formed
in the reaction A ? D? b) how many transition
states are there? c) which step is the
fastest? d) is the reaction A ? D exothermic or
endothermic?

5
Problem 14.58 (cont) a) how many intermediates
are formed in the reaction A ? D? valleys,
2 b) how many transition states are there?
peaks, 3 c) which step is the fastest? the
one with the smallest Ea, C ? D d) is the
reaction A ? D exothermic or endothermic? Since
ED gt EA, endothermic
6

• Problems 14.82
• Cyclopentadiene (C5H6) reacts with itself to form
  dicyclopentadiene (C10H12). A 0.0400 M solution
  of C5H6 was monitored as the reaction
• 2 C5H6 ? C10H12
• proceeded. The following data were collected.
• t (s) C5H6 (M)
• 0.0 0.0400
• 50.0 0.0300
• 100.0 0.0240
• 150.0 0.0200
• 200.0 0.0174
• What is the order of the reaction?
• What is the value of the rate constant?

7
The graph of 1/concentration versus t is the one
that results in the best linear fit. Therefore,
the reaction must be second order. For a second
order reaction
The best straight line through the points is y
25.2 0.163 x so k 0.163 M-1s-1.
8
Problems 14.87

• The following mechanism has been proposed for the
  gas-phase reaction of chloroform (CHCl3) and
  chlorine
• Step 1 Cl2 (g) ? 2 Cl (g) (fast
  equilibrium)
• Step 2 Cl (g) CHCl3 (g) ? HCl (g) CCl3
  (g) (slow)
• Step 3 Cl (g) CCl3 (g) ? CCl4 (4) (fast)
• What is the overall reaction?
• What are the intermediates in the mechanism?
• What is the molecularity of the elementary steps
• What is the rate-determining step?
• What is the rate law predicted by this mechanism?
• We add the elementary steps to give Cl2 (g)
  CHCl3 (g)? HCl (g) CCl4 (g)
• The intermediates are the species that are made,
  then consumed Cl and CCl3
• The first step is unimolecular and the others are
  bimolecular
• The rate determining step is the slowest one,
  Step 2

9
Step 1 Cl2 (g) ? 2 Cl (g) (fast
equilibrium) Step 2 Cl (g) CHCl3 (g) ? HCl
(g) CCl3 (g) (slow) Step 3 Cl (g) CCl3
(g) ? CCl4 (4) (fast) e) The rate is determined
by Step 2, so Rate k2ClCHCl3 but
the concentration of Cl is determined by the
equilibrium in the first reaction, ie, when
k1Cl2 k-1Cl2 Cl
(k1Cl2/k-1)½ given a rate law of Rate
k2(k1/k-1)½ Cl2½ CHCl3 or, setting
the combination of rate constant factors to
k Rate kCl2½ CHCl3