Chapter 13 - Rates of Reaction Chemical Kinetics

1
Chapter 13 - Rates of ReactionChemical Kinetics

• End-of-Chapter Problems Pages 577-589
• Sapling Homework, Chapter 13, due 11/25/2013

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End-of-Chapter Problems - Chapter 13 pages 567-579

• 1 - 12 16 17 18 20 23 25 26 33
  34
• 35 36 37 39 45 49 51 53 55 57
  
• 59 60 65 67 68 71 75 83
  84
• 87 89 117 125

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Lecture Outline

• I. Background on Rates Mechanisms
• II. Rates
• III. Rate Law (Variables T, Mech M)
• IV. Order
• Determining order a) Varying M, measuring
  rates
• Determining order b) Integrated rate
  expression graph
• 1st Order
• 2nd Order
• 0th Order
• Order Summary
• V. Temperature Arrhenius Equation
• VI. Mechanisms
• Introduction, examples and catalysts

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I. Background on Rates Mechanisms

• Chemists study reactions. Some of what we study
  
• - Major Minor Products
• - Reactants
• - Completion
• - Effects of Catalysts
• - Separation and Purification of Products
• - Effects of Variables on Rxn Speed (rate) on
  Products
• - Rates of the Reaction
• - Mechanism of the Reaction
• This chapter deals with rates mechanisms of
  reactions.
• Mechanism Step-by-step progress of the chemical
  reaction.
• Rate How fast the reaction proceeds (usually
  ? M / Time)

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I. Background on Rates Mechanisms

• Study of rates is useful since the results will
• 1) Indicate how to manipulate factors to
  control the reaction
• 2) Lead to the mechanism of the reaction
• 3) Indicate time needed to get a given amount
  of product
• 4) Indicate amount of product in a given amount
  of time
• Study of mechanisms important. With the
  mechanism we can
• 1) Predict products of similar reactions
• 2) Better understand the reaction
• 3) Accurately manipulate the reaction for a
  desired result
• 4) Organize and simplify the study of organic
  chemistry
• OH
  I
• Example CH3-CH-CH3 H I-
  -------) CH3-CH-CH3 H2O

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I. Background on Rates Mechanisms

• Main Factors which influence reaction rate
• Concentrations of Reactants - Rates usually
  increase as reactant concentrations increase.
• Reaction Temperature - An increase in temperature
  increases the rate of a reaction.
• Presence of a Catalyst (not all rxns have
  catalysts)
• A catalyst is a substance which increases the
  rate of a reaction without being consumed in the
  overall reaction.
• The concentration of the catalyst or its surface
  area (if insoluble) are variables which influence
  the rate.
• Some catalysts are incredibly complex - like
  enzymes and others are quite simple H
  H2O CH2 CH2 ------) CH3-CH2-OH H
• Type of Reactants
• Surface Area of an Insoluble Reactant

7
II. Rates

• Reaction Rate either the increase in M of
  product per unit time or the decrease in M of
  reactant per unit time ?M / ?T
• Note X moles X / Liter
• Example H Catalyst
• Sucrose H2O --------------) Glucose
  Fructose
• Rate rate of formation of either product.
• Rate ? M of glucose / ?sec ?glucose
  / ?sec
• or
• Rate rate of disappearance of either
  reactant.
• Rate - ?sucrose / ?sec (- since want a
  rate)
• In order to obtain rate, we need a way to measure
  ?M of a reactant or product with respect to time.

8
II. Rates

• Example 2 N2O5 -----) 4 NO2 1 O2
• If we want to equalize the rates then
• Rate ?O2 1/4 ?NO2 - 1/2
  ?N2O5
• ?t ?t ?t
• - divide by balancing coefficients when we
  equalize rates.
• Various Rates can be determined 1)
  instantaneous rate at a given time 2) average
  rate over a long period of time or 3) the
  initial rate rate at the beginning of the rxn
  ie rate at t0.0 (this is used the most).
• On next slide the ?O2 versus time is plotted
  for a reaction.
• Note 1) how the rate changes with time 2) that
  rate is the tangent at a given point on the curve.

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II. Rates
10
II. Rates
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II. Rates

• Calculate the Average Rate for
• I- ClO- -----) Cl- IO-
• Given the I- and time in seconds, then what is
  the average rate?
• Time (s) M I-
• 2.00 0.00169
• 8.00 0.00101
• - Rate - ?M / ?T
• - Rate - (0.00101 - 0.00169) M / (8.00 - 2.00)
  s 6.8x10-4 M / 6.00s
• - ave Rate 1.1 x 10-4 M/s

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II. Rates

• Need to obtain the change in M of a given reagent
  per change in time could follow any parameter
  related to concentration.
• Examples of what one might follow to obtain
  rates
• - a change in pressure (if gas produced or
  consumed in the rxn)
• - a change in pH (if acidity changes in the rxn)
• - a change in absorbance of electromagnetic
  radiation (EMR)
• Usually measure absorbance of Visible or UV EMR
  at a given ? - caused by a change in reactant or
  product concentration.
• A ?bc at given ? (wavelength) This is Beers
  Law from Ch 121 (know)
• A absorbance use spectrophotometer to
  measure has no units.
• ? molar absorptivity a constant _at_ given ?
  has units of M-1cm-1
• b pathlength of EMR through sample usually
  1.00 cm cuvette used.
• c concentration in M
• A plot of A versus M _at_ given ? will yield a
  straight line and the equation
• A Ebc intercept If follow ?A then
  can convert to ?M get rate.

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III. Rate Law rate k x Am x Bn
for A B

• Rate Law relates the rate to temperature
  concentration.
• Rate law is given in terms of REACTANTS only
  (convention).
• k rate constant handles the temperature
  variable.
• The exponents are the order handle the
  concentration variables.
• General form of the rate law for a A b
  B c C d D
• rate k x Am x Bn
• - Order for A is m order for B is n
  Overall order is m n
• - m n are determined experimentally.
• - k, also determined experimentally units
  depend upon overall order.

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III. Rate Law - Rate Constant Units

• Note Assume time is in seconds (s). Rate k
  Ax
• Solve for k plug in units k Rate / Ax
• (this may be useful for one of the online HW
  problems)
• Overall Rxn Order, x Units for k
• zero Ms-1
• first s-1
  
• second M-1s-1
• third M-2s-1

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III. Rate Law

• Example 2 NO2 1 F2 ----) 2 NO2F
  Rate k NO2n x F2m
• In the laboratory, the overall rate was found to
  be second order.
• n m 2 Possibilities n2 m0
  n0 m2 n1 m1
• Experiments demonstrated that n1 and m1 How?
  By running the reaction at least three times
  1 getting rate at certain initial
  concentrations of NO2 F2 2 getting rate
  when keeping NO2 the same doubling F2 3
  getting rate when keeping F2 the same
  doubling NO2.
• They found that doubling NO2 doubled the rate
  doubling F2 doubled the rate so, both
  coefficients had to be 1.
• Rate k NO2n F2m k NO21 F21

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III. Rate Law - Rate Constant

• Determination of the rate constant, k. You are
  given
• 1) aA bB ? cC dD
• 2) Rate kA0B1 (0 1 were determined
  experimentally)
• 3) M of A B 2.00 moles/L Rate 2.50 x
  10-2 M/s
• Determine the value for k give complete rate
  expression.
• Rate kA0B1 k Rate
• A0B1
• k Rate 2.50 x 10-2 M/s 2.50
  x 10-2 M/s
• A0B1 2.00 M0 x 2.00 M1
  1 x 2.00 M1
• k 1.25 x 10-2 s-1
• Rate (1.25 x 10-2 s-1) A0 B1
  (completed rate expression)

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III. Rate Law

• Rate Law for a reaction is found experimentally
  except for a single step in a mechanism
  (elementary reaction). Assume rxn is NOT
  elementary unless told that it is a one step in
  mechanism.
• Given 1NO2 1CO -----) 1NO 1CO2
• By experiment, the rate law was found to be
• rate kNO22CO0 or rate kNO22
  ( Note CO0 1 )
• The order WRT each reactant the overall order
  are
• 2nd order WRT NO2 0th order WRT CO
  2nd order overall

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IV. Order

• The concentration variables are handled by the
  exponents - the order.
• The orders are determined experimentally except
  for one case An
• elementary reaction.
• Elementary reactions are one step reactions which
  are the individual steps in a mechanism. (For
  an elementary reaction only the balancing
  coefficients determine the order.) - Important
• Example 1 for a multistep reaction 1 CH2Br2
  2 KI ---) 1 CH2I2 2 KBr
• If experimentation found that m n were both
  first order then
• rate k CH2Br2 1 KI1
• Example 2 for an elementary reaction 2 O3
  ---) 3 O2 (told it is elementary)
• No need for experimentation order comes from
  balancing coefficients
• rate kO32

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IV. Determination of Order

• Order - from units of k If you are given the
  units of the rate constant for a reaction, then
  you will know the overall order (slide 14). Not
  too common.
• Order by Method 1 - from altering M Measure
  initial rates keeping one reactant constant and
  change the concentration of another observe the
  rates calculate order as illustrated in the next
  few slides.
• Order by Method 2 - from integrated rate
  expression Use calculus integrate the rate
  expression between the limits of time 0 time
  t. By plotting out the variables of these
  integrated rate expressions you can determine the
  order. This will be shown in the lecture, and
  you will be doing this in the kinetics lab.

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Order by Method 1 - from altering M Measure
initial rates keeping one reactant constant and
change the concentration of another observe the
rates calculate order as illustrated in the next
few slides.
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IV. Determination of Order by varying M

• Example 1 Determine the order for rate
  expression for
• 2N2O5 ---) 4NO2 1O2 rate k
  N2O5m
• Exp 1 Rate 4.8x10-6 Ms-1 at 1.0x10-2 M
  N2O5
• Exp 2 Rate 9.6x10-6 Ms-1 at 2.0x10-2 M
  N2O5
• Order Note that when N2O5 doubles, the rate
  doubles. Since
• rate a N2O5m rate doubles when N2O5
  doubles, the value of
• must be 1 the order is 1.
• - rate a N2O5m rate doubles when
  N2O5 doubles, then
• go from 1m 1 to 2m 2 m has to be
  1
• Rate law rate k x N2O51

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IV. Determination of Order by varying M

• Summary
• EFFECTS of doubling reagent M while keeping
  others constant
• Rate remains the same 0th order M0
• Rate doubles 1st order M1
• Rate quadruples 2nd order M2
• Rate increases eightfold 3rd order M3

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IV. Determination of Order by varying M

• Example 2 2 NO Cl2 -----) 2 NOCl -
  Calculate order of Rxn
• Exp Initial NO Initial Cl2 Initial Rate,
  Ms-1
• 1 0.0125 0.0255 2.27x10-5
• 2 0.0125 0.0510 4.57x10-5
• 3 0.0250 0.0255 9.08x10-5
• Rate kNOmCl2n
• a) calculate n From 1 2 - double Cl2 keep
  NO constant
• rate increases by factor of 2.01 n 1
• b) calculate m From 1 3 - double NO keep
  Cl2 constant
• rate increases by factor of 4.00 m 2
• Rate kNO2Cl21
• 2nd order wrt NO 1st order wrt Cl2
  3rd order overall

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• Order by Method 2 - from integrated rate
  expression
• Integrate the rate expression between the limits
  of time 0 and time t. By plotting out the
  variables of these integrated rate expressions
  you can determine the order. You will be doing
  this in the kinetics lab.

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IV. Determination of Order by Integrated Rate
Expression

• Summary on use of logarithms
• Log involves s to the base 10 Log 10x
  x
• Ln Natural log uses s to the base e Ln ex
  x
• Ln A/B Ln A - Ln B (or Log)
• Ln A x B Ln A Ln B (or Log)
• Ln Ab b Ln A (or Log)
• To obtain either log or ln use the appropriate
  calculator function.
• Log 2.1x10-4 - 3.68 (note significant figure
  change - see below)
• (-4.0000000. 0.32 -3.68 cut off at
  first doubtful digit)
• To remove Ln Log use the inverse ex 10x
  functions on cal.
• Inverse log 3.00 or 103.00 1.0 x 103
• Inverse ln 3.00 or e3.00 20.

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IV. Order Integrated Rate Law - First Order Rxns

• 1) 1st Order Reactions aA -----) Products
  If 1st order, then
• -?A/?t kA1 (rate expression)
• This plot for first order data only gives minimal
  information

A
Time
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IV. Order Integrated Rate Law - First Order Rxns

• 1) 1st Order Reactions aA -----) bB -?A/?t
  kA1
• if we integrate from time t to 0, we get the
  following
• Y mX b
• lnAt - kt lnAo or
  lnAt/Ao -kt
• where At M of A at time t Ao M
  of A at t 0
• - A plot of lnAt versus t gives a straight line
  (Y mX b)
• b Slope (m) - k
• Note Only linear for 1st order

lnAt
Time, t
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Example of an integrated rate plot for a 1st
order reaction
Slope -k rise/run
Must be 1st order since plot of lnN2O5 vs t is
linear. Can get k from the slope.
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IV. Order Example using 1st order integrated
equation

• Example 2N2O5 ---) 4NO2 O2 rate k
  N2O51 (1st order)
• Given k 4.80x10-4s-1 N2O5to 1.65x10-2
  M what is N2O5 at 825 s?
• ln At - k x
  t ln Ao
• ln N2O5 - 4.80x10-4s-1 x 825 s
  ln 1.65x10-2
• ln N2O5 - 0.396
  - 4.104
• ln N2O5 - 4.500
• Take inverse ln of both sides INVERSE ln
  N2O5 N2O5
• INVERSE ln -4.500 e -4.500
  0.0111
• N2O5 0.0111 M

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IV. Order Integrated Rate Law - First Order Rxns

• Half-Life (t1/2) of 1st Order Reaction
• t1/2 time it takes for Ao to decrease to 1/2
  initial M ½Ao
• ln At /Ao -kt ln 1/2Ao /Ao
  -kt1/2
• ln 1/2 -kt1/2 -0.693 -kt1/2 t1/2
  0.693 / k
• Note 1) Time for 1/2 to disappear is
  independent of A for 1st order reaction. 2)
  This is an easy way to calculate 1st order rate
  constant, k.
• Example If t1/2 189 sec for 1st order
  decomposition of 1.0 mole of H2O2, then how much
  H2O2 will be left after 378 sec?
• Note 378/189 2 Goes through
  two half lives
• 1.0 mol ---) 0.50 mol ---) 0.25 mol

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IV. Order Integrated Rate Law - First Order Rxns

• Example Given a) k 3.66x10-3s-1 for
  decomposition of H2O2 and b) H2O2o 0.882 M.
  (Note Reaction must be 1st order examine units
  for k)
• Calculate
• 1) t1/2
• 2) How much will be left after one half-life?
• 1) t1/2 0.693/k
• t1/2 0.693 / 3.66x10-3s-1 189 s
• 2) M of H2O2 cut in half in one half-life
  (t1/2) will go from 0.882 to 0.441 M in 189 s.

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IV. Order Integrated Rate Law - Second Order Rxns

• 2) Second Order Reactions
• - Assume that aA -----) Products is 2nd
  order
• Rate - ?A / ? t k A2
• Integrate rate expression from time t to 0 get
  following
• 1/At k t 1/Ao So, a plot of
  1/At vs t should give a straight line with
  slope k and y intercept 1/Ao
• t1/2 1 Note Now t1/2 depends on
  initial M
• k x A0
• Note can tell if reaction is 2nd order from
  1/A vs t plot.

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IV. Order Integrated Rate Law - Second Order
Rxns Example
Plot of lnNO2 vs t is not linear not 1st
order.
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IV. Order Integrated Rate Law - 0th Order

• 3) 0th Order Reactions Assume A ---) B
  is 0th order
• Rate -kA0
• Rate -k
• - Integrated Rate Equation for a 0th order
  reaction
• At -k x t A0
• - a plot of At versus t will give a straight
  line
• - Again, if you let At 1/2 Ao then t
  t1/2
• - t1/2 A0 / 2k

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IV. Order Integrated Rate Law - Summary
? Rate when double M None
Double Quadruple
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IV. Order Integrated Rate Law - Summary
A? B ?A/?t kAn
Note slope k in each case
At
0th Order n0 At - kt Ao
Time, t
1st Order n1 lnAt - kt lnAo
1/At
2th Order n2 1/At kt 1/Ao
Time, t
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V. Temperature

• A collision needs to occur before a reaction can
  take place, the rate constant ( rate) of the
  reaction depends upon the
• 1) collision frequency (temperature)
• 2) number of collisions having enough energy for
  rxn (Ea)
• 3) orientation of particles upon collision
• Ea energy of activation minimum energy of
  collision in order for the reaction to take
  place.
• Ea ?H can be represented by Potential Energy
  Diagram can draw for one step or for several
  steps in a mechanism.

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V. Temperature Reaction Rate A)
Potential Energy Diagram for an Elementary
Reaction
?H
(Exothermic)
What is Ea for reverse reaction?
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V. Temperature Reaction Rate Arrhenius
Equation

• Arrhenius Equation relates rate constant (k),
  temperature (T), energy of activation (Ea in
  J/mole), orientation factor.
• k A e-Ea/RT R gas constant use R
  8.31 J/(Kmole)
• Take ln of both sides ln k -(Ea/R) 1/T
  ln A
• Y m X b
• Measure k at several temperatures and make plot
  of ln k
• versus 1/T. Slope of the curve - Ea/R
  (will give Ea)
• Note A is a constant includes orientation
  factor.
• Note Page 559 contains a form of the Arrhenius
  equation which may be useful for some online HW
  questions.

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V. Temperature Reaction Rate Arrhenius
EquationData below from 4 experiments - detn of
rate constant, k, at 4 temperatures for a rxn
ln A
o
Slope -Ea/R Can now determine Ea
ln k
o
o
o
1/T (in K-1)
ln k -(Ea/R) 1/T ln A
From k Ae-Ea/RT Y m X
b Use R 8.31J/(K.mol)
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VI. Mechanisms A) Introduction

• Mechanism step by step progress of chemical
  reaction.
• Most likely mechanism is determined
  experimentally from a study of rate data.
• The mechanism consists of one or more elementary
  reactions which add up to give you the overall
  reaction.
• Species which is generated then consumed in the
  mechanism is called an intermediate Species
  which is added, consumed then regenerated is a
  catalyst.
• Step with largest Ea (slowest step) is called the
  rate determining step governs overall reaction
  rate.

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VI. Mechanisms B) Example 1 - Information

• Overall Rxn O3 2NO2 -----) O2 N2O5
• Suggested two-step mechanism (from
  experimentation)
• Step 1) O3 NO2 -----) NO3 O2
  (slow)
• Step 2) NO3 NO2 -----) N2O5
• rate k O31NO21 - from slow first step
• Notes a) Two elementary reactions (NOTE
  balancing coefficients orders in an
  elementary rxn)
• b) Steps add up to give overall rxn
• c) NO3 is an intermediate (produced used
  up)
• d) There is no catalyst
• e) Slowest step governs the overall rate
• mechanism is useful will give us a) practical
  data, b) rate law, c) theoretical data, d)
  understanding of reaction

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VI. Mechanisms B) Example 2 - Calculate Rate
Expression

• Determine a) general rate expression b)
  complete rate law from the following mechanism
  Note can directly get the order for an
  elementary rxn from the balancing coefficients.
• 1) 1I2 2Io (fast equilibrium)
• 2) 2Io 1H2 2HI (slow)
• a) Overall Rxn from addn of steps 1I2 1H2
  2HI
• General rate law rate kI2xH2y
• b) Complete rate expression from mechanism
• From step 2 (rxn rate slow step rate)
  rate k2 Io2 x H21
• From step 1 Keq Io2/I21 Io2
  KeqI21 Substitute into above
• rate k2 Keq I21 H21
• rate k I21 H21

44
IV. Mechanisms C) Catalysts

• Catalyst A chemical which speeds up a reaction
  without being consumed in the reaction.
• - They operate by lowering the Ea for the rate
  determining step.
• - One example is Pt which speeds up the
  following rxn
• CO 1/2 O2 -----) CO2
• - Pt can be used in catalytic converter for
  your car exhaust.
• Most famous catalysts are proteins called
  enzymes.
• - Enzymes extremely specific biochemical
  catalysts that allow complex reactions to take
  place in living systems under mild conditions.
• - Enzymes are very complex, well designed, and
  usually have molecular weights in the tens of
  thousands.
• - Their mode of operation uncovered only 60
  years ago.

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VI. Mechanisms C) Catalysts
A catalyst speeds up the rxn by lowering the Ea
provides a different mechanism with a lower Ea
NewEa
46
VI. Mechanisms C) Catalysts
47
Final ExamTable

• Chapter 13 Equations
• Integrated Rate Equation Half-Life
• At - kt Ao Ao/2k
• lnAt - kt lnAo 0.693/k
• 1/At kt 1/Ao 1/kAo
•  
• R 8.31 J/Km 0.0821 La/Km k A e-Ea/RT
•  
• Water
• ?Hvap 40.7 kJ/m ?Hfus 6.01 kJ/m
• S water 4.18 J/goC Kb 1.86 oC/m Kf
  0.512 oC/m
• Selected values for ?Hof and ?Hbond energy
  in kJ/mole